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# CIVIL ENGR SYSTEMS CVEN 322

Texas A&M

GPA 3.53

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This 24 page Class Notes was uploaded by Carolyn Kuhn on Wednesday October 21, 2015. The Class Notes belongs to CVEN 322 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/226123/cven-322-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15

Refuse hauling transhipment nodes Z Page 1 of 2 5 B Problem 511 8 Three cities A B and C produce refuse of 100 tonsday 50 tonsday and 30 tonsday respectively The refuse can be shipped in several directions including to each other or to two transfer stations D and B using local trucks Once the refuse is shipped to the transfer stations at D or B it is transferred to highway trucks and shipped to one of three land lls F G or H The land lls have the following maximum capacities per day F 100 tonsday G 50 tonsday and H 60 tonsday The following unit costs per ton are charged for any and all refuse that comes into through or to the cities transfer stations or land lls It is a quothandlingquot charge applied to any material accepted City A 4ton City B 6ton City C 4ton Transfer Station D 3ton Transfer Station E 4ton Land Fill F 8ton Land Fill G 12t0n Land Fill H 7ton Thus the 100 tonsday entering City A will incur a 400 charge simply because the refuse entered City A The costs to transport material between locations are as listed A B C D E F G H City A 2ton 3ton 4ton City B 2ton 3ton SSSton 4ton City C 3ton 7ton 9ton Transfer 3ton 1 Oton 8ton 12ton Station D Transfer 3ton 6ton 4ton 1 3ton Station E SEE TABLE ABOVE FOR THE NUMBERS SHOWN ON THE MAP BELOW httplowerytamueducven3 22homeworklp5l lhtm 6252008 Refuse hauling transhipment nodes i Page 2 of 2 19 H Total hauling costs per ton are shown in the table and on the schematic above Set up and solve the LP solution to determine the least cost way to get the refuse from the cities to the land lls Note that it is possible to haul refuse between the cities as shown and also between the transfer stations if that proves to be more economical httplowerytamueducven322homeworklp51 lhtm 6252008 ClaSS 13 Page 1 of2 53 M Tars Ream Ti39ME germs A no swarm Print out the following page and bring it to class with you Learning Objectives Class 13 After today39s lecture and after working the homework problems the student should be able to solve problems such as Waste dis osal for laroe combined area You should also see how problems such as 5 gravel hauling and integer problems are solved using linear programming Topics covered in today39s class Waste disposal for large combined area Given A contractor is planning a job that will require a large amount of gravel and sand The estimates for the necessary gravel and sand are Gravel 90000 tons Sand 10000 tons There are two pits from which material can be obtained and the plan is to haul from these two pits and to separate screen the material on the job for whatever material is needed Analysis shoWs that the material at each pit has the following composition Pit A Pit B Gravel 20 10 Sand 80 90 It costs 80ton for material and hauling from pit A and 2 ton from pit B Required Solve this as a linear programming problem to min ize the delivered cost Integer solutions WW To solve a problem where you want everything to be an integer run the problem using the Gomery Cut Algorithm not on autoselect with the following lines added at the end of the model BOUNDONSUM http lowerytamueducven322 Classes Class 1 3Class l 3 htm 6232008 Class 13 Page 2 of2 200 and on the next line ALLINTEGER The BOUNDONSUM variable should be set the the approximate sum of all the answers for your variables LE if you think you will be making 300 one bedroom apartments 100 two bedroom apartments and 400 three bedroom apartments perhaps try the value of boundonsum 800 the sum of all your answers X1X2X3 BOUNDONSUM is used to assist the program in making a reasonable guess as to where the answer lies It is used only to speed up the solution and is not critical that you get it correct However the closer you can get it the fewer dead ends the computer will go down and the quicker the program will run In any case it is required that you put something in there If you put in BOUNDONSUM 2 and the true nal sum of all your answers is 2000 he will still get the solution he will just say quotidiotquot quotfoolquot quotjackassquot a bunch of times The Gomery Cut Algorithm can also be used when you have some variables that you want to be integer and others that should be real numbers by using the statement INTEGERvariablel variable2 httplowerytamueducven322ClassesClass 1 3 Class 1 3 htrn 6232008 Laying paving and binder Page 1 of 1 12 CD Problem 5 17 Given In order to keep to a construction schedule I must place 1000 tons of pavement a day Because of other projects in the area I will be unable to purchase all of my pavement from one batch plant In fact I will have to purchase from three plants in the area because of the high demand Plant G can supply me with up to 275 tons day total of either Topping or Binder my choice Thus you can get up to 275 tons of whatever you want 275 tons of Topping and no Binder or 100 tons of Topping and 175 tons of Binder or 2 tons of Topping and 273 tons of Binder or 10 tons of Topping and no Binder just anything you like but no more than 27 tons per day Plant H can supply up to 350 tons day of either Topping or Binder same idea as for Plant G Plant I can supply me with up to 595 tonsday of either Topping or Binder same idea as for Plant G I have two pavers one laying Topping and one laying Binder The Topping Paver lays 575 tonsday and the Binding Paver lays 425 tonsday which as you might imagine is the exact daily amount required to construct the road The cost to transport paving material from Plant G to the Topping Paver is 32ton Cost from Plant G to the Binder Paver is 36ton from Plant H to Topping Paver 30ton from Plant H to Binder paver 33ton I to Topping paver 3 lton and I to Binder paver 35t0n Required Solve for the minimum cost of acquiring the necessary material httplowerytamueducven3 22homeworklp5 l 7html 6252008 Car Barn distribution for a day Page 1 of 1 Problem 5 21 g l Given A rapid transit system serving a metropolitan area has ve bus stations in the suburbs Each morning busses are dispatched to these stations from three bus barns near the center of town Data for bus availability at each bus barn and bus requirements for each station along with the costs of dispatching a bus from any bus barn to any station are given below Station Busses Required 1 30 2 2O 3 10 4 10 g9 Bus Barn Busses Available 1 40 2 2O Dispatching Costs to station gt 1 2 3 4 5 From bus barn l 40 30 20 30 60 From bus barn 2 20 20 4O 50 70 From bus barn 3 60 40 50 30 30 Required 1 Solve the problem to nd the least cost method of dispatching the required busses 2 Solve problem a assuming that there were now 50 busses available at bus barn 1 instead of 40 ie that there were now more busses available at the bus barns than were needed at the stations 3 Solve problem a assuming that the number of available busses at bus barn 1 is only 30 ie that there were now fewer busses available in the bus barns than were desired at the stations http lowerytamu educven3 22homeworklp5 21 html 6252008 Machine assignment problem Page 1 of 1 22 Problem 5 33 Given A company has four machines that can accomplish three jobs Each job can be assigned to one and only one machine The objective is to assign the jobs to the machines in the most costef cient manner Costs for each job machine combination are given in the following table Machine 1 Machine 2 Machine 3 Machine 4 Job 1 19 23 28 31 Job 2 7 14 16 19 Job 3 10 15 20 22 Required Solve for the minimum cost httplowerytamueducven322homeworklp533html 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 1 of 9 TABLEAUS Print out the following page and bring it to Class with you HOW TO SET UP HAND SOLUTIONS TO LINEAR PROGRAMMING PROBLEMS Say you have been presented with the following LP problem Min Z 2R 8Y ST 1 R lt 20 red ones quot 2 Y gt 14 yellow ones 3 5R lOY 150 Note that basically the quotsolutionquot to this problem as far as we are concerned is R something and Y something where R and Y always have to give legal results and hopefully at some time optimal results These quotsomethingsquot can be positive numbers or zeros but never negative Note that there are three constraints subj ect to39s to the problem each of which is an equation of a line in two dimensions forming quotboundariesquot to the solution Further the objective anction is also a line Since we have previously seen that the solution to such a problem falls on the boundaries of the constraints it becomes evident that if we could nd a legal starting position for R and Y even though not optimal and then move that point to a boundary then move that point to a boundary intersection then crawl up with respect to the slope of the objective function that boundary intersection until we get to another boundary intersection and continue this process we would nally reach the optimum value of R and Y This can be done using a quotTableauquot Even neater is the fact that this works in three dimensions R Y and x3 or 500 dimensions R Y x500 Solving the above problem using a Tableau INEQUALITY 1 says quotDO NOT use more than 20 of the red ones since we only have 20 available quot Thus we try to make INEQUALITY 1 into an EQUATION by ADDING a SLACK variable to it and see how it feels 1RSl20 Great Because with 1 representing how many red ones we didn39t even need ie the number still left in the box after we found the optimum solution to our problem EQUATION I says 1 How many red ones you used in getting the optimum solution How many red ones are still left in the box Slackl 20 Le when the boss comes in and says quotYou didn39t use more than 20 red ones did youquot we reply quotNo sir In fact we still have 81 of those little suckers left in the box quot Next INEQUALITY 2 says quotBe SURE to use AT LEAST 14 of the yellow ones since our storage room is ill and we absolutely MUST get rid of at least 14 of them to make room for other things I quot Thus we try to make INEQUALITY 2 into an EQUATION by SUBTRACTING a SURPLUS variable from it and see how that would work httplowerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 2 of 9 i gt itir 2g 2 Y S2 14 Say Y 20 then S2 6 WORKS GOOD Because Y represents how many yellow ones we used in getting the optimum solution and S2 represents how many MORE than the 14 minimum we were told to use that we actually ended up using Ie when the boss comes in and asks quotWere you sure to use up at least 14 of those rotten yellow ones like I told youquot we reply quotYes indeed sir in fact we actually used S2 more than the minimum that you speci edquot LE SurplusZ 6 Next EQUATION 3 is already an EQUATION so to heck with it 3 5R 10Y 150 We now have formed a set of EQUATIONS to work with as opposed to our previous INEQUALITIES and can thus use various math methods such as Gaussian Elimination or Cramer39s Rule for solving for the intersections and boundaries given by these equations If you have forgotten how these methods work see httowwwsosmathcommatrixsvstem l svstem l html or httn mathworidwolframcomCramersRu lentml Now we must some select some initial legal point R and Y at which to start the solution to get an initial although nonoptimal answer This is not an easy task Where in a diamond in threeD space is a point inside that diamond Thus I propose that we just always start out saying quotToday no workquot No kidding Initially the initial solution will be Today do no work make no pro t incur no costs make no widgets use no red ones and no yellow ones no no no Even though we both agree that the quotdiamondquot does not touch 000 we also admit that we have no way of nding a legal point inside of it Obviously the problem of an initial legal point becomes even more impossible for a problem with 50 or 500 variables Thus the initial proposed solution will always be xi 0 R0 YO Rltz Now let39s try putting these values back into our modi ed EQUATIONS to make sure we won39t be shooting ourselves in the foot If the modi ed equations themselves become illegal then we have no hope of quotimprovingquot the solution since even our initial solution is invalid 1 R 81 20 m 2i my That39s ok since when R 0 S1 will 20 and that s legal g gt m i g C 2Y s214 That39s not so great because when Y 0 S2 must l4 and no rogramming variable can EVER be negative It would have no meaning So let39s simply add anothef ariable to the EQUATION to permit our proposed initial solution to be legal at the quotdo no workquot point it we add an arti cial variable Al which will represent how badly we ignored the boss39s demand that we n at least 14 of the yellow ones I2 6 it 2YS2 g 14 f Now let s try it again Say Y 20 then S2 can 6 and Al will 0 That works ok Say Y 14 then SZ can 0 an A2 ill 0 That works ok WWW httplowerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 3 of 9 Say Y 0 then 32 can o anm4 That39s GREAT Z Now remember S2 Surplus Variable 2 represents how many above and beyond the minimum 14 yellow ones we used above and beyond our boss39s demands while Al Arti cial Variablel represents how many fewer than his demand of 14 we used in our optimum solution thereby incurring his wrath Now we39ll try to get equation 3 to work under all conditions 7 W 35R10Y150 39 6 Trying R 0 and Y 0 we see that 0 150 Rats So let39s use the same trick as before and add an arti cial variable to the equation 3 5R 10YA2 150 Say R 10 and Y 10 then A2 0 Good enough Say R 0 and Y 0 then A2 150 Great Thus the rules of the game are 5 52 3 2 Whenever you have gt subtract a SURPLUS variable and add an ARTIFICIAL variable S 3 Whenever you have add an ARTIFICIAL variable A i A 2 Also a summary of just what these things MEAN 1 Whenever you have z add a SLACK variable l Slack variables If a constraint inequality lt demands an upper limit on the usage of some resource the slack variable represents the unused amount of that limited resource once the optimal solution is found if any exists 2 Surplus variables If a constraint inequality gt demands a minimum usage of some resource the surplus variable represents that amount above and beyond the minimum of that resource that you used in nding the optimal solution if any 3 Arti cial variables Used in either gt or constraints to be able to legally start the solution at a quotno work will be donequot point It39s purpose is to get the solution legally started at a simple initial point namely R Y X3 0 However we must try and force all arti cial variables to zero in the nal solution since arti cial variables represent how badly you have ignored your boss s wishes and disobeyed his direct orders Now since we added all those extra slack surplus and arti cial variables how are we going to make the solution realize our preferences 16 how will the solution know Whether we like leaving those ARTIFICIAL VARIABLES and SLACKS and SURPLUS VARIABLES in the solution or are we greatly displeased with their presence in the solution Well here39s what we will do we will tell him that SLACK variables areigae if he leaves them in the solution that39s ok since I really don39t care pro twise or cost wise whether there are some red ones left in the box or not I was only told not to use too many of them anyway And I also really don39t care if he uses 3 more yellow ones than the minimum of 14 or whether he uses 30 more than the minimum of 14 required yellow ones Thus SURPLUS variables are also free Actually I have already paid for them elsewhere in the objective function http1owerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 4 of 9 Zia However I really must make clear to him that Arti cial variables are truly expensive items For example if he gives me back a solution that has A1 2 that means the bosses instructions were not carried out and I may lose my job That would say I only used up 12 yellow ones rather than the 14 he instructed 39 So what I will do is tie a really BIG cost to Al so he will do his dangd st t drive it out of the solutio as soon as possible Thus my objective function becomes MinZ2R8YOSlOSZMA1MA2 Where M one quadrillion dollars Also since I don39t know HOW to minimize I will do this instead MaxZ2R8YOSl Os2MA1 MA2 Note there are rules which would work to minimize something but rather than learn how to minimize AND maximize we will just learn how to maximize If anyone ever asks you to minimize something simply reverse the signs in the objective function and maximize it instead Note that only the signs of the objective function are reversed not the signs on the subject to constraints Note also that whether you are minimizing OR maximizing when you set up your tableau the quotbig M39squot will be negative This is because if you are maximizing you will SUBTRACT MAl to greatly penalize your pro ts If you are minimizing you will ADD MA1 to greatly penalize your costs which when you reverse your objective function signs will again give the arti cial coef cients as negative A QUICK SUMMARY OF WHAT YOU DID IN ORDER R lt 20 Not an equation R 20 An equation but doesn39t allow R 14 as originally permitted 1 20 An equation permitting R 14 but perhaps not legal at R 0 don39t go to work today 0 Legal and permits R 0 Good Y gt 14 Not an equation Y 14 An equation but doesn39t allow Y 20 as originally permitted Y 2 14 An equation permitting Y 20 but perhaps not legal at Y 0 don39t go to work today 0 2 14 Sure enough not legal W vY S2 A1 14 Good 5R 10Y 150 Already an equation but not legal for R O and Y 0 do no work today 0 0 re enough not legal w 5R 10Y 2 15 Adding the new variables required into the objective function MINZ2R 8YOSlOSZMA1MA2 GENERAL RULES FOR WHO IS IN THE SOLUTION 1 All SLACKS are IN the solution 2 All ARTIFICIAL VARIABLES are IN the solution httplowerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 5 of 9 BASIC RULES OF GETTING A SOLUTION WITH A TABLEAU 44quot 7 1 Calculate Zj the sum of the Cj values down the left column times the corresponding coef cient under the x1 or x2 or x3 or 51 etc column This is basically the slope of the boundary with respect to the variables x1 x2 x3 Please note that it39s almost impossible to explain this by writing it down LISTEN IN CLASS 2 Calculate Cj Zj This is basically the slope of the boundary with respect to the objective function 3 The largest value of Cj Zj will tell you whether a better solution is available and give you which Entering Variable has the largest slope and therefore will improve the solution the fastest If there are no positive Cj Zj then the best solution has been found You are nished The Entering Variable is the variable which will replace the leaving variable on the next Tableau thereby kicking one of the other poorly performing variables out 4 Once you have found the Entering Variable calculate the Right Hand SideEntering Variable coef cient The smallest positive NON ZERO NONNEGATIVE value of RHSEnter is the Leaving Variable The Leaving Variable is doing you the least good in attempting to improve the solution and therefore will be removed in the next Tableau to make room for the Entering Variable 5 The intersection of the Entering Variable and the Leaving Variable is called the pivot element 6 To get the next tableau re write the previous tableau replacing the Leaving Variable and its coef cient with the Entering Variable and its coef cient 7 By algebraic manipulation make the pivot element 1 Note that whatever you do to the pivot element to make it into the number 1 you will have to do that same operation to all other elements on that row Seems only fair Write all of these new numbers on the pivot row on the next Tableau 8 Make all 19 terms above the pivot element and all 37 terms below the pivot element equal to 0 using any multiple OF THE PIVOT ROW legal algebraic operations only and write these new rows in the new tableau Whatever you do to any element to make it a zero must be done to all the other elements in that row Note that this is just like the Gaussian Elimination procedures you learned in math 9 After you get the new tableau written down you MUST MAKE SURE that all the RHS terms are positive If not change the sign of every term on that row This is because otherwise you are saying that the current solution for something is NEGATIVE which is never permitted in linear programming Secondly you must make sure that all the cross terms are 1 Le if S2 is in the solution make sure that the value which intersects S2 horizontally with 2 vertically is 1 If it is a 6 then divide that entire row by 6 making it a 1 Or if it is 35 divide that entire row through by 35 to make it a l 10 Proceed with this same operation until there are no positive Cj Zj values at which time the optimum solution has been found 11 Note that although the method is similar to Gaussian Elimination or Cramer s Rule because of the book keeping done around the outside of the equations you have several other restrictions that you must follow a All Intersection terms must be made to 1 That is to say you MUST get 1000000000000 at the intersection of all variables which are in the solution in all tableaus If you don39t make it so Thus if X is IN the solution in any tableau ie if it is currently in the SolutionVariable SV column check that the quotintersectingquot number on the quotXquot row directly BELOW the quotXquot at the top is a 1 If Y is IN the solution in any tableau ie if it is currently in the SolutionVariable SV column check that quotintersectingquot number on the quotYquot row directly BELOW the quotYquot across the httplowerytamueducven3 22howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 6 of 9 top is a 1 If Al is IN the solution in any tableau ie if it is currently in the SolutionVariable SV column check that quotintersectingquot number on the quotAl quot row directly BELOW the quotAl quot across the top is a l b All Righthand side RHS numbers MUST be positive If they are not make them so This is because this number IS the solution for the variable on this row and negative numbers have no meaning in Linear Programming ANSWERS 1 For each and every tableau including the rst the value of the variables currently in the solution ie the variables listed on the left hand side of the tableau can be determined by simply looking at the RHS 2 The current value of any variable not listed in the tableau is zero 3 The current value of the objective function for any tableau is listed as the last number under the RHS column EXAMPLE PROBLEM M z 4 1 3 2 3 WWW ax X X X ST x1 x2 lt 20 x2 x3 gt 4 Xl X3 10 Thus as per the rules above and the objective function becomes MAXZ4X13x2x3 081082 The INITIAL solution since we have decided to quotdo no work todayquot is From the rst equation X1 0 X2 0 so 1 20 From the second equation X2 0 X3 0 S2 0 so A1 4 From the third equation X1 0 X3 0 so A2 10 Putting these quotdo no work todayquot numbers into the objective function leads us to realize a loss of Z 14M Who is currently IN THE SOLUTION S 20 A1 4 and A2 10 All others are 0 and are not quotinquot the solution Then the tableau will be set up as follows SV RHS RHSENT http1owerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 7 of 9 Zquot Z Cj 2 where Cj is the coef cient for each variable SV is the name of the Solution Variable Zj is the slope of the Solution Variable CjZj is the quotrelativequot slope between the Solution Variable and the Objective Function RHS is the Right Hand Side of the equation and also the current solution for the Solution Variable quotin the solutionquot and RHSENTER is the slope of the Solution Variable relative to the Entering Variable W t L if Est w sea We ng Filling in the rst values c 4 R 3 1 0 0 M M sv it X1 if x2 x3 81 82 A1 A2 RHS RHSENT 0 s1 9 1 1 1 0 0 29 20120 1 A1 310 3 1 1 0 1 1 0 4 314951 M A2 v39qvk 0 0 0 1 10 iii1102 Eta H yj win g9 Zj M M 0 0 M M M l4m Cj 4 3 1 0 0 0 Zj M M M W M a a E ma 1 teen a 5 a a a a at Now at this time tableau l the current proposed solution is 81 20 A1 4 A2 10 xl x2 x3 S2 zero and the current value of the objective function is quotlose a cool 14 million dollarsquot mainly because you didn39t follow your boss39s orders and there are still some arti cial variables in your solution Note that all quotIntersectingquot terms are 1 and all RHS Right Hand Side terms are and MUST BE positive The reason all RHS terms must be positive is because those are the current answers for 81 Al and A2 and as such must be positive Negative answers have no meaning in linear programming problems Carrying out the calculations for Zj Cj Cj and RHSENT gives the values shown above You calculate Zj the sum of the Cj values down the left column times the corresponding coef cient under the XI or x2 or x3 or s1 etc column Thus Zj in the x1 column 01 M0 Ml M Zj in the x2 column l Ml MO M Z in the x3 column 00 M l Ml 0 etc Thus for the next tableau the entering variable found above will be x1 since he has the largest relative slope CjZj The leaving variable will be A2 since he has the smallest nonzero non negative RHSENT Note that the pivot element has been marked with asterisks 1 and the entering and leaving variables have also been marked OK Now let39s see if we can get a better solution out of the next tableau Should be possible since Mister xl has said that his Cj Zj is positive and therefore if we will let him into the solution he will improve things for us Also we have kicked out the bum A2 since he was the smallest RHSENTER value we had Remember Do not try to remove anyone whose RHSENTER is zero or negative you will end up outside the feasible solution region Also we will be making the pivot element l and all fifteen or so terms above and below the pivot element if there are any equal to zero Actually we got kind of lucky since the pivot element was already 1 and one of the values directly above him which was to be turned into a zero was already zero Then subtracting the pivot row from the top row will turn the term directly above the pivot element into a zero giving the next tableau NOTE You MUST use the pivot row in combination with other rows in your attempt to make terms above and below the pivot row 0 This is necessary because of the bookkeeping going on around the edges of the tableau httplowerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 8 of 9 h D 1331 9 1 0 0 M M X3 A1 RHSENT l 0 101 10 1 1 l 0 100 INF 4 4 M M Cj 3 5 4 M M M M Zj Thus at the second tableau the current value of the objective function is a loss of only 40 shy of 4 million dollars The solution shown consists of Sl 10 Al 4 x1 10 with X2 x3 S2 A2 zero since they are not in the solution Note also that we carefully made sure of the following after our mathematical manipulations all cross terms are 1 all RHS are positive the pivot element is 1 all terms above and below the pivot element are 0 Note that the Cj Zj values for S 1 x1 and A1 in the tableau above are zero That makes sense because since they are already IN this solution so they couldn39t be expected to improve the solution if we were to quotlet them inquot They are already in Note also that the Cj Zj coef cient of A2 is minus millions He is saying quotif you let me back in I will hurt you GOODquot Note that all quotIntersectingquot terms are l and all RHS terms are positive OK as seen above the entering variable is x2 and the leaving variable is Al Thus the next tableau looks LEQAI Cj 4 3 l 0 0 M M SV x1 x2 x3 S1 82 A1 A2 RHS RHSENT 0 1 0 0 0 l 1 l l 6 6l6 3 x2 0 l 1 0 l l 0 4 4 1 4 4 x1 1 0 l 0 0 0 l 10 100 INF NOTE Don39t use x2 as the leavin z 4 3 1 0 3 3 4 52 variabi H Cj 0 0 2 O 3 3 4 Zj M M W i w Mquot Note that all quotIntersectingquot terms are l and all RHS term the leaving is S1 The next tableau looks 3 1 0 M M SV x2 x3 S2 Al A2 RHS RHSENT S2 0 1 l 6 x2 1 0 l 10 x1 1 0 l 10 Zj 1 00230 1 CJ M M lt Note No more p051tive httplowerytamueducven322howtolpHowtolphtm 6252008 HOW TO SET UP EQUATIONS LINEAR PROGRAMMING PROBLEMS Page 9 of 9 it lellllllll ll Notethat all quotlntersectin quot terms arel and all Right Hand Side terms are osmve The RHSterms MUST be posmye because they A the solutxons propose at each step or the solu Ion and negatwe Values are not germ1ttecl fog l1near rogrammIn JSOIutIons they make no sense Also no more posmve C3 ZJ eXISt thus he solutlon lS comp ete WHO N THE nal solutlon therefore1s x1 10 x2 10 S2 6 1 x3 A1 A2 zero The current value of the object1ye functlon 15 70 whlch IS the maXImumdpro t oss1b1e for th1s groblem Now s1nce l was suszosed to be Ijnlnlm1z1n costs but d1dn39t know how an there ore pretended 0 be max1rn121n6ro t Instea I now admlt that an tell the boss that the mlmmum poss1ble cost for th1s problem w111 be Wthh W111 occur when Xl1 X210 2 6 an 81 X3 A1 A2 zero httplowerytamueducven322howtolpHowtolphtm 6252008 Blank Tableau W W Sol vac K 33 0 pm 0 D9 O 2 KW 93 http1owerytamueducven322h0meworkLpblanktableau1 htm Page 1 of 1 a 6252008 Blank Tableau 2 C 45352 a o m am 1 I f S 39 XE xif a 35 2 a 2 gas var RHS Enter Q OO mqu gig ii EQ WEQWVE Eig 29 fgt EO Cgt 063 E i 4 O 4 Zj M 5 415 0 Zj quot4qu j http10werytamueducven3 22h0meworkLpb1anktableau1 htrn 37252 Page 1 of 1 6252008 Blank Tableau 0 Sol RHS Var RHS Enter 2 CJ zj http lowerytamu edu cven3 22 homeworkLpblanktableaul htm Page 1 of 1 3 6252008 Blank Tableau 0139 So RHS Var RHS Enter 2139 CJ39 zj httplowerytamueducven3 22homeworkLpb1anktableau1 htm Page 1 0f 1 6252008 Class 16 Page 1 ofl 353 Print out the following page and bring it to class with you Learning Objectives w Class 16 After today39s lecture and after working the homework problems the student should be able to o Solve linear programming problems by hand Topics covered in today39s class See main page httplowerytamueducven322Classes Class 1 6Class 1 6th 6252008 Lowery39s Old 422 Exams Page 1 of 2 5 w LOWERY39S OLD 422 EXAMS n2 Please note that I am happy to share the type of questions asked of previous semester39s students However please also note that sadly I will NOT be able to sit down with each of you individually one at a time and show you how to work all 132 of the problems listed Sorry but that just isn t possible If you want to work these problems for practice fine Get with your friends go over to someone39s house and pound them out but please don39t ask me to solve them for you Please also note that the course changes from year to year Thus this year we may have economics on quiz A while last year quiz A covered optimization Sometimes material is completely removed from the course and something else takes its place For example Critical Path Methods were removed from 422 when engineering economics was added Thus don39t get upset if you don39t know how to do something on one of these quizzes that you never heard of It39s possible that we just no longer cover it It39s also possible that you didn39t come to class that day or slept through that class in which case I39m not willing to spend two hours teaching it to you individually Sorry Get a kinder soul in the class to show you QuizA Summer of 1996 Quiz13 Summer of 1996 EinalExarn Summer of 1996 QuizA Fall of 1996 Quisz Fall of 1996 FinalExam Fall of 1996 an Spring of 1997 QQiLB Spring of 1997 Final Exam Spring of 1997 Q iLA Fall of 1997 QuizB Fall of 1997 EinalExnm Fall of 1997 Quiz A Spring of 1998 Quiz B Fall of 1998 Final Exam Spring of 1998 Quiz A Fall of 1998 Quiz B Fall of 1998 EinalExam Fall of 1998 Quiz A Spring of 1999 Quiz B Spring of 1999 Final Exam Spring of 1999 Quiz A Spring of 2000 Quiz B Spring of 2000 Final Exam Spring of 2000 Quiz A Spring of 2001 httplowerytamueducven422quizzesindexhtm 2202003 Lowery39s Old 422 Exams Page 2 of 2 m Quiz B Spring of 2001 Final Exam Spring of 2001 Quiz A Spring of 2002 Quiz B Spring of 2002 Final Exam Spring of 2002 http10werytamueducven422quizzesindexhtm 2202003 Time Independent Production Models 3 Determine What scaffolding components to make for the following month given the data below Side Rails Floors Braces Connections Limitsmonth Ft 0 P IPC 50 ft 20 ft 10 ft 50000 ft Used Ft of 2X2xl4 Angle Iron 30 ft 50ft 12 ft 8 ft 30000 Used 1 bolts 20 30 4 10000 Pounds of Welding 2 1 4 3 1000 Labor 10 6 2 1 3200 Profit 80 50 12 8 Minimum Required 30 50 Additionally marketing requires that you make at least 2 sides for every oor manufactured

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