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by: Jada Daniel

EngineeringReliability MEM361

Jada Daniel
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INTRODUCTION TO ENGINEERING RELIABILITY Lecture Notes for MEM 361 Albert S D Wang Albert amp Harriet So a Professor Mechanical Engineering and Mechanics Department Dreer University Philadelphia PA 19104 Chapter Introduction I 1 CHAPTER I INTRODUCTION Why Engineering Reliability In a 1985 Gallop poll 1000 customers were asked what attributes are most important to you in selecting and buying a product On a scale from 1 to 10 the following were the results from the poll Performance quality 95 Lasts a long time dependability 90 Good service 89 Easily repaired 88 Warranty 84 Operational safety 83 Good look 77 Brand reputation 63 Latest model 54 Inexpensive price 40 It is interesting to note that the top five attributes are related to Engineering Reliability the study of which is the essence of this course I 1 Basic Notions in Engineering Reliability Eng39lleering Product An engineering product is designed manufactured tested and deployed in service The product may be an individual part in a large operating system or it may be the system itself In either case the product is expected to perform the designed functions and last beyond it s designed service life Product Quality It is a quantitative measure of the engineered product to meet it s designed functional requirements including the designed service life Example 11 A bearing ball is supposed to have the designed diameter of 10mm But an inspection of a large sample off the production line finds that the diameters of the balls vary from 99lmm to 1013mm although the average diameter from all the balls in the sample is very close to being 10mm Now the bearing ball is an engineering product it s diameter is one quality measure The diameter of any given ball off the production line is uncertain although it is likely to be between 99lmm and 1013mm Example 12 A color TV tube is designed to have an operational life of 10000 hours Aftersale data shows that 3 of the tubes were burnt out within the first 1000 hours 65 were burnt out within 2000 hours Here the TV tube is an engineering product and the operating life in service is one quality measure For any one given TV tube off the production line it s operating life the time of failure during Chapter Introduction I 2 service is uncertain but there is a 3 chance that it may fail within 1000 hours of operation and a 65 chance that it may fail within 2000 hours of operation Random Variable A random variable denoted by X is one which can assume one or more possible values denoted by x but at any given instance there is only a chance that Xx In this context the quality measure of an engineering product is best described by one or several random variables say X Y Z etc Discussion The diameter of the bearing balls discussed in Example ll can be described by the random variable X according to the sample taken X can assume any value between x99lmm and x10l3mm If a much larger sample is taken from the production line some diameters may be smaller than 99lmm and some diameters may be larger than 1013mm In the extreme case X can thus assume any value between 0 and ac Similarly the operating life of the TV tube discussed in Example 12 can also be described by a random variable say Y From the statement therein for a given tube the chance that Y S 1000 hours is 003 and that Y S 2000 hours is 0065 Probability Function Let X be a random variable representing the quality measure of a product and it can assume any value x in the range say 0lt x lt oc Furthermore for Xx there is a certain associated chance or quotprobabilityquot this is denoted by fXx or by fx Note that fx is the probability that the value of X is exactly x Discussion In Example 11 the random variable X describes the diameter of a bearing ball From the sample data we see that the probability for Xlt99lmm or Xgt10l3mm should be very small while the probability for X10mm should be much higher this is because the bearing ball is designed to have a target diameter of 10 mm In this context fx may be graphically displayed as follows fX fx 05 I gt xmm x 1000 1013 desrgn target If fx is known a number of questions related to the quality of the bearing balls can be rationally answered For instance the percentage of the bearing balls having the diameters X g x is obtained as FX s x Fx IX fxdx 991 Here Fx represents 1 the probability that a given bearing ball is less than or equal to x or equivalently 2 the percentage of the bearing balls with diameters less than or equal to x Clearly for a given bearing ball the probability that its diameter is larger than x is Rx lFx Note Based on the sample given Fx denotes graphically the area under the fx curve from 991mm to x while FX10l3 denotes the total area under the entire fx curve Since the diameter of any given bearing ball is less or at most equal to 1013 mm the probability that the diameter of a given bearing ball being less than or equal to 1013mm is 100 or FX10l3 l Chapter Introduction 1 3 Note the difference in notation between fx and Fx and their meanings fx is termed the probability density function while Fx is termed the cumulative distribution function We shall discuss these functions and their mathematical relationships in Chapter 11 Probability of Failure Ifthe quality of a product X is measured by it s time of failure during service then the value of X is in terms of time t Let the range of t be 0lt Km and the associated probability for X is ft Discussion In Example 12 the operating life of the TV tube may be described by the probability function ft such as shown below 1 0 rm 0 1 05 t 1000 des1gn target Here the TV tube is designed for a life of 10000 hours of operation the chance for a given tube to last 10000 hours is better than any other tvalues Based on the sample data there is a 3 of failure up to t 1000 hours thus we have gt th1s Fth Ft from 003 The above is indicated graphically by the shaded area under the fx curve in the interval 0 S t S t Here note the relation between ft and Ft Product Reliability Let the quality of a product be measured by the timetofailure probability density function ft the probability of failure up to t t is given by the cumulative distribution function Ft Then the probability for non failure before t t is FX gt t 1 Ft 0c ftdt tgtllt The term FXgtt which is associated with the probability density function ft represents the probability of survival or known as the reliability function A precise de nition of the latter will be fully discussed in Chapter IV Discussion In Example 12 the data shows that 3 failed before tS1000 hrs and 65 before tS 2000 hrs Hence in terms of the reliability function we write R100010i0O ftdt 097 R2000I quot ftdt 0935 2000 Chapter Introduction I 4 Ifwe want to know the service life t for which no more than 5 failure or 95 or better reliability we can determine t from the following relation Rti ftdt 095 Clearly one attempts to answer all the questions regarding product reliability and this can be done when the mathematical form of ft is known In fact the knowledge of ft or the pursuit of it is one of the central elements in the study of engineering reliability I 2 Probability Statistics and Random Variables In the preceding section we used terms such as random variable probability of occurrence say Xx sample data etc which form the basic notions of engineering reliability How these notions which are sometimes abstractive in nature all t together in a realworld setting is a complicated matter especially for the beginners in the eld As in nature the occurrence of some engineered event is often imperfect indeed it may seem to occur in random but when it is observed over a large sample or over a long period of time there may appear a de nitive mechanism which causes the event to occur If the mechanism is exactly known the probability for the event to occur can be inferred exactly if the mechanism is not known at all sampling of a set of relevant data observations can provide a statistical base from which at least the nature of the mechanism may become more evident The latter is of course keenly dependent on the details of data sampling and on how the sample is analyzed Example 13 The number obtained by rolling a die is a random variable X In this case we know all the possible values of X the integers from 1 to 6 and the exact mechanism that causes a number to occur Thus the associated probability density function fx is determined exactly as fxl6 forx l 2 6 Note for example the probability that the number from a throw is less than 3 is given by Fxlt3 fl f2 l6l6 26 13 Similarly the probability that the number from a throw is greater than 3 is given by Fxgt3 f4 f5 f6 36 12 The probability that the number from a throw is any number x S 6 is given by FXS 6 f1f2 F6 1 Note In this example X is known as a discrete random variable since all the possible values of X are distinct and the number of all the values is finite The distribution of fx is said to be uniform since fxl6 for all x l 2 6 Example 14 Now let us pretend that we do not know any thing about the die By conducting a sampling test in which the die is rolled N100 times and each time the integer x on the die Chapter Introduction 1 5 recorded the following data is obtained from an actual experiment of N100 throws Xvalue x 1 2 3 4 5 6 timesxoccurn 17 14 16 20 15 18 The above is said to be a quotsamplequot of size N100 It is comparatively speaking a rather small sample even in fact only the integers from 1 to 6 can actually be observed we cannot be certain integers other than 1 to 6 could not appear rem ember we pretended not to know any thing about the die However we can infer from the sample quite closely what is actually happening namely we observe that the number 1 appears 17 times out of 100 throws the number 2 appears 14 times out of 100 throws and so on Hence we can estimate the probability density function fx as fl17100 f214100 f316100 f420100 f515100 f618100 We see that the estimated fx is not uniform over the range of X39 but they vary slightly about the theoretical value of 16 A graphical display of the above results is more revealing km 17 3920 167 theoretical value Vi 1 2 3 4 5 6 It is generally contented that all estimated fx would approach the theoretical value of 16 if N is sufficiently large say N1000 Qlou may want to experiment on this The relation between the sample size N and the theoretical probability function is another central element in quotstatisticsquot The above example illustrates the statistical relationship between a test sample and the theoretical probability distribution function fx for X X being generated by a specific mechanismrolling a die for N times This example casts an implication in engineering ie in most cases the theoretical fx is unknown and the only recourse is to find estimate fx through a test sample along with and a proper statistical analysis of the sample Example 15 Suppose we roll two dice and take the sum of the two integers to be the random variable X Here we know the exact mechanism in generating the values for X First there are exactly 36 ways to generate a value for X39 and X can have any one of the following 11 distinct inters i 2 3 4 5 6 7 8 9 10 11 and 12 For instance the number 2 is generated by the sum of 1 and 1 in one throw while the number 3 is generated by the sum of 1 and 2 39 but there is only one chance to obtain X2 while two chances to obtain X3 12 and 21 Hence the probability for f213639 and f3236 In fact the probabilities associated with each of the 11 values are f2f12136 f3f11236 f4f10336 f5f9436 f6f8536 f7636 A graphical display of fi is shown below Chapter Introduction I 6 fi 636 636 336 136 2 3 4 5 6 7 8 9 10 11 121 Here X is also a discrete random variable but its probability density function fi is not uniform It is however a symmetric function with respect to X7 For X7 the theoretical probability is 636 the largest among the 11 numbers Discussion Again if we do a statistical sampling by actually rolling 2 dice N times we may obtain an estimated fi function In that case we need a very large sample in order to approach the theoretical fi as displayed above Central Tendency In most engineering settings the random variable X o en represents some measured quantity from identical products for example the measured diameters of a lot of bearing balls off the production line constitute just such a case In general the measured values tend to cluster around the designed value ie the diameter thus the central tendency To evaluate this tendency is clearly important for design or quality control purposes The following example illustrates the evaluation of such a tendency Example 16 A master plumber keeps repaircall records from customers in his service area for 72 consecutive weeks 71 73 22 27 46 47 36 69 38 36 36 37 79 83 42 43 45 45 55 47 48 60 60 60 49 50 51 75 76 78 31 32 35 85 58 59 38 39 40 40 41 42 42 54 73 53 54 65 66 55 55 56 56 57 49 51 46 54 62 62 54 62 63 64 67 37 58 58 61 62 52 52 Here let X be the number of repaircalls per week which seems uniformly random While one sees that the smallest X value is 22 and the largest is 85 the sample really does not provide a definitive value range for X Furthermore since no definitive mechanisms could be identified as to how and why the values of X are generated the true probability distribution of X could never be determined Hence instead of looking for the mechanisms the sample data can be analyzed in some way to show its central tendency which may in turn be used to estimate the probability distribution function fx for X Here we follow a simple procedure as described in the following First we note that there are 72 data values in the sample roughly within the range from 21 to 9039 so we divide this range into 7 equal quotintervalsquot of 1039 namely 2130 3140 4150 etc Second for each Chapter Introduction I 7 interval we count from the sample the number of X values that fall within the interval For instance in the first interval 2130 there are 2 values 22 27 in the second interval 3140 there are 13 values 36 38 36 36 37 31 32 35 38 39 40 40 and 37 and so on After all is counted in the 7 intervals the following result is obtained intervals 21 30 31 40 41 50 5160 6170 7180 8190 X Values 2 13 15 22 11 7 2 in interval In this manner we can already observe that fewer values fall into the lower interval 2130 or into the upper interval 8190 but more values fall into the middle intervals especially in the central interval 5160 With the above interval grouping we may estimate the probability for X to fall inside the intervals Instead of treating X we introduce a new variable I representing the value intervals the values of I are the integers l to 7 since there are 7 intervals Thus the probability density function of I fi can be approximated as follows fi272 f21372 f31572 f42272 f51 172 f6772 f7272 A bar chart for the above is constructed as shown below it is termed a histogram for the sample data considered fi fx 2072 e tted fx 1072 10 20 30 40 50 60 70 80 90 calls The above bar chart displays some important features for the weekly repaircalls Namely it suggests that the most probable number of the repair calls occurs at i4 the 4th valueinterval39 or 5060 calls er week Secondly the shape of the histogram provides another clue as to the form of the estimated probability distribution function fx Note that the quotfittedquot fx shown in the figure is just a qualitative illustration details of sample fitting will be further discussed in Chapter III Discussions We note that the histogram obtained above is not unique For one may take more or fewer intervals value intervals in the range from 21 to 90 In either case one may obtain a somewhat different histogram for the same sample39 often one may even draw a quite different conclusion for the problem under consideration This aspect in handling sample data will be examined further in later chapters Sampling Errors As has been seen most engineeringrelated events are unlike rolling dice Rather the mechanisms that generate the random variable X are not exactly known Moreover Chapter Introduction I 8 the values of X may not be distinct and the range of their values may not be de ned exactly either In such cases the probability density fx can be estimated from test samples However questions arise as to the proper size of the sample possible bias that may have been introduced in taking the sample and the manner the sample is analyzed This necessitates an evaluation of the con dence level in estimating fx We shall defer the discussion of this subject in Chapter III 1 3 Concluding Remarks This chapter provides an overview of the intrinsic elements that embody the subject of engineering reliability At the heart is the interrelationship linking the random variable X the mechanisms which generate the values of X the statistics of the sample data related to X and the determination and or estimation of the probability distribution function fx The fundamentals of probability ie the mechanisms that generate the random variable X and the properties of various probability functions are investigated more critically in Chapter II Sample statistics and methods for fitting sample data some known probability distribution functions and sampling error estimate and related subjects are discussed in Chapter III The basics in reliability and failure rates are included in Chapter IV Chapter V presents techniques for reliability testing while Chapter VI discusses applications to some elementary product quality control issues Throughout the text simple but pertinent examples are used to illustrate the essence of the important points and or concepts involved in the subject of concern A modest number of homework problems are included in each chapter the students are urges to do the problems with a clear logical reasoning rather than seeking the formula and plugging in just for an answer Chapter Introduction Summary As a beginner it is useful to be conceptually clear about the meaning of some of the key terminologies introduced in this chapter and to distinguish their differences and interrelations The random variable X is generated to occur as an event by some mechanisms that are or are not known When it occurs X may assume a certain real value denoted by x and x can be any one value inside a particular range39 say 0 S xlt 00 In one occurrence there is a chance probability that Xx that chance is denoted by fx here Xx means X equals exactly x Since x denotes value in the range of X values fx is treated mathematically a distribution function over the value range Thus fx is the mathematical representation of X39 fx has several properties and these will be discussed in Chapter II In one occurrence the probability that X S x is denoted by PxlX Sx or simply Fx39 here Fx is the sum of all fx where X Sx Similarly PxlXgtx denotes the sum of all fx where Xgtx sometimes it is also denoted by Rx andor by lFx In this course we sometimes mix the use of the symbols between PxlX S x and Fx39 between PxlXgtx and Rx and lFx etc This can be a source of confusion at times If the exact mechanism that generates X is known it is theoretically possible that the exact valuerange x along with the theoretical fx is also known39 if the mechanism is not known exactly one can only rely upon statistical samples along with a proper statistical analysis methodology in order to determine fx A certain Quality of an engineered product can be treated as a random variable X x is then the measure of that quality in one such product picked in random Often than not the exact mechanisms which generate the product quality X are not completely known39 hence sampling of the quality and statistical analysis of samples become essential in determining the associate fx function for X Engineering reliability is a special case where the random variable X represents timetofailure such as service lifetime of a light bulb39 for obvious reasons it is necessary to obtain the timetofailure probability ft for say the light bulb Assigned Homework 11 Let the random variable X be defined as the product of the two numbers when 2 dice are rolled List all possible values of X by this mechanism Determine the theoretical probability function fx Sketch a barchart for fx similar to Example I539 Compute F25 explain the meaning of F2539 Compute R15 explain the meaning of ROS Show that the sum of all possible value of fx equals to one Partial answer there are 18 values for X39 f6l939 f25l36 12 A coinbag contains 3 pennies 2 nickels and 3 dimes If 3 coins are to be taken from the bag each time their sum is then a random variable X I9 L Lquot Chapter Introduction List all the possible values of X by this mechanism Determine the associated probability distribution fx Plot the distribution in a graphical barchart Show that the sum of all possible value of fx equals to one Partial answer there 56 combinations in drawing three coins but only 9 difference values 020 and 025 are among them Optional for extra effort Let the random variable X be the sum of the three numbers when 3 dice are rolled I Complete the theoretical probability distribution fx for X I Show your results in a barchart I Comment on the form of the barcharts obtained by rolling 1 2 and 3 dice respectively There are 216 possible outcomes in rolling 3 dice they provide only 16 values from 3 to 18 f1027216 f1321216 one die gives a uniform fx 2 dice yield a bilinear fx 1n Example 16 the exact mechanism that generates repair calls X is not known but the sample provided can be used to gain some insight into the probability distribution function fx I Now by using the classinterval 6 calls instead of 10 calls Redo a histogram for the sample I Discuss the difference between your histogram and the one obtain in Example 15 Optional for extra effort The Boeing 777 is designed for a mean service life of 20 years in normal use Let the service life distribution be given by ft where t is in years and the form of ft looks like the one shown in Example 12 I Sketch ft as a function of t ears and locate the design life 20 years on the taxis I If a B 777 has been in serve for 10 years already what is the chance that the craft is still fit to fly until for another 6 years This is a case of quotconditionalquot probability 110 Chapter II Fundamentals II 1 CHAPTER II FUNDAMENTALS IN PROBABILITY II 1 Some Basic Notions in Probability Probability of an Event Suppose that we perform an experiment in which we test a sample of N quotidenticalquot products and that n of them fail the test If N is quotsuf cientlyquot large N gt co the following de nes the probability that a randomly picked product would fail the test PXpnN 0SnSNN gtoo 21 Here X denotes the event that a product fails the test it is a random variable because the picked product may or may not fail the test thus the value of PX represents the probability that the event X fail the test does occur Clearly the value of PXis bounded by 0 S PX S 1 22 The Non Event Let X be an event with the probability of PX We de ne X the non event of X meaning that X does not occur Then the probability that X occurs is given by PX 1 PX 23 The relationship between PX and PX can be graphically illustrated by the socalled Venn Diagram as shown below PX PX In the Venn diagram the square has a unit area the shaded circle is PX and the area outside the circle is PX If PX is the probability of failure PX is then the probability of survival or the reliability Event and Non Event Combination In a situation where there are only two possible outcomes such as in tossing a coin one outcome is a head and the other a tail X is a random variable with two distinct values say 1 and 0 the associated probability distribution are then f1 PXp and fO PX q It follows from 23 that f0 f1 q p 1 Chapter II Fundamentals II 2 Example 21 In tossing a coin the head will or will not appear we know that the probability for the head to appear is PXp 12 and that for the head not to appear is PX q l p 12 Similarly in rolling a die let X be the event that the number 1 occurs Here we also know that PX p 16 and the probability that I will not occur is PX q lp 56 In the above we know the exact mechanisms that generate the occurrence of the respective random variable X In most engineering situations one can determine PX or p from test samples instead Example 22 In a QC quality control test of 500 computer chips 14 chips fail the test Here we let X be the event that a chip fails the QC test39 and from the QC result we estimate using 21 PX p E nN 14500 0028 Within the condition of the QC test we say that the computer chip has a probability of failure 7 0028 or a survivability ofq 0972 Discussion In theory 21 is true only when N gt oc The p0028 value obtained above is based on a sample of size N500 only Hence it is only an estimate and we do not know how good the estimate is There is a way to evaluate the goodness of the estimate and this will be discussed in Chapter III Combination of Two Events Suppose that two different events X and Y can possibly occur in one situation if the respective probabilities are PX and PY The following de ned PXr Y the probability that both X and Y occur and PXU Y the probability that either X or Y or both occur X n Y is termed the intersection of X and Y XU Y is termed the union of X and Y a graphical representation of these two cases is shown by means of the Venn diagrams X Y XUY In each diagram the outline square area is 1X1 representing the total probability the circles X Chapter II Fundamentals II 3 and Y represent the probabilities of the respective events to occur The shaded area on the left is X n Y in which both X and Y occur the shaded area on the right is XU Y in which either X or Y or both occur The union is mathematically expressible as PXU Y PX PY PXn Y 24 which can be inferred from the Venn diagram Note that the blank area outside the circles in each case represents the probabilities of non eventquot that is neither X nor Y will occur PX U Y39 1 PXU Y Independent Events If the occurrence of X does not depend on the occurrence of Y or vice versa X and Y are said to be mutually independent Then PXnY PXPY 25 Expression 25 is an axiom of probability and it cannot be shown on Venn diagram Example 23 In rolling two dice let the occurrence of 1 in the first die be X and that in the second is Y In this case occurrence of Y does not depend on that of X39 and we know PXPY16 It follows from 25 and 24 respectively that PX Y 1 appears in both dice 1616 136 PXU Y 1 appears in either or both dice 16161361136 Discussion The fact that PX U Y1136 can also be found as follows there are in all 11 possible combinations in which 1 will appear in either or both dice 11 12 13 14 15 16 21 31 41 51 61 while there are a total of 36 possible outcomes Conditional Probability If the occurrence of X depends on the occurrence of Y or vice versa then X and Y are said to be mutually dependent We define PXY probability of X occurrence given the occurrence of Y PYX probability of Y occurrence given the occurrence of X It follows the axiom 25 that PX n Y PXY PY PYXPX 26 Example 24 Inside a bag there are 2 red and 3 black balls The probability for drawing a red ball out is PX25 and that to draw another red ball from the rest of the balls in the bag is PYX14 Thus to draw both red balls consecutively the probability is PXr YPYX PX1425110 Example 25 An electrical system is protected by two circuit breakers that are arranged inseries When an electrical surge passes through at least one breaker must break in order to protect the system if both do not break the system would be damaged In a QC test of the breakers individually the probability for the breaker not to break is PX00239 Chapter II Fundamentals II 4 and if two are connected inseries the probability of failure to break of the second given the failure to break of the first is much higher PYXPXY0l The probability that the system fails by an electric surge is when both breakers fail to break PX YPYX PX 010020002 The probability that at least one fail to break when they are inseries is PX U YPXPY PX Y 002 002 0002 0038 Discussion If failuretobreak of one breaker does not affect the other the failure probability of the system is when both fail to break PX n YPY PX 002o02o0004 Mutually Exclusive Events In a situation where if X occurs then Y cannot and Vice versa then X and Y are said to be mutually exclusive In the Venn diagram X and Y do not intersect So PX Y 0 27 It follows from 24 and 26 respectively that PXU Y PX PY 28 PXYPYX0 Discussion To illustrate mutually exclusive or nonexclusive events consider the following example for a deck of poker cards the probability of drawing an quotacequot of any suite is PX452 and that of drawing a quotkingquot is PY452 These are two independent events but mutually exclusive in a single draw since if an quotacequot is drawn it is impossible to draw a quotkingquot Thus PX Y 0 PXUY PXPY 852 Alternatively given the occurrence of an quotacequot the probability of drawing a quotkingquot in a single draw is PYX 0 Now in a single draw the probability of getting a quotheartquot is PZl352 the chance of getting an quotace of heartquot is then PX ZPX PZ 452l352 152 The chance of getting either an quotacequot or a quotheartquot is the union ofX and Z PX u Z PXPZPX n Z 4521352152 1652 Note that to get an quotacequot X and a quotheartquot Z is not mutually exclusive while to get an quotacequot and a quotkingquot Y are mutually exclusive Combination of N Events In a situation where N possible events X1 X2 X3 XN can occur their intersection or union cannot be obtained However if they are independent events then their intersection is Chapter 11 Fundamentals H 5 PX1 0X20 X3 nXN PX1 PX2 PX3 PXN 29 Note that 29 represents the probability that all N events occur Similarly X39l X Z X393 X39N are the respective nonevents and their intersection is PX 1 n X 2 0 X3 n X NPX 1PX 2 PX 3 PX N 210 And the above is the probability that none of the N events occur As for the union of the Nevents PX1 U X2UX3 U XN it represents the probability that one or more or all of the N events occur Since Pone or more or more events occur Pnone occurs 1 we can write PX1 U X2 U X3 U XNPX 1 PX 2 PX 3 PX N1 211 Note that 211 is the total probability of all possible outcomes thus it is a unity The terms PX i i12N in 211 can be replaced by PX i 1 PXi i12N 212 Hence the union of all the Nevents can be expressed in the following altemate form PX1UX2UX3 UXN11PX11PX2 as 1 PXN 213 A Special case ifPXi p for all i1N then the intersect in 29 becomes PX1nX2rX3 nXN pN39 And the union in 213 becomes PX1UX2UX3 UXN 11pN The example below illustrates such a special case Example 26 A structural splice consists of two panels connected by 28 rivets QC finds that 18 out of 100 splices have at least one defective rivet If we assume defective rivets occur independently and the probability of being defective is p what can we say about the quality of the rivets Here let Xi i128 be the event that the ith rivet is found defective in one randomly chosen splice and PXi p The probability for one or more or all rivets to be found defective in one splice is the union of all Xi i128 PX1 UXZUX3 UXZS But QC finds the probability of a splice to have at least one defective rivet is 18 100 hence PX1UX2UX3 szg11p28 018 Chapter 11 Fundamentals 11 6 Solving we obtain 7 00071 We can say that about 7 out of 1000 rivets may be found defective Discussion In this example QC rejection rate of the splice 018 is given but the probability of being defective of a single rivet is not By using the definitions of intersect and union of multiple events we can estimate the probability of being defective for a single rivet 7 Conversely if p is given we can use the same relations to estimate the rejection rate of the splice 11 2 The Binomial Distribution A special case of Nevents leads to the socalled binomial distribution it is also referred to as the Bernoulli Trials Speci cally if the events X1 X2 XN are independent as well as statistically identical PXnp and PX n lp q for n 12N Then we can always write PXn PX n p q 1 for n12N 214 It follows that q p N1 215 Note that 215 is a binomial of power N upon expansion we have CNOq NCN1q Nlp CN2q NZp 2 CNiq Nip i CNNp N 1 216 where CNi N Ni i i 01 2 N 217 It turns out that each term in the binomial expansion 216 has a distinct physical meaning CNOq N qN is the probability that none of the Nevents ever occur fi 0 Cqu N 1p is the probability that one of the Nevents occur fi 1 CNNp N p N is the probability that all of the Nevents occur fi N and In particular CNiq N39ip i are the probabilities that i out of N events occur fi The expression Chapter 11 Fundamentals ll 7 CNiq N39ipi fi 218 is known as the binomial distribution representing the probability that of the Nevents there will be exactly i events i 1 2 N that will occur Note that 216 is the total probability for all possible outcomes f0f1f2fN1 219 It can in turn also be rewritten as f1f2fN1f01qN 220 which is the probability for one or more events to occur or the union of all events The binomial distribution 218 is also known as the Bernoulli trial the meaning of the trial is illustrated in the following examples Example 27 Suppose that a system is made of 2 identical units A and B Under a certain prescribed operational conditions the failure probability of each unit is p For the system one of the following 4 situations may occur during the prescribed operation Afail and Bfail39 Afail Bnot fail39 Anot fail Bfail39 Anot fail Bnot fail The associated probabilities for the four situations are respectively pp pq qp and qq Thus we have f0 probability of none failure qq fl probability ofjust one failure pq qp 2pq f2 probability of two failures pp Check that the total probability of all possible outcomes is 100 fof1f2 q 2 2m 1 2 q p 2 1 The above follows the binomial distribution for N2 qp2 1 Example 28 In rolling a die repeatedly what is the probability that the 1 appears at least once in 4 trials Here the same event the appearance of 1 is observed in N4 repeated quottrialsquot the random variable of interest is the number of times the observed event occurs this can be any number i 0 1 2 3 or 4 This is a trial case known as the Bernoulli Trial Now for f0 fl f2 f3 and f4 we can write fO no no no no q fl yes no no no pq3 no yes no no pq Chapter 11 Fundamentals H 8 no no yes no 7113 no no no yes p113 f2 yes yes no no 7qu yes no yes no 7qu yes no no yes 7qu no yes yes no 7qu no yes no yes 7qu no no yes yes 7qu f3 yes yes yes no 7311 yes yes no yes 7311 yes no yes yes 7311 no yes yes yes p31 4 f4 yes yes yes yes p The total probability of all outcomes is thus f0f1f2f3f4 q 4 4q 3p 6q 2p 24q 7 31 4 1 17 4 1 Again it follows the binomial distribution Discussion For engineering products a system or a single component may be under repeated and statistically identical demands Say in each demand the failure probability is p 16 while that for nonfailure is L 56 Then the probability that failure of the system or component occurs at least once in 4 repeated demands is given by f1f2f3f4 1 fO 1 q 4 1 564 518 The above result can be obtained by applying 218 to 220 directly The Pascal Triangle The coefficients in the binomial expansion of power N in 217 can be easily represented by a geometrical construction known as the Pascal Triangle if N is not very large 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 18285670562881 zzzzzzzzz OONCNUIAUJNt O Example 29 Suppose the probability of a light bulb being burnt out is 7 whenever the switch is turned on In a hallway 8 light bulbs are controlled by one switch Compute f0 fl f8 when the switch is turned on Chapter 11 Fundamentals H 9 Using the Pascal Triangle for N8 we can quickly write K0 qg fl 8q7p K2 28 6122 f3 56q5p3 f4 70q4p4 f5 56q3p5 f6 28q2p6 fD 811177 f8 pS The Poisson Distribution While the Pascal triangle becomes cumbersome to use when N is large say Ngt20 there is a simple expression for 218 when N is large and p is small say Ngt20 andp ltlt1 fi Np i expNp i i 012 N 221 Expression 221 is known as the Poisson Distribution It is an approximation of the binomial distribution 218 when N is large and p small Example 210 Suppose that the probability for a compressor to pass a QC test is 109 If 10 compressors are put through the QC test compute the various probabilities of failure fi i0l 10 In this case N10 and p 0 l39 the binomial distribution 218 and the simplified Poisson distribution 221 will be used39 the following are the respective results f0 fl f2 f3 f4 f5 f6 f7 f8 f9 f10 by2l8 0349 0387 0194 0057 0011 00015 by22l 0368 0368 0184 0061 0015 00031 H H 0 0 Discussion The exact binomial distribution 218 gives a maximum at f00349 and fl038739 the Poisson approximation 221 gives f0fl0368 For the rest the two distributions are rather c ose The Poisson approximation would yield better results if N is larger or 7 smaller In this example the N 10 value is not large enough and p 0 l is not small enough thus the difference II 3 Properties of Discrete Random Variables The random variable X is discrete when it s values xi i 1N are distinctly de ned and N is nite If the probability that X takes the value xi is fXi fxi has the following important properties The axiom of total probability 2 fxi 1 Z sums over i 1 N 222 Here the anction fxi is formally termed as the probability mass function of X or pmf for short It is called mass function in the sense that the probability fxi is exacted at Xxi much like the lumpedmass representation of particle dynamics in physics the expression in 222 which equals to unity resembles the total of all the lumped masses Chapter 11 Fundamentals 11 10 The partial sum for nSN FXn Z fXi Z sums overi 1 n 223 It is termed the cumulative mass function of X or CMF for short Note that 0 FX nl forl SnSN The Mean of X The mean of fXi is de ned as u Z Xi fXi39 Z sums overi 1 N 224 The Variance of X The variance of fXi is de ned as 62 Z Xiu2 fXi39 Z sums over i 1 N 225 By utilizing 224 the variance de ned in 225 can be alternately expressed as 62 Z Xiz fXi Hz 2 sums over i 1 N 226 The standard deviation The standard deviation is 6 as de ned in 226 The above properties are illustrated by the following example Example 211 Suppose the values ofX are given as 0 1 2 3 4 5 and the associated pmf are f00 f1116 f214 f338 f414 and f5116 Compute the mean variance and standard deviation for X We compute the total probability by checking the sum 0 116 38 14 116 139 the mean ofX by applying 224 p 001116 214 338 414 5116 3 the variance of X by applying 226 02 02 01211622143238421452116 732 1 the standard deviation is computed as o i1 Discussion It is geometrically revealing if the pmf is represented by a bar chart versus x and the CMF by stair case chart as shown in the figures on the next page Note that the sum of the pmf bars lumped masses is a unity and that the center of gravity of the total mass is at xu 3 Note that the mean is not the average of all of the values ofX Chapter 11 Fundamentals 11 11 It is also interesting to note that the mass moment of inertia about the mean is 02 Which equals 139 the radius of gyration of the total mass about mean is 0 Which is the standard deviation This analogy in physics is sometimes helpful A me 12 14 I 39 gt X1 0 1 2 3 4 5 A CMF 10 1216 1116 05 516 116 39 0 I l I I I gt X1 0 1 2 3 4 5 Note In the upper figure the length of the bar at each xi represents the value of fxi in the lower figure the staircase rise at each xi equals the value of fxi The charts provide a geometric view of the pmf and CMF respectively Example 212 For the binomial distribution fi given by 218 it s mean can now be found by using 224 p E i C1li1p1l391p1 Np E sums overi 1 N The variance is found by using 225 02 E lp2 CNi1p N39i i Np1p E sums over i 1 N The proof of the above results for p and 02 requires manipulations of the binomial terms in 218 Chapter II Fundamentals H 12 the details of which will not be included here interested readers may consult the text by E E Lewis 2nd Edition p23 Recall that the Poisson distribution 221 is a simplified version of the binomial distribution when N gt 0c and p ltltl The mean u and the variance 02 for the Poisson distribution are readily obtained from the above H 62 Np Thus the Poisson distribution is a oneparameter exponential function see 221 f0 pi e39Hi The Expected Value Suppose that a weighting function gxi is accompanying fxi whenever the random variable X assumes the value of xi Then the expected value of X with respect to the weighting function gxi is de ned as Eg 2 gxifxi Z sums over i 1 N 227 Here the weighting function gxi may be understood as follows 0 Each time xi is taken by X a value of gxi is realized But for X to assume xi there is the probability fXi hence the expected realization is gxifxi39 o Eg is the cumulative realized value for all possible values of X Example 213 Suppose in Example 211 the accompanying weighting function is gxiocxi where i0 to 539 and on is a real constant Then the expected value Eg is given by 22D Eg oc0 011162143384145116 ocp 30c Applications of the expected value formula in 227 will be detailed in chapter VI Summag of Sections 11 2 and 11 3 The following table summarizes the essential elements in the discrete random variable X discussed in Sections 112 and 113 FunctionProperties General Discrete Special Case Bernoulli Trials Binomia Poisson values ofX xi ilN i 0 l 2 N pmf fog fiNltNi npini fimpii eNP CMF Fxn2fxi over ln fn2fi over 0n Chapter II Fundamentals ll l3 mean of X it uExifxi39 over lN tFNp variance of X 02 02Exi2fxiu2 02Nplp 02Np nonevent reliability RXnl FXn f01pN f0eNP II 4 Properties of Continuous Random Variables A random variable X is continuous when it s value X are continuous distributed over the range of X For definiteness let the range of X be olt lt X lt oc then the probability that X takes the value X in the range is a continuous function fX Here fX is formally termed the probability density function or pdf for short The pdf must satisfy the aXiom of total probability X I fX dX 1 228 70C The cumulative distribution function or CDF for short is de ned as X FX I x dx 229 70c FX represents the probability that value of X X The Mean u and The Variance 62 0f fX are Hl XfX dX 230 62 l xii fX dX 231 The variance in 231 can alternatively be eXpressed as 62 l X2fX dX H2 232 The Proof of 232 is left in an eXercise problem at the end of this Chapter Discussion A pdf which is common engineered product quality variation is shown graphically on the next page This fx resembles a belllike curve39 the head of the curve is with diminishing fx as X increases While the tail is with diminishing fx as X decreases The maximum of fx occurs Chapter II Fundamentals H 14 at xxm Ode which can be determined by setting dfxdx0 The total area under fx is the integral in 228 Which must equal to l39 the centroid of the area under the fx curve represented by the integral 230 is located at xp Which is the mean ofX the area moment of inertia with respect to the axis xp and represented by 231 or 232 is the variance 2 Finally the radius of gyration of the area under fx revolving about the axis xp is given by to it is the standard deviation of X Hquot 69 OCV VN lt V X mode The Expected Value given gxi accompanying fXi the expected Value of X with respect to gx is given by Eg I w gx fx dx 233 Note when gxx Egp39 when gxx2 Eg 02 2 Thus the expected value of X is it s mean39 the expected value of X2 is 02 2 The latter is readily seen from 232 The Median of X The median of X is when X xm such that Xm FX m I fx dx 12 234 00 From the geometric point of View Xxm separates the total area under the fx curves into two halves The Mode of X The mode of X is when X mode such that it corresponds to the maximum of fx as discussed above Note The mean u median xm and mode xmode are distinctly defined quantities each has it s own physical Chapter II Fundamentals H 15 meaning But the three quantities become the same if fx is a symmetric respect to the mean xmode xm u Skewness of fix The pdf is a skewed distribution when it is not symmetric with respect to the mean then in general Xmode 7E Xmi u A measure for the skewness of fX known also as the skewness coef cient is given by CC 3 sk 3 x H fX 1 235 0 IfX is discrete then 235 can be expressed as sk 163Z xi 3 fog Z sums over 1N 236 It may be shown that when sk gt 0 fX is a left skewed curve Xmodelt Xmlt u sk lt 0 fX is a right skewed curve Xmodegt Xmgt u sk 0 fX is a symmetric curve or Xmode Xm u A graphical display of the leftskewed and rightskewed curves is shown below fX left skew right skew I Example 214 LetX be a continuous random variable with its values defined in the interval a x b Suppose that the corresponding pdf is a constant fxk This is a case of uniform distribution Now in order for fx to be a bona fide pdf it must satisfy the total probability axiom b J39 kdx kba 1 a This yield the value for k k lba Chapter II Fundamentals H 16 The cumulative function Fx is obtained by integrating fx from a to x FX Xaba The mean and variance can be easily obtained p ab2 02 ba212 Example 215 Let the random variable T be defined in the valuerange of 0 t lt 0c and the pdf in t e orm ft Ae39M Here the pdf is an exponential function It s CDF is easily integrated as Ft 1 eM The mean and the standard deviation are also easily obtained 0 p lA Example 216 The lifetime time to failure of a washing machine is a random variable Suppose that it s pdf is described by the function ftAte05t 09m where A is a constant andt is in years We examine the following properties of this pdf a The CDF t t F0 I A t e 0quot dt 4AI 050 e 0quot d05t 4A1 1 050 e39 05 0 0 where the integration is carried out by integration by parts b The pdf satisfies the axiom of total probability Foc l we have x c 0 5 t IftdtI Ate dt1 0 0 Upon carrying out the integration we obtain A 14 c The mean of the pdf which is also called the mean time to failure or MTTF for short H I t At 6 0quot dt 14 2053 4 Chapter II Fundamentals H 17 where the integration is carried out using an integration table d The variance is given by 2 0C 6 I0 t2At e 0 dt 2 14 3054 16 8 e The standard deviation is hence o xg A graphical display of the pdf and the CDF in this example is shown below f The skewness coefficient sk of the pdf is given by the integral sk 163 It 3 A t O39St dt With 0 xg and u 4 it can be shown that sk gt 0 so the pdf is a leftskewed curve as can be seen in the plot Discussion Plots of the pdf andor CDF provide a visual appreciation of the life distribution of the washing machine We see that the failure probability rises during the first two years in service about 25 of the machines will fail by the end of the second year ie F2m 025 Similarly the mean timetofailure MTTF is 4 years and we find F40594 so nearly 60 of the machines will fail in 4 years lfthe manufacturer offers a warranty for one full year tl then Fl009 or 9 of the machines is expected to fail during the warranty period Chapter II Fundamentals H 18 Another Note Often one has to integrate complex integrals in order to obtain explicit results An integration table at hand is always helpful At times some integration can become extremely tedious to the extent that it may not be integrated explicitly Such difficulty can be circumvented by a suitable numerical method of integration One should not view such difficulty as an inherent feature in this course Summary This chapter introduces l the basic notions in probability and 2 the mathematical definitions of the properties of a probability distribution function It is essential to be conceptually clear and mathematically proficient in dealing with these subjects In the former we should be clear about the following I The event X the probability of X to occur is denoted by PX39 the probability of X not to occur is PX lPX I If event X is a random variable X can have a range of possible values denoted by x if value of x is continuous or by x if value of x is discrete The probability of Xx is denoted by PXx fx similarly the probability of X x is denoted by fxi I fx is called pdf since x is continuous fxi is called pmf since x is discrete39 fx is the probability when X is exactly equal to x I Fx is called CDF and Fxi CMF Fx is the probability when X S x39 it is the area under the fx curve up to the specified value of x I For multiple events X Y Z or X1 X2 X3 etc make clear the physical meanings of their intersection and union as well as the meanings of dependent independent and mutually exclusive events I If occurrence of X depends on the occurrence of Y the probability of X to occur is PXY39 this is known as the conditional probability But if X and Y are independent events PXYPX39 and PYXPY as well I For identical and independent events the binomial distribution applies The simpler Poisson distribution is a degenerated binomial distribution when N is large and 7 small For identical but mutually dependent events the binomial distribution will not apply As for the properties of a probability function fx note the definition of I The total probability axiom39 the distribution mean variance and the skewness I The value of X may be discrete or continuous handle them with care I Elementary integration skill will be helpful Chapter 11 Fundamentals 11 19 Assigned Homework 21 Suppose that PXO32 PY044 and PX U Y058 Answer the following questions with proof a Are X and Y mutually exclusive b Are X and Y independent events c What is the value of PXY d What is the value of PYX Partial answer a no b no PXYO40939 PYX05625 22 Suppose that PAO4 PA U B08 and PA B02 Compute 801333 b PAB C PBA Partial answer PBO639 PBAO5 23 In a QC test of 126 computer chips from two different suppliers the following are the test results Pass the QC test do not pass Supplier 1 80 4 Supplier 2 40 2 Now let A denote the event that a chip is from supplier1 and B the event that a chip pass the QC test Then answer the following questions a Are A and B independent events b Are A and B independent events c What is the meaning of PAU B CD What is the value of PAU B Hintz PA84126 PB120126 and PA n B80126 24 Use the Venn diagram and show the following equalities I PY PY X PY X X and Y may be dependent events I PY PYX PX PYX PX X and Y may be dependent events I PY PY PX PY PX ifX and Y are independent events 25 An electric motor is used to power a cooling fan39 an auxiliary battery is used in the event of a main power outage Experiment indicates that the chance of a main power outage is 06 And when the main power is on the chance that the motor itself fails is pm 025x1039339 when the auxiliary battery is on the chance of the motor failure is pb 075x10 3 Determine the probability that the cooling fan fails to function Hint let X be the event of main power outage Y the event of fan failing to operate39 note that Y depends on X and X39 and PYPYXPXPYX39PX39 applies Note also PYXpb and Hyman Answer PY0253x10393 26 The values of the random variable X are 123 the associated pmf is fxi Cxi3 i13 a Find the value for C39 b Write the expression of Fxi39 c Determine p and o and sk39 d Plot fxi and Fxi graphically Partial answer C13639 2722 0050639 sk 162 Chapter 11 Fundamentals 11 20 27 A single die is rolled repeatedly 6 times each time the 6 is desired a Use 218 to compute the pmf fi i06 is the number of times the 6 appears b Plot a barchart for fi i06 indicate the largest fi value c Determine the mean and variance of f i d Redo the computation of fi by the Poisson equation 221 comment on the difference 28 Show the details how Equation 225 is reduced to 226 similarly show how Equation 231 is reduced to Equation 232 29 In a QC test of a lot of engines 3 failed the test Now 8 such engines are put into service what is the probability of the following situations a None will fail b All will fail c More than half will fail d Less than half will fail Partial answer a 0784 b nearly 0 210 The probability of a computer chip being defective is p0002 A lot of 1000 such chips are inspected a What is the probability that 01 or more of the chips are defective b What is the probability that more than 05 of the chips are defective c What is the mean expected number of the defective chips Hint Use the Poisson distribution for pltlt1 and Ngtgt1 Partial answer a Pngt1 1 f0 f1 0594 211 Suppose that the probability of finding a flaw of size x in a beam is described by the pdf fx 4xe392X 0 S x S oc x is in microns 10396 m a Verify that fx satisfies the total probability axiom Determine the mean value of the flaw size distribution c If a flaw is less than 15 micron the beam passes inspection what is the chance that a beam is accepted Partial answer b 1 micron c PXlt15 08 212 A computer board is made of 64 kbit units the board passes inspection only if each kbit unit is perfect On line inspection of 1000 boards finds 60 of them unacceptable What can you say about the quality of the kbit unit Hint let p be the probability that a kbit unit is imperfect and then find the value of p 213 Optional for extra effort A cell phone vendor assures that the reliability of the phone is 99 or better the buyer would consider an order of 1000 phones if the reliability is only 98 or better The two sides then agree to inspect 50 phones if no more than 1 phone fails the inspection the deal will be made a Estimate the vendor s risk that the deal is off b Estimate the buyer s risk that the deal is on Hint For the vendor p 001 estimate the chance that more than 1 in 50 inspected fail For the buyer 7 002 estimate the chance that no more than 1 in 50 inspected fails Chapter II Fundamentals H 21 ChapterIII Data Sampling IIIl CHAPTER III DATA SAMPLING AND DISTRIBUTIONS In the preceding chapters some elementary concepts in probability andor random variables are introduced and these are illustrated using a number of examples whenever possible We note that the central element in these examples is the pertinent probability distribution function pmf or pdf for the random variable X identi ed in the particular problem In general the pertinent pmf or pdf may be determined or estimated by one of the following two approaches a Probabilistic Approach If the exact mechanisms by which the random variable X is generated are known the underlying pmf or pdf for X can be determined on the basis of a certain probability theory In Chapter II we have illustrated this approach using simple examples such as rolling dice or ipping coins In addition a somewhat more complex mechanism involving the socalled Bernoulli Trial was shown to lead to a class of pmfs known as binomial distribution b Data Sampling Approach Engineering issues such as the lifetime of a product in service or the expected performance level of a machinery often involve intrinsic mechanisms that are not exactly known the underlying pmf or pdf for the identi ed random variable cannot be determined exactly Then the alternative is to estimate the pmf or pdf using techniques involving data sampling In Chapter I we demonstrated brie y how a statistical sample could provide an estimate leading to the underlying pdf for the identi ed random variable But in that process there are certain techniques and the details of which need be thoroughly discussed Thus this Chapter discusses the basic elements in the data sampling approaches along with their connection to some of the wellknown probability distribution functions III 1 Sample and Sampling At the outset let us introduce the following terms Population A population includes all of its kind Namely when a random variable X is de ned all of it s possible values constitute the population The underlying pdf of X must be de ned for each and every element in the population it is referred to as the true pdf of X Clearly for a continuous random variable the population size is infmite Sample A sample is a subset of the population Thus the size of the sample is nite even if the population is in nite Elements in the sample represent only partial possible values of X In general more than one sample may be taken from the same population If a sample contains N elements it is referred to as a sample of size N ChapterIll Data Sampling Ill2 Sampling This refers to the quotcreationquot of sample data in order to estimate the underlying pdf of the random variable X Depending on the sample is taken the estimated pdf may not be close to the true pdf of the population this is especially the case when the sample size is small compared to that of the population Hence proper techniques in sampling become important Moreover one would also want to have a degree of con dence in the sampling technique as well as the estimated pdf Random Sampling This refers to sampling techniques that guarantee each possible value in the population will have an equal chance of being sampled Such techniques ensure a closer agreement between the estimated pdf and the true pdf Sampling Error This refers to the difference between the estimated pdf from a sample and the true pdf of the population Logically the sampling error can be easily assessed if the true pdf of the population is known But the true pdf may never be known in some cases hence the error in sampling can be estimated only based on some statistical reasoning III 2 Sample Statistics Random Sample Consider a random sample of size N denote the data in the sample as Xi ilN Assume each Xi is selected in random the sample pdf is then a uniform distribution fXi UN for ilN 31 The sample Mean according to 224 is then us Z XilN lNZ Xi Z sums over 1N 32 The sample variance and skewness are respectively 692 1N2 xi us2 2 sums over 1N 33 skS 1N63 2 xi us3 2 sums over 1N 34 Note that 32 is actually the averaged value of the sample Xi it is based on the assumption that each Xi has the same chance being sampled However if N is not large this assumption can bias the variance it is likely that at least one value outside Xi may have been excluded in the sampling To admit this possibility the sample variance in 33 is often modified in the form 692 1N12 xi us2 2 sums over 1N 35 The expression 35 will be used in all subsequent discussions and in all homework problems ChapterIll Data Sampling Ill3 Note In most engineering situations one cannot assume a uniform pdf for a sample eg the one shown in 31 without a justification The following example is a case in point Example 31 Seventy 70 cars are picked randomly from the assembly line for QC inspection For each car the braking distance from the running speed of 35 mph to a complete stop is recorded The following table lists the raw data obtained in the QC tests referred to as the sample xi il70 Braking distance in feet 39 54 21 42 66 50 56 62 59 40 41 75 63 58 32 43 51 60 65 48 61 27 46 60 73 36 38 54 60 36 35 76 54 55 45 71 54 46 47 42 52 47 62 55 49 39 40 69 58 52 78 56 55 62 32 57 45 84 36 58 64 67 62 51 36 73 37 42 53 49 Here the braking distance is treated as the random variable X theoretically it39s possible value could range from 0 to infinite feet But the quotsamplequot has only 70 data N7039 so the underlying pdf can only be approximated In the sample the smallest and largest values are shown in bold face39 the sample range isR 8421 63 ft As a firststep estimate we assume that the sample has a uniform pdf lN39 then the sample mean and variance can be calculated from 32 and 35 respectively HS 523 ft 092 1685 ft2 os 1298 ft Sample Histogram Ifthe sample size is large say N gt 50 a histogram can be constructed which may reveal the true distribution character of the sample If constructed properly the histogram may guide us to estimate the pdf of the sample To illustrate consider Example 32 Let us return to the sample data in Example 3l First we group the 70 data points in 7 class intervals numbered from i l7 Class interval i l 2 3 4 5 6 7 Value range 2029 3039 4049 5059 6069 7079 8089 Next we do a tally by recording the number of the data that fall inside each of the class intervals For instance there are 2 data points in the 2029 interval39 ll data points in the 3039 interval 16 in the 4049 interval etc We then define the occurrence frequency by dividing that number by N 70 resulting in the corresponding occurrence frequency 270 1170 1670 etc These results are tallied below Chapter111 Data Sampling 1114 i Interval number 1 2 3 4 5 6 7 Interval limits 2029 3039 4049 5059 6069 7079 8089 data in interval 2 11 16 20 14 6 1 Occurrence frequency 270 1170 1670 2070 1470 670 170 Now we treat the classinterval number i i17 as a new random variable with the real integers 1 to 739 the unit of the integer is A which equals 10 feet in this case The underlying pmf fi is then estimated as fi270 f21170 f31670 f42070 f51470 f667039 f7170 A graphical representation of the above in a bar chart known as the histogram is shown below fX 2070 1570 1070 570 m 10 20 30 40 50 60 70 80 90 Discussion The histogram is the pmf of the discrete random variable representing the class interval i whose value range is 17 We can use 224 and 226 to compute the mean and the variance of the pmf fi and convert them in term of the unit ofx feet hence 5286 ft and F132 ft Note that the results for u and 0 compare fairly well with 5523 ft and Us 1298 ft that were computed earlier in Example 3139 but the histogram reveals a distribution that is nonuniform in fact the histogram suggests a symmetric belllike distribution with the maximum at about x55 feet Clearly to assume a uniform distribution for fx would be quite imprudent In general the histogram can often provide a useful clue as to the mechanism by which the sample is generated In this example the mechanism stems probably from the engineering design of the brake and the targeted designed braking distance is probably around 55 feet or so The histogram can also serve as a basis for tting the sample with a continuous distribution function fx Note The histogram obtained above is based on 7 classintervals with the interval width A10 ft The question is often asked as to why 7 interval What if 5 intervals say with A15 ft are used instead Or what if 15 intervals with A5 ft are used instead Actually the histogram corresponding to A15 ft is shown as follows ChapterIll Data Sampling Ill5 10 20 35 50 65 80 95 The shape of the above looks like a halfsine curve rather than a bellshaped curve39 but it s mean is 530 ft and the standard deviation is 1275 ft which are about the same as before Similarly if A5 ft is used the resulting histogram is shown below fX 157 107 m 10 20 30 40 50 60 70 80 90 Here the shape of the histogram is rather rugged although the mean and standard deviation 5293 ft and 1275 ft respectively are again within good agreement as before The question is given a sample what classinterval should one select in order to obtain a proper histogram The Sturges Formula From the preceding we see that selection of the classintemal width affects the shape of the resulting histogram the in uence is a matter related to the sample size N the value range R and the sample data within the range Yet a proper selection for the classinterval width is essential for obtaining a relevant histogram shape and thus the distribution function that is to be tted In 1926 H A Sturges oiTered an empirical formula for the optimum selection of classinterval size A Chapter111 Data Sampling Ill6 AR 1 33 log10N391 36 Discussion Returning to Example 32 we find from the sample N70 andR 84 2163 ft According to 36 we calculate A m 9 Recall that we have used A10 as the first try which yielded good result Ranking Statistics In many situations only a small sample is available say N lt 20 such a sample would be too small to yield a useful histogram In that case the method of ranking statistics is often employed instead The essence of ranking statistics is to estimate the CDF instead of the pdf for the given sample Of course once the CDF is found the pdf follows The procedures in the ranking statistics are illustrated by the example below Example 33 Prooftests of 14 turbo engines provide the timeto failure data in hours 103 113 72 207 82 97 126 117 139 127 154 127 199 159 Here the sample size is only N14 It is not useful for histogram construction Instead the 14 data points are first ranked in ascending order as follows Rankingi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Timetofailti 72 82 97 103 113 117 126 127 127 139 154 159 199 207 The above ranking of the data may be interpreted by one of the following three assumptions 1 The equal rank assumption Each of the N data points has the same ranking order This implies that the underlying probability of occurrence for each data is the same that is fti1N for all i1N It then follows that the CDF each ranked data is see 223 Fti iN for i1N The above states that the value of the CDF corresponding to the ith ranked data ti is simply iN When paired with ti the following is obtained Timet0failti 72 82 97 103 113 117 126 127 127 139 154 159 199 207 Fi1N 071 143 214 286 357 428 50 571 643 714 786 857 929 10 Since ti are ranked in ascending order the paired data are readily plotted as shown below ChapterIll Data Sampling Ill7 A Fi 10 75 395 Fitted CDF Ft 25 0 l I I I I 25 50 75 100 125 150 175 200 225 t 1118 In the above plot the solid line is the fitted CDF Ft based on the 14 data points Discussion It is said that the equalrank equation is somewhat biased especially when N is small Note from the above data that the cumulative probability is 71 for t g 72 hours while 100 for t207 This implies that there is a 71 chance for some data smaller than 72 while no chance for any data larger than 207 Thus for a small sample the equalrank assumption may overestimate the CDF at the tail end of the population and underestimate the head end of the population 2 The meanrank assumption This assumption allows the possibility that one or more virtual data may be present in between the ith and ilth data Specifically the existing ith data is the mean of it s two neighboring virtual data In particular there is at least one data below the lowest ranked data and there is at least one data above the highest ranked data By a statistical reasoning the details of which are omitted here the value of Fi corresponding to the ith data is computed by FiiN1 for i1N If we use the above formula in the previously example we find Ft72667 instead of 71 and Ft207 9333 instead of 100 These results are deemed more realistic than that found using the equal rank assumption Consequently a better fit for Ft may be realized 3 The medianrank assumption This assumption is similar to the mean rank assumption but each of the existing data is regarded as the median instead of the mean of it s two neighboring virtual data Again by a similar statistical reasoning the CDF for the ith data is given by Ftii03N04 for i1N This approximation gives Fti values lower than that by the meanrank assumption near the tail portion of the Fti curve but higher in the head portion of the curve Note All three representations yield essentially the same results for samples with large size say Ngt50 However for reason of uniformity the meanrank assumption will be used exclusively in all future discussions and in all the homework problems ChapterIII Data Sampling III8 III 3 Parametric Probability Distributions In the preceding section we have introduced several methods for evaluating the distributional character of the underlying probability functions 7 fx or Fx 7 given a sample data for instance the character of fx may be revealed by a histogram or the character of Fx may be revealed by a plot based on ranking statistics But we have yet to t a speci c set of data with a speci c probability function In statistics and probability literature a number of known distribution functions for the pdf or CDF have been employed in various applications A common feature in these lnctions is that their forms are explicit mathematically and each may contain one or more free parameters in particular the properties of these functions eg the mean variance skewness etc are thus expressible in terms of the free parameters This gives a multitude of choices to t a given sample data with one of these functions In this section a number of the wildly used lnctions will be discussed in detail The Normal Distribution The normal distribution also known as the Gaussian distribution is arguably the most wellknown and widely used distribution function in many practices The pdf contains two 2 free parameters and is expressed in the following form 2 fx exp7171 aw 37 1 bxE where a and b are the two free parameters of real value For 37 to be a bona de pdf we require fx 0 for all values of x and it must satisfy the total probability condition 7 C 1 1 xa 2 7 Fox H706 b VZ n exp 77 T 1 dx 7 1 38 It is easy to check that fx 0 is satis ed by 37 and the proof of 38 can be accomplished by the method of residues from the theory of complex variables the reader is here challenged to complete the proof The mean and variance of the pdf 37 are found from using 230 and 231 respectively MI 73 exp i l2 Lia2 ldx 39 0c x 2 1 2 39 E 310 627J3970cbX2n exp 2 b dx Chapter111 Data Sampling 1119 Integration of 39 and 310 requires formula derived from the method of residues Here we shall omit the details of the integration and state only the results ua 6b Consequently 37 is in fact expressible explicitly in terms of its mean and standard deviation x exp 7171351 ocltxltoc 311 1 NZ n A graphical display of 311 as a function of x is shown below Note that fx is always symmetric with respect to its mean value LL its overall shape resembles a belllike curve whose spread or band width varies with the standard deviation 7 The gure shown above illustrates two normal functions having the same u but each has a different 6 value Note that a large 6 value yields a large spread of fx and vice versa The Standardized Normal Distribution One short coming of the normal distribution function is that the pdf 311 cannot be integrated explicitly to obtain a closeform expression for the CDF Fx the alternative is to integrate 311 numerically To do so however the numerical values of u and 6 are needed In situations where u and 6 are not at all known the difficulty can be alleviated by a mathematical transformation of the variable x to the variable 2 zx u 312 It then follows that dx 6 dz 313 From the above 311 is transformed to a function of z pz satisfying the following equality Chapter1H Data Sampling Ill10 I X 2x FX fX dX dz J pz dz 314 70c 70c The above invokes the fact that the cumulative probability FX is the same as z provided that X and z are onetoone related by 312 Differentiation of the integral expressions in 314 yields fX pZdZdX pZl6 315 Combining the above with 311 and 312 we obtain 1 1 PZ N TZZl oltltzltoc 316 Note that pz does not explicitly contain any free parameters so is its CDF z 1 1 2 Z TeXP l z dz 317 Lodzn 2 Consequently a numerical evaluation of z in 317 can be carried out through the value range of z without the need of having the speci c values of u and 6 The function pz in 3 16 or the function z in 317 is known as the standardized normal distribution By standardization it means that all normal functions in the regular form 311 can be transformed into 316 and the functional shape of pz in 316 is a belllike curve as shown below 92 ChapterHI Data Sampling Illll Note that the normalized pdf is symmetric about 2 0 with HZ 0 and oz l the maximum occurs at z 0 with the numerical value of q0l211M2 m 04 Note also pz decays rapidly to approaching zero for z gt 3 or z lt 3 As for the CDF Z it is obtained by integrating pz numerically over the zvariable say from 00 to z Note that z rises from 0 to l as 2 increases from oo to 00 see the gure shown below practically z is nearly 0 for z S 3 and nearly 1 for z Z 3 due to symmetry of pz z 05 at z 0 0132 A 39UI39cxkl39oo39on A Conversion Table for gzl As mentioned the value of z can be obtained by numerically integrating 317 if the value of z is given A comprehensive table which lists values of z in the domain of 750ltzlt50 is included in Appendix IIIA at the end of this chapter Note that the value of z varies with the increments of 001 and for each 2 value four signi cant digits are retained in value of z This table can also be used to determine the inverse of z Speci cally if the value of z is given say zF then the inverse of z is de ned as z 391F and the latter can be found from using the same table in Appendix IIIA More extensive mathematical tables list values for z in the range of 6 S 2 S 6 and with up to 15 signi cant digits some electronic calculators have builtin statistical functions with options for z as well as the inverse of z Use of the conversion table in Appendix IIIA will be illustrated in the examples below Example 34 Let the timeofwearout of a cutting knife be normally distributed with the mean and standard deviation of 28 hrs and 06 hrs respectively Determine Chapter111 Data Sampling 11112 a The probability of a knife wearing out within 15 hrs Here the random variable is the timeofwearout t39 The CDF is given a normal distribution with ILL 28 hrs and 0 06 hrs In the standardized form Ft ltIgtz with ztuo Now for t15 hrs ztuo157280672167 This means that FtS 15ltIgtzS216739 and we are interested in finding the value ofFt S 15 or lt1gtz S 2167 By using the table in Appendix IllA we locate 2 values in the first column at z21 and then go horizontally to at z21639 here we are into the Ffield and find F001539 at z21639 similarly we find F001500 at z217 By means of a linear interpolation we estimate F001512 at 2 2167 Thus we have FtS15ltIgtzS 2167 001512 Thus the probability of the cutting knife to wearout within 15 hrs is 1512 b Time to replace the cutting knife when there is 75 probability of wearout Here ltIgtzFt075 Again from using Appendix IllA we scan the Ffield to find a value closest to 075 We see that when 2067 F0748639 and when 2068 F07517 By a linear interpolation we obtain 2 06745 for z075 From ztuo t72806 06745 we find t32047 hrs Thus the time to replace the knife is 32047 hrs Example 35 The tensile strength of a metal wire is known to be normally distributed with p 1000 lbs and 040 lbs In a proof test 100 wires are loaded up to 950 lbs in tension39 estimate a How many wires will pass the test For x950 we have 2 xuo 950100040 12539 from Appendix IllA we find the value of ltIgt12501056 Thus the cumulative failure probability up to 950 lbs of load is F9501056 Thus about 11 wires will fail the test or 89 wires will pass the test b If the first 15 weakest wires are to be screened out at what load should the wires be prooftested Here the cumulative failure probability is 15 out of 10039 or z015 From Appendix IllA we find the corresponding z1037 Thus the proof load x should be x zo H 1037401000 9585 lbs Central Population With the standardized normal functionq3z one can readily see the following relationship z z 1 318 From the graphical display of the pdf pz on the next page one can see that the area under the curve from iz to z is equal to z z owing to symmetry or by means of 3 18 this area is also equal to 12 z This area is commonly called the central population or the Yield denoted as Y 1 2d3z 319 Check that for 21 Y683 for 22 Y9545 and for 23 Y9973 etc The meaning of ChapterIll Data Sampling Illl3 the yield will be lrther discussed in Chapter VI A PZ Central population elm2 1qgtz A LA V Example 36 The diameter of a bearing ball is normally distributed with u10mm and 002mm What is the central population whose diameter is in between 99mm and 101mm Solution From using zxuo we find 2 10l1000205 or z05 From Appendix IllA we find lt1gt 0503085 Hence the yield Yl2ltIgtO50383 383 of the balls will have the diameter in between 99mm and 101mm Fitting A Sample to the Normal Distribution Suppose a sample of size N is initially given and we wish to t the sample with a proper distribution function In general there are two alternate routes to followed depending the size N of the given sample rank sample data Xilt Xi1 i1N if Nlt50 experience suggests choice 00mm a Proper ofdistributi on h1stogram use ranking method to fit the FX ofa known distribution h1stogram shape suggests choice of distributi on Chapter111 Data Sampling 11114 From the above ow chart we see that for a sample of size Ngt50 it is possible to construct a histogram whose distribution characteristics can then suggest a proper choice of the distribution function for the sample But when Nlt50 it is impractical to do a histogram then experience often plays a role in the choice of a distribution function to be fitted with the sample In either case the general guideline is that a the tted anction should preserve the physical properties of the sample with minimum error and b the tted anction is mathematically convenient to apply in subsequent reliability and related analyses Here we use one specific example to outline the steps in fitting a given sample to the normal distribution function Example 37 Consider the sample in Example 33 where the timestofailure of 14 turbo engines are recorded Here the sample is not large enough for a histogram construction We first rank the sample data in ascending order Rankingi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Timetofailti 72 82 97 103 113 117 126 127 127 139 154 159 199 207 Before fitting the sample to the normal distribution function let us just compute the sample mean and variance using 32 and 35 us 13014 hrs 052 15512 053938 hrs Normal Probability Plot To fit the sample formally to the normal distribution we apply the mean rank formula to approximate the cumulative probability function Fti Fti iN1 i 114 Since Fti can be put in the standardized form ltIgtzi with zitiuo it follows that zi lt1gt 1iN1tics uG The above is a linear equation between ltIgt 1iN1 and tio However still with u and 0 are so far unknown But for each i1N we know zi Fti iN139 and with the aid of Appendix IllA we can determine the inverse zi 1iOT1 for i114 ltIgtzi Fti 07 13 20 27 33 40 47 53 60 67 73 80 87 93 ltIgt 1iN1 15 11184 62 44 25 07 08 26 43 62 85 111 15 We then pair the values of zi 1iN1 with ti and plot each pair in the 21 verses ti frame as shown on the next page the 14 open circles are from the paired data ChapterIll Data Sampling Ill15 A straight line labeled yat b is drawn through the paired data points Here y corresponds to 21 and t corresponds to ti and it is readily identified that a1o b po Note In the plot below we have added a vertical axis labeled Ft whose scale corresponds to ltIgtz39 the fitted straight line is not in the Ft verses t frame F00 ll 391iN1 999 3 yatb 977 2 O 841 1 gt D D 0 5 0 gt o D O 159 1 gt 10 lt A lt gtlt gt 023 2 6 6 0014 3 7 W t 50 8ymo 150 200 250 hrs 93 111325 172 To nd u we note that when tu 20 From Appendix IllA ltIgt00539 or F05 Hence a horizontal line is drawn from F05 or from 20 to intersect the straight line yatb39 from the intersect a vertical line is drawn downward to intersect the taxis at tn 1325 hrs hence the value for p p 1325 hrs To find 0 We note that when tpro zil39 hence from Appendix IllA we find 1371 Fpio 0159 and ltIgt1 Fpo 0841 Thus a horizontal line is drawn from FO16 or zl to intersect the straight line yatb39 then from the intersect point a vertical line is drawn downward to intersect the taxis at about 93 hrs The distance from t93 t0 the distribution mean is equal to 0 namely on 1325 93 395 hrs ChapterIII Data Sampling III16 Alternatively a horizontal line from F084 or 22H is drawn to intersect the straight line39 from there a vertical line is drawn downward to intersect the taxis at about 172 hrs The distance from tl72 to the distribution mean 1325 is equal to 039 or 017271325395 hrs Discussions In this example the values for p and 0 1325 and 395 found in fitting the sample to the normal function appear to be fairly close to the sample mean 513014 hrs and standard deviation os 3938 hrs obtained earlier The straight line in the graph can provide answers to many related questions For instance within 80 hours of operation there is a 10 failure probability for the engine We find this result from drawing a vertical line upward from t80 to intersect the straight line and then a horizontal line to intersect the Faxis at F m 01 In the graph the vertical axes Ft and 171 are put sidebyside to show their transformation relationships only ID 1 is linearly related to t39 and both are in arithmetic scale Ft is not linear with t39 and F is in the transformed scale through the use of Appendix IIIA With F axis in the transformed scale the sample data can be paired with Fti versus ti directly without having to compute 13 1 This will save time and avoid mistakes in computation Appendix IIIB provides such a plotting paper that can be copied off for ready use In the probability plot the straight line yatb was drawn through the 14 paired points by eye balling This approach is highly subjective and it can result in error A more rigorous way to determine the straight line is by the method of linear regression or the least square The method is analytical39 no plotting is necessary This is explained in detail in the next section The Least Square Method In the probability plot the values of yi 1iN1 are paired with ti In a perfect situation all the paired points yi ti should fall onto the straight line yatb if they are not an error is present between the data and the tted line For each paired point the error is expressed by yiyti Thus we de ne the squared error function as s 2 yr Yti2 l 2yi ati b2N Z sums over i1N 320 The idea here is to nd the line y atb such that S is the minimum Hence we set BSaaBS6b0 321 Upon carrying out the minimization process we nd x x aH byax 322 X X39 x where Chapter111 Data Sampling 11117 1 x Ztl y 2y1 1 1 XXN Ztitl W Zylyl 1 Ky itiyl The minimum of S is then given by 3min1r2W 1W 323 where W 7lty 2 r2 324 E y39 The quantity r2 is known as the correlation factor whose value lies in between 0 and 1 When r2 gt 1 the paired data points is said to correlate well with a straight line t when r2 gt 0 the correlation is not good at all Since the straight line represents graphically the tted normal function the constants a and b are related to u and 6 as u ba 61a Example 38 We now return to Example 3739 and treat the sample data there by the method of least square From the tabulated data in that example we first calculate the following 1 1 X 2ti13014 yFZyi0 1 1 xx 2titi 1837757 yy FZyiyi069 1 t Xy N2 iyi30915 Substitute the above into 322 we obtain a002145 and b27917 Consequently we obtain Gla 4662 and uba13015 The correlation factor is computed from 324 r2 096 The correlation is rather good as we can see from the plot Discussion We list the estimated values for the mean and the standard deviation using the various methods 11 13014 hrs 0 3998 hrs 11 1325 hrs 0 395 hrs 11 13015 hrs 0 4662 hrs By the lststep sample calculation By eyeballing from plot By the least square fit ChapterIII Data Sampling III18 We see that the mean is rather stable regardless the method used but the standard deviation can vary considerably Clearly the least square method provides the best results As the latter requires a considerable computation a computer routine can be easily programmed however Sampling Error When a sample of nite size is taken randomly from a population of in nite size there is uncertainty that the sample may not have the same statistical characteristics as that in the population But in practice we often use the sample to represent the population and to estimate the underlying probability distribution function for the population thus question often arises as to how much con dence do we have in the estimated function To answer this question properly however it requires a considerable development in the theories of statistics Here we will discuss only brie y some of the basic notions in sampling sampling error estimate or con dence levels in sampling Theoretical Error Suppose a sample of nite size N is taken randomly from a population and it is tted to say a normal function thus the sample mean and standard deviation us and GS are obtained Owing to the size of the sample both us and GS are likely to be different from the true u and 6 of the population However us and GS will approach the true u and 6 under two extreme conditions 1 when N is large or N gt co and 2 if samples of size N are repeatedly taken from the population In the latter case the estimated sample means and standard deviations are themselves random variables for instance the sample means us may be described by the pdf fps According to the central limit theorem in statistics fps tend to be normally distributed if the number of samples is large and moreover the mean of fps tends to be the mean of the true mean of the population u while the variance of fus tends to be equal to 62N Clearly in the limit that N gt oc the variance of fus gt 0 The term 6N12 is a theoretical measure of how much the sample us deviates from the true u of the population The dif culty here is of course that the true u and 6 of the population is in fact unknown Estimated Error When the true 6 of the population is unknown one can estimate the sampling error by using GS of the sample in place of 6 In that case the meaning of GSN12 may be best explained by the following example Take Example 37 The sample size is Nl4 and the computed by the least square method sample mean psl3015 hrs and standard deviation 054662 hrs The estimated sampling error is then oSN1Z 466214m 1246 hrs This error represents the range of deviation of the estimated sample mean In other words if samples of similar size are taken repeatedly from the population the means from all the samples will fall inside the range of 13015i1246 hrs Thus by extrapolation the true mean it of the population also falls inside this range ChapterIII Data Sampling IIIl9 Note that the smaller the sample size N the larger the estimated sample error In the above example the sample error is rather large This is due to the sample size Nl 4 which is rather small Theoretical Con dence Level The re ned way to assess sampling error is to establish a con dence level Here we return to the premise that if multiple samples are taken from a population the sample mean us is a random variable and it s distribution is normallike according to the central limit theorem So we normalize the random variable us as in 312 ZsHs39H5Nml 325 In the above u and 6 are still the true mean and standard deviation of the population both unknown Nonetheless the standardized normal pdf qzs and CDF zs can be evaluated without knowing the values of u and 6 In particular the mean of 25 is at zero and the standard deviation is 1 In fact the pdf qzs is shown below ll ltpzs Central population 1 0c ZOC Zoe Now consider the central population between izoc where at is a small number less than 1 And at represents the probability that zS falls outside the range of izoc Conversely 1oc represents the probability that 25 falls inside the range of izoc Hence 0c is called the risk factor that zS falls outside izoc39 and 1oc is called the con dence level that zS falls inside izoc The range between izoc is called the con dence interval associated with ac Note that if at is speci ed the value of zaq3 1oc2 can be found via Appendix IIIA and there is no need to specify the true u and 6 of the population Example For a 80 confidence level we set loc0839 so oc02 The confidence level is found via ChapterIll Data Sampling Ill20 Appendix IIIA Z02lt1gt 1oc2 lt1gt 101128 or i Z02128 For a 99 confidence level we set o001 from Appendix IllA we find lt1gt 1oc2 258 the confidence interval is Z001258 We return to 325 to obtain the confidence level for the sample mean us Hi it i ZaGNm Estimated Con dence Level Since the exact pdf of the population is unknown we can still obtain a pair of estimated con dence intervals by substituting us and GS of the sample in place of the true u and 6 of the population pi s i z0c rsNV2 326 The above is read for a loc con dence level the mean of the population falls inside the estimated interval Hi The con dence interval for the variance 62 of the population can be similarly established and a pair of estimated con dence interval for 6 can be obtained by substituting us and 65 from the sample in place of the true u and 6 of the population Here we omit the mathematical details in the derivation and present only the nal result at as r z cs2N1m 327 Example 39 Again take Example 37 Let the confidence level be oc02 80 confidence then hi S i zOc 65m Z 13015 i 128 466214m 13015 r1595 hrs at US i zOc oS2Nlm 4662 i 128 466226 2 4662 r 1170 hrs Similarly for 99 confidence 06001 we have hi 13015 i 258 466214m 13015 r 3215 hrs at 4662 i 258 466226m 4662 r 2359 hrs Discussion For higher confidence the ranges of pi and or are wider III 4 Other Parametric Distribution Functions In many engineering applications the normal distribution is not always the best choice for tting a given sample For example the tensile strength distribution of some materials can be skewed to the right of its mean the timetofailure distribution for fatigue failures in machine parts can span from a few thousands hours to millions of hours resulting in a very skewed pdf Of course if the sample size is suf ciently large one can always construct a proper histogram and nd a clue as to the skewness in the sample s distributional character and to t a skewed distribution with the r 330 In turn it leads to 1 1 1 lntP c 2 gt f c1t 61V t exp776 331 ChapterHI Data Sampling Ill22 At this point we formerly replace HI and 6 by two new parameters as follows pt lnt0 and 65 030 332 Consequently the pdf gt of T in 331 is expressed as 1 gt 007 Tlexp 71423 2 333 Not that the gt is not a normal lnction It s mean and variance can be determined following 230 and 231 omitting the details of the derivation these are given by 7 2 27 2 032 032 334 utitoexp nOz stito e 0e 0l The general characteristics of f39 which is normal and gt which is not are illustrated by the graphs shown below fI normal distribution K I 6T GT oc g T I l gt oc 395 5 gt nonnormal K 3 39 39 gt an 0 l to t 4 D c 4 Mt t Note that gt is always skewed to the left of the mean while f39c is symmetric about it s mean as it is normal The physical difference between them is that the taXis in gt is in real time arithmetic scale while the EaXlS in f39c is in logarithmic time scale Note for instance ChapterIll Data Sampling Ill23 the location for the mean of f39c is at ut which corresponds to t0 in the realtime taxis see 332 the mean of gt is located at toexpn022 according to 334 The lognormal function gt is characterized by the two free parameters t0 and no In essence to shi s the distribution along the taxis while the value of no dictates the distribution s scatter band or the overall shape It can be shown numerically that 1 when no 2 1 gt approaches an exponentiallike function 2 when no gt0 gt then becomes a normallike in particular gt coincides with f39 with their respective mean and the variance also coinciding and 3 in the range of 0lt m0lt 1 gt is always a leftskewed lnction These features are schematically shown in the sketch below gt no u 01 03 1 H 39 L 7 Discussion The lognormal function offers a greater freedom than the normal function in fitting a set of data whose distribution may or may not be symmetric The lognorm al function can be exponential leftskewed or normally distributed depending only on the parameter 0J0 Use of the lognormal function is as simple as that of the normal function the extra step is to transform the variable t to Ilnt As f E is normal it can then be expressed in the standardized form in order to facilitate numerical computations as it has been done before The Standardized Log Normal Function The lognormal function gt in 333 cannot be integrated explicitly to obtain a closed form for the CDF Gt But through tlnt gt is related to f39 in 328 which is normal Hence the CDF of f39 F39c can be expressed in the standardized normal form via the variable z z39c Jt6c lntt0n0 335 Note that the pdf pz has the standardized normal form in 316 and its CDF z can be evaluated numerically as before In short the value of Gt equals to that of z and the value of z is rendered through the use of 335 in conjunction with Appendix IIIA Example 310 Fatigue failure of a rotating shaft is fitted with the lognormal function gt as given in 333 The characterizing parameters are obtained as to 5000 hrs and mo 02 Chapter111 Data Sampling 11124 Observations a The random variable here is t timetofailure in hours b We note that 0ltOJOlt139 the pdf should be slightly leftskewed c The standardized normal CDF is ltIgtz with zlnttO0JO We may then compute the following quantities 1 The expected failure time the meantimeto failure MTTF toexpoJOZ2 5101 hrs 2 The variance of gt otZ tOZexpoJ02 expm021 10169x106 hrs2 ot 1030 hrs 3 The design time td say for 1 probability of failure or Ftd 001 For ltIgtz001 we find z232 from Appendix IllA from lntdtO0JO232 so td 3144 hrs 3 1f the inspection period is set at 3000 hrs what of failure is expected before the inspection 7 We compute z ln3000500002 2554 from Appendix IllA ltIgt2554 000539 Thus we expect a 0539 failure probability before inspection Fitting A Sample to Log Normal Distribution Given the sample data xi i1N let it be tted with the lognormal function in the form of 333 11 1 1m 2 fx 0302 XeXP T Here the quantities to be determined are the two free parameters x0 and 030 Using 335 we transform the variable x to z by z lnx lnx0oa0 Then the pdf pz is a standardized normal and its CDF is z which is equal to the CDE of fx Fx The inverse z is 1F z lnx lnx0oa0 336 We see that 1F is a linear function of lnx Hence in tting the sample xi i1N we follow the same procedures outlined earlier for tting the normal function Speci cally a Rank xi in ascending order b Compute lnxi c Compute Fxi i N 1 d Find 1iN1 Via Appendix IIIA e Plot 1iN1 versus lnxi f Obtain x0 and 030 from the plot Example 311 The failure times xi in days for 20 heating devices are recorded and are listed in ChapterIll Data Sampling Ill25 ascending order from 1 to 20 The corresponding values for lnxi Fxi iNl and lt1gt39liNl are then determined accordingly these are summarized in the table below 1 FxiiN1 xi lnxi lt1gt11N1 1 00476 26 09555 16684 2 00952 32 11632 13092 3 01429 34 12238 10676 4 01905 39 13610 0 8761 5 02381 56 17228 07124 6 02857 71 19601 0 5659 7 03333 84 21282 04307 8 03810 88 21748 03030 9 04286 89 21861 01800 10 04762 95 22513 00597 11 05238 98 22824 00597 12 05714 113 24248 01800 13 06190 118 24681 03030 14 06667 119 24765 04307 15 07143 123 25096 05659 16 07619 127 25416 07124 17 08095 160 27726 08761 18 08571 219 30865 10676 19 09048 224 31091 13092 20 09524 242 31864 16684 Plotting of the above data can be done in two different ways 1 pair ltIgt 1iNl versus lnxi and 2 pair Fxi versus lnxi Of course the scale for ltIgt 1iNl is arithmetic while the scale for Fxi is not The following plot displays both axes in the plot ChapterIll Data Sampling Ill26 A Fi A q3391iN1 999 3 977 2 814 1 05 0 159 1 023 2 0014 3 0 mean of lnX lnXO In the plot a straight line is then drawn through the paired points This line denoted by ya b where y 1F lnx in the plot represents the fitted CDF function ltIgtz graphically The free parameters of gt however are determined from the plot as follows 1 The mean of ltIgtz is found by drawing a horizontal line from 1FO to intersect the straight line39 and from the intersect point drawn a line vertically downward to intersect the lnxaxis in this case it is at about lnx2239 it thus yields the parameter x0 9025 days Alternatively we can draw a horizontal line from F0539 this is the same as that from 1710 0 2 The slope of the straight line ya b is then found in the 1407 versus lnx frame hence al36 According to 336 al000 So m0 0735 Here 000 is greater than 05 but less than 139 so the pdf gt should be considerably leftskewed As for the mean and the variance of the gt they are computed by px0exp00022ll82 days and by 02 xozexp0002exp0002l 100l5 day The standard deviation is thus 010 days Discussion The constants in ya b can of course be found by using the leastsquare method discussed earlier In that case we set yilt1gt 1Fi and iilnxi For this example the least square fit yields the following results a 135 b 2969 and r2 0955 ChapterIII Data Sampling III27 Consequently we find x0expba9018 days and 0J0la074 In addition the correlation factor r2 0955 indicating a relatively good fit Note A ready made lognormal plotting paper is provided in Appendix IIIC39 both Fxi and 171 F axes are included With this form the plot can be made in the F versus lnx frame without the need of computing ltIgt 1F This will save a considerable amount of time Weibull Distribution Function The Weibull distribution anction is due to Walodie Weibull who in 1931 established a probability distribution function for the strength of textile yarns Similar to the lognormal function it is very versatile in tting a wide range of engineering data including the skewed and normal like distributions But unlike the normal or the lognormal functions the Weibull function can be expressed in close forms for the pdf and the CDF and there is no need to perform numerical integration This unique attribute provides a considerable mathematical ease in probability and0r reliability analysis of complex systems The random variable X is said to be a Weibull distribution if its pdf is a 2parameter anction of the form fx m9x9m391expx9m OS X lt oc 337 Here m is called the shape parameter and 9 the scale parameter both have values greater than The characteristics of the Weibull pdf in 337 depend on the values of m and 9 In general the role of m is similar to that of 6 in the normal distribution or no in the lognormal distribution it determines the shape of fx The role of 9 is similar to that of u in the normal or to in the log normal distributions it shifts the fx along the xaxis In particular it can be readily shown that 1 The Weibull becomes an exponential function when m1 2 It approaches the normal anction when mm 4 3 It tends to be leftskewed when 1ltmlt4 and 4 It tends to be rightskewed when mgt5 The above features of the Weibull function are illustrated graphically below ChapterIII Data Sampling III28 Upon integrating 337 the CDF of the Weibull function takes the simple close form FX l eXpXGm O S X lt 0C 338 The mean and the variance of the Weibull function are found from using 230 and 231 n 9F11m 52 92r12m F11m2 339 Here FC is the Gamma function de ned by the integral FC 139v C lexpy dy 340 In 340 C is real and is always greater than 1 C is not necessarily an integer however About the Gamma Function Given the value of Q which is always greater than 1 the integral in 340 can only be integrated numerically Numerical tables listing values of FQ as a function of Q are available in many mathematical handbooks some calculators have such a function as well The graph shown on the next page is obtained by numerically computing the Gamma function FQ in the range of 1 S Q3 2 with the increment of 01 ChapterIll Data Sampling Ill29 NC 10 11 12 13 14 15 16 17 18 19 Q Use of the Gamma function graph shown above The graph shown above is convenient for a quick estimation of the value of Hg given the value of Q For example from the graph we find l l5m 0886 For C gt 2 the following recurrence relation of the Gamma function is useful T C C UUC U For example the value for F45 may then be computed as follows r45 45 1r35 45 1351r25 45 1351251r15 35 x25 x 15 x0886 1163 Discussion The Weibull function can fit data in a variety of forms exponential normal leftskewed and rightskewed Note The parameter 0 is neither the mean nor the median of the distribution a case in point when xe Fe0632 Example 312 Suppose that the tensile strength distribution of a cotton yarn is described by a Weibull function with m5 and 91 Kg Then from 337 the pdf is given by fx 5x4expx5 0 S x lt 0c Then the mean and variance of fx are calculated by using 339 along with the Gamma function and its recurrence formula 340 p 0r115 0918 Kg 02 02r125 7r1152 00443 Kg2 The standard deviation is Chapter111 Data Sampling 11130 0 021 Kg The graphical form of fx is plotted and shown as follows fX 20 16 12 08 04 Discussions As it can be seen from the graph above the mode maximum of fx occurs at x61 while the mean is at xp0918 The distribution is slightly skewed to the right of the mean not quite a normal distribution At xe the CDF value is F60632 which is greater than 05 Clearly the Weibull parameter 9 does not correspond to the mean the mode or the median of the distribution Fitting Samples to Weibull Distribution Let the sample Xi i1N be tted with the Weibull distribution function in the form of 337 then it is necessary to determine the parameters m and 9 from the sample The tting procedure is similar to that for the lognormal distribution function discussed previously From 338 we rst de ne the reliability anction RX as RX 1 FX expX9m 341 We then take the logarithmic of the inverse of 341 twice and obtain ln1n1R m1nX mn9 342 We see that ln1n1R is a linear function of lnX We then return to the sample and do the following 1 rank the values of the data Xi i1N in ascending order 2 approximate the CDF FiFXi by the meanrank formula ie Fi iN1 ChapterIll Data Sampling 11131 3 calculate for each data point the value of lnln llFi or lnlnlRi 4 plot lnlnlRi versus lnXi either graphically or by means of linear regression If carried out graphically the paired data points provide a linear t in the lnlnlRi versus lnxi frame and the slope of the line equals to m To determine 9 we observe from 342 that at lnlnlR0 lnxln9 hence a horizontal line is drawn from lnlnlR0 to intersect the straight line and a vertical line is drawn downward to intersect lnx axis at lnxln9 The above tting procedure is illustrated by the example worked out below Example 313 Consider the sample of 20 data points given in Example 3ll39 and the sample is to be fitted with the Weibull function Following the procedures outlined above the sample and the various related quantities are then computed and tabulated as follows Rank i Fii1l1 RflFi Xi lnxi lnlnlRQ 1 00476 09534 26 09555 30202 2 00952 09048 32 11632 23018 3 01429 08571 34 12238 18698 4 01905 08095 39 13610 15544 5 02381 07619 56 17228 13022 6 02857 07413 71 19601 10892 7 03333 06667 84 21282 09027 8 03810 06190 88 21748 07349 9 04286 05714 89 21861 05805 10 04762 05238 95 22513 04360 11 05238 04762 98 22824 02985 12 05714 04286 113 24248 01657 13 06190 03810 118 24681 00355 14 06667 03333 119 24765 00940 15 07143 02857 123 25096 02254 16 07619 02381 127 25416 03612 17 08095 01905 160 27726 05057 18 08571 01429 219 30865 06657 19 09048 00952 224 31091 08550 20 09524 00476 242 31864 11133 We note that the fitting must be conducted in the lnlnlRQ versus lnxi frame ie use the paired data listed in the last two columns of the table However note also the onetoone relationship between Fi and lnlnlRil39 the plot can in fact be conducted in the Fi versus lnxi frame as well In particular at lnlnlRi0 Fi 0632 This F value corresponds to lnxlne The plot shown on the next page is done in both the lnlnlRQ versus lnxi frame and the Fi versus lnxi frame In the plot however the straight line ya b is drawn through the paired points by eye balling here y lnlnlR and g lnxi ChapterIll Data Sampling Ill32 The slope of the line is found to be al65 which equal to the parameter m The parameter 0 is found by drawing a horizontal line from y0 to intersect the straight line and then a vertical line downward to intersect the axis at lnx24839 since lnxlne we find 91194 days A 1 F 1 iln 1n lR 999 2 y ya b 934 1 632 0 gt 308 1 127 2 048 3 018 4 1nX gt 0 1 2 1110 3 4 The mean and variance of the distribution are calculated from 338 with the aid of the Gamma function chart displayed on p Ill28 Thus H eFlll65 ll94x089 1063 days and 02 02r12165r111652 4432 The standard deviation is o 665 days Discussions In plotting the Weibull distribution we note that the parameters In and 0 must be determined in the lnlnlR versus lnx frame both axes are in arithmetic scale only then the slope of the fitted straight line is the parameter m The Faxis is added alongside the lnlnlR axis only to provide a timesaving alternative that the fitting can also be conducted in the F versus lnx frame without having to compute the values for lnlnlRD A Weibull plotting paper is provided in Appendix IllD with both the F and lnlnlR ordinates A ain one must be aware of that the slope of the fitted straight line must be determined in the lnlnlR vs lnx frame By the Method of Linear Regression A better fit may be obtained if the least square fit procedure is used to determine the parameters In and 6 In that case we set yi lnlnlRD and ilnxi and the fitted line is denoted as ya b39 following the detailed steps outlined in Example 37 the values of the constants a and b are then obtained Then ChapterIll Data Sampling Ill33 ma and ln0ba For the above example the leastsquare method yields ml66 and 01236 days The mean and the standard deviation are respectively 1 112 days and F689 days In addition the correlation factor 39 2 is r 0964 It is noted that the Weibull fit is slightly better than the lognormal fit see Example 311 the correlation factor of the former is 0964 for while 0955 for the latter The Extreme Value Distribution Functions In addition to the normal lognormal and Weibull distribution functions discussed previously many distribution functions of different forms are in use for engineering reliability problems For instance for aws in materials or noises in machines that occur in random and in large numbers the Extreme Value Distributions have been very use ll in practice The pdf39s of the maximum and minimumvalue distributions are in the general form fx 16 eiXu exp eiXu oc lt x lt oc 343 Here both functions contain the same two parameters u and D The maximum value function takes the negative of the ambiguous sign in the exponential power and the corresponding CDF is obtained by integrating 343 Fx exp eXu olt lt x lt cx 344 The minimum value function takes the positive of the ambiguous sign and the corresponding CDF is obtained as 1 eXp exu oc lt X lt X The distribution mean and variance of the respective anctions are expressible in terms of the parameters u and z p u i 0577226 62 n2 26 346 The maximumvalue anction is left skewed it s mean is located at pa1057722 the mode occurs at xmode u resulting in Fu 03679 The minimumvalue anction is right skewed it s mean located at uu057722 but the mode occurs also at xmodeu resulting in Fu06321 Both distributions have the same standard deviation given in 346 and their respective skewness coefficients are ski1 1396 The skewness is independent of the values of u or 9 Note that the pdf and CDF of the extremevalue anctions are expressed in explicit forms there is no need to use any numerical integration scheme or any tabulated tables to evaluate the values of the CDF only if the parameters u and D are given ChapterIll Data Sampling Ill34 The respective pdf39s are graphically shown blow in the frame of fx versus xu fx Xu 00 4 3 2 0 9 2G 3G 4G Discussions The maximumvalue distribution is commonly referred to as the Gumbel distribution it is also known as the double exponential distribution The latter name is due to the fact that the CDF in 344 can be expressed explicitly in terms of a single variable mxu resulting in a double exponential function as F0J expl eXIX03 The minimumvalue distribution in 345 can be similarly expressed in the same double exponential form as shown above by letting 03 xu In addition if x be replaced by lnx u by 1116 and by lm the CDF of the minimumvalue function in 345 reduces to Fxl exp xem which is the form of the Weibull function Thus the minimum value function is just another form of the Weibull function Also Note The range of the random variable X in the minimumvalue distribution is focltxltoo while that of the transformed random variable y in the Weibull form is 0ltyltoo Example 314 During aircraft landing the maximum tensile force in lbs in a key fastener in the landing gear is described by the maximumvalue function with u8000 lbs and 61500 lbs Let us examine the following issues a The mean and standard deviation of the distribution function using 346 wsooo 057722x1500 88661bs 01t2150026 1924 lbs ChapterIII Data Sampling III35 b Suppose that the maximum force distribution in the fastener during each landing is the same as that during the previous landing then for N successive landings the maximum force distribution FNx in the fastener is given by FNx FxN exp e39x39u N exp e39X39Tl In the above expression 7 u 8 lnN Discussion We see that the maximum force distribution in the fastener in N successive landings is still an maximumvalue function it is with the same but with u replaced by nu lnN ie the parameter 7 depends on and increases with N For example if N10000 landings then n80001500lnl 000021815 lbs Thus during 10000 landing the maximum force distribution function has the same functional form but with the parameter 8 still equals to 1500 lbs while the parameter u increases from 8000 lbs to 21815 lbs Furthermore the mean of the latter distribution is uN22673 lbs and the standard deviation UN is the same as that of the single landing 1924 lbs ie the shape of distribution remains unchanged with The fact that the mean maximum force increases with lnN implies that in repeated applications there is a greater chance for a higher tensile force to occur the larger the N the more so A Related Situation In a unit volume of material the detectable size of flaws can be described by the maximumvalue function it implies that the fact that the size distribution of the detectable flaws scales with the material volume 111 5 The Weakest Link Model Involving Extreme Value Functions In the previous section we see that repeated applications e g in N landings of the maximum Value distribution result in another maximumvalue distribution ie the shape parameter 9 is preserved regardless of the number of applications but the mode parameter 11 and hence also the mean uN will shi t0 the right as N increases Similarly repeated applications of the minimumvalue distribution will result in another minimumvalue distribution whose shape parameter 9 is preserved but the mode and the mean shi t0 the left as N increases As the minimumvalue distribution is just another form of the Weibull distribution the same characteristics are observed with the Weibull distribution as well The Weibull weakest link model refers to a chain of N links as shown below xlt w4vv w w gt x Suppose each link would fail by the applied tensile force x and x is describable by the CDF Fx ChapterIII Data Sampling III36 Then the reliability of the chain of N links is RNX 1 FXlN 347 Or the failure probability of the chain is FNx l l FxN 348 Let Fx be the CDF of the Weibull function Fx l expx9m It follows that FNx 1 expx9Nm 349 where 9N eN1m 350 Thus the failure distribution of the chain FNx is also a Weibull function with the same shape parameter m as that of the individual link only the scale parameter is now 9N 9N1m The latter decreases with lN1m So N shifts the scale parameter to the left ie It reduces the mean of fx Discussions Though the weakestlink model expressed in 348 can accommodate Fx of the individual link having any distribution form only the Weibull function would preserve the distribution shape parameter for a chain of Nlinks In many engineering situations the Weibull distribution which is very versatile itself is often the choice for reliability applications as it is demonstrated in the above example As for the weakestlink model it finds application in many engineering problems Examples include fibers of a longer length are weaker in tension than those of a short length glass rods of a larger diameter are weaker under bending andor torsion than those of a smaller diameter N electrical circuit arranged in series would have a shorter life in service if N is large longer life if N is small39 etc Somewhat more complicated situation is when Nlinks arranged InParallel In such case the resulting distribution function would not retain the shape of the links distribution even if the latter is a Weibull function In general the distributional shape for a system of Nparallel links would spread more than that of the individual link ie the standard deviation would increase while the mean would decrease from that of the single link Example 315 Suppose the tensile strength of a 10inch long fiber is described by the Weibull function with m7 and 68 kg Then the CDF for the 10inch long fiber is Fx 1 expx87 21 Now if the design load xd is defined by Fx xd 001 it follows that xd 415 kg That is under 415 kg tension there is 1 chance that the 10inch long fiber would fail b For a fiber having 20inch length and assuming that the weakestlink theory applies the resulting distribution is still a Weibull with m7 but with 92 8217725 kg That is the CDF for two 10 inch long fibers arranged inseries is FZSx1expx7257 Chapter111 Data Sampling 11137 Then under the same design load of 415 kg as before the failure probability of the 20inch long fiber is FZSX 4151 exp4157257002 So there is a 2 chance that the 20inch long fiber would fail before 415 kg tension c If two 1inch fibers are bundled inparallel we may assume that the load x applied to the bundle is shared equally by the two fibers before any one of them fails However there are three possible failure conditions 1 both fibers fail at the strength xZ39 2 one fiber fails at the strength xZ while the other fails at a strength in between xZ and x and 3 the reverse role as in 2 Thus the CDF for the bundle of two fibers sz can be written as the reader should independently verify the following szx Fx2Fx2 2Fx2Fx Fx2 1n the above Fx is the CDF for the single 10inch long fiber 1f the bundle of two fibers is under the design load x415 kg the probability of failure of the bundle is then F2p415 F4152F4152 2F4152F415F4152 We can easily compute F415 1 exp41587 001 F4152 1 exp207587 79xio5 Hence F2p415 79x1052 279x10395 001 79x10395 1574x10396 Note The probability of failure for 2 fibers in a bundle is much less than the single fiber under the same load as it should39 ie the bundle is stronger than the single fiber This is obvious However if the applied load is 2x415 83 kg the failure probability of the bundle is then sz83 F415F415 2F415F83F415 144x10392 In this case the applied load in each fiber before failure is 415 kg39 but the probability of failure of the bundle under 83 kg is 00144 which is higher that of the single fiber under 415 kg which is 001 So the averaged fiber strength in a bundle of N fibers is weaker than that of the single fiber But to derive the distribution function for a bundle of N fibers would be mathematically tedious especially for large N say Nltlt339 those details are beyond the scope of this chapter ChapterIll Data Sampling Ill38 Summary The main purpose of this chapter is to introduce aspects of statistics in place of probability In engineering settings a random variable is almost always generated by some unknown mechanisms Hence it is not possible to determine the exact values andor the valuerange of X nor is it possible to figure out the exact probability distribution for X The alternative is to collect all or nearly all possible values of X in the field and this is called the population of X From the population one then devises a set of techniques based on the theories in statistics to figure out the probability distribution function of X In general the population is large for practical reasons a smaller sample is taken instead It is in this context one uses the same statistical techniques to figure out the probability distribution of the sample Though it is hoped that the latter is a close approximate of the former This chapter thus contains several working topics 1 the statistical nature of a sample that is taken from a population 2 use of the sample to approximate the underlining probability distribution of the population 3 confidence evaluation of the probability distribution estimated from a sample 4 the physical characteristics of some well known probability distribution functions including the normal lognormal Weibull and the extreme value functions 5 techniques for fitting a sample to the normal lognormal and Weibull functions In studying this chapter it is essential to be conceptually clear andor become pro cient in the following areas I The meaning of the population of X it contains all or nearly all of the possible values of X it can be finite or infinite in size I The meaning of a sample it is a subset of the population the size of a sample is almost always finite N Random sampling it is a sampling technique that attempts to insure that the sample taken possesses the same probabilistic characteristics as those of the population The physical characteristics of the normal distribution why do we standardize the normal function How to use the numerical table in Appendix IIIA And how to fit a sample to the normal function The physical characteristics of the lognormal distribution what is the similarity and difference between the normal and the lognormal How to fit a sample to the lognormal function The physical characteristics of the Weibull distribution the need to use of the Gamma function How to t a sample to the normal function Note the versatility and the mathematical ease in the Weibull function Use of the graphical papers in Appendices IllB C and D to fit the normal lognormal and Weibull distributions respectively The meaning of the fitted straight line also note the double vertical scales used in each of the plotting papers and know their relationships Details of the linear regression the least square method in fitting a sample analytically without using the plotting paper Keep in main the correlation factor in this method Chapter111 Data Sampling 11139 Assigned Homework 31 Data for the response time in seconds are collected from 90 electrical n a quality inspection program These are listed below 148 146 149 142 135 134 142 170 156 158 159 159 161 125 131 166 158 143 180 132 155 160 129 151 148 161 167 136 150 147 152 137 166 144 129 180 155 146 162 148 164 155 146 162 148 164 155 165 154 153 146 157 165 159 147 138 166 159 146 161 156 138 157 148 139 162 149 126 153 143 130 158 143 133 139 156 148 153 159 140 127 130 172 148 166 137 168 177 162 133 a Calculate the sample mean sample variance sample standard deviation and sample skewness b Use the Sturges formula and construct a proper histogram Does the histogram suggest a normal function a 515086 050127339 sks0052739 b distribution is almost belllike 32 Fit the sample in the previous problem with the normal distribution by means of the plotting paper provided in Appendix IllB Qlou may down load a copy from the web site Then provide answers to the following a The mean variance and standard deviation39 b The probability that a relay has at least 15 sec relaytime c The probability that a relay whose relaytime is outside the range of 15 i 01 sec 33 The following are data from testing 16 cutting knifes for their useful life in operating hours on a cutting machine 210 082 280 251 315 273 455 500 427 266 480 169 358 199 460 211 a Calculate the sample mean standard deviation and skewness b Fit the sample to a normal function and determine the distribution parameters from the plot39 c If five 5 knifes are used for a 120minutes mission on a cutting machine what is the probability that none of the knifes would fail the mission d What is the probability that at least one of the knifes would fail the mission 34 Dynamic compressive strength of an automobile chock absorber is normally distributed with p 1000 kg and 040 kg Then a If a shock absorber is tested under 900 kg compressive force what is the probability of its failure b For 99 reliability a chock absorber should not be loaded beyond what compressive load a FXS 900 kg 00062 35 Fit the sample in Problem 33 to a normal distribution by means of the least square method39 list the values of the distribution parameters p and o and the correlation factor 3085 61466391392 096 p 36 Fit the sample in Problem 33 to a lognormal distribution by means of Chapter111 Data Sampling 11140 a The plotting paper provided in Appendix IllC list the values of the distribution parameters b The least square method list the values of the distribution parameters and the correlation factor c Compute the mean and standard deviation of the fitted function a t0281 03005737 b r2 0906 c 331o207 37 Repeat Problem 36a b and c this time fit the sample in Problem 33 to a Weibull distribution m218 0356 r2 095 315 0153 38 For problem 35 evaluate the confidence limits pi and c5r for 80 90 and 99 confidence levels for 80 pi 3085i039 oi 1215i0284 39 Service life of a machine tool is described by the lognormal distribution with t0100 hrs and 3005 Determine a The meantim etofailure and the standard deviation of the distribution b The service life at 10 failure probability c The failure probability within the first hour of service a 1133 hrs and 604 hrs b 5273 hrs c 0 310 Service life of a bearing ball follows a Weibull distribution with the shape parameter m25 Field data shows that 10 of the bearing balls fail with in 2years of service Determine a The other parameter 0 b The meantimetofailure MTTF of the bearing ball c The failure probability of the bearing ball within 6month of service and d The service life for 1 failure probability b MTTF437 yrs d 078 yrs 311 Tensile strength of a 10 long graphite fiber is described by the Weibull distribution with m07 and 00 8kg Determine a The strength distribution for a 40 long fiber b The strength distribution for two 40 long fibers arranged in parallel c The probability of failure of the 240quotlong fiber arranged in parallel when loaded to 15kg in tension a for single 40 fiber set N4 b for a bundle of 240 long fiber use F2bx F25x2F25x2 2F25x2 F25x F25x2 c substitute x 15kg into the above ChapterHI Data Sampling Ill41 APPENDIX III A Values of ltIgtz in the Standardized Normal CDF for 500 S 2 S 000 Table to be handed out in class ChapterHI Data Sampling Ill42 APPENDIX III A Continued Values of ltIgtz in the Standardized Normal CDF for 000 S 2 S 500 Table to be handed out in class ChapterHI Data Sampling III43 ChapterHI Data Sampling III44 APPENDIX III B Plotting Paper for Normal Function 0888 0401 0228 0122 0082 0030 0014 ChapterHI Data Sampling III45 APPENDIX III C Plotting Paper for LogNorm a1 Function ChapterHI Data Sampling III46 APPENDIX IIID Plotting Paper for Weibull Function Fm 1nln1R ChapterIV Failure Rates IV1 CHAPTER IV FAILURE RATES AND RELIABILITY MODELS The term reliability in engineering refers to the probability that a product a system or a particular component will perform without failure under the specified condition and for a specific period of time Thus it is also known as the probability of survival To quantify reliability a test is usually conducted to obtain a set of timetofailure sample data say ti i1N The sample can then be fitted to a probability density function ft or to a probability cumulative function Ft The reliability function is defined as Rt 1 Ft Hence the behavior of Rt is conjugate to that of Ft the cumulative probability of failure in time But failure of an engineering product or system may stem from such random factors as material defects loss of precision accidental overload environmental corrosion etc The effects on failure of the these random factors are only implicit in the collected data ti i1N and it is dif cult to ascertain which factor is predominant and when it is predominant from using Ft Another way to look at the failure behavior in time is to examine the failure rate Failure rate is the time rate of change of the probability of failure Since the latter is a function of time failure rate is also a function of time But in terms of failure rate one can obtain physical information as to which factor is controlling the failure behavior andor when it is controlling the failure behavior Example 41 A TV producer tested 1000 sets in an accelerated reliability evaluation program In that program each set is turned onandoff 16 times each day to mimic a typical TV usage for a week Based on a failuretoperform criterion failure data are obtained for the first 10 days of test day1 day2 day3 day4 day5 day6 day7 day8 day9 day10 18 12 10 7 6 5 4 3 0 1 Here we define the failure rate as the quotprobability of failure per day denoted by M i1 10 For the first day i1 M1181000day39 For the second day i2 M21210001812982 day For the third day i3 M3 1310001812 13970day Note that the failure rate for day1 is based on a total of 1000 TV sets in which 18 failed during the day39 the failure rate for day2 is based on a total of 100018 sets and for day3 a total of 100018 12 sets etc In this way we can obtain Mt up to t10 days A plot of Mt versus t is displayed on the next page Clearly for this procedure to yield reliable Mt the number of the TV sets tested each day must be large relative to the number of failures in that day However we also note that the time required in gathering the data is only 10days a relatively short time period compared to what might be needed to ChapterIV Failure Rates lVZ generate a set of timetofailure data ti ilN xi 103day t day Physical signi cance of Mt The above plot of shows that the TV failure rate is initially high but it decreases with time Such a decaying behavior is known as infant mortality or wearin phenomenon It implies that early failure is caused by quotbirth defectsquot that are present in the product before it is put into service those products that have survived the wearin period are deemed to have fewer defects at birth statistically speaking The plot can be easily fitted by a smooth function Mt for 0ltt In fact as it will be shown next Mt is analytically related to ft the timefailure probability density function In many engineering situations it may be more tim esaving and less expensive to datafit Mt and than ft IV l Relation Between Failure Rate and Failure Probability Failure Rate The relation between the failure rate function Mt and the failure probability density function ft can be readily established by examining ft graphically as shown below i ft L gt Rt 1 Ft Ft At gt g 4 At Since the area under ft from 0 to t is Ft while the area from t to cx is Rt the area shaded from t to tAt is noted by ftAt which represents the fractional probability that failure occurs within At The latter occurs when the product has actually survived the time period from 0 to t ChapterIV Failure Rates lV3 Hence the probability that failure occurs within At is a conditional one ftAtRt and the rate of change of that probability at the time t is Mt ftAtRtAt ftRt 41 The above establishes the relation between Mt and ft In fact one can obtain ft from Mt To this end we note that ft dFtdt dlRtdt dRtdt Eq 41 can then be written in the form Mt dRtdtRt The in turn yields di erential relationship Mtdt dRtRt Integrating the above from 0 to t and noting that ROl we obtain Rt I t exp I E d 4 2 Then from 1 t ft Mt eXp xi M c d39c 43 Noting that R 00 0 one can readily verify the following 00 00 p MTTF t ft dt Rt dt 44 0 0 Discussion The failure rate data the bar chart in Example 41 can be fitted nicely by the function Mt 002 r056 From 43 the corresponding failure function ft is obtained as ft 002 t39056exp004545t044 Similarly from 42 the reliability function Rt is obtained as Rt exp004545t044 And the meantim etofailure is obtained using 44 cc ChapterIV Failure Rates IV4 MTTF p J39 exp004545t044dt 0 Example 42 A company produces videocassette recorders In order to formulate a pricing warranty and aftersale service policy a reliability evaluation program is carried out39 the test finds that the failure rate is somewhat a constant 7ttl875hour Thus from 42 the reliability function is obtained Rtexpt 875 0 t in hours And from 43 the timetofailure probability function is obtained ft expt87508750 From using 44 the mean of ft or the MTTF is given by p 8750 hours In this case we note that a constant failure rate leads to an exponential function for ft Significance of this relationship will be discussed next The Bath Tub Curve of Mt For many engineered products the failure rate function Mt has a timepro le much like the crosssection of a bath tub such as shown below Mt infant youth aging wearin 001151 rate wearout t1me This curve is a ubiquitous character of all living things for instance the human life expectancy Service life of many engineered products have much in common As illustrated above the curve may be broadly classi ed in three time zones infancy youth and ageing each may correspond to a distinctive failure mode The infancy or wear in period is generally short with a high but decreasing rate such as in the case discussed in Example 41 This mode of Mt may be due to defective parts defects in materials damages due to handling out of manufacturing tolerance etc such factors can have ill effects early in life To correct the situation a number of measures can be taken design improvement stricter material selection tightened quality control just to mention a few If such ChapterIV Failure Rates lVS measures are insufficient a proof test may be instituted The latter refers to products under go a specified period of simulated tests in hope that most early failures are weeded out Another measure may also be instituted known as redundancy which is built into the product to provide a fail safe feature The youth period is exacted by a constant rate this occurs with products that do not contain fatal defects or that have survived the infancy period The ratevalue is generally the lowest and in some cases it maintains a long and at behavior such as shown below Mt gt time This constantrate mode is generally due to random events from without rather than by inherent factors from within Such events are beyond the control during the periods of design prototype development manufacturing etc The constant rate period is often used to formulate the pricing warranty and servicing policies of the product the latter is of particular importance in commerce As it will be shown later product with a constant failure rate has the unique attribute that its probability of failure is independent of the products past service life this aspect aids mathematical ease in modeling repair frequency sparepart inventory maintenance schedule etc The aging or wear out period is associated with increasing failure rate it is attributed to material fatigue corrosion contact wear etc modes often encountered in mechanical systems with moving parts such as valves pumps engines cutting tools bearing balls wheels and tires just to mentioned a few In some products the youth period is absent or relatively short while the wear out period is long such as depicted below Mt gt time ChapterIV Failure Rates lV6 For products with rapidly increasing failure rates it requires corrective measures regularity of inspection maintenance replacement etc Thus the central concern in the wearout period is to be able to predict the probable service life with a suitable model so that a prudent schedule for preventive maintenance can be formulated Generally speaking the wearin mode is a quality control issue while the wearout mode is a maintenance issue The random failure or constant rate mode on the other hand is widely used as the basis for product reliability considerations Some of the key features in this regard are discussed in the next section IV 2 Reliability Models Based on Constant Failure Rates Reliability Model for a Single Unit Let the failure rate of a certain product be constant say 7tt7to the corresponding reliability lnction is given by 42 Rt eXp hot 45 And the corresponding probability density lnction is given by 43 ft Aoexp hot 46 The mean MTTF and the standard deviation of ft are given by u MTTF 17 0 47 6 u lho At the meantimetofailure H1Lo the reliability value is Ru eXpl 0368 or Fu 0632 Note that the number 0632 is also the value of F0 in the Weibull distribution see III31 but 0 is not the mean of the Weibull function Example 43 A device in continuous use has a constant failure rate AOO02hr Then the following may be computed a The probability of failure Within the first hour of usage Pt1 F11 exp 002x1 198 b The probability of failure Within the first 10 hours Pt10 F101 exp 002x10 181 c The probability of failure Within the first 100 hours ChapterIV Failure Rates lV7 Pt100 F100 l exp 002x100 865 d The probability of failure within the next 10 hours if it has already been in use for 100 hours This is conditional probability situation since the device has already survived 100 hours Now let XlFlOO be the probability the devise survives 100 hours and YFllO be that fails within 110 hours see the sketch below Then the answer to d is found from the conditional probability of Y d givenX already occurre PYD PX n YPX 17010 F1001 F100 f0 lt X1F100 100 110 YF110 I Since Fl lO l exp 002xl lO 8892 and F100 865 we find t PYX 0889208651 0865 181 Discussion The answer to d is identical to that in b Thus being of constant failure rate the device has no memory of prior usage Single Unit Under Repeated Demands When a product is called to service by a demand there is a probability p for failing to respond to the demand If N such demands are called during the time period t we de ne the average number of demands per unit time as m Nt 48 Assuming failure of the unit during each demand is an independent event the reliability of the unit subjected to N repeated demands is see Chapter III the binomial distribution RN 1pN In particular if pltlt1 and Ngtgt1 the above is reduced to the Poisson distribution RN e39NP emp t 49 Now let ho mp 410 ChapterIV Failure Rates IV 8 Here X0 mp represents the equivalent failure rate for the unit under repeated demands And 49 becomes identical to 45 the constant rate reliability function Hence the reliability of a product under repeated demands is a case of constant failure rate when pltltl and Ngtgtl Example 44 Within oneyear warranty period a cell phone producer finds that 6 of the phones sold were returned due to damage incurred in accidental dropping of the phone on the ground A simulated laboratory test determined that when a phone is dropped on a hard floor the probability of failure is l in 1039 or 702 Based on this information the engineers at the phone manufacturer made the following interpretation a Let the time unit be year Then for a single unit Fl cumulative probability of being damaged within 1 year is 6 Or Rl 094 The above can also be interpreted as 6 phones out of every 100 were damaged per year39 Or 94 phones out of every 100 did not suffer any damage during the year b Then let m number of demands drops on hard floor per phone per year From 49 Rl l Fl exp mpt exp mx02xl 094 Solving the above we obtain m03l drop per phone per year Interpretation On the average there are 31 drops per 100 phones per year Or alternatively the mean time of drops for one phone is MTTF l03l 323 yrs Discussion Given m03l a factor stemming from customers habits the phone producer can only redesign the phone by making it more impact resistant or more robust this mean to decrease the value ofp If for instance p0l then Rl exp 03lx0lxl 09739 or Fl 003 It cuts the returning rate from 6 to 3 per year Single Unit Under Step Wise Constant Rates A unit often operates at different levels of performance during a typical operating cycle If at each performance level the unit fails with a constant rate it can then be treated as a step wise constant failure rate problem For instance the electric motor used in a household heatpump system is called when the room temperature is low it is shut off when the room temperature is raised to the preset high During a typical service cycle say 24 hour the motor may be called N number of times the failure rate profile may be depicted schematically such as shown below Mt Chapter1 V Failure Rates 1V9 This pro le provides the following information for an evaluation of the motor s reliability N the number of starts demands per service cycle 24 hours c time fraction when the motor is in running state during the service cycle lc time fraction when the motor is in standby state during the service cycle p probability of failure when the motor responds to a operation call start hr failure rate per hour when motor is in the running state and ks failure rate per hour when motor is in the standby state Under the constant rate condition the combined or equivalen failure rate A0 for the motor in service can be expressed as he kd c hr 1CLS In the above Xd mp m being the number of calls per unit of time N24 calls per hour And the reliability function of the motor is given by RC eXP39 Mt 412 It is clear that 4 12 is accurate only if the service time t is much greater than the length of one single cycle 24 hours 411 Example 45 An electric blower is used in a heating system The manufacturer of the blower has provided the following failure rate data 7 failure probability on demand 00005 per call 7L1 failure rate per hour when blower is in running 00004hr 7L5 failure rate per hour when blower is in standby 000001hr During the a typical 24 hours in the winter months the following data is obtained from the heater s operation recording of calls time of call time of stop running running time 1 047 am 101 am 023 hr 2 141 am 207 am 043 hr 3 253 am 304 am 018 hr 4 355 am 413 am 030 hr 5 443 am 505 am 037 hr 6 558 am 619am 035 hr 7 650 am 714am 040 hr 8 746 am 807 am 035 hr 9 855 am 908 am 022 hr Chapter1V Failure Rates 1V10 10 949 am 1005 am 027 hr 11 1049 am 1101 am 020hr 12 1152 am 1208 pm 027 hr 13 1259 pm 111pm 020hr 14 149 pm 204 pm 025 hr 15 252 pm 311pm 032hr 16 358 pm 405 pm 012 hr 17 441 pm 459 pm 030 hr 18 543 pm 602 pm 032 hr 19 637 pm 700 pm 038 hr 20 737 pm 758 pm 035 hr 21 837 pm 855 pm 030 hr 22 929 pm 952 pm 038 hr 23 1035 pm 1047 pm 020 hr 24 1137 pm 1153 pm 027 hr Total calls Total running time 696 hrs N24 m24241hr Time fraction c69624029 Thus the combined failure rate for the blower is Ac mp cAr 1c S 1x00005029x00004071x000001623x10394hr With Ac the reliability of the blower in a month service 720 hours is given by R720 exp 0000623x720 064 Failures of A Maintained Unit It occurs in many engineering situations that a single device in continuous use can be regularly maintained so that the device can function inde nitely If the unit is of a constant failure rate it is possible to estimate the number of repairs andor replacements needed to maintain the device over a long period of continued service Now let pnt be the probability that exactly 11 repairs or replacements are needed over the time period t Note that pnt must satisfy the conditions that for t 0 there is no failure pOO 1 pr O 0 for n gt 0 412 At any time period t gt 0 however pnT must also satisfy the total probability condition 2 pnT 1 sum over 11 012 oc 413 1low consider the time interval from t to tAt the probability that zero repair will occur before t is pOt and the probability that zero repair will occur before tAt is pOHA Note then in order for pOtAt to occur we must have pOt in the rst place Since the device has a constant failure rate say lo the failure probability during At is XOAt while the nonfailure probability is 1 XOAt Consequently we can write Chapter1V Failure Rates lVl l p0tAt p0T1 AOAt Realrange and we obtain the di erential relation Ap0tAt hop0t Integrating the above over the range from t0 to t and noting the initial conditions in 412 we nd the probability for zero repair within the time period of t pOt ex130M 4 14 The result in 42 is not surprising for it means that there is no failure before t or it is simply the reliability of the unit for the time period t To nd the expression for pnt however we consider the probabilities that 1 n repairs occur already over t so no more repair occurs during At 2 n1 repairs occur over t one repair occur during At though At gt 0 In short we write pr tAt pr tl AOAt pLltAOAt The above can be rewritten in the differential form Apr t At mo pnt ho pLlt 415 Integration of 415 from 0 to t yields the integral expression for pnt t pnt A0 exp7 Ot lo pk 11 exp7 O39c d39 416 Equation 416 is a recursive relationship it allows for the determination of pnt successively for n123 For instance for n1 we substitute the result in 414 into 4 16 and carry out the integration obtaining p lt hot exp7 Ot 417 For n2 we in turn obtain 1320 40020 eXIX401 4 18 A general expression for pnt is given by ChapterIV Failure Rates IV 1 2 pnt kotnn eXp hot 419 We note that 419 is in the form of the Poisson distribution for the random variable 11 see Eq 22l whose mean and variance have the same value see Example 2 l 1 HT 6112 tot 420 In the above un is known as the mean number of repairs needed over the time period t Mean Time Between Failures The quotmean time between failuresquot or MTBF of the maintained unit is de ned as MTBF tun 1 ho The above is identical to the MTTF of the unmaintained unit see 46 and 47 With the distribution for n in 419 the cumulative probability that more than N repairs are needed over the designated time period t is given as follows 00 N PngtNt Z kotnn eXp7tot l ZEtOtnn eXp7tot 422 nNl n A more precise interpretation of the terms in the above equation is as follows a the term which sums from n 0 to nN represents the probability that up to N repairs will occur during the time period t 2 the term which sums from nNl to n gt oc represents the probability that more than N repairs will occur during the time period t Together their sum represents the total probability during the period t including n0 123 00 see Eq 413 Example 46 A DC power pack in a computer is in continuous use and it has a constant failure rate AO04 per year If a spare is kept on hand in case of the power pack failure what is the chance of running out of the replacement spare within a 3months period Solution Here the designated time period is tl4 year the mean number of failures in 3 months is Aot0l The probability that more than 0 or at least 1 failure occurs within tl4 is calculated from using 422 for N0 P 0t l exp lot 0095 Or there is roughly 10 chance that the spare will be used within 3 months Discussion If 2 spares are kept in hand the chance of running out of the spares within 3months can be calculated also from 422 By setting Nl ie n is more than 1 or at least 2 we obtain 421 Chapter1V Failure Rates 1V13 Pngtlit 1 1Aot exp lot 000468 Hence the chance to use 2 spares in 3month is less than one half of 1 Example 47 Field data have shown that truck tires fail due to random puncture on the road If the meantimebetweenfailure MTBF of a tire is 1500 Km then a truck with 10 wheels must carry some spare tires a What is the chance that at least 1 spare will be used on a 100 Km trip b What is the chance that at least 2 spares are needed on a 100 Km trip c How many spares should be kept in order to have more than 99 assurance that it will not run out of spares on a 100 Km trip Solution Puncture of tire is a random event hence it can be treated as a case of constant failure rate In this example the MTBF1500 Km is known from field data thus we can approximate the constant failure rate of a single tire as AO111TBF 11500km Since the truck has 10 wheels the 10 tires will accumulate a total of t10x1001000 Km over a trip of 100 Km Hence lot 10001500 0667 Thus a The probability that at least 1 tire is punctured is Pngt0t 1 exp tot 0487 b The probability that at least 2 tires are punctured is Pngt1t 1 17tot exp 7tot 0144 c For more than 99 reliability or less than 1 failure chance N spares should be kept and the chance of running out of N spares must be less than 1 PngtNt 1 1f N2 it takes more than 2 punctures to fail the 100 Km run and that chance is Pngt2t 1 1A0t7tot22 exp 7tot 003 The probability of running out of 2 spares is 3 or the reliability is 97 1f N3 more than 3 punctures would fail the 100 Km run and that chance is Pngt3t 117t0t Aot22 000 exp tot 000486 The probability of running out of 3 spares is 0486 Hence for 99 reliability or better N3 spares should be kept IV 3 Time Dependent Failure Rates ChapterIV Failure Rates IV 1 4 The general characteristics of the failure rate function Mt resemble the pro le of a bathtub consisting 3 time periods each is associated with a distinct failure mode In the infancy wearin period the rate is initially high but decreases with time the youth period then follows with a constant rate it is in turn followed by the ageing wearout period where the rate increases with time In the previous section we have discussed several reliability models that are based on the constant rate or the youth period assumption in this section we examine the cases where the failure is timedependent The Wear In Mode of Failure The wearin mode is characteristic to products with initial defects stemming from productdesign to manufacture and handling When put into service the products may initially experience a high rate of failure caused mainly by the inherent defects as the early failures occur the failure rate then gradually reduces The test data for the TV sets discussed in Example 4l exhibits just such a wearin behavior in fact the data can be fitted to a decreasing time function for Mt Mt at b In the above a and b are positive and real constants Having found Mt one can obtain the reliability function Rt and the failure probability density function ft see 42 and 43 respectively Then a number of questions related to product reliability may be rationally answered Example 48 A circuit has the failure rate described by Mt005tm per year tin years Here Mt has the form of 423 The associated reliability function Rt and the failure probability density function ft are obtained by integrating 42 and 43 respectively Rt exp 01015 ft 005 exp 01t12t12 With the above the following may be computed a The reliability for 1 year use Rlexp 010905 Fl95 b The reliability for the first 6month use R05 exp 01 I05 093 c The fraction of failed circuits in 3 years F3lR3lexp0l l3016 84 still in use d A circuit has been in service for 1 year the reliability for it to be in service for another 6month R05l l F05l F051 F15 F1R10115 009509050022 423 ChapterIV Failure Rates IV 1 5 RO511 0022 098 Discussion The result in d shows that the circuit in question had a fewer defects and thus lasted 1 year without failure this particular circuit when used for another 6 months has a higher reliability than a randomly picked new circuit the reliability of the latter is the answer in b The above also demonstrates that if early failures can be eliminated from a lot of the circuits before putting them into service the remaining circuits will have a higher reliability This gives rise to the concept of prooftest a practice often used in offline quality control The Concept of Proof Test Prooftest is usually conducted with products having an initial high failure rate followed by a relatively short wearin period The test is designed to subject the products under simulated service conditions for a short period of time in the process most products with fatal defects will be weeded out while products that have passed the test will yield a higher reliability Suppose that the failure rate function Mt of a product has a short wearin period such as shown in the sketch below Here as the case in general the associated wearin time period tp is not well de ned but it can be estimated Beyond this period it is assumed that the failure rate function begins with the constantrate mode Mt I l quot wear1n I l tp If a prooftest is conducted for the time period tp the fraction of failure and the fraction of survival can be estimated from the Rt curve as sketched below The upper shaded area represents the fraction of failure while the lower shaded area represents the fraction of survival Rt A early failures Rtp39c Chapter1V Failure Rates 1V16 Now for the product that survived the prooftest let 1 tp be the actual service life Thus the associated reliability is given by Reap Rap map exp 10 1 M d 1explotp Mod 1 Upon rearranging tp 5 R39ctp eXplt k d 1 gt 0 424 P It can be readily shown that R39ctp is much greater than Rt t being 1tp Discussion Example 48d is a case related to prooftest We can now use 424 to compute the desired result In fact we find R051098 using 424 which is the same as that found differently in Example 48d The Wear out Mode of Failure The wearout mode is characteristic to products that have been in service for some length of time usually well into the constantrate period Occasional over loading and adverse environment may have induced damages in the product and the effect of fatigue then sets in Consequently the failure rate begins to rise Fatigue failures in durable products are common in practice and the problem must be adequately addressed within the context of reliability In fact timetofailure probability distribution that can be described by the normal lognormal or the Weibull functions are associated with the wearout failure mode This is brie y discussed below Normal Function and Failure Mode The timetofailure probability described by the popular normal distribution is associated with a failure rate function that is of the wearout mode We can show this starting from the standardized normal CDF z where z is the transformed time 2 t u and u and 6 are the mean and standard deviation of the normal probability density function ft see 311 to 217 for details The corresponding reliability function is the given by Rt1q3z It follows from 41 that the failure rate function is Mt 1 121116 eXp2221q3z 425 With the help of Appendix IIIA a plot of Mt versus time can be obtained using 425 Chapter1V Failure Rates 1V17 Mt 156 106 0 t 39 u Zc u c u 46 H26 It is seen that Mt rises sha1p1y when the time t exceeds the MTTF ie beyond t gt u Example 49 Field data of a certain brand of tires show that 90 of the tires on passenger cars fail to pass inspection between 22 to 30 kmiles 1f the timeto failure probability of the tires can be described by a normal distribution a What is the failure rate when a tire has 20 kmiles on it b What is the failure rate when a tire has 25 kmiles on it Since 90 of the tires fail between 20 and 30 kmiles ie the central population is 90 we can write 220k 005 Z30k 095 Using Appendix IIIA we find 220k 20p0 165 Z30k 30p0 165 Solving the above for p 25 kmiles and 0 303 kmiles Having found the values of 0 and p in normal pdf ft the corresponding failure rate function Mt is then computed from 425 Thus a For a tire having 20 kmiles on it 7Lt200l4 per kmiles and b For a tire having 25 kmiles on it 7Lt2508 per kmiles Since 7L increases with t miles failure of the tire has the wearout mode Discussion For a product in service knowledge of its failure rate helps to formulate a rational repair or replacement schedule In the above example a failure rate of 014 per kmile at the 20 kmile life may be more tolerated than one of 08 per kmile at the 25 kmile life Log Normal Distribution Its Failure Modes The lognormal timetofailure distribution is explicitly expressed in 333 its behavior is that of leftskewed function the degree of skew is dictated by the parameters 030 When the value of no gt 1 the pdf gt becomes an exponential ChapterIV Failure Rates IV18 function when no gt01 gt approaches the normal function Thus the failure rate function corresponding to no gt 1 is one of constant rate mode the failure rate function corresponding to no gt 01 is one of wear out mode Hence we see that no can affect the failure mode According to 41 and with the substitution of 333 we obtain a general expression for the failure rate function 1 xtm OV21T Tlexp 71752 2lqz 426 In the above z is the standardized CDF of gt via the transformation see Eq 335 z lntt0o0 427 Example 410 Failure of a brand of shock absorbers used in passenger cars is described by the log normal function Field data shows that 90 of the shock absorbers fail between 120 kmiles and 180 kmi es What is the failure rate of the shock absorber at t 150 kmiles Solution Given gt being lognormal the parameters t0 and 030 can be determined from the field data The standardized CDF of gt is ltIgtz where z is expressed in 427 From the field data we have ltIgt2120 005 and ltIgt2180 095 Using Appendix IIIA and then 427 we find 2120 lnl20tOoo 0l645 2180 ln180tOoo 01645 From there we solve for lognormal parameters t0 and mo tO 147 kmiles and 030 01232 Then the failure rate function is given by 426 At tl50 kmiles 7L150049 per kmile Compare this value to 7Ll200009kmile and Mto 7Ll470044kmile39 we see that Mt increase rapidly once tgt to In this case 030 01232 which is close to being 0 139 so the distribution is almost normal like Discussion In general the failure rate for lognormal is of the wearout mode the rate will increase sharply once t is greater than to But if 30 1 the lognormal reduces to exponential the failure rate is then constant in time see Chapter III section III4 Weibull Distribution and Failure Modes Following the same procedures as before the failure ChapterIV Failure Rates IV 1 9 rate function for the Weibull distribution can be obtained from using 337 and 41 M m0t0m391 428 The behavior of Mt thus depends on the values of the Weibull parameters 9 and m In particular when 0ltmltl Mt is a decreasing function of t representing the wearin mode when ml Mt is simply a constant representing the random failure mode or the youth mode when mgtl Mt is an increasing function of time representing the wearout mode We can see this in the following example Example 411 A hearing aid has the timetofailure described by the Weibull distribution ft m0t0m391 expt0m The Weibull CDF is Ftlexpt0m39 the reliability function is Rtexpt0m The corresponding failure rate function is then found from using 41 Thus we have Mt ftRt m0t0m391 Thus if the shape parameter m05 and the scale parameter 0180 in days then Mt 00373 05 per day This is a decreasing time function representing the wearin failure mode If we increase the value of m to 15 we obtain Mt 000062 tos This is an increasing function representing the wearout mode Similarly if we set ml Mt 10 which is a constant Discussion The parameter In in the Weibull function controls the shape behavior of ft hence the reliability function Rt as well as the failure rate function Mt The figures shown below illustrate the shape behaviors of ft Rt and Mt for m05 10 20 and 40 0 m l Example 412 For the hearingaid considered in the previous example the failure rate function is Mt00373t3905 ChapterIV Failure Rates lV20 where t is in days and the associated failure mode is that of wearingin The plot below illustrates the behavior of Mt as a function of t A 03726 L V 1 day It is seen that the product is inherently defectsladen as the failure rate before the first day of service is more than 3 per day Suppose that the manufacturer conducts a prooftest on the hearingaids so as to screen out the infancy failures and the new pdf for the timetofailure of the screened hearing aids is again a Weibull function but with ml5 and 6180 days Then according to 428 the new failure rate function is At621x104t05 The mode of the new failure rate is now an increasing function of t representing the wearout mode And it can be shown that the failure rate of the screened product at dayl is only 00621day However the failure rate is rapidly increasing as it goes beyond 100 days 7 00621 50 100 day39 Model for Bath Tub Like Failure Rates From the above it is seen that the Weibull function can describe the wearin mlt1 constantrate m1 and wearout mgt1 modes Thus it can be used to describe the entire bathtub curve with a combined failure rate function in the general form Mt ma9at9ama391 mbGbt9bmb391 meectGcmc391 429 where 0ltmalt1 mb1 and mc gt1 also 9a lt 9b lt Sc Note that the rst term in 429 is decreasing with t which is in the infantile mode while the second term is constant in t hence in the youth mode the last term increases with t representing the wearout mode Proper choice of the Weibull parameters m s and 939s in 429 will yield a smooth bathtub curve for Mt This is illustrated by the example below Example 413 Suppose that the failure rate function for a cutting knife behaves like a bathtub curve39 its wearin period is described by 7ttmetem391 with m02 and 610 days the constant rate mode by 7ttllOOday and the wearout mode by 7ttmetem391 with m25 and 6120 days ChapterIV Failure Rates lVZl The combined failure rate function is thus Mt 01262t3908 001l58x10395t15 per day 7 per day combined wearin 7 0 const rate wear out Discussion The failure rate behavior of the cutting knife does not have a significant constantrate period the wearin period is short about 25 days While the wearour period is long VI 4 Failure Rates for Systems of Multiple Units Almost all engineering systems are a complex combination of multiple units or subcomponents in general each of unit or subcomponent has its own failure rate when functioning in the system one or more of the units can fail which may or may not cause the system failure Of course failure of a unit or a subcomponent can reduce the reliability of the system even if the system is still safe Analysis of the system s failure rate can often be conducted by representing the system as a combination of two elementary models the in series and in parallel models The In Series Models Let the system S contain N subcomponents X with i1N connected in series as a chain 9 9 9 quotquot Let fit be the timetofailure pdf of Xi then the corresponding failure rate is AirfitRit see 41 where the reliability function Rit is given by 42 t Rit exp lO Mom 430 Here the N subcomponents are arranged in series failure of one or more of components will ChapterIV Failure Rates lV22 cause failure of the whole system Thus according to 29 the failure probability of the system is represented by the intersection all subcomponent failures Xi PSsy PX1UX2UX3U o o o UXN Alternatively the system reliability requires the nonfailure X39i or the reliability RNt of each and every subcomponent thus using 2 10 Rsy PX391nx392nx393n o o o nX N If each subcomponent failure is independent from any other subcomponent failure then the system reliability can be reduced to RsyO R1t39R2t39R3t39 39 39 RNO 431 Upon substituting 430 into 431 we obtain l t l t l t Rsy eXp 0 X1tdt 39eXp 0 X2tdt 39 39 39 39eXp 0 XNtdt 63XpJOt k12 39 39 7tNdt eXpJ0t ksytdt 432 In the above expression hsyt is the system failure rate which is defined by AS t Elia sum over i lN 433 Discussion The inseries model is often referred to as the weakestlink model as first discussed in Chapter III Section Ill5 In particular if Xi X for all i lN 433 reduces to Asy tNAt Most engineering systems are more complex than the inseries model but the model can at least provide a lowerbound estimate for the system s failure rate Note that the reliability lowerbound of the system cannot be better than the reliability of the poorest subcomponent in the system Example 414 N identical subcomponents are arranged in series the timetofailure pdf for each of the subcomponents is described by the Weibull function with the parameters In and 6 Determine a The system failure rate and b The timeto failure pdf for the system Solution The failure rate function of the subcomponents is given by 428 Atmetem1 Chapter1V Failure Rates 1V23 Hence the system failure rate is found follows from 433 Asyt Nm0t0m391 mGNt0Nm391 Where 9N eNlm Note that the system failure rate function has the same form as that of the subcomponent so the system timetofailure pdf is also a Weibull function but with the parameters In and 0N In essence the inseries arrangement of N units shifts the parameter 0 of the unit to 0N the latter is much reduced depending on the value of N While the parameter In remains unchanged regardless the value of N Example 415 A computer circuit board is made of a total of 67 components in 16 different categories The failure rate of each of the 16 categories is listed in the table below the first column lists the 16 categories the second column indicates the number of components n in each category39 the third column lists the component failure rate 7L constant and the forth column is the cumulative failure rate n7 derived from the n components in each category Component type Number of units n Unit failure rate 7L Cumulative failure rate n7 type1 capacitor 1 00027 X10396hr 00027 x10396h type2 capacitor 19 00025 00475 resistor 5 00002 00010 flipflop 9 04667 42003 nand gate 5 02456 12286 diff receiver 3 02738 08214 dual gate 2 02107 04214 quad gate 7 02738 19166 hex inverter 5 03196 15980 shift register 4 08847 35388 quad buffer 1 02738 02738 4bit shifter 1 08035 08035 i inverter 1 03196 03196 connector 1 43490 43490 Wiring board 1 15870 15870 solder connector 1 02328 02328 Total units N 67 Sum of all failure rates 21672 x1039 6hr Here the sum of all failure rates in the fourth column is the lowerbound system failure rate Asy21672x10396hr Based on the lowerbound estimate the system reliability is Rsyexp21672x10396 t And the system meantimetofailure is MTTFl7Sy 46142 hrs ChapterIV Failure Rates IV24 Since Asy is a constant the system failure is of the random mode39 and the timetofailure probability of the system is described by the exponential function The In Parallel Model and the Bundle Theorv Suppose that the system S contains N sub components denoted by Xi i1N and are arranged in parallel as depicted below input output gt gt v In this case failure of one subcomponent may or may not cause failure of the system Indeed this is a system with a degree of redundancy or with a failsafe feature In general a certain load sharing mechanism is built into the inparallel system namely the load carried by one subcomponent that fails will be quotsharedquot by the unfailed ones according to the loadsharing mechanism In addition if the failure probability of the unfailed ones depends on that of the failed ones it will result in a quotconditional probabilityquot situation 00 Hence in order to evaluate the system reliability for the in parallel models the following input information is needed in addition to the reliability lnctions Ri s of the subcomponents a the loadsharing mechanism in the in parallel arrangement and b the conditional failure probability between any two subcomponents This can often result in mathematically complex formulations However if there is no speci cally builtin loadsharing mechanism and if failure of one sub component does not depend on that of the others mathematical complexity in the inparallel model can be greatly simpli ed In fact the system reliability can be readily found as Rsyt 1 1R1tl1R2tl1R3tl39 39 39 lRNtl 434 The above result is based on the assumption that the system is safe as long as one subcomponent is safe under the system loading In fact the term 1R1t1R2t1R3t1RNt in 434 is the probability that all N components fail Note that Rsy in 434 is better than or at least equal to the best of the Ri s Hence 434 is in the upper bound for systems of N subcomponents regardless of their arrangement Example 416 The reliability of a pressure valve during a specific service period is RO08 If two such valves are arranged in parallel evaluate the system reliability for the service period Solution The system reliability is readily given by 434 if failure of one valve does not affect the ChapterIV Failure Rates lV25 failure of the other Rsy l l08l08 096 Discussion If three valves are arranged in series the system reliability can be improved Rsy l l08l08l08 0992 Thus parallel structure improves system reliability Example 417 Suppose the valve in the previous example have the timedependent reliability given by RU exPlkotl Then the reliability of the system with two valves in parallel is Rsyt 1 1 exp7tot2 2exp7tot exp27tot If N valves are arranged in parallel and it requires at least M MltN of the valves be reliable durin a particular application then the system reliability can be determined by the quotbinomial distributionquot see section HZ for reference to binomial distribution Rsyt 2 cNi Rti 1RtN39i sum over i M N Series Parallel Combination Model Many engineering systems are con gured in a network of inseries and in parallel units each unit may contain subcomponents that are in turn arranged in series andor inparallel If the reliability function or the failure rate anction of each component in the units is given one can evaluate the following for the system a The lower bound for the system Rsy using 432 b The upper bound or the system Rsy using 434 andor 0 The exact system Rsy using the method of network reduction technique this is to be discussed in Example 4 18 below Example 418 A system is made of 7 subcomponents arranged in a network as shown below For the specific service duration the reliability value of each subcomponent is known as indicated Chapter1V Failure Rates 1V26 The system is a relative simple network of inseries and inparallel units the lowerbound and upper bound for the system reliability can be readily computed using 431 and 434 respectively the exact system reliability Rsy can be evaluated by the network reduction technique a The lowerbound by 431 Rsy1b 093 082 0752 026244 b The upper bound by 434 Rsym 1 1093 1082 10752 09999975 c The exact by the network reduction technique The inparallel unit from B to D is replaced by a single equivalent component with R31 1 1 0752 09375 The inseries unit from A to B to D is replaced by a single equivalent component with RABD 09 08 09375 0675 The inseries unit from A to C is replaced by a single equivalent component with RAc 09 08 072 The inparallel unit from A to O is replaced by a single equivalent component with RAO 1 10675 1072 0909 The system reliability is determined by the inseries unit from 1 to O Rsy 09 0909 08181 Discussion This network has 2 levels of inparallel units one is from B to C and the other is from A to 0 Such a system is said to have a much greater degree of redundancy than the allinseries system The exact reliability in this case is much higher than the lower bound yet it is also substantially lower than the upper bound The Bundle Theory The bundle theory due to Daniels 1945 refers to a system of many subcomponents that are arranged in parallel Originally Daniels considered a loose bundle of N identical threads If the bundle is loaded by a tensile force T then the tensile load on each thread is assumed to be x TN and when one of the threads fails the remaining Nl threads would share the bundle load Nx equally so the tension on each thread will be increased to NxNl increase in tension may in turn increase the failure probability for each of the surviving threads If one or more of the threads fail again the rest of threads continued to shared the bundle load equally this of course may further increase the probability of failure of the surviving ones Clearly the failure probability of the bundle depends on the thread strength distribution and the assumed load sharing mechanism that the applied bundle load is always shared equally by the surviving threads This assumption helps to reduce the complexity of the problem if the bundle of threads are twisted into a rope or cast into a binder matrix material the loadsharing mechanism ChapterIV Failure Rates lV27 would be much more complex Now let the tensile strength of the thread be the random variable X Then the probability that a typical thread fails at or before XS X is denoted by FX PXS x 435 Now let the random variable TB be the failure load of the bundle then XBTBN represents the quotaveragedquot thread tension at the outset when no thread has failed before TB Note that X is the strength of the thread while TB is the strength of the bundle When XB reaches X X also reaches X if no thread has failed X reaches TBNl when one of the threads fails Similarly X reaches TBNi when i of the threads fail The probability that the bundle fails at the average thread strength X3 is denoted by FBX PXB S X 436 Thus given the thread strength FX Daniels theory leads to the bundle strength FBX N FBX Z Z 1NnN FNXr1r1FNXr1r2r2 FXrnr1r2 rn 437 nl r In the above the 1nner sum 1s taken over r r1 r2 rn and r1 r2 rn are 1ntegers equal or greater than 1 their combination is subject to the condition r N 438 For a bundle of only 2 threads N2 the running number n can only be 1 and 2 So for nl there can be only r1N2 for n2 there can be only r1l and r2l Accordingly we obtain from eXpanding 437 the following eXpression for the CDF of the bundle strength FBX 2F2X FX FX2 439 EXpansion of 437 for a bundle of 3 threads N3 is similar but is considerably more tedious Details are le in one of the assigned eXercises Beyond N3 eXpansion of 437 becomes unmanageable However when the value of N becomes larger say Ngt10 Daniels showed that FB in 437 reduces to the CDF of a normal distribution In that case the normal parameters in FB namely HE and 6B can be eXpressed in terms of the parameters in the pdf of the individual threads the pdf of the threads need not be normal That part of the derivation however is outside the scope of this Chapter ChapterIV Failure Rates IV28 Example 419 Suppose the tensile strength X in GPa of a single fiber is given by the CDF Fx PX x 1 expx87 Then at 1 of failure probability the maximum applied fiber stress is determined from Fx 001 1 expx87 Solution of the above yields x 415 GPa Now if two fibers are bundled together loosely the maximum applied bundle stress at 1 of failure probability is determined via 439 FBx 001 21exp2x871expx87 1expx872 The above yields x m 402 GPa Discussion From this example the strength of loose bundle is weaker than that of the single fiber Alternatively the probability of failure of a bundle of N loose fibers under TNx is actually greater than that of the single fiber under the load of x Note The CDF in 439 for the bundle of 2 fibers has actually been obtained earlier in Chapter 111 Example 313 case c In the latter the random variable Xx is the total load on the bundle39 the load on the single fiber is thus x2 when no fiber fails the load on the bundle and on the surviving fiber is x when one fiber fails Failure Rate of A Loose Bundle If the random variable XB in 436 and 437 is a time variable one can simply replace X by t to obtain ft or FBt the corresponding failure rate function for the bundle ABC is then obtained from using 41 X130 dFBtdt11FB01 fBtRBt 440 ChapterIV Failure Rates IV29 Sum m ary This chapter introduces the reliability function RtlFt where Ft is the probability of failure during the time period 0 to t An alternative expression for Rt is through the failure rate function Mt as expressed in 42 The reasons for expressing Rt in term of Mt instead of ft or Ft are several a The timeprofile of Mt reveals the physical modes in which failure occurs namely the wearin mode the constantrate mode and the wearout mode knowledge of the failure mode helps to devise proper reliability enhancement measures b It is generally less timeconsuming in collecting test failurerate data than the timetofailure data and c It is mathematically simpler in formulating reliability models in terms of failure rates especially for systems of multiple units In studying this chapter and when working on the assigned exercise problems pay attention to the following areas I The physical meaning of and the mathematical relations between the timetofailure pdf ft and the failurerate function Mt see 41 to 44 I The mode of Mt whether it is in the wearin mode constantrate mode or wearout mode Failure mode dictates the approach taken to address the reliability concern for instance the wearin mode is mainly an offline qualitycontrol issue the wearout mode is a repairreplacement issue and the constantrate mode is a salewarrantyinventory issue I The timedependent wearin and wearout modes Each is attributable to distinctive causes briefly the former is related to early failures due to birth defects while the latter is related to longterm failures due to accumulative damages fatigue both are intrinsic factors I The timeindependent constantrate mode Physically it implies that the product inherent no major defects and has incurred no appreciable damages failure occurs due to random extrinsic factors such as accidental over loading service misuse etc I The reliability models discussed in this chapter Most are based on c0nstantrates only for reasons of mathematical simplicity these include unit under repeated demands unit of stepwise rates and regularly maintained unit I Models for tim edependent rates This topic is narrowly discussed but pay attention to the concept of prooftest and the failure modes implied in the familiar normal lognormal and Weibull functions I Systems of multiple units of constant rates These are modeled only by assuming failures of the units are independent events again for reasons of simplicity the bundle theory discussed brie y at the end of the chapter is an exception only to illustrate the mathematical complexity involved in modeling such systems ChapterIV Failure Rates lV30 Assigned Homework 41 Fit the test data given in example 41 to a failure rate function a Comment on the mode of failure b Compute the value of Rl539 c Compute the meantim etofailure 42 Given Mt kt k is apositive constant and t 0 a What is the associated failure mode b Determine the associated ft Ft and Rt in terms of k39 c Sketch ft Ft and Rt on a graph paper if t is in hours and k 0 lhr2 43 The pdf for the timetofailure in hours of a system has the following form ft 0001 exp0001t a Determine the corresponding failure rate function b What is the associated mode of failure c What is the reliability Rt when t100 hours d What is the value for the MTTF 44 The reliability function of a machine is described by Rt exp004t0008t2 tin years a What is the corresponding failure rate function b What is the associated mode of failure c What is the design life at which the reliability is better than 90 failure probability a Mt 0040016t c tl907 yrs 45 A device that controls a machine has a constant failure rate of 07yr39 in order for the machine to function normally for a long time the device must be repaired at once upon failure Then a What is the probability that the device fails during the first year of operation b What is the probability that the device fails during the second year of operation c What is the probability that there will be at least one repair in 3 years of operation b 5034 c 8775 46 The failure rate of a circuit is constant in time with the value of 004day If 10 circuits are put in continuous and independent use a What is the probability that there be none failure during the first day of use b What is the probability that there be exactly one failure during the first day of use 0 What is the probability that there be more than one failure during the first day of use d If the circuit is repaired upon failure What is the probability there be more than one failure in 3 days a 67 c 616 d 3374 47 The landing gear of an airplane has the constant failure rate Ap0000001hr when in parking M00000ll I when taxiing on the runway and Af 00000001hr when in ight During takeoff it has the Chapter1V Failure Rates 1V31 failure probability p1 0000139 during landing it has the failure probability p 00002 Typical ight schedule for this plane in a 24hour day is 4 takeoffs and landings the average taxing time before takeoff and after landing is 30 minutes the average flying time between takeoff and landing is 25 hours a Develop an expression for the combined failure rate similar to 411 b Estimate the reliability of the landing gear in a 30day service c How often in hours should the landing gear be maintained as new for 99 reliability 48 The timetofailure pdf of an electric clock is normally distributed with 115 years and 008 years a Determine the design life for 99 reliability b What is the reliability that the clock will run for 3 years c What is the failure rate if the clock has successfully served 3 years without failure d What is the probability that the clock will run for another 3 years without failure a t 3144 yrs b R3 9938 c 713 0022yr d F33 1063 49 A motionsensing floodlight is turned on when it senses a motion in the dark39 it is lit for 5 minutes then turned off if there is no more motion detected On the average the floodlight is turned on 12 times during each night The failure probability of p00005 when it is turned on39 the failure rate of 00001hr when it is lit and 000001hr when it is off a What is the combined failure rate for a 24hours cycle b What is the reliability for 30 days of service c How many spares should be kept for the period from November 1st to March 315t for 99 reliability a Mom 00002638hr b 173 c 4 spares 410 A cutting tool has the MTBF2000 hours Assume random failure a What is the reliability for a 500hour mission b What is the chance that at least one repair is needed during the 500hr mission c What is the probability that more than one repair is needed during the 500hr mission d What is the probability that more than two repairs are needed during the 500hr mission a R500 7788 b Pngt0500 2212 c Pngt1500 265 c Pngt2500 022 411 A relay circuit has an MTBF of 08 years Assuming random failure a What is the reliability for one year of service b If two relays are arranged in parallel what is the reliability for one year of service c If five relays are arranged in parallel what is the reliability for one years of service Do this part by expanding 434 for N5 a R1 2865 or F1 7135 b F21509 412 A widget has the timetofailure probability described by a Weibull distribution with MTTF of 5 days and standard deviation of 12 days If 10000 widgets are prooftested for 1 day before they are put into actual service a Determine the failure rate function for the virgin widgets plot the function for 0lt t lt10 days b Determine the failure rate function for those passed the prooftest39 Chapter1V Failure Rates 1V32 c What is the expected number of failures during the prooftest d What the number of failures during the first day of the wedges in actual service Hint determine numerically first the parameters In and 0 for the Weibull pdf b M1 00015day c F1 003 3 will fail proof test d P1lttlt2 081 or about 81 failures out of 10000 413 A designer assumes a 90 probability that a new machine will fail between 2 and 10 years in service a Fit a lognormal distribution to this design b Compute the MTTF c Obtain the associated failure rate function d plot the function for 0lt t lt10 years a t0447 0J0049 b MTTF 5 years 414 A unit consisting of three identical components with the reliability R it can be connected either in a quotstarquot or in a quotdeltaquot as shown below A A 6 3 C 3 6 B a C By means of the inseries andor inparallel models show that the reliability of the quotdeltaquot is better than the quotstarquot B RA7RRR3 415 A network of 4 identical components with the reliability R0 are arranged in configuration A and B as shown below Derive the system reliability for configurations A and B respectively ChapterIV Failure Rates IV33 Configuration B is better than A RBRA 2R022R02 416 Optional Derive the CDF FBX of bundle strength for a bundle of 3 loose fibers given the CDF Fx of fiber strength Use 437 and let N3 FBxFx33F3xFx23FxF3x226F3XF3x2Fx ChapterV Testing Vl CHAPTER V RELIABILITY TESTING One principal reason for investigating product reliability is to insure andor improve the quality of the product at par with the expected service life but a fullscale reliability investigation can be costly As a minimum adequate time to failure andor failure rate database is needed in order to determine the pertinent reliability function for the product in question In general such database is generated either by a test program involving new products randomly picked offline or by inservice fieldtests or both In either instance the data so gathered re ects the quality of the product after it is already made off the production line furthermore the cost or the time needed to complete such a fullscale testing program can be prohibitive For this reason product engineers have devised tests that may 1 improve the quality before it is made or 2 require limited offline tests andor 3 accelerate the time of the tests provided that the data so gathered still capture the reliability characteristics of the product V 1 Reliability Enhancement Reliability enhancement is practiced during product design and development stage before full scale production In this stage product prototypes are constructed based on initial designs and put in a certain simulated testrun Generally the first round of prototype testing tends to fail early and frequently if so diagnostic actions are then taken to identify the modes andor mechanisms of failure of course other possible deficient factors inherent in the initial design may also be identi ed design modification is subsequently implemented in the next round of prototype construction and it is expected to enhance the product quality as well as reliability Clearly the enhancement can be more or less effective depending on the efficiency of the diagnostic and designmodification cycles A frequently practiced approach is to testrun the prototypes under over load andor severe environment conditions The idea of the former is similar to that of proof testing it helps to identify factors that cause product infantile mortality The latter is related to product quality that may degrade faster in severe service environment Both ideas are a way to accelerate the product timetofailure process The specifics in carrying out a reliability enhancement program often depend on the individual product in question besides to discuss any diagnostic or designmodi cation details is beyond the scope this chapter In what follows we brie y describe some of the techniques used in reliability enhancement tests The Duane Plot To a large extend infantile product failure is traceable to defects introduced in the design and development stage it is hence essential to debug any possible defect that can be corrected in the design modification cycle The Duane Plot is an empirical method to achieve enhanced reliability through the debuggingmodification iteration the general procedure is as follows In the first round one or more prototypes are constructed and testrun to failure Suppose that the causes for failure are diagnosed and are traced to deficiencies in the design a designmodification then follows to eradicate or minimize the deficiencies a new round of prototypes are built and test ChapterV Testing V2 run again If additional failures are discovered and the causes are ascertained a new round of design modification is then carried out The testdesignmodification cycle can be repeated several rounds until the design is rid of all possible defects The Duane plot is an analysis tool to minimize the number of testdesignmodification cycles or the number of prototypes required in each test cycle Now let 139 be the accumulated test time during successive prototype testruns and let 111 be the number of failures observed during the period from 0 to 5 Though the value of 11 increases with E the frequency of failure decreases with each designmodi cation cycle Hence in general the quantity T E E representing the averaged failure rate during the period from 0 to 5 will decrease with E In particular T E E and I maintain a linear relationship in the loglog scale 1n1139c39c ocln39c 1 51 In the above or and l are coef cients de ning the straight line The linear behavior depicted in 51 empirically found by J J Duane 1964 is quite pervasive over a variety of products in the design and development stage but it is not clear why it exhibits a linear behavior Use of 51 is illustrated in the following example in particular the meaning of the coef cient at will become evident Example 51 A computer circuit is developed to interface a complex mechanical device Failure of the circuit occurs during testrun debugging and design modification are made and new round of testruns are conducted In this case 8 cycles are repeated with one prototype tested in each cycle The following is data for nearly 240 hours of testruns in which 8 failures had occurred failures 7 l 2 3 4 5 6 7 8 cumulative test time 5 11 39 82 178 797 1131 2084 2391 hrs The Associated Duance Plot ln1111 versus 1111 lnn 01 11 5 C ChapterV Testing V3 Signi cance of plot We observe that the data points in the plot follow a straight line with the correlation factor r20988 In fact the slope of the line is found to be on 7 0654 The meaning of on is interesting it is known as the reliability growth coefficient Graphically it is the slope of the straight line and its value lies within the interval 071 Physically the value of ion is a measure of the efficiency of the designmodification cycle When oc0 the line is flat or the number of failures is proportional to the cumulative time 539 this means that no reliability enhancement is achieved In that case two possible explanations exist a Failure occurs not due to defects in design or b Designmodification cycle is inefficient or defects are not eliminated Usually the latter is the case When oc gt l the line is vertical it means that the number of failures is a constant independent of the length of testrun Hence the first round designmodification has already reached perfection In this example the reliability growth coefficient is 0654 it indicates that the designmodification is relatively effective Furthermore the last two data point appears to be unnecessary as the first 6 data points would have been sufficient to provide the same straight line or the same ocvalue any more growth testruns would not add much improvement but only adds more cost and time Environmental Stress Test Engineering materials such as polymericbased plastics are prone to degradation by environmental agencies the latter can include factors of temperature change humidity and vapor invasion longterm cyclic loading rusting due to oxidation ultra violet light exposure etc Generally environmental effects are insides in nature and they accumulate over a long period of time product reliability enhancement against these environmental effects can bring the level of product quality higher especially during the product infancy period A procedure known as environmental stress test is commonly employed during the product design and development stage Usually one or more selected environmental variables are introduced into the reliability growth test along with normally applied mechanical loading For instance temperature conditioning may be superimposed onto mechanical stressing in prototype test runs timetofailure can be accelerated by testing under a higher temperature The latter is known as the time temperature correspondence behavior Thermal cycling and cyclic over loading are other ways that can be superimposed onto normal stressing during the test runs In addition to possible time saving environmental stressng along with the normal loading provides a combined effect on product failure mechanisms and failure modes Sometimes the combined effects are additive most often they are coupled depending on the specifics of the problem at hand V 2 Censored Reliability Test Let us consider the following situations An electric motor is designed with a running life in the hundreds of millions of cycles If the motor runs at 3600 1pm it would take 193 days to reach 100 million cycles ChapterV Testing V4 An automobile tire is designed for a life of about 40000 miles in normal use In a simulated testrun the tire accumulates 40 to 60 miles per hour failure of a tire under such test is not expected to occur for at least 600 to 1000 hours Censored reliability test is a technique used to gather timetofailure data from test specimens some or most of which are suspended from the test before failure occurs Sometimes data from suspended tests may also come expectedly due to sudden interruption of the test apparatus or due to power outage Though a censored data point is not a timetofailure point in the true sense it nevertheless represents a timepoint at which the specimen has lasted some amount of time without failure Thus censored data can be statistically significant if handled properly Censored Ungrouped Data Let t1 t2 t3 tf tN be a sample of N ranked timeto failure data The i3911 data tf with a sign signi es that the test was censored suspended at ti without failure Now if ti was not censored we can t the data with a probability distribution anction ft or Ft using say the meanrank method Fti iN1 Since Rt1Ft we have Rti N1iN1 52 From the above we can write 11011 N2iN1 53 Combining 52 and 53 we have Rti N1iN2iRti1 54 In 54 Rti1 is the reliability at ti1 Rti is that at ti Hence the quantity N1iN2i is the reliability between ti1 and ti given the reliability Rti1 Rtiti1 N1iN2i 55 With 55 54 isjust a statement based on 25 ROi Rtiti1Rti1 56 Now in the event that censoring takes place at ti ie no failure takes place at ti then there is no change of reliability between ti1 and ti Then it follow that Rti Rti1 or ChapterV Testing V5 Rtiti1 l 57 Between 55 and 57 we can state the following if censoring is not taken at ti Rtiti1 is given by 55 and if censoring is taken at ti Rtiti1 l as shown in 57 If there are more than one censored data taken either before or at ti the reliability Rti can be expressed in the general form ROD ROME 1 Rti1ti2 Rti2t13 Rt10R0 58 Note that the terms Rtiti1 etc must take the appropriate values calculated either from 55 if not censored or from 57 if censored In 58 R0l as it should In this way one can obtain Rti versus t from the sample with one or more censored data Of course from Rti versus t one easily obtains Fti versus t which can be tted with a properly selected parametric function be it the exponential normal log normal or Weibul etc Example 52 Ten electric motors underwent life testing three of the motors were censored indicated by a sign below The failure times t in hours are ranked as 27 39 40 54 69 85 93 102 135 144 Solutions Here il10 and censoring took place at i369 We first calculate the quantities Rtiti1 using 55 or 57 depending on whether or not censoring takes place at ti We then calculate Rti according to 58 for all il10 These result are summarized in the table below 1 ti ROME 1 ROD 1 27 0909 0909 2 39 0900 0818 3 40 1000 0818 4 54 0875 0716 5 69 0857 0614 6 85 1000 0614 7 93 0800 0491 8 102 0750 0368 9 135 1000 0368 10 144 0500 0184 Let us spot check the calculations listed in the table above At i2 t23939 the data is not censored and we calculate using 55 Rtiti1 Rt2t211012102209 ChapterV Testing V6 Use 58 and we calculate Rtz RtZt1Rt1t009x0 9090 8 l 8 At i3 t340 the data is censored Rti does not change from t39 to t40 so Rt3Rt20 8 l 8 At i4 t54 the data is not censored so Rtiti1l0l410240875 Rt4 00874xl000x0900x0909 0716 A plot of the calculate Rti versus ti is shown below ce nsore d data b t 0 50 100 150 hours Discussion In the above plot there exists an extra data point at t00 and ROl and we have attempted to link only the uncensored data points by drawing a piecewise linear line Note that the censored points are left outside the line The line represents the approximated Rt function which can always be fitted by an analytical function Even though the censored point are not a part of the line on the graph the effect of censoring is inherent in the calculation of Rti For instance if the three censored points were actually failures the Rt line would have gone down to intersect the taxis before t150 hours with the censored points Rt is still positive finite m 02 at t150 hours An alternative is to draw the line pass through all the points censored or not Then a slightly different Rt function will result In practice the former approach not to include the censored points is preferred since the fitted line tends to underestimate rather than overestimate the Rt function to underestimate the reliability is safer Censored Groupned Data When the reliability test sample is large say Ngtgt100 and the failure time may extend to several decades of the timeunit used say ti ranges from 1 to 1000 hours it is a common practice to group the data in nite number of time intervals say 7 to 10 intervals ChapterV Testing V7 refer to Sturges formula in Chapter III and the grouped data are then fit to a probability distribution function The question here is what to do if there are censored data in the sample First let the class intervals be selected in the following way t0 t1 t2 ti tn where t0 is the initial time say to0 the first interval is from tO to t1 the second interval is from t1 to t2 and so on Suppose mi data points begin at the start of the i3911 interval from ti1 to ti di data points fail within the interval and ci data points are randomly censored during the interval Then the conditional probability of survival at the end of the interval is approximated by Rtiti1 1 dimi 05ci 59 Note that at the beginning of the ilth interval there will be mm midici items under test With 59 valid for all iln the estimated reliability Rti is given by 58 again Example 53 206 turbine disks were tested for failure Timetofailure data are tabulated in 16 intervals each may be 100 to 200 hrs in length During each of the intervals failure may or may not occur in some intervals there may be censored data The rough data are tabulated in the first 5 columns of the table shown below i tim einterval mi di ci ti Rtiti1 Rti 1 0 200 hrs 206 0 4 200 10000 10000 2 200 300 202 1 2 300 09950 09950 3 300 400 199 1 11 400 09948 09898 4 400 500 187 3 10 500 09835 09735 5 500 700 174 0 32 700 10000 6 700 800 142 1 10 800 09927 09664 7 800 900 131 0 11 900 10000 8 9001000 120 1 9 1000 09913 09580 9 10001200 110 0 18 1200 10000 10 12001300 92 2 5 1300 09776 09366 11 13001400 85 1 13 1400 09873 09247 12 14001500 71 0 14 1500 10000 13 15001600 57 1 14 1600 09800 09062 14 16001700 42 1 14 1700 09714 08802 15 17002000 27 0 5 2000 10000 16 20002100 22 1 2 2100 09524 08384 Note The time intervals shown above are not the same length the first and the 6th interval are 200 hours each the 15th interval is 300 hours the rest of the intervals are 100 hours each The reason for the uneven class intervals is probably due to limitations encountered during testing Basis for calculation the censored data in each interval are assumed taken randomly hence the calculation of Rtiti1 can be based on 59 the calculation of Rti is based on 58 The calculated values for Rtiti1 and Rti are listed in the last two columns of the above table For a spot check at i10 ti1000 hour From 59 the calculated Rt10t9 is 12922509976 from 58 the calculated R10 09976XR9 09976x0958 09366 ChapterV Testing V8 A plot of the reliability function Rti versus ti is shown below ll Rt data tted 08 s extrapolated 07 t 0 500 1000 1500 2000 hours Discussion Of the sample of 206 data there are only 13 failures with 172 data censored By the end of 2100 hours 22 items are still under test Thus the fitted Rt function is based on life data from t0 to t2lOO hours only The plot however can be extrapolated for longer times if needed to as shown by the dotted line in the plot There may have been too many censored data in the early time intervals see column ci in the above table Usually the number of censors should be more or less constant throughout the time intervals V 3 Accelerated Reliability Testing Condensed Time Test Many products are not operated continuously in real time Home appliances for instance are used only a fraction of the day but are expected to be operational for many years of use Actually over the life of an appliance its actual on time is much smaller than its off time To assess the reliability of these products lifedata collected in the usual way can be expansive and timeconsuming In such cases reliability tests are often conducted in condensed time meaning that the test is conducted under continuous operation without any offtime Clearly the condensedtime method is useful only if the onoff switching has no or little effect on product failure thus a veri cation of this fact is important Generally the veri cation is an integral part of the reliability test itself In this regard some interesting insights may be gained in the following example Example 54 Timeto failure data of a ashlight bulb is generated in the laboratory where the bulb ChapterV Testing V9 is lit by a constant 6volt DC source Data from two sets of tests are collected a Condensedtime tests where 26 bulbs are lit continuously till burnt out39 and b Cyclic onoff tests where 27 bulbs are turn0n for 30 minutes and turnoff for 30 minutes and so on till burnt out The following is a tabulation of the respective data in calendar hours from the two tests CondensedTime test Cvclic OnOff test 72 82 161 177 87 97 186 186 103 111 196 208 113 117 219 224 117 118 224 232 121 121 241 243 124 125 243 258 126 127 262 266 127 128 271 272 139 140 280 284 148 154 292 300 159 177 317 332 199 207 342 355 376 Statistical analysis of the above data using the quick firststep estimation technique see Chapter 111 provides the following results 12842 hours 03132 hours for the condensedtime test39 257 11 hours 05590 hours for the cyclic onoff test Since the ontime in the cyclic onoff test is just onehalf of that in the condensedtime test the parameters u and o in the onoff test should be reduced by half 12855 hrs 02795 hrs for the cyclic onoff test Thus the two sets of data have nearly the same mean 12842 versus 12855 but their respective standard deviations are somewhat different 3132 versus 2795 namely the cyclic onoff data is slightly less scattered than that of the condensedtime data It is not clear if there is any physical reason for this difference But based on the quick analysis it suggests that the cyclic onoff test has no or little adverse effect on the flashlight bulb s operational life distribution the condensedtime test can be used as an acceptable reliability test method Discussion The above data may be better fitted by a known distribution function such as the normal or the Weibull39 and use of the leastsquare method would further yield a better fit Over Stress Test Many products are in continuous use without any signi cant offtime Failure of such products is usually caused by longterm deterioration known as fatigue during their service life To access their reliability life tests involving over stressing may be employed in ChapterV Testing Vl 0 order to shorten the timetofailure For the method to be useful however overstressing must not signi cantly alter the product failure mode and mechanisms By overstressing it is meant that the severity of loading is elevated higher than the designed load so failure may occur in short timeunderload But mechanical load is not the only over stressing agency for instance electronic components may be tested at elevated temperatures in order to hasten the incidence of failure Similarly steel pipes in nuclear power plants may be exposed to extreme neutron irradiation that can increase the brittleness of the steel thus causing brittle failure To apply overstressing in reliability test it is not necessary to require that the failure mode be the same at all the stress levels The key idea here is to obtain a relationship between the stresslevel and the product lifetime with some level of con dence The following example illustrates the essence in overstress testing Example 55 A steel crankshaft is designed for the designstress 0d and the designlife td Life tests are then conducted on the crankshaft at three stress levels 01 02 and 03 which are all higher than the design stress 0d At each of the three test stress levels life distribution data is gathered and the results are displayed in a 0 vs t plot as shown below stress level test stress levels design load 6d V timet ofailure Discussion In the above life distributions the shaded normallike curves at the three test stress levels are plotted with test data39 but the life distribution curve for the design load Cd is projected by extrapolation without any test data The point here is that at the higher stress level test time needed to generate data is relatively short compared to that needed under the designstress level 0d In the plot shown above two parameters are of interest the mean u and the standard deviation 0 The distribution means at the three teststress levels are fit into a decreasing function say ut shown by the solid line the dash line is the projection of ut towards the distribution mean at the design stress 0d ChapterV Testing Vl 1 1n the same plot it is also noted that the life distribution curve is less scattered at the higher stress level than at the lower stress level The physical reason is that at a higher stress level the most dominant mechanism usually emerge and cause failure in a particular mode39 at lower stress levels more than one mechanisms may be simultaneously in effect and one of them may ultimately become the dominant failure mechanism Hence among many specimens tested there may occur more than one failure mode when tested at a lower stress level The SN Curve Fatigue test on structural materials often takes the form of cyclic loading about a meanstress level In that case timetofailure data are plotted with the meanstress S versus the number of load cycles N most often the log of N the line that connects the means at the different S levels is commonly known as the S N curve Fatigue Limit The rate of decrease of the SN curve in time signifies the ability of the material against fatigue39 if the curve is steep it indicates the material being fatiguesensitive if it is relatively flat it indicates otherwise Extension of a relatively at SN curve establishes the socalled fatigue limit meaning the material is fatigue free if tested under that stress limit Example 56 Flashlight bulbs considered in Example 54 were designed to operate under 6v DC power In an accelerate test the bulbs were subjected to 9 12 and 15 volts respectively39 12 bulbs were tested at each voltage level and the timetofailure data in hours are tabulated below w 44 15 8 56 19 9 58 23 9 59 25 10 60 28 11 61 30 11 62 32 11 63 34 12 64 37 12 70 37 13 74 39 13 88 41 15 Solution As was done in the previous example the above 3 sets of data are analyzed using the first step estimation for the respective distribution parameters p and o and these are tabulated as follows 9volts 12 volts 15 volts Mean 6325 3000 1116 Std Dev 1586 822 203 From the above we see the trend that both the distribution mean and standard deviation decrease with the test voltage39 in fact the rate of decrease in both parameters is rather sharp Physically a rapidly decreasing mean indicates that timetofailure is accelerated with increasing rate while a decreasing standard deviation indicates that failure is possibly dominated by a single mechanism ChapterV Testing V12 In order to obtain the distribution mean and standard deviation for the ashlight tested under 6volt we use the above overstress test results to make an extrapolation for the each of the parameters To do so we plot both 11 and 0 versus the three overstress test voltages as shown in the figure below A H A 526 6 15 25 132 100 15 std dev 50 mean 5 39 0 0 I I I 5 3 6 9 12 15 V The above plots show that the mean is a slightly nonlinear function of the testvoltage while the standard deviation is nearly a linear function By extrapolating the plotted curves toward the 6volt level the values of p and 0 under 6volt are estimated as follows u132 hours o26 hours Compare the above results with the respective values of u12842 hrs and o3232 hrs tested under 6volt in Example 5439 we see that the accelerated test provides a reasonable extrapolation for p and 0 under 6volt Discussion In the above the nonlinear plot of the distribution mean can be replotted in terms of lnu versus the test voltage voltage 9volts 12 volts 15 volts lnu 415 340 271 Then a straightline relation is obtained as shown in the plot below ChapterV Testing Vl 3 111u49 p exp49 1343 hours By extrapolating the straight line to 6volts we find lnu 49 or u 1343 hours In practice the condensed time and over stressing concepts are often used in combination in accelerated life tests it usually can provide further reduction in test time ChapterV Testing V14 Sum m ary When a product fails to perform the designed functions in life it is often difficult or at least timeconsuming to pin down the real cause or causes Most often a sufficiently large database is needed in order to render a rational estimate for the reliability of the product This chapter briefly introduces the concept of reliability enhancement it is an effective approach to eliminate or at least minimize the defects that are inherent in the product designanddevelopment stage Methods of censored test and accelerated test are all aimed at savingtime in gathering the needed statistical data I The Duane Plot is an empirical method that can be applied in a reliability enhancement program The essence of the method is illustrated in Example 51 Distinguish the difference between the ungrouped censored and the grouped censored tests The former is illustrated in Example 52 the latter in Example 53 Be familiar with the way the test data are analyzed in each case The condensed time is applied to products that are operated with some ontimes and some oftimes And the method is reliable only if the onoff switching operation does not significantly cause failure or accelerate the failure rate See Example 54 for some details Overstressing is another approach to accelerate product failure The general provision is that the failuremode and failure mechanisms caused by overstressing should not be much different from that caused by the product s designstress See Examples 55 and 56 for some details Assigned Homework 51 Use the test data provided in Example 51 a Apply the leastsquare fit to the Duane plot39 b Verify that OL0654 and r2 0988 52 By debugging a computer software failurecausing defects are found and then corrected at 14 89 243 681 1172 and 2293 hours Make a Duane plot and estimate the reliability growth coefficient on on m 7065 53 The wearout times of 9 emergency road ares are 170 206 213 214 227 256 270 277 and 2974r minutes numbers with a superscript refer to ares distinguished by accident a Use the ungrouped censored method to make a plot for the reliability function b Determine the probability of wearout within 24 hours for a randomly picked flare 54 Grouped data uncensored for the timetofailure in unit of 103 hours of an electrical circuit are given as follows 0 t 6 5 6 t 12 19 12 t 18 61 18 t 24 27 24 t 30 20 30 t 36 17 a Make a plot for the reliability function b Estimate the reliability of the circuit for the design life of 10000 hours 55 Repeat Example 56 by fitting the data using the normal distribution plotting paper39 compare the results with that obtained in the example ChapterVI Quality Control VIl CHAPTER VI PRODUCT QUALITY MEASURES Quality and reliability are often synonymous in the views of consumer For product engineers however there are distinct measures in each case in term of design development manufacture and service performance of the product As we have learned in Chapters IV and V reliability refers specifically to the probability of survival of a product or system within a certain service life and that possibility can be improved through some reliability enhancement programs including quality control In broad terms product quality is associated with a the ability to incorporate and optimize all the design parameters to meet the specified performance targets including product reliability and b the ability to reduce variability in meeting the performance targets Generally speaking the former is central to quality assurance in the early design and development stages while the latter is essential to quality control during the manufacturing and infreld service stages The owchart below illustrate the intertwined relationship between quality and reliability Reliability Improvement Cycle product jnception product development manufacture msemce product life 7 design amp modification online QC field service A This chapter discusses only some of basic notions pertaining to on line quality control QC Other aspects in the reliability improvement cycle will be included in a subsequent course on quality control in much broader terms VI l On Line Quality Control Desi Target and Quality Variability In engineering a product is almost always designed with one or more quality indices X and each is tied with a design target Ei In the most ideal situation the actual product should all be meeting the targets X Ei However during the process from design to manufacturing many random factors may be introduced and the quality of the product can inherent various degrees of scattering Thus each of X can become a random variable For simplicity let there be only one index X and one target X39c And online inspection of the product for the index X often yields a normal like distribution fX with the mean u near or around the target 5 and with some degree of scatter that can be measured by the standard deviation 6 such as shown schematically in the figure below ChapterVI Quality Control VI2 ll x 6 c lt gtlt gt l LSL US39L 7 lt u gt 5 ta1get In the above gure if the distribution mean coincides with the target the quality index X is said to be on target otherwise it is off target As for the variation of X from its mean value it is usually judged by a set of independent speci cation limits labeled USL upper speci cation limit and LSL lower speci cation limit respectively The signi cance of the various elements shown in the gure is as follows a We would want u E39 that is the mean quality is on target b We would want 6 be small so X is clustered around the target 5 Thus through quality control one attempts to place u on target and reduce 6 to an acceptable small value This however requires a identi cation of the physical factors that in uence the values of u andor 639 and b the implementation of corrective actions in order to meet the limits on these parameters in the quality indeX distribution Generally speaking factors in uencing u are inherent in the designdevelopment stage most are easily identi ed though some may be traceable to manufacturing precision or the lack of it Of course these factors must be identi ed accurately so that relevant corrective measures can be devised in the design modi cation cycle On the other hand factors that in uence 6 are more random in nature some are traceable to materials variability others may be due to variables encountered during manufacturing and handling processes These variables are usually dif cult to identify and not easily controlled Methods and techniques for controlling the mean and the standard deviation are subjects belonging to product robust design to be brie y discussed later in this chapter AcceptanceRejection Limits In the previous gure the acceptance limits USL and LSL are set for purpose of acceptreject evaluation Thus given the quality distribution fX or FX those products with X below the LSL and above the USL will be rejected thus the fraction of rejection is FLSL lFUSL or those that are accepted known as the yield is Y FUSL FLSL 61 In particular if the distribution mean is on target u39c the speci cation limits can be set in the following manner USLuA and LSLu A ChapterVI Quality Control Vl3 A fX reject LSL 4 A A Here 2A is known as the acceptance gate widt and the white area between the specification limits represents the fraction of product yield If the normal fX is standardized see chapter 111 for details through 2 Xu6 we can readily obtain LSLu6 A6 and USLu6 A6 Hence the yield in 61 can be expressed as Y l 2 A6 62 In this case since ut the fraction of yield depends singly on the distribution standard deviation 6 Note that setting 2A is independent of the manufacturing process itself so the online quality control is based sorely on the value of AG The 3 6 Criterion The conventional online quality control follows the socalled 3 6 criterion Speci cally the product quality distribution fX meets the conditions pm and 6 S A3 63 Under this criterion the quality of the product will pass the QC with Fraction yield 2 l 2 3 9973 or Fraction rejected S 2 A6 or S 2 3 0027 The 6 6 Criterion Most modern electronic parts eg computer chips are required to meet more stringent quality speci cations The 6 6 criteria requires that ut and 6 S A6 64 In this case with 6 0000000001 the quality ofthe part will pass QC if and only if Fraction rejected S 2 6 0002 ppm or ChapterVI Quality Control VI 4 Fraction yield 2 1 0002x106 Example 61 The manufacturer of a bearing ball performs an online inspection and finds that the bearing ball diameter can be described by a normal distribution with the mean no and 00 Although no is on target the value of 00 is unacceptable An online QC program is then instituted which screens out the bearing balls with diameters outside 14500 from the mean Now examine the following questions a What is the yield after the QC screening b For the bearing balls passing QC what is the new diameter distribution c What does the QC actually achieve Answers a The yield after QC is obtained via 62 Y l 2437156000 l 2006681 86638 b Assume that the diameter distribution of the 86638 yield is a truncated normal function its mean remains the same as upo due to symmetry while the standard deviation changes by a factor of K oKoO thus 1 1 xu02 fx ex 7 mompi2 K60 Ix H0Ilt1560 fX0 lXHOgt1560 For fx to be a pdf the following must be satisfied 0 K60 I fx dx 1 390 K60 The above in turn yields K074263 and o 07426300 c The QC screening has narrowed the spread of the original diameter distribution39 ie the original diameter scatter 00 is reduced by about 25 Shon Term Process Capability Index The 36 criterion is often used as a standard index for measuring the effectiveness of product processing capability for a batch of products during on line QC The short term Process Capability Index Cp is thus de ned by C1 N36 65 If the quality of the batch meets the 36 criterion exactly that is 1 and 6 A6 then Cp 1 if the quality of the batch falls below the 36 criterion 6gtA3 then Cplt1 if the quality exceeds the ChapterVI Quality Control Vl5 3 6 criterion 6lt A3 then Cp gt1 The index CI is frequently used in industrial circles to measure the processing capability in short term since the product shortterm yield can be readily expressed in term of Cp Y l 2 3Cp 66 Note If the quality of the product meets the 66 criterion exactly then the shortterm process capability index is Cp 2 Long Term Process Capability Index Product quality can vary from batch to batch over long period of time and online QC for many batches the variation in quality can be characterized by the shift of the distribution mean u from the design target I and sometimes also by the change of the standard deviation 6 from one batch to another As an example suppose several batches of a product are impacted online quality distribution of each batch is shown in the following graphical sketch 5 N target 7 loss of precision 7 7 over calibration mn target 7 It is noted that The initial batch of the product is on target ut Loss of precision shits the mean to the left of E in the second batch Over correction shifts the mean to the right of E in the third batch Loss of precision shifts the mean back on target in the fourth batch ChapterVI Quality Control VI6 Suppose the above represents a quality variation cycle and the cycle repeats itself again and again during a long period of processing it is then considered a case of long term processing characterized by the long term process capability index Cpk Note Over long period of time it is possible that the overall quality distribution mean uk is still on the target 539 but the overall standard deviation may be increased to 6k or worse the over all uk is olT target and the standard deviation is increased to 6k In the first case uk39c the long term process index is de ned as Cpk A3ck 67 In the second case H451 the long term processing index is modi ed as Cpk 1kA361Q 68 where k absuk39cA 69 In both cases the long term yield is given by Y 1 2 3Cpk 610 The Taguchi Processing Capability Index Another form for the processing capability index takes into account the longterm effects due to shifting of the mean from the target as well as possible increase of the standard deviation Specifically the effective longterm distribution standard deviation 6m is defined as 61192 6192 Mk 02 611 It follows that the processing capability index is given by Cpm A3cm 612 Note that in the definition of Cpm the longterm quality distribution fx need not be normal As it will be discussed in Section VI3 the expression in 611 is associated with the Taguchi loss axiom Example 62 The diameter of a driving shaft is designed to be 10 cm with the acceptance tolerance ofi 001 cm From online QC over a long period of time it is found that 15 of the shafts exceeded the USL while 004 fell below the LSL Find the long term processing capability index Solutions Here we have 5 1000 cm and A 001 cm Assume normal distribution we have ChapterVI Quality Control Vl7 lt1gt100170K5K1715 0985 lt1gt 99970Kch 004 00004 Using the table in Appendix III and solving for K and UK we obtain K 10002 cm 0K 00036 cm We see that the mean of the distribution is slightly off target 10002 vs 1000 cm For the longterm processing index Cpk Using 610 we compute k 1000210001 0239 and the long term processing capability index is given by 68 Cpk l 020013x00036 0741 Since Cpk is less than 1 the longterm quality of the shafts is not meeting the 30 criterion For the Taguchi processing index Cpm The longterm variance is calculated first by using 611 om2 000362 10002 7 102 1696 x10 5 om 000412 Then from 612 we have Cpm 0013x0004l2 08094 The Taguchi index is also less than 1 Discussion Both Cpk and Cpm measure the effect of longterm deviation of the product quality from the design target39 in practical use the two should not be compared Example 63 A TV producer uses a large quantity of 509 resisters The acceptance tolerance is i259 Two suppliers have provided samples of the resisters and an inhouse QC on 30 randomly selected resisters in each supplier provided the following data Sample 1 4847 4849 4866 4884 4914 4927 4929 4930 4932 4939 4943 4949 4952 4954 4969 4975 4978 4993 4996 5003 5006 5007 5009 5042 5044 5057 5070 5077 5087 51879 Sample 2 4767 4770 4800 4841 4842 4844 4864 4865 4868 4885 4917 4972 4985 4987 5007 5075 5060 5063 5090 5102 5105 5128 5133 5138 5143 5160 5170 5174 5206 52339 Analysis of the above data provides the following points for consideration a Both samples fall within the specification limits of 50 i 259 both are acceptable b The sample data are fitted to a normal distribution via the leastsquare method yielding For sample 1 p 4970939 0 0849 r2 0962 For sample 2 p 5010939 0 1599 r2 0956 ChapterVI Quality Control VI 8 c Assume that the samples supplied were randomly selected from longterm stocks the long term processing capability indices are For sample 1 Cpk 0873 Cpm 0934 For sample 2 Cpk 0503 Cpm 0523 On the basis of the above analysis the following considerations are relevant a The resisters from both of the suppliers fit the normal distribution well b Both meet the acceptance tolerance 4750 to 52509 c Samplel is more offtarget than sample2 d Sample2 is more variant than samplel e Supplierl is a better choice based on the processing index Cpk or We shall revisit this problem latter in chapter based on a different set of considerations VI 2 Systems of Multiple Parts The forgoing discussions were focused mainly on a single item part subject to a single set of speci cation limits Real engineering systems however may contain many parts and each part may be subjected to more than one set of speci cations In many cases the total number of speci cations imposed on a system of multiple parts can grow rapidly The following are some basic notions pertaining to online QC for systems of multiple parts System Yield Suppose that a system is made of N parts each part is subjected to a certain speci cation limits Now let X i1N be the event that the i3911 part fails to meet its speci cation then the probability that all the N parts fail to meet their respective speci cation is given by the intersection of all X PX1nX2nX3n nXN 613 For simplicity assume that X p for all i1N and that the events are independently of each other then the system of N parts passes all the individual speci cation is given by PX 1nX 2rX 3 nX N PX 1PX 2 PX N 1 p N 614 The expression in 614 is actually the system yield based on the abovementioned assumptions Y lpN 615 IfN is large andp is small p ltlt1 615 can be reduced to Y e39NP 616 Example 64 A computer manufacturer found that the circuit boards used in the computer meet the 30 criteria a What can we say about the quality of the board ChapterVI Quality Control VI9 b If the board contains N chips what can we say about the quality of the chip Analysis a Based on the 30 criterion the yield of the board is 9973 b Now suppose there are N chips in the board each chip can fail with the probability 7 Then for large N and small 7 it follows from 616 and the result in a that the yield of the board is Y e39NP 09973 Thus we find from the above 7 ln09973N Now suppose that each board contains N100 chips then 7 ln o9973100 2704x106 This implies that there are to be 27 chip failures per million or 27 ppm Note Here one may consider that 7 re ects the technology level of chip production39 to improve the yield of the board the value of 7 should be reduced Then consider the following situation Suppose the same technology is used ie p 2704x10396 in making the chips then a board that contains N10000 chips has the yield of only Y elOOOO000002704 0763 That is certainly not a desirable outcome Suppose a circuit board contains N10000 chips but the desired yield of the board must meet the 30 criterion ie Y09973 This can be achieved only by improving the quality of the chips ie to reduce the failure probability p which is found by setting the yield of the board at Y 09973 e10000P From that we find 7 ln0997310000 02704x106 This translates to 027 failures per million chips Discussion In modern electronic technology a system say a computer often contains millions of chips consequently the reliability of the chips must be very high in order to insure a high yield of the system computer The Sigificance 0f the 6 6 Criterion In modern times engineering systems such as electronic appliances can contain millions of parts In order to assure such systems with a high degree of reliability the probability of any one part not meeting its speci cations must be measured in terms of a few ppm or less In this context the 66 criterion is often instituted in on line QC that the shortterm quality variation must be such that GSA6 This also implies that the shortterm processing capability index Cp220 Alternatively the fraction rejected on a shortterm basis should be equal or less than p S 2 6 0002 ppm 617 ChapterVI Quality Control VIlO Thus if a system contains N104 parts and each part meets the 66 criterion the system yield on the shortterm basis is given by Y e39NP e000002 99998 However when longterm effects are taken into consideration the system yield is expected to reduce somewhat Let the longterm processing capability index for each part be Cpkl5 the corresponding failure to meet speci cation for each part increases to p 2 3x15 6796 ppm Compared to 617 the longterm quality of the parts deviated much from the 6 6 criterion consequently the corresponding system yield is reduced to Y e39NP e006796 9343 Discussions The above example illustrates the close relationship between system complexity and system yield on both the shortterm and longterm bases When the system is complex ie whenN is large a tight acceptance specification limit such as the 60 criterion may be required for each part then and only then a higher level of system reliability can be achieved Implement the 60 methodology has been a new trend in advanced manufacturing the ISO standards are example are based on the 60 concept For a through coverage of the latter is beyond the scope of this chapter V 3 Loss Due to Quality Variation When a product does not meet the speci cation limits it must be rejected or if not the chances are that it will not perform well within the designed functions In either case this will incur a certain monetary loss to the producer How to estimate the loss due to product quality variation is usually proprietary in business and the model used is a highly guarded secrete Nonetheless the general approach in loss estimate is based on the product quality distribution and the rejectionacceptance criterion In this connection there are two commonly used models the apparent estimation model and the intangible estimation model A typical model for the former is the socalled goal post model one for the latter is the Taguchi model The Goal Post Loss Estimate Model Let L be the cost per product manufactured Then any one such product rejected represents a loss of at least Lo Now if the pdf fx is a measure of the product quality and it must meet the speci cation limits say LSLuA and USLpA then the probability that a product being rejected is given by 2 A6 according to 62 Hence the expected loss per product manufactured is given by EL L02 A6 618 Altematively one can formally de ne a loss function Lx for the product as follows LX 0 for LSL lt x lt USL 619 ChapterVI Quality Control Vlll Lx L0 for xlt LSL and xgtUSL Then weighing against the product quality distribution fx the expected value for product loss is given by the following integral EL j m Lx fx dx 620 0C Substituting the loss function Lx of 619 into 620 and carrying out the integration yield the exact result as expressed in 618 The above can be graphically visualized in the following sketch A fX y1eld reject releel LSL 4 D A A A u A Lx L 0 I I re39ected I I rejeCted J acceptance gate loss no loss 1055 X I I L Example 65 A bearingball producer adheres to the 30 for online QC The cost for each bearing ball produced is 025 Based on the goalpost model what is the expected loss per bearing ball produced due to quality variation In this case the expected loss per product produced is simply EL 0252lt1gt73 00675 cents Or the loss is about 027 of the product cost Example 66 A car manufacturer produces a driving shaft for 350 a piece The length of the shaft is designed to be exactly 100 inches but it is acceptable within the tolerance of i0l inch On line ChapterVI Quality Control VI12 inspection of the shaft finds that 15 of the shaft was rejected because it exceeds the USL and 004 of the shaft was rejected because it is below the LSL Using the goalpost model estimate the expected loss per shaft due to quality variation Solution In this case the total rejection is 15 004 154 so the expected loss is simply EL 350 x 00154 539 per shaft produced Discussion Of course the above estimate is based on the implied assumption of mass production of the shaft Since the length of the shaft is targeted at 100 inches we can actually infer that the shaft length distribution is a normally function Thus with the online inspection data we can write I LSLipo I 9997100 00004 1 USLipo I 1001100 17 0015 0985 By using Appendix IIIA we find 9997100 33 and 10017po 217 From the above we obtain it 10002 and o 0036 The length of the manufactured shaft is off target by 002 The Taguchi Loss Estimate Model In the goalpost loss estimate model the product that is acceptable does not contribute to any loss only the rejected ones do and they each incur the same loss no matter how badly it is rejected In the case discussed in Example 66 above even for those that are off the design target no loss would be incurred as long as the product meets the acceptance limits The Taguchi model Genichi Taguchi proposes that loss can be caused both by quality scatter measured by 6 and the degree of the quality that is of target measured by the difference between distribution u and the target 5 even if the quality actually meets the speci cation limits loss could be incurred simply because it is not on target The rationale behind this model is that there are some intangible effects that cause the oiT target product not performing to the designed level A common form of the Taguchi loss function is de ned as Lx Lo 0 W412 621 Note that 621 contains the absolute value of XT which measure the deviation from the target and it is weighted with respect to the quality acceptance gate A Substituting 621 into 620 and denoting kL0A2 we have ChapterVI Quality Control Vll3 EL j kX 02 fX 1 14 xu2fxdxu rj xufxdxu r2j foodx Carrying out the integration over the range of X typically from oo to 00 and simplifying we obtain EL LoA2 62 H UZ 622 It is noted that 622 is obtained without specifying the specific form of fX only the mean u and the variance 62 of fX need to be specified Hence fX needs not be normally distributed Note also that the quantity 62u39c2A2 in 622 is related to the Taguchi processing capability indeX Cpm as de ned in 612 The Taguchi loss function in 621 is a quadratic function of X the origin of which is at the target 5 and the quadratic range is between LSL and USL It implies that loss will be incurred as soon as the product quality deViates from the target 139 even it is within the acceptance limits A graphical representation of the loss function LX is shown below 11 L00 a parabolic curve Example 67 Let us return to Example 66 and use the Taguchi model in 622 to estimate the loss Here we have 5 100 39 A 01 u 10002 and o 0036 hence 622 yields EL 350001 00362 10002 1002 5936 Point to ponder 539 based on the goalpost model Or 5936 based on the Taguchi model ChapterVI Quality Control VI14 Which estimate is right Example 68 Let us return to Example 63 The two samples of 30 resisters were each fitted by the normal function and the following were the results For sample 1 u 49709 0 0849 r2 0962 For sample 2 u 50109 0 1599 r2 0956 The longterm processing capability indices were calculated as For sample 1 Cpk 0873 Cpm 0934 For sample 2 Cpk 0503 Cpm 0523 From the above we see that both samples meet the specification though sample1 is deemed of having the better quality due to higher Cpk or Cpm Now if we use the goalpost model to calculate loss estimate then either sample would incur any loss However if the Taguchi model is used the calculated loss on per 39 cost basis for each sample wo For sample 1 EL 013 per 39 For sample 2 EL 041 per 219 Based on loss estimate the Taguchi model points to supplier1 as a better choice Smaller is Better and Larger is Better Models The Taguchi model degenerates into two special cases known as the smaller is better and the larger is better respectively In the former case the model implies that it is better if the measure for the product quality is smaller An example of this case is the noise level of automobile engines the lowerspeci cationlimit LSL for the noise level should be as low as possible or LSL gt0 the upperspeci cationlimit USL for the noise level on the other hand is set at some nite acceptable level Thus by setting 1 LSL gt 0 and A USL the Taguchi loss function in 621 reduces to LX LOXZUSL2 623 Then the associated loss for the smallerisbetter case is given by EL L0 USL2 I X2fx dX 624 0 Graphically the smaller is better model is illustrated as shown below ChapterVI Quality Control Vl15 MK MK re39ect L x a gt LSL USL In the larger is better case it is better if the product quality measure is as large as possible An example is the impact resistance of the automobile bumper Here the USL is the ideal target and it is better that 139 gt co the LSL is set at some nite acceptable level In this case the Taguchi loss function in 621 reduces to Lx L0LSL2x2 625 Then the associated Taguchi loss is EL LoLSL2 j x2 fX dx 626 0 The larger is better model is illustrated graphically as shown below rej ect Example 69 The purity of a chemical solution is measured by the of contaminants in the solution Thus it is better if the contaminant in the solution is as small as possible Online inspection of a particular batch of the solution finds that 05 of the solution sampled exceeded the USL and the contaminant distribution in the solution can be described by the pdf fx 100 eW Where on is a parameter characterizing the distribution function Given the cost of the solution at 1000 per pound What is the expected Taguchi loss ChapterVI Quality Control VIl6 Discussion amp Answer In the above two parameters are not explicitly specified on and USL In this case the pdf has the exponential form and the CDF is readily obtained as Fxl e39w Hence the condition F 00 l is automatically satisfied regardless the value of on Now we use the inspection result that 05 of the solution exceeded the USL FxUSL 1 7 em 0995 From the above we obtain USL 52980c Then by applying 624 for smallerisbetter we obtain EL 0712 per pound Note In obtaining EL the parameter occancelled out automatically the value of ocremains still undetermined Example 610 The tensile strength of a coating material is described by the Weibull function fx m0x0m391 expx0m where m4 and 0500 MPa In a certain thermal coating situation the lowest tensile strength limit is 100 MPa The production cost is 3000 per pound Determine a The expected loss based on the goalpost loss model39 and b The expected loss based on the Taguchi loss model Solution This is a larger is better case as coating with tensile strength lower than 100 MPa would fail the specification From the Weibull pdf given above we first find the CDF Fx l expx0m With m4 and 0 500 MPa the fraction of rejected material is Fx 100 l exp1005004 00016 21 Based on the goalpost loss model the expected loss is EL 00016x30 0048 cent b The loss function for largerisbetter is given by 625 Lx LO LSL2x2 The expected Taguchi loss is found by integrating 626 EL 4254 cents per pound Discussions In the above two examples integrals containing exponential functions had to be ChapterVI Quality Control VI17 evaluated in closeform In Example 69 integration by parts was needed in Example 610 a transformation of the variable x was necessary An integration table would often be handy otherwise a numerical integration routine can be helpful as well Example 611 Air bags used in automobiles are designed to fully in ate within 25 milliseconds usec upon impact but airbag full in ation sooner or later than 25 psec is unsafe to the passenger The Highway Safety Regulation stipulates that the permissible time limits for airbag full inflation are 25i0375 psec A producer of air bags finds that the inflation times of their product are normally distributed with the mean of 25 sec on target and the standard deviation of 05 psec In order to meet the Safety Regulation the producer instituted a screen test on the air bags to remove those with in ation times outside the upper and lower limits p i150 a What is the yield of the air bags those passing the screen test b What is the standard deviation of those passed the screen test Solutions The in ation times of the asproduced air bags are normally distributed the pdf is given e ow f0 1027Emlexp tH220l where o 05 and 25 psec respectively Note that the fullin ation time is ontarget psec but the standard deviation of 05psec may be too high In particular within one standard deviation the yield is only Y12ltIgt 03750505468 However the screening test has removed the ones outside the limits of pi 150 Then from 62 the yield from the screen test is Y 1 2ltIgt 15 08664 For those passing the screen test assume that the in ation time is still normally distributed so the new pdf has the form 1 A ft for 7150 lt t lt p15o t 0 for t ltp715o and t gt p15o In ft A is scaling constant that can be determined from the following fwf 0dr 1 J39Mil39saAfawt 08664A 47156 The above yields A 1154 Note that the mean of ft will remain as p 25 psec but the standard deviation of the new f2k t is reduced to 0 4 074260 i 03713 psec Since the Safety Regulation limits are 25i0375 psec we see that the screened air bags provide a yield to Y12ltIgt03750371306876 Discussion The integral associated with the variance 62 can only be evaluated numerically To do so we first transform the variable tto C by introducing C t po It then follows that ChapterVI Quality Control Vl18 15 2 5quot2 2A6 W2 no Qzexp g221dg The integral involving C is then evaluated numerically VI 4 Robust Design A Brief Introduction In product quality control one often places attention to a product yield and b projected long term loss If the quality is ontarget the product yield is determined by the parameter Ao as in A6 see 62 Whereas the value of A is usually preset by the designer along with the target 5 it leaves the reduction of 6 as the only option to increase product yield and at the same time the projected longterm Taguchi loss is reduced see 622 Factors in uencing 6 are intrinsic in nature and they can originate from many random sources Hence it is usually dif cult to reduce 6 beyond a certain limit On the other hand factors that in uence the bias u t are relatively few and they can be identi ed physically known as noises In most cases noises can be controlled in the early design and development stages The essence in robust design is to understand the physical nature of the noises and to correct them in the product designdevelopment stages The nal outcome is to keep the quality mean on target and to reduce the quality variance to as small a value as possible Noise Behaviors Let X the product quality measure Suppose that the noise factor A affects the outcome of X If the in uence mechanism of A on the value of X is known a mathematical relationship or a function xA can be formulated on a deterministic basis In general there exist three distinctive behaviors of xA the linear behavior the softening behavior and the hardening behavior The following examples illustrate the key features in each of these three behaviors Example 612 Torque is required to open the cap of a peanut butter jar39 but the required torque is related to the tightness of the cap The cap tightness is often a random variable stemming from a variety of factors the cap s dimension the material used the ambient temperature etc Let the torque required to open the jar be denoted by X39 the noise be the tightness of the cap denoted by A For simplicity assume that the tightness is in uenced by the stiffness of the material used only Then a certain relationship between X and A exists denoted by the function xA In the following we examine the behavior of this function in the context of robust design 11 Linear Behavior If xA is a linear function the slope of the straight line depends only on the stiffness of the material used Then consider two cases a the cap is made using a softer material39 and b the cap is made using a stiffer material The respective xA functions are illustrated graphically as shown below ChapterVI Quality Control Vll9 f stiff materlal it material 4 A gt Based the xA lines shown above a robust design may begin as follows The design target for the torque needed to open cap is denoted by I and it is placed on the Xaxis indicated by an arrow Note that it is undesirable if the torque is too high or too low from the target Let the design start by selecting a distribution of the tightness with its mean at positionl39 the function xA then yields the related torque distribution with the mean at position3 if the softer material is used while at position5 if the stiffer material is used Between these two designs position5 is on target while position3 is below the target39 but the design at position3 yields a smaller scatter than that at position5 A different design let the cap tightness be designed with the distribution placed at position2 made tighter then the cap with the softer material yields a torque distribution at position4 while the cap with the stiffer material lands at position6 Between these two designs the one at position4 is on target39 the one at position6 is above the target Furthermore the design at position4 yields a smaller scatter than that at position6 Thus the design at position4 is a better design39 it is on target and it has a smaller scatter 12 Softening Behavior If the dependence of X on A is a softening relationship then xA is a concave downward curve as illustrated graphically below ChapterVI Quality Control VlZO X stiff material g 6 soft material A gt In this case we note that shifting A along the horizontal direction will affect both the mean and the variance of X in the vertical direction Specifically shifting A to the right will result in an increased mean and decreased variance for X regardless the material used soft or stiff However increase in the mean of X is more rapid if the stiffer material is used while decrease in the variance of X is more rapid if the softer material is used A robust design in this case would look for a material with A that will result in optimum design that is the mean of X is on target and the variance of X is acceptably small 13 Hardening Behavior If the dependence of X on A is a hardening relationship ie the XA curve is concave upward then by shifting A to the right it will increase both the mean and the variance of X39 note that any increase in the variance is undesirable And this undesirable effect is more pronounced if the stiffer material is used A robust design in this case would not consider a hardening behavior of XA This effect of shifting A to the right on the increase of mean and variance of X can be readily seen in the plot shown below X stiff material soft material ChapterVI Quality Control VlZl Discussion In the above we have demonstrated the essence of robust design using three different behavior functions of only a single variable the linear softening and hardening of XA In real World design situations the behavior function of X may depend on a multiple of parameters XA B 39 there robust design is a science by itself involving linear andor nonlinear dynamic programming ChapterVI Quality Control Vl22 Assigned Exercises 61 A process is found to have a shortterm Cp 095 and a longterm Cpk09 What is the shortterm yield What is the longterm yield The longterm yield Y 9931 62 A part must meet the 50 criterion in each of 10 independent specifications determine the probability that the part fails to meet each of the specifications then estimate the final yield Parts failed to meet the specifications p 06038x1039639 final yield Y 9999 63 The target quality of a product is set at being 10 Offline sampling finds that the product quality distribution pdf fits the following function fx 004xe3902X 0 x lt 0c a Is the product quality on target b If the specification limits are 10i5 what is the probability of not meeting the specification c What is the Taguchi loss if it costs 500 per product out of specification Ans a yes 10 b 4633 c o 707 and EL 10 per product 64 An offline inspection of a sample beer cans shows that 05 of the cans failed under the compression load of 10137 lbs while 03 sustained the load of 10252 lbs The design target of the can s compressive strength is 10200 lbs a Assume that the compressive strength is norm ally distributed determine the mean and variance b Is the quality of the cans in terms of their compressive strength on target c If the acceptance criterion is 102i025 lbs what is the expected yield d What is the Taguchi process capability index e If it costs 005 to produce a can what is the expected loss based on the goalpost criterion f What is the expected longterm loss based on the method of Taguchi a p 101925 lbs c Y 7242 d Cpm 036439 e EL 065 cent 65 Voltagedrift in a computer circuit is allowed within the limits of i08 volt When the circuit is connected to a battery with a known voltagedrift distribution function the probability that the voltagedrift allowable not be exceeded must be evaluated Now a If the voltagedrift distribution in volts of the battery is x 34X1 X2 for absx lt1 fx 0 for absx gt 1 What is the probability that the circuit will encounter a voltage outside the allowable b If each time the voltage exceeds the allowable the battery has to be replaced and it costs 100 for a replacement what is the expected Taguchi loss per battery a p 56 b EL 3125 66 The solderjoints in a computer chip must have a diameter within 4i001pm It is known that for a batch of solder joints the diameter distribution is normal and the mean is on target ChapterVI Quality Control Vl23 a What is the shortterm standard deviation 0 if the solder joint meets the 60 criterion b If there are 10000 solder joints on a chip What is the yield of the chips a o 000167 pm b Y 9998 In a longterm production period p moved off target by 0005 mm and 0 increased 10 c Determine the longterm processing capability index for the solder joints d Determine the probability of failure to meet the 60 criteria for the solder joints e Determine the longterm yield of the chips which contain 10000 solder joints c cpk 1724 d p 2lt1gt517 e Y e10000P 67 Lifetime of an electric motor is normally distributed There is a 5 chance for the motor to fail before 2000 hours and a 15 chance for it to last beyond 4000 hours In a device the motor is used as a power source with the guarantee to provide at least 3000 hours of operation however should the motor fail before 3000 hours it must be immediately replaced The cost for replacing the motor is 1000 Determine the following with relevant reasoning and calculations a The mean u and the standard deviation 0 of the motor s operational life distribution b The probability that the motor fails before 3000 hours of operation c The expected Taguchi loss by the largeisbetter model a p 3225 hrs 0 7449 hrs b p 3821


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