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# FINITE MATH & ITS APPLIC MA 162

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This 136 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 162 at University of Kentucky taught by A. Sathaye in Fall. Since its upload, it has received 11 views. For similar materials see /class/228133/ma-162-university-of-kentucky in Mathematics (M) at University of Kentucky.

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The Linear Programming Begins Ma 162 Spring 2009 Ma 162 Spring 2009 February 2009 a 72 mm lne qualit We discuss inequalities in two or more variables a An inequality in one variable looks like 2x 3 g 5 and is solved by rearranging it so only the variable appears on the left hand side a S 1 o This can also be done graphically as follows Convert it to an equation and solve it Thus 2z351eadstoz1 o On the number line7 plot the point z 1 and notice that all points to the left of it satisfy the inequality and the ones on the right don t o The interval 00 1 on the number line looks like Tn infinily 11 VVe Verify test Values 5 O and x 2 to decide that this interval consists of the solutions and the other part of the number line does not 0 The set of solutions is said to be the feasible set of the inequalities used 0 If We similarly handle another inequality say 3x 10 Z 4 then the solution to the associated equation 3x 10 4 is x i2 and the interval 27 00 is deduced as before 20 Tu infinity n 9 If We try to solve both 29 3 g 5 and 39 10 2 4 together then We get the intersection of the two intervals But this can be a s explaind thus a We solve both associated equations plotting their solutions on the number line 42 17 0 By using test points on each interval say a 73 O 2 We pick up the ones which satisfy all the inequalities This gives the feasible set 72 1 a An inequality like 2 23 g 4 is the next topic As before we rst conVert it to the equation 2 23 4 We note that this is a line and we know how to plot it It is not di icult to see that the plane is split into two halves so that on one side of the line the inequality is true while on the other side it is not Thus haan plotted 2 23 4 we see that at the origin 00 0 the inequality is satis ed So we choose as the feasible set the half plane containing the origin 2 In the picture only the rst quadrant is shown since inequalities 2 Z O 7 are D i 2 3 4 typically going to be part of our conditions o If we have nnore than one inequalities then we solve t separately and take the common L em pert Here is the solution for ogowgo wlyglzxwygz the rst o uedrent Here are the separate regions for the two inequalities followed by the sonnbineol region As before the rst two inequalities nneen we only draw things in 3x4y 12 x2y22 Cambinaiv 3 m 2 25 E 1 E 1 1 4 1 D5 2 as U 1 2 3 4 D 2 A E am Summary of The Graphical l o On a common graph paper7 draw the equations corresponding to each inequality7 and mark the regions indicated by each inequality using a directional arrow In our course7 the assumption always puts the region in the rst quadrant The directional arrow is usually decided by using a test point For any inequality at least one of the three points 07 07 17 07 07 1 is always a good point to use 0 Take the common part of the plotted regions 0 Calculate and list all the corner points Make sure that the chosen corner points actually satisfy all ineualities 3 Some corner points7 or even some whole lines may be lost7 meaning they do not have any points in the region 7 Summary C ntinued The aim of sketching and marking the corner points is to solve an optimization problem for a linear function on the resulting region Review various examples in the section 32 where a practical situation leads to a set of inequalities and a linear function to be optimized ie maximized or minimized The rst step is to clearly name the variables and write down the inequalities and the function to be optimized Then you sketch the region carefully7 provided that you have only two variables More variables are handled in the next chapter using the Simplex algorithm Then you list all the corner points of the region in order and evaluate the function at each of the corners The optimum value of the function is among the values at the corner points7 with one exception7 which we discuss next ll Further Ctnnments If the region is not bounded7 then the function may not have a maximum or a minimum This can be decided by checking along the edges of the polygon running off to in nity Why does the graphical method work We give a brief explanation below Parametric lines Consider a line in the plane7 say7 y 3x 5 It passes through a point 17 8 We want to study the line near this point So we take a point on this line Whose z coordinate is 1 t and notice that its corresponding y coordinate shall be y31t583t In fact7 all points of this line can now be described by the parametric equations z1ty83t This is called the parametric form of the line It is useful to be able to calculate such a form near any convenient point l l lanation Continued n Thus for the same line we could also have started with a point 72 71 and concluded a different parametric form z72ty713t 9 Functions on Parametric lines Consider a function say fz y 3x 43 We can analyze how it behaves on our parametric line by plugging in the parametric form Thus we have fz y 35 1575 0 This shows that as t increases so does the function value Remember that the parameter t was the change in the z coordinate from the point 1 8 Thus as we let x coordinate increase on our line the function value increases lanation continued a Conclusion Thus on a line in parametric form7 a linear function increases or decreases with the parameter depending on the coef cient of the parameter For the above line7 if we consider a different function7 say gz y 3x 7 y 27 then we see that g1 t7 8 3t 31 t 7 8 3t 2 and this simpli es to gl t8 32 73 a Thus7 a linear function on a line is either constant at all points or increases in one of the two available directions and decreases in the opposite direction 0 Consider the plane region of feasible points that we can plot for our problem Where would a linear function become maximum on such a region If we take any point in the interior of our region7 then we can draw a little line segment through the point which is still entirely in the region Now if our function is not constant on the line7 it would be increasing in one of the two directions and thus would not be maximum at our given point If by luck7 we had chosen a line segment on which the function happens to be constant7 we can choose a different segment through the same point and make the same argument We could not have the same function constant on the second segment as well7 for it is clear that then the function would be identically constant on the whole plane Think why and our problem has a triVial answer every point is a maximum point ll Thus7 our maximum point has to be on the boundary It could be on a boundary segment or at a corner Note that if it is on a segment7 but not at a corner7 then the function has to be constant on the whole boundary segment for otherwise we get a contradiction as above This is why it is enough to only check the corner points for locating a maximum This also explains that if we nd two maximum points which are corners7 then the line joining them must form a boundary line7 ie they must be adjacent points on the boundary polygon If the region is unbounded7 then it has boundary lines running o to in nity and we may nd that the maximum point may not exist in the sense that it has to be a point of in nity on one such boundary line 7 Some Sample Prol ems a Problem 322 by graphing The problem is to maximize the pro t P 2x 153 subject to I Z 07 y 2 07 3x 1 43 g 1000 and 6x 3y 1200 We rst sketch the lines and nd their common point Then we decide on the region a Note that in the picture loelow7 the inequality 3x 1 43 g 1000 corresponds to the line BC and the inequality 6x 1 33 g 1200 matches the line AB Their regions both point towards the origin7 since the origin satis es both of them The axes are automatically included with regions pointing towards the rst quadrant The corners are 00 0 A200 0 B120 160 C0 250 The values of 2x 153 at these are 0400480375 0 So the maximum is at 3120 160 With maximum value 480 5n mu 15m 2m x 9 Consider the problem similar to B28 Suppose that 2 2y S 2 52 S Etogether With 2 Z Oy Z 039 The max the function 62 91 2 on the resulting regio y Th aXimum Value of the function is imum Value of noccurs 2 and We rst convert all inequalities to equations and plot after nding common points The equations are 22y2y52 52Oy0 The corners are lO8Q5QO1OOV The Values of 62 91 2 at these are 395 l 15 2 83173113 0 So the maximum is at 89 59 and the maximum Value is 373 9 Consider the problem similar to 1329 Suppose that y S 5967 y 2 3x and z4 y5 S 1 together With z 2 07 The maximum value of the function 96 7 on the resulting region occurs as A AA and y A A AV The maximum value of the function is 0 We rst convert all inequalities to equations and plot after nding common points The equations are y5zyy3zyz4y51yz07y0 Note that the last tWo do not contribute to the picture A 3 2 T e corners are 1 0 0 434 201776017 The a o y at these are 07245 48 8017 47059 So the maximum is at 4574 and the maximum value isy4VSv U nzmnena 1 12 More 011 Priaility Ma 162 Spring 2009 Ma 162 Spring 2009 April 2009 a 72 n q a ProlJability in Real Life Experimental Probability In real life situation the sample spaces are huge and we don t even know them very well So it is important to be able to identify events properly Moreover usually we cannot calculate a probability but can only estimate it For instance suppose we consider the population of the united states and consider the possibility that each person is either for the Stimulus Package or against it A or undecided We can imagine a sample space S consisting of pairs person opinion where the opinion is F or A or U We will get a sample space S of size 3 where n is the number of people in the united states Of course this n has to be understood as consisting of only those persons whose opinion is relevant to the decision process It makes sense to de ne the event F to be the set of all pairs with opinion equal to F ll Example Then the probability of a person being for the stimulus package is nF3 0 There is no way to actually carry out the necessary survey in a reasonable time or even in a meaningful way So7 statisticians resort to testing a few sample cases This is why we have the word sample space a They will select a few random persons7 say 01 in number and ask how many of them are for the stimulus package If they get a number f7 then they will estimate the probability PF fd 0 Of course7 this answer depends on the experiment of surveying and is only expected to be equal to or close to the true probability Such problems appear in section 72 o Statisticians try to estimate the probability of the answer being correct by general theories and this is the basis of the estimated con dence levels or margin of error ll Prtlalility Rules o If we know or are con dent about the probabilities of certain events it is possible to calculate the probabilities of related events Here are some rules based on the set counting formulas and the conviction that the probability of any event E can be interpreted as PE PE U F PE PF 7 PE F a We say that two events E F are mutually exclusive if they cannot happen together or in other words PE F O In this case we get PEUF PE PF PEC 17 PE This could be also explained that either E or EC is always true so PEUEC 1 and hence 1 PE PEC since E and EC are clearly mutually exclusive ProlJability Distrilutitii o A simple event is an event with a single sample point a In Mathematical Probability we are acting as if all the sample points are equally likely In real life this is far from the case 0 A more realistic situation is that di erent sample points or di erent simple events might have di erent probabilities Then the probability of an event is de ned to be the sum of the probabilities of its costituent sample points Consider an example of a class of 40 students with 4 10 18 6 2 grades of A B C D E respectively If we think of the ve grades as our sample space then the probability of getting A is not 02 but 01 Thus the probability of getting a poor grade D or E is 4 02 The table or formula describing the probabilities of all simple events is called a probability distribution ll ProlJability Talc ula ticnis a What is the probability of drawing a red card higher than 9 from a standard deck of 52 cards Don t count aces high 0 Answer The sample space clearly has 52 elements Our event of a red card higher than 9 can be enumerated thus In each suit there are 4 such cards and since we have two red suits7 the total count by multiplication principle is 2 4 8 So PE 5 01539 a What would be the probability of drawing a face card from a standard deck Answer If F is this event7 then nF 3 4 so PF We are not counting aces as face cards What is the probability of not getting a face card Answer Easiest way to handle this is 1 7 g 7 Ccnitinued Probabilit lulatitns Consider the situation where we draw three cards from a standard deck without replacement Now the sample space has 7527 3 227 100 sample points What would be the probability of getting at least one face card Answer If we call the event E and try to count we soon realize that we have to make cases getting one or two or three face cards The calculation is clearly complicated However7 the count of choices without face cards is easy It is just 7407 37 the number of ways of choosing two cards from the forty non face cards So 7 C40374039387 C523 525150 85 So PE17 04471 05529 PEC 04471 Poer Han o If you are playing a 5 card poker What is the probability of getting a straight 0 Answer Recall that a straight is de ned as ve cards in a sequence which are not from the same suit The sample space has nS C52 5 2 598 960 We count the straight hands thus We can start with a 1 2 3 10 as the lowest card and ll up a straight Note that an ace is both a top and a bottom card Thus we have 10 cases of starting numbers to handle For any choice of the starting number there are 45 1024 ways of lling up a sequence when we don t worry about the not same suit77 condition Now we remove the four sequences in the same suit to get 1020 Thus the total number of straights is 10 1020 10200 0 The desired probablity is then 10200 00039 2598960 I i Another Poer Hand What is the probability of getting three of a kind77 which is de ned as three cards with the same rank and the other two cards mismatched of di ernt ranks We have thirteen di erent ranks and for each rank7 there are four di erent three of a kind triples of that rank Thus a total of 52 triples are possible To pick the remaining two cards7 we choose two of the remaining ranks and these are C127 2 choices Moreover7 we choose one of the four possible cards from these ranks Thus we have a total of 52 C127 2 4 4 54912 The desired probability is then 00211 tniditimial Prilalility Suppose there are events A7 B under consideration Suppose we know PA If we later know that B has happened This naturally a ects the sample space7 reducing it to those sample points belonging to the event B What is the new probability of the event A it is easy to see that in a mathematical set up7 the answer is simply nA This is called the conditional probability of A given B and shall be denoted as PAlB We note n l H B ME mAnm PAlB 133 or PA B PAlBPB A ample of acuiditicnial prtlalility 3 Suppose we are playing a ve card poker Suppose that we have drawn an ace and a 9 What is the probability of getting a straight now 0 Answer It is 0 We have to choose three more cards and no choice will ll up a straight joining these two cards 0 Now suppose we have drawn an ace and a 10 of di erent suits What is the probability of a straight now 0 Answer Now it is possible to ll in a jack7 queen and king to nish a straight The sample space now consists of picking 3 cards from the remaining 50 cards7 or nS 7507 3 The three cards which will yield a straight can be made in 4 4 4 64 ways by choosing one of the four jacks7 one of the four queens and one of four kings Thus the answer is 64 7 4 C5073 1225 l l 00032653061 of tniditimial Prtlalility a Quite often7 in order that a certain event happens7 another event has to happen rst For instance7 to get to class on time7 you need to get up on time with enough time to get ready and then you need to get in a reasonable traf c 0 Given that the probability of getting up on time is 80 and then the probability of catching a good traf c pattern is 70 then the probability of getting to class on time is calculated as the product 08 07 056 or 56 o What happens if you are a little late in getting up7 say the probability of this is the remaining 20 Perhaps7 then the probability of getting a good traf c pattern is only 40 Then the probability of getting to class on time is changed to 02 04 008 or only 8 ll Since to get to class you need to get up and ght with the traf c you can now calculate the combined probability of getting to class on time as 056 008 064 or 64 Note that we simply added the two probabilities since the two events are mutually exclusive If you get up in time you have the rst probability and if you get up late you have the second Let us formalize the above calculation as follows De ne events E and L as getting up early and getting up late De ne events G and B as getting a good traf c pattern and getting a bad traf c pattern Thus to get to class on time you must have the event E H G or L H C so the probability desired is that of the event En 0 UltLn 0 Note that the probability of getting a good traf c pattern PG is not given to us at all mainly because it changes based on the timing ll Ccnitinued Discussimi We do know that The probability of G given E is 70 While probability of G given L is only 40 Our formulas say that then The probability PE G PE PGlE is then calculated as 08 07 056 Similarly PL G 02 04 008 Finally7 the two events E G and L G are mutually exclusive7 so their probabilities are added Food for thought You can imagine a further possibility Where you can get up so late that the traf c pattern does not matter Think What could happen to the probability then You should also calculate the probability of missing the class due to bad traf c lent Events If A B are two events we say that they are independent if the probability of A is the same Whether B occurs or not In other words PA PAlB We recall that PA B PAlB PB and thus the condition for independence comes out to be PA B PA PB 0 Of course in real life situations the numbers are not likely to be exactly equal Also both sides being numbers less than 1 we need to have a stricter de nition of considering them nearly equal 0 We accept the convention that the two events A B are independent if the corresponding percentage probabilities di er by less than 05 0 Consider the following situation In a factory 85 of the 260 managers and 185 of the 570 non managers got laid o a Are the events A quotbeing a managerquot and B quotbeing laid o quot independent 0 We estimate from the given data that PA 260 While PltBgt 39 830 Also PA n B 0 We check the condition if 85 260 270 Q 7 39 The LHS evaluates to 01024 or 1024 The RHS evaluates to 01019 or 1019 The di erence is 005 This is pretty close to being independent However7 keeping further digits7 we get a slightly bigger number and may consider them dependent o This illustrates how a personal judgement and convention can I tra11sactilt311s Ma 162 Spring 2009 Ma 162 Spring 2009 March 2009 a 72 mm 0 Interest is rent on borrowed money We use the notation P for principal This is the amount of money borrowed or lent a We say T is the interest rate7 if it is the agreed interestrent on one dollar per year Of course7 we change the name of the currency as appropriate The rate is often quoted as something like 7 which means T 00 Thus the words per cent77 mathematically mean the fraction m a We shall typically let t denote the period of lending in years and thus the interest I accumulated in t years on a principal of P dollars at a rate T is given by the formula I PTt it acuitinued 3 Thus after the t years7 the total amount owed is the original principal P plus the interest I and thus has the formula Accumulation A P Prt P1 rt 0 If we know three of the four quantities A7 P7 T t then we can nd the fourth We should learn to recognize what is given and what is unknown 3 Example 1 If you invest 77091 at 8 simple interest7 how much will your investment be worth in 15 months 0 We note that T 8 or T 008 Also P 77091 We are given if 15 months which must be converted to years and thus t g 125 We are looking for A7 the net accumulated value of the investment Hence A 770911 008 125 848001 or 848 0 Example 2 If you invest 112752 and after 18 months it is worth 1229007 What simple interest rate7 expressed as a percentage and rounded to 01 did you receive 0 We are given P 1127527 t g 15 and A 1229 We want T We recommend solving the formula A P1 Tt for T and then evaluating it Thus A 1229 1 112752 1 The answer comes out T 006000189206 We multiply by 100 to make a per cent rate and report T 6 after rounding 0 Example 3 If you invest 152088 at 6 simple interest7 after how many months7 rounded to 0017 will your investment be worth 165700 0 We are given P 1520887 T 0067 A 1657 and asked to nd if As before7 we solve our formula for t and evaluate A 1657 t 1 152088 1 r 006 39 This gives t 1491680255 Be sure to multiply by 12 to make months So the answer is 1790016306 or 179 months after rounding A M e Cbmplic d Example 0 Example 4 Homer won a prize in the lottery of 3000 payable 1500 immediately and 1500 plus 4 simple interest payable in 260 days Getting impatient Homer sells the promissory note to Moe for 1440 cash after 170 days Using a nominal 360 day year nd the simple interest rate rounded to 001 earned by Moe o This is a simple interest problem of nding T but needs careful set up If Homer were to patiently wait the 260 days he would earn 260 A P1 rt15001 004 1543333333 dollars a From Moe s perspective this is his A after a lending of 1440 for a period of 260 7 170 90 days Thus for Moe the calculated interest rate as in Example 2 is A 7 1 154333 7 1 7 P7 02870370361 287 t W Such high rates are not uncommon for short term lenders l l 0 Suppose you lend somebody 100 for a period of one year at 10 interest rate You will receive an accumulated payback of 110 at the end of the year But if you demand a repayment in six months you will be entitled to receive 105 Now suppose you lend this total amount back to the borrower then using the usual formula with P 105r 010 t 05 we get 1051 010 05 11025 dollars It is easy to see that the net formula is 1001 1 2 Of course you don t really want to carry out the transaction just demand the money This is called the accumulation by compounding every six months or twice a year Thus a greedy lender can claim more money by simply imagining a transaction 7 The lOIHPOUHCl hite o If one gets greedier and imagines compounding the interest m times a year then it is easy to see that in each of the m periods7 we get the accumulation by multiplying the starting principal for that period by 1 and thus the full formula for the interest after one year is T m A P 1 7 m o It is useful to develop some new notation Assume that we are compounding m times each year Thus in t years7 we shall have mt periods of compounding and we de ne Periodic interest rate 239 and Total term of loan in periods n tm This gives us the formula for accumulated amount when we compound m times a year as T mt APlt1igt P1239 m This Greed has at Limit 0 Continuing our example of lending 100 for one year at a rate of 10 If we compound it m times a year then we have the formula Am 100 1 a We can calculate the accumulation for different values of m l m l 1 l 11 l 51 l 101 l 1 Am 1 11000 1 11046717 1 11050627 1 11051162 1 0 Thus though increasing it is not growing very fast Indeed using techniques of algebra it is possible to show that the limit of the quantity Am as m goes to in nity is a famous function of mathematics namely T m lirn P 1 7 Pexpr mace m Thus even if we imagine in nite compounding our accumulation for the above P 100 T 10 is only 100 exp010 1105170918 7 st Formulas a To summarize we have the formula that for principal P annual rate T period t years and compounded m times a year we have Am P1 where 239 L n tm m We describe the idea of in nite compounding as continuous compounding The accumulation if we compound continuously is given by the formula AC Pexprt Typically we just write A for accumulation but mention the value of m in words A amples of mpound lnt 0 Example 5 If you invest 500000 at 9 compounded bi weekly7 how much will your investment be worth in 8 years a We have P 50007 T 0097 t 8 The meaning of the phrase bi weekly is that it is compounded once every two weeks or m 26 using a nominal year of 52 weeks We have 239 0003461538462 and n 8 26 208 Thus A 50001 0003461538462208 1025940275 9 Warning It is crucial to learn good calculator techniques here7 since if you don t keep enough accuracy for 239 then the power calculation introduces a lot of error and multiplication by a large P makes a very inaccurate amount One should try not to copy down intermediate results7 but store and reuse them for better accuracy 7 9 Examples 0 Example 6 How much did you invest at 8 compounded bi Weekly if 15 years later the investment is worth 97000007 o If the investment is P then our formula gives 1526 W 97000 P 1 which can be solved for P as 008 71526 P97000ltlt1gt a This evaluates to 2926971472 It is an excellent idea to double check that this value of P does generate the 97000 ie 008 1526 2926971472 1 97000 Within reasonable accuracy The computer answer is 9699999712 l l a Often7 lending terms are described by di erent rates and di erent number of compoundings per year It is necessary to be able to compare them to decide which is a better rate a One way to do this is to nd an e ective rate reg which is de ned as a simple interest rate which will give the same yield as the given scheme 0 Thus7 if we invest one dollar at T annual rate compounded m times a year7 then our net yield is 1 and if reg is to be the e ective rate7 then we have T m lt1ggt 1Te so we have the formula W re lt1gt 71 a Example 7 Bank A is o ering an interest rate of 660 compounded monthly while bank B is o ering an interest rate of 669 compounded quarterly What are the e ective rates of the two banks expressed as percents and for the investor which bank o ers the bettter rate 0 We apply the formula for the e ective rate to get 12 The W for bank A is 1 71 006803356 4 and the Tag for bank B is 1 7 1 006859714600 0 The reported answers should be 680 and 686 respectvely with bank B declared as haVing a better rate 0 Note that if the problem was about borrowing from the bank instead of investing then bank A would be a better choice 7 o Next7 we study the concepts of progression or a sequence and a series or their sum Of special interest are the Arithmetic series and the Geometric series a must study for all students of mathematics Afterwards7 we tackle a problem of annuity Such problems have three types a We discuss how to borrow a large sum and pay back with periodic payments mortgage 0 We discuss how much money can be withdrawn on a periodic basis from set up funds which are earning interest until drawn sinking funds 0 We also dicuss how to build up future reserves by periodic saVing Understanding Annuities Ma 162 Spring 2009 Ma 162 Spring 2009 March 2009 a 72 n q a a We recall some terms and calculations from elementary algebra o A nite sequence of numbers is a function of natural numbers 12 n Thus the formula ak 2k1for k12 10 describes a sequence 3 5 7 9 11 13 15 17 19 21 a We may also let a sequence run out to in nity as in 1 5 Here the sequence can also be described as Wheren12 a A sequence may also be called a progression Two progressions are important the Arithmetic Progression and the Geometric Progression AP and GP 0 Arithmetic progression This is a sequence which has a starting number a and successive numbers are obtained by adding a number 01 called the common di erence Thus7 its n th term is a n 7 1d Example Take a 37 d 4 The sequence is 37111519 734n71 The n th term can be better written as 4n 7 1 Geometric progression The geometric progression has a starting number a and successive terms are obtained by multiplying by a common ratio T Thus7 its n th term is arm l Example Take a 2 and T The sequence is 1 1 2 2717571739 7 2 717quotquot Note that the n th term is better written as 1 2M4 a We need the formula for the sum of terms in AP 0 The sum of the AP a7ad7a2d7 711n71d 7L is called an Arithmetic Series and is written as 211 k 7 1d k1 0 Its sum is given by the formula ak1dnwnltanildgt39 M3 2 2 k 1 An alternate way to remember it is number of terms average of the rst and the last term a We need the formulas for the sum of terms in GP 9 The sum of the GP 111 1T2 aTW l 7L is called a Geometric Series and is written as 2ark 1 161 6 Its sum is given by the formula i zrwim a TV 71gt 117 Hz 161 T71 lir o If M lt 17 then we can make sense of the formula even for an in nite GP and write SWAP alt1irgt c Annuity a What is an annuity An annuity is a combination of investments or payments 0 For conveniencewe assume the following conditions which are valid in most practical situations A xed amount R is invested exactly m times a year This gives exactly m periods in a year and each is th part of the year Each payment is made at the end of its period 0 The payments are made for a period of t years and thus the number of payments is exactly mt n For each period7 the interest rate is the same T annual and thus in each period7 the interest earned by 1 dollar is exactly 239 This is called the periodic rate Basic Annuity Formula 0 With the notation as explained above7 how much money will be accumulated by making a periodic investment of R dollars at the end of each of the n periods when the periodic rate is 239 and the interest is compounded in each period The answer comes out as a geometric series Here is how we reason it out o The payment at the end of the rst period is compounded for n 7 1 periods and hence becomes worth R1 239 1 a The payment at the end of the second period is compounded only for n 7 2 periods and becomes worth R1 239 2 Continuing7 the very last payment is worth R1 00 R In other words7 it acquires no interest Adding up in reverse7 we have 12 717 1i 71 m 7 s RR12 R1iltHgt R Present lue of an Annuity o Often7 the periodic investments are just payments like mortgage against borrowed funds What is the relation between the periodic payment R and the borrowed amount P7 when the interest rate is T and the payment is m times a year As usual7 we let 239 be the periodic rate and n the number of periods or the total number of payments We think like the lender and argue that he makes a single payment of P dollars and expects it to grow to the same sum as the periodic payements by the borrower over the same number of periods This gives us the equation P1i SR 1Zl 1 Z and thus the formula 7 1 717 171 lt gt 211 m 239 an JJL 1 Using the Annuity Formulas We now have the basic formulas needed to answer all questions about periodic investments or payments Example of a Trust Fund If a trust is set up so that you take 6 years to travel and pursue other interests Suppose that you will make bi weekly withdrawals of 27 000 from a money market account that pays 400 compounded bi weekly How much should the fund be Answer Imagine the trust fund to be a lender and your withdrawls as mortgage payments to you Thus7 we use the formula 1 7 1 39 W P R l 2 Z HereR20007 i2 andn266156 The formula yields 2771951659 or 2777 19517 Kamples of Annuities Sinking Fund This means a fund set up with periodic investments to be sunk or used up at the end of the n periods 3 Example You plan on buying equipment worth 307 000 dollars in 3 years Since you rmly believe in not borrowing7 you plan on making monthly payments into an account that pays 400 compounded monthly How much must your payment be 0 You have to nd out the value of R7 but know that S7 the expected accumulation is 307 000 with t 3 and T 004 Moreover m 12 from the word monthly and hence i 003333andn12336 0 Using S EggE we get 239 1 7 1 a Thus7 the reported answer is 78572 which actually yields 3000002 l l R 30000 lt gt 7857195502 ntinued Examples 0 About Accuracy In the above calculation the evaluation of 1 71 7 1 000333336 71 239 7 0003333 is involved If you calculate this and divide into 30000 you need to keep many digits of accuracy Try various approximations to see how to get the most accurate answer to the penny You will nd that you need to keep at least four accurate decimal places the the rst answer Thus as a general principle in these problems you should not copy down intermediate answers but store and recall them so that maximum accuracy is maintained Further Examples of Annuity As another example7 consider this problem If you can afford a monthly payment of 1010 for 33 years and if the available interest rate is 410 What is the maximum amount that you can afford to borrow You note that R 10107 239 and m 12 with t 337 so that n 12 33 396 0 But you don t want S7 the future accumulation You want the money now7 to be paid back over the years So7 you use the formula for P7 the present value Thus7 you evaluate if 1 2 239 P R 1010 2168603683 21902897 Note that due to the large numbers involved7 your fraction needs 10 digit accuracy Thus7 the hardest part is always to gure out which formula is appropriate l l Lecture lt ltJ11ti1111aticn1 of 36 Ma 162 Spring 2009 Ma 162 Spring 2009 February 11 2009 a 72 mm eqiiaticnis Now we describe the main use of matrices for solving systems of linear equations In this lecture7 we would mainly consider systems where the number of equations equals the number of variables 0 A linear system of n equations in 71 variables can be descibes by a single matrix equation of the form AX B For example The equations 2x 7 3y 196 7 2y 5 can be written as l i Z llil lil eqiiaticnis Now we describe the main use of matrices for solving systems of linear equations In this lecture7 we would mainly consider systems where the number of equations equals the number of variables a A linear system of n equations in 71 variables can be descibes by a single matrix equation of the form AX B For example The equations 2x 7 3y 196 7 2y 5 can be written as l i Z llil lil eqiiaticnis Now we describe the main use of matrices for solving systems of linear equations In this lecture7 we would mainly consider systems where the number of equations equals the number of variables a A linear system of n equations in 71 variables can be descibes by a single matrix equation of the form AX B For example The equations 2x 7 3y 196 7 2y 5 can be written as l i Z llil lil A 11111198 lt lt11ti1111elt1e o The solution7 in turn can also be described as IX C which reduces to X 0 Tth7 the solution to the above system is x 713 y 79 and can be written as 5 9H3 113 a a 72 mm 1 111111198 lt lt11tinueltii o The solution7 in turn can also be described as X C which reduces to X C Thus7 the solution to the above system is x 7131 79 and can be written as 5 Hit iii 0 The equationsxiyiz5xy2202xy321 can be written as 17171 x 5 1 12 y 0 213 2 1 1 111111198 lt lt11tinueltii o The solution7 in turn can also be described as X C which reduces to X C Thus7 the solution to the above system is x 7131 79 and can be written as 5 Hit iii 0 The equationsxiyiz5xy2202xy321 can be written as 17171 x 5 1 12 y 0 213 2 1 1 111111198 lt lt11tinueltii o The solution7 in turn can also be described as X C which reduces to X C Thus7 the solution to the above system is x 7131 79 and can be written as 5 Hit iii 0 The equationsxiyiz5xy2202xy321 can be written as 17171 x 5 1 12 y 0 213 2 1 The 11min iiiiez t of soiutim o The solution7 as before can be written as IX C which reduces to X O Thus7 the solution to the above system is x 491 2 2 7393 and can be written as 4 y 2 z 73 a a 72 n q a 1 0 0 x 0 1 0 0 0 1 The main idea of solution o The solution7 as before can be written as X C which reduces to X C Thus7 the solution to the above system is x 4 y 22 73 and can be written as 1 0 0 x 4 0 1 0 y 2 0 0 1 z 73 Both the solutions above can be described by the following simple philosophy Let the original equations be AX B Where we assume that A is a square 71 gtlt n matrix7 X is the column of 71 variables and B denotes the right hand sides We nd an n gtlt 71 matrix M such that AM MA LL The main idea of solution o The solution7 as before can be written as X C which reduces to X C Thus7 the solution to the above system is x 4 y 22 73 and can be written as 1 0 0 x 4 0 1 0 y 2 0 0 1 z 73 Both the solutions above can be described by the following simple philosophy Let the original equations be AX B where we assume that A is a square 71 gtlt n matrix7 X is the column of 71 variables and B denotes the right hand sides We nd an n gtlt 71 matrix M such that AM MA LL The main idea of solution o The solution7 as before can be written as X C which reduces to X C Thus7 the solution to the above system is x 4 y 22 73 and can be written as 1 0 0 x 4 0 1 0 y 2 0 0 1 z 73 Both the solutions above can be described by the following simple philosophy Let the original equations be AX B where we assume that A is a square 71 gtlt n matrix7 X is the column of 71 variables and B denotes the right hand sides We nd an n gtlt 71 matrix M such that AM MA LL se De ned o Multiplying both sides of the equation AX B by M on the left we get MAX MB which becomes IX MB and yields the solution X MB a 72 mm se De ned o Multiplying both sides of the equation AX B by M on the left7 we get MAX MB which becomes IX MB and yields the solution X MB 0 Thus7 it would be good to have a mechanism for nding such a matrix M when possible o Multiplying both sides of the equation AX B by M on the left7 we get MAX MB which becomes IX MB and yields the solution X MB 0 Thus7 it would be good to have a mechanism for nding such a matrix M when possible 0 We de ne the inverse of a square matrix A to be a square matrix M such that MA AM I Multiplying both sides of the equation AX B by M on the left7 we get MAX MB which becomes IX MB and yields the solution X MB Thus7 it would be good to have a mechanism for nding such a matrix M when possible We de ne the inverse of a square matrix A to be a square matrix M such that MA AM I The matrix M can be shown to be uniquely de ned by A7 when it exists and is called the inverse of A Multiplying both sides of the equation AX B by M on the left7 we get MAX MB which becomes IX MB and yields the solution X MB Thus7 it would be good to have a mechanism for nding such a matrix M when possible We de ne the inverse of a square matrix A to be a square matrix M such that MA AM I The matrix M can be shown to be uniquely de ned by A7 when it exists and is called the inverse of A The matrix A is said to be invertible or non singular if its inverse exists and it is said to be non invertible or singular otherwise F111lti1111g13he Inverse 2 X 2 case a b A c d then we have a very simple answer Let A detA ad 7 be a 72 mm Finding the 111V erse 2 X 2 a b A 7 i c d i then we have a very simple answer Let A detA ad 7 be 0 Then A is invertible if and only if A 7Q O a 72 mm Finding the 111V Arse 2 X 2 i it a b A i c d i then we have a very simple answer Let A detA ad 7 be 0 Then A is invertible if and only if A 3A 0 o Moreover7 if A 7E 0 then the inverse of A is the matrix i deb Aica39 Finding the lnv arse 2 X 2 i 7i a b A l c d l then we have a very simple answer Let A detA ad 7 be 0 Then A is invertible if and only if A 3A 0 o Moreover7 if A 3A 0 then the inverse of A is the matrix 1dib Aica39 2 73 172 Then A 2727173 71 and hence the inverse is Mi li 3 o Thus7 for our rst example above A o It is easy to check that M solution a 72 mm 1 713 5 791stheold 11latilt11 lt11t111uelti1t o It is easy to Check that M 1 solution 713 5 791stheold a 72 mm drillaticni ontinued 1 713 o It is easy to check that M 5 79 is the old solution As a side note7 we observe that in this case the inverse M is the same as A7 or AA A2 I Such matrices are said to be unipotent 0 Important notation When the inverse of A exists7 it is denoted by the convenient notation A l Do not ever write in place of A lg it is both illegal and meaningless A11 obse I We solved the equation AX B above as X MB A lB for a speci c 2 X 2 matrix A Note that B did not enter the calculation until the product MB a 72 mm An obse 0 We solved the equation AX B above as X MB A lB for a speci c 2 gtlt 2 matrix A Note that B did not enter the calculation until the product MB 0 Thus we observe that if A is invertible7 then the equation AX B has a unique solution X A lB An obse We solved the equation AX B above as X MB A lB for a speci c 2 gtlt 2 matrix A Note that B did not enter the calculation until the product MB 0 Thus we observe that if A is invertible7 then the equation AX B has a unique solution X A lB This should be compared with the statement If ab are numbers and if a 3A 0 then the equation ax b has a uniques solution x a An obse We solved the equation AX B above as X MB A lB for a speci c 2 gtlt 2 matrix A Note that B did not enter the calculation until the product MB Thus we observe that if A is invertible7 then the equation AX B has a unique solution X A lB This should be compared with the statement If ab are numbers and if a 3A 0 then the equation ax b has a uniques solution x a What happens if A 0 Let P 1 2 2 4 and consider the equations PX Q where Q 1 We solved the equation AX B above as X MB A lB for a speci c 2 gtlt 2 matrix A Note that B did not enter the calculation until the product MB Thus we observe that if A is invertible7 then the equation AX B has a unique solution X A lB This should be compared with the statement If ab are numbers and if a 3A 0 then the equation ax b has a uniques solution x a What happens if A 0 Let P 2 4 1 2 and consider the equations PX Q where Q 1 We inVite you to check that when U 2u this system has in nitely many solution7 but it has no solution when U 3A 2u l l 0 Now we discuss the general inverse the formula is not as convenient as in the 2 gtlt 2 case So we give a procedure a 72 mm 0 Now we discuss the general inverse the formula is not as convenient as in the 2 gtlt 2 case So we give a procedure 0 Suppose we are trying to nd the inverse of a matrix VLX KL39 Start with the augmented matrix AH and row reduce it7 ie nd its RREF 0 Now we discuss the general inverse the formula is not as convenient as in the 2 gtlt 2 case So we give a procedure 0 Suppose we are trying to nd the inverse of a matrix 7 Lgtlt7 L39 Start with the augmented matrix All and row reduce it7 ie nd its RREF o The matrix A is invertible if and only if the RREF becomes I M for some n gtlt 71 matrix M Moreover7 this M is the desired A l Now we discuss the general inverse the formula is not as convenient as in the 2 gtlt 2 case So we give a procedure Suppose we are trying to nd the inverse of a matrix A AW Start with the augmented matrix All and row reduce it7 ie nd its RREF The matrix A is invertible if and only if the RREF becomes I M for some n gtlt 71 matrix M Moreover7 this M is the desired A l If one of the pivots is on the right hand side of the separator bar7 then the matrix A is non invertible or singular unple of in second example 0 We now illustrate the procedure on our 3 X 3 matrix in the a T 72 mm A aruple of 11 a We now illustrate the procedure on our 3 gtlt 3 matrix in the second example a Start with 1714100 AI11 2 010 21 3 001 a 72 mm 1 ample of 11 0 We now illustrate the procedure on our 3 gtlt 3 matrix in the second example a Start with 1 71 4 1 0 0 Ail112 010 2 1 3 0 0 1 0 When we do row transformations R2 7 R1 R3 7 2R1 and R3 3R27 we get 1 i1 71 1 0 0 0 2 3 i 71 1 0 0 0 12ii12 732 1 This is REF 7 l l se mntinued 0 Now we go on to make RREF o The Operations 1 2R3 R2 3337 R1 R37 1132 R1 R2 produce the RREF 100M 2 71 010M 5 73 0017173 2 a 72 mm 13989 911 511111911 0 Now we go on to make RREF o The operations 2R3 R2 7 3R3 R1 R3 R2R1 R2 produce the RREF 1 0 0 1 2 i1 0 1 0 1 5 i3 0 0 1 1 i1 i3 2 0 Thus the desired inverse is 1 2 i1 A l 1 5 i3 i1 i3 2 0 Verify that our answer is correct Thus 1 2 1 5 1 73 a 72 n q a 1 73 2 2 1 0 Verify that our answer is correct Thus 1 2 71 5 4 1 5 73 0 2 71 i3 2 1 3 a We note that when we nish the work of converting AH we can either nd the inverse or determine that the inverse does not exist 0 Verify that our answer is correct Thus 1 2 71 5 4 1 5 73 0 2 71 i3 2 1 3 a We note that when we nish the work of converting AH we can either nd the inverse or determine that the inverse does not exist 0 If A has an inverse7 then the equations AX B always have a uniques solution X A B Verify that our answer is correct Thus 1 2 71 5 4 l 5 73 0 2 71 i3 2 1 3 We note that when we nish the work of converting All we can either nd the inverse or determine that the inverse does not exist If A has an inverse7 then the equations AX B always have a uniques solution X A lB The main drawback of this method is the calculation of the RREF which can be lengthy Testing lime 0 Thus it would be useful to know if we are likely to nd an inverse before doing the full work a 72 mm Testing 111 o Thus7 it would be useful to know if we are likely to nd an inverse before doing the full work 0 Luckily7 we already have such a tool Let A be an n X 71 matrix We can convert All to REF It is easy to see that we have exactly 71 pivots Testing 111 Thus7 it would be useful to know if we are likely to nd an inverse before doing the full work Luckily7 we already have such a tool Let A be an n gtlt 71 matrix We can convert All to REF It is easy to see that we have exactly 71 pivots The inverse exists if and only if all the pivots are on the left hand side of the searator bar It is instructive to see that for the matrix 1210 2401 The operation R2 7 2R1 gives REF with pp 13 Since 2 2 we have no inversel Lelt ture Simplex algm ithm 1 Ma 162 Spring 2009 Ma 162 Spring 2009 February 2009 a 72 mm Old problem with a new method 0 We reca the problem we solved using the graphic method Maximize P 2x15y st 3m4y g 1000 61133y g 1200 37 y Avinash Sathaye 1 Ia 162 Spring 200E h IRtI iCE S Old problem with a new method 0 We recall the problem we solved using the graphic method Maximize P 2x15y st 3x4y 3 10007 6x3y 12007 as y o This was the sketch of the feasible region Avinash Sathaye l la 162 Spring 200E h latfiCE S Old problem with a new method 0 We recall the problem we solved using the graphic method Maximize P 2x15y st 3x4y 3 10007 6x3y 3 12007 as y o This was the sketch of the feasible region x o The corner points were 007 0 1412007 0 B120 160 and C07 250 By Checking the function values at the four corner points we found the maximum value at B Avinash Sathaye l la 162 Spring 200E l latrices February 2009 2 11 A unple mntinuecl 0 We begin by setting up a problem table which will become useful later 3 HHS 4 1000 3 1200 15 19030332 a 72 mm A ample mntinued 0 We begin by setting up a problem table which will become useful later 36 y RHS 3 4 1000 6 3 1200 2 15 0 Note that x y 2 0 is not listed7 since it is always assumed We are listing inequalities which are always assumed to be of the form 3 with the last row giving the function coef cients L ample mntinued 0 We begin by setting up a problem table which will become useful later 36 y RHS 3 4 1000 6 3 1200 2 15 0 Note that x y 2 0 is not listed7 since it is always assumed We are listing inequalities which are always assumed to be of the form 3 with the last row giving the function coef cients 0 We next convert these to equations and write a proper augmented matrix called a simplex tableaux Simplex taibl 39 a 0 We introduce new variables called slack variables so that an inequality 3x 4y 3 1000 is replaced by 396 4y u 1000 With the understanding that u 2 0 Sirnilarly7 the second inequality becomes 6x By v 1200 With v 2 0 Simplex taibl 39 a 0 We introduce new variables called slack variables so that an inequality 3x 4y 3 1000 is replaced by 396 4y u 1000 with the understanding that u 2 0 Similarly7 the second inequality becomes 6x By v 1200 with v 2 0 The maximized function declaration is rewritten as 7296 7153 P 0 Note the changes of signs Simplex taibl 39 a 0 We introduce new variables called slack variables so that an inequality 3x 4y 3 1000 is replaced by 396 4y u 1000 with the understanding that u 2 0 Similarly7 the second inequality becomes 6x By v 1200 with v 2 0 0 The maximized function declaration is rewritten as 7296 7153 P 0 Note the changes of signs 0 This sets up an initial tableaux x y u v P RHS 3 4 1 0 0 1000 6 3 0 1 0 1200 72 715 0 0 1 0 Bar i Solution o The simplex tableaux always has a set of unit columns columns of the Identity matrix which exactly ll up an identity matrix after possible rearrangement n a m i Solution o The simplex tableaux always has a set of unit columns columns of the Identity matrix which exactly ll up an identity matrix after possible rearrangement o The corresponding variables on top of these columns are said to be basic variables and form a basis for the current tableaux The number of basic variables must be the same as the total number of rows i Solution The simplex tableaux always has a set of unit columns columns of the Identity matrix which exactly ll up an identity matrix after possible rearrangement The corresponding variables on top of these columns are said to be basic variables and form a basis for the current tableaux The number of basic variables must be the same as the total number of rows Corresponding to the basis7 we have a basic solution to the current tableaux It is obtained by setting the non basic variables to zero value and reading off the solutions of the basic variables from all the equations i Solution The simplex tableaux always has a set of unit columns columns of the Identity matrix which exactly ll up an identity matrix after possible rearrangement The corresponding variables on top of these columns are said to be basic variables and form a basis for the current tableaux The number of basic variables must be the same as the total number of rows Corresponding to the basis7 we have a basic solution to the current tableaux It is obtained by setting the non basic variables to zero value and reading off the solutions of the basic variables from all the equations Thus7 for our rst tableaux above7 the basis is u v P and the basic solution is x y u U 07 0 l l li liill i ttli11 of the Simplex table 1x 0 Note that our original variables 3c y have values 00 and thus7 this corresponds to the corner point 0 of our original feasible region a a 72 n q a lVlCH jll C thHl of the Simplex table 1x Note that our original variables x y have values 0 0 and thus77 this corresponds to the corner point 0 of our original feasible region If we solve the last equation for its basic variable P7 then we have P 2x 15 and this says that increasing the values of x or y will increase the value of P Note that this is signi ed by the negative numbers 72 715 in the tableaux 0 Note that our original variables x y have values 0 0 and thus77 this corresponds to the corner point 0 of our original feasible region o If we solve the last equation for its basic variable P7 then we have P 2x 15 and this says that increasing the values of x or y will increase the value of P Note that this is signi ed by the negative numbers 72 715 in the tableaux 0 So we try to make one of thern7 say 36 as a basic variable This means the 36 column has to become a unit column Where shall it have its 1 a You may try and make the pivot at any available non zero entry in the 36 column However7 sorne choices will produce negative entries in the last column This would mean that the new basic solution will have negative values for some basic variables This is not allowed7 since by our set up all the variables have non negative values You may try and make the pivot at any available non zero entry in the 36 column However7 some choices will produce negative entries in the last column This would mean that the new basic solution will have negative values for some basic variables This is not allowed7 since by our set up all the variables have non negative values There is a simple test which will guarantee that we won7t get in trouble Check the ratios obtained by dividing non zero entries in the x column into the corresponding entries in RHS and choosing the one which gives a minimum value You may try and make the pivot at any available non zero entry in the 36 column However some choices will produce negative entries in the last column This would mean that the new basic solution will have negative values for some basic variables This is not allowed since by our set up all the variables have non negative values c There is a simple test which will guarantee that we won7t get in trouble Check the ratios obtained by dividing non zero entries in the x column into the corresponding entries in RHS and choosing the one which gives a minimum value 0 In our example the choices are and 200 respectively Note that we do not use negative entry 72 The smallest ratio is 200 in the second row 7 1111p R i quot and below 0 Thus7 we use the 21 entry 6 t0 Clean out all entries above E39 5 r n a m 7 nup h 5 quot o Thus7 we use the 21 entry 6 to Clean out all entries above and below 0 The row oeprations R3 7 132 R1 7 132 and nally Rz produce the following new tableaux a a 72 mm nup R i quot o Thus7 we use the 21 entry 6 to Clean out all entries above and below 0 The row oeprations R3 7 R2 R1 7 1132 and nally RZ produce the following new tableaux O x 1 U U P RHS 0 52 1 712 0 1 12 0 16 0 200 W 400 7 1111p h 5 quot o Thus7 we use the 21 entry 6 to Clean out all entries above and below 0 The row oeprations R3 7 R2 R1 7 1132 and nally RZ produce the following new tableaux x 0 1 0 0 Note that the new basis is now 36 u P and hence the new basic solution is 2 52 12 u 1 0 0 v P 16 0 13 1 712 0 712 RHS 400 200 400 x y u v P 200 0400 0 400 bntinilecl Pi 0 We note that in terms of x y variables we are at a point 200 0 which was our point A Thus we have marched from O to A on our feasible graph We also have improved our function value from 0 to 400 0 We note that in terms of x y variables we are at a point 200 0 which was our point A Thus we have rnarched from 0 to A on our feasible graph We also have improved our function value from 0 to 400 0 We notice the negative number 7 in the function row and this tells us that it would help to put y in the set of basic variables Explicitly7 the last equation is P g 7 g 4007 so y wants to get bigger bntinuecl Pi 0 We note that in terms of x y variables we are at a point 200 0 which was our point A Thus we have rnarched from 0 to A on our feasible graph We also have improved our function value from 0 to 400 0 We notice the negative number 7 in the function row and this tells us that it would help to put y in the set of basic variables Explicitly7 the last equation is P g 7 g 4007 so y wants to get bigger 0 We check the pivot ratios for the second y column 400 i 160 d 200 7400 52 7 an 12 7 39 Thus the new pivot must be at 11 position 319111111 and end I The pivot operations are R 7 R1 and R3 R1 and produce 95 y u v P RHS 0 1 25 715 390 160 1 0 0 715 A 0 120 0 15 397 1 480 E a r o a m Tieaiiiil and end 0 The pivot operations are R2 7 Rl and R3 R1 and produce x y u v P RHS 0 1 25 715 0 160 1 0 715 1445 0 120 0 0 15 310 1 480 o The new basic is Clearly x y P and the new basic solution is xyuvP 1201600 0480 Note that this matches our point B in the graph of feasible points

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