An Introduction To Management Science: Quantitative Approaches To Decision Making, Revised, 13th Edition Solution Manual

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ContentsPrefaceChapter1.Introduction2.An Introduction to Linear Programming3.Linear Programming: Sensitivity Analysis and Interpretation of Solution4.Linear Programming Applications in Marketing, Finance and Operations Management5.Advanced Linear Programming Applications6.Distribution and Network Models7.Integer Linear Programming8.Nonlinear Optimization Models9.Project Scheduling: PERT/CPM10.Inventory Models11.Waiting Line Models12.Simulation13.Decision Analysis14.Multicriteria Decision Problems15.Forecasting16.Markov Processes17.Linear Programming: The Simplex Method18.Simplex-Based Sensitivity Analysis and Duality19.Solution Procedures for Transportation and Assignment Problems20.Minimal Spanning Tree21.Dynamic ProgrammingAppendix A: Building Spreadsheet Models

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Chapter 1IntroductionLearning Objectives1.Develop a general understanding of the management science/operations research approach to decisionmaking.2.Realize that quantitative applications begin with a problem situation.3.Obtain a brief introduction to quantitative techniques and their frequency of use in practice.4.Understand that managerial problem situations have both quantitative andqualitative considerationsthat are important in the decision making process.5.Learn about models in terms of what they are and why theyare useful (the emphasis is onmathematicalmodels).6.Identify the step-by-step procedure that is used in most quantitative approaches to decision making.7.Learn about basic models of cost, revenue, and profit and be able to compute the breakeven point.8.Obtain an introduction to the use of computer software packages such asMicrosoft Excelin applyingquantitative methods to decision making.9.Understand the following terms:modelinfeasible solutionobjective functionmanagement scienceconstraintoperations researchdeterministic modelfixed coststochastic modelvariable costfeasible solutionbreakeven pointSolutions:

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1.Management science and operations research, terms used almost interchangeably, are broaddisciplines that employ scientific methodology in managerial decision making or problemsolving. Drawing upon a variety of disciplines (behavioral, mathematical, etc.), managementscience and operations research combine quantitativeand qualitativeconsiderations in order toestablish policies and decisions that are in the best interest of the organization.2.Define the problemIdentify the alternativesDetermine the criteriaEvaluate the alternativesChoose an alternativeFor further discussion see section 1.33.See section 1.2.4.A quantitative approach should be considered because the problem is large, complex, important,new and repetitive.5.Models usually have time, cost, and risk advantages over experimenting with actual situations.6.Model (a) may be quicker to formulate, easier to solve, and/or more easily understood.7.Letd= distancem= miles per gallonc= cost per gallon,Total Cost =2dcmWe must be willing to treatmandcas known and not subject to variation.8.a.Maximize10x+ 5ys.t.5x+ 2y40x0,y0b.Controllable inputs:xandyUncontrollable inputs: profit (10,5), labor hours (5,2) and labor-hour availability (40)c.

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P rofit :Labor Hours:5/unit forx2/ unit fory$10/unit forx$ 5/ unit fory40 labor-hour capacit yUn con trol lable InputsP roduct ion Quant itiesxandyCon troll abl eInputP roject ed P rofit andcheck on product iont ime const raintO u tpu tMax10x+5ys.t.10xy+540xy00Mathe maticalModeld.x= 0,y= 20 Profit = $100(Solution by trial-and-error)e.Deterministic-all uncontrollable inputs are fixed and known.9.Ifa= 3,x= 13 1/3 and profit = 133Ifa= 4,x= 10 and profit = 100Ifa= 5,x= 8 and profit = 80Ifa= 6,x= 6 2/3 and profit = 67Sinceais unknown, the actual values ofxand profit are not known with certainty.10.a.Total Units Received =x+yb.Total Cost = 0.20x+0.25yc.x+y= 5000d.x4000 Kansas City Constrainty3000 Minneapolis Constrainte.Min0.20x+ 0.25ys.t.x+y=5000x4000y3000x,y011.a.at $20d= 800-10(20) = 600

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at $70d=800-10(70) = 100b.TR=dp= (800-10p)p= 800p-10p2c.at $30TR= 800(30)-10(30)2= 15,000at $40TR= 800(40)-10(40)2= 16,000at $50TR= 800(50)-10(50)2= 15,000Total Revenue is maximized at the $40 price.d.d= 800-10(40) = 400 unitsTR= $16,00012.a.TC= 1000 + 30xb.P= 40x-(1000 + 30x) = 10x-1000c.Breakeven point is the value ofxwhenP= 0Thus 10x-1000 = 010x= 1000x= 10013.a.Total cost = 4800 +60xb.Total profit= total revenue-total cost= 300x-(4800 +60x)= 240x-4800c.Total profit= 240(30)-4800 =2400d.240x-4800 = 0x= 4800/240 =20The breakeven point is20students.14.a.Profit= Revenue-Cost= 20x-(80,000 + 3x)= 17x-80,00017x-80,000= 017x= 80,000x= 4706Breakeven point = 4706b.Profit = 17(4000)-80,000 =-12,000Thus, a loss of $12,000 is anticipated.c.Profit=px-(80,000 + 3x)= 4000p-(80,000 + 3(4000))= 04000p= 92,000p= 23d.Profit= $25.95 (4000)-(80,000 + 3 (4000))= $11,800

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Probably go ahead with the project although the $11,800 is only a 12.8% return on the total cost of$92,000.15.a.Profit= 100,000x-(1,500,000 + 50,000x)= 050,000x= 1,500,000x= 30b.Build the luxury boxes.Profit= 100,000 (50)-(1,500,000 + 50,000 (50))= $1,000,00016.a.Max6x+ 4yb.50x+ 30y80,00050x50,00030y45,000x,y017.a.sj=sj-1+xj-djorsj-sj-1-xj+dj= 0b.xjcjc.sjIj

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Chapter2An Introduction to Linear ProgrammingLearning Objectives1.Obtain an overview of the kinds of problems linear programming has been used to solve.2.Learn how to develop linear programming models for simple problems.3.Be able to identify the special features of a model that make it a linear programming model.4.Learn how to solve two variable linear programming models by the graphical solution procedure.5.Understand the importance of extreme points in obtaining the optimal solution.6.Know the use and interpretation of slack and surplus variables.7.Be able to interpret the computer solution of a linear programming problem.8.Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linearprogramming problems.9.Understand the following terms:problem formulationfeasible regionconstraint functionslack variableobjective functionstandard formsolutionredundant constraintoptimal solutionextreme pointnonnegativity constraintssurplus variablemathematical modelalternative optimal solutionslinear programinfeasibilitylinear functionsunboundedfeasible solution

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Solutions:1.a, b, and e, are acceptable linear programming relationships.c is not acceptable because of22Bd is not acceptable because of3Af is not acceptable because of1ABc, d, and f could not be found in a linear programming model because they have the above nonlinearterms.2.a.84480BAb.BA84480c.BA84480Points on lineare only feasiblepoints

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3.a.BA0(0,9)(6,0)b.BA0(0,60)(40,0)c.BA0(0,20)(40,0)Pointson line are onlyfeasible solutions4.a.BA(20,0)(0,-15)

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b.BA(0,12)(-10,0)c.BA0(10,25)Note: Point shown wasused to locate position ofthe constraint line5.BA0100200300100200300abc

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6.7A+10B= 420is labeled (a)6A+ 4B= 420is labeled (b)-4A+ 7B= 420is labeled (c)80404080BA206010020600100-20-40-60-80-100(c)(a)(b)7.0BA1505010020025050100

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8.A0200-200-1001002001331/3(100,200)B9.BA0200100-100-200200300(150,225)(150,100)100

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10.012345B123456Optimal SolutionA= 12/7,B= 15/7Value of Objective Function = 2(12/7) + 3(15/7) = 69/7AA+2B=6(1)5A+3B=15(2)(1)×55A+10B=30(3)(2)-(3)-7B=-15B=15/7From (1),A= 6-2(15/7) = 6-30/7 = 12/7

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11.0100B100200Optimal SolutionA= 100,B= 50Value of Objective Function = 750AA= 100B= 8012.a.(0,0)12345B123456Optimal SolutionA= 3,B= 1.5Value of Objective Function = 13.56(4,0)(3,1.5)A

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b.(0,0)123B123456Optimal SolutionA= 0,B= 3Value of Objective Function = 18A78910c.There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).13.a.02468B2468Feasible Regionconsists of this linesegment onlyAb.The extreme points are (5, 1) and (2, 4).

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c.02468B2468Optimal SolutionAA= 2,B= 414.a.LetF= number of tons of fuel additiveS= number of tons of solvent baseMax40F+30Ss.t.2/5F+1/2S200Material 11/5S5Material 23/5F+3/10S21Material 3F,S0

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b.c.Material 2: 4 tons are used, 1 ton is unused.d.Noredundant constraints.15.a.SF

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b.Similar to part (a): the same feasible region with a different objective function. The optimalsolution occurs at (708, 0) with a profit ofz= 20(708) + 9(0) = 14,160.c.The sewing constraint is redundant. Such a change would not change the optimal solution to theoriginal problem.16.a.A variety of objective functions with a slope greater than-4/10 (slope of I & P line) will makeextreme point (0, 540) the optimal solution. For example, one possibility is 3S+ 9D.b.Optimal Solution isS= 0 andD= 540.c.DepartmentHours UsedMax. AvailableSlackCutting and Dyeing1(540) = 54063090Sewing5/6(540) = 450600150Finishing2/3(540) = 360708348Inspection and Packaging1/4(540) = 135135017.Max5A+2B+0S1+0S2+0S3s.t.1A-2B+1S1=4202A+3B+1S2=6106A-1B+1S3=125A,B,S1,S2,S3018.a.Max4A+1B+0S1+0S2+0S3s.t.10A+2B+1S1=303A+2B+1S2=122A+2B+1S3=10A,B,S1,S2,S30

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b.02468B2468Optimal SolutionAA= 18/7,B= 15/7, Value = 87/710121410c.S1= 0,S2= 0,S3= 4/719.a.Max3A+4B+0S1+0S2+0S3s.t.-1A+2B+1S1=8(1)1A+2B+1S2=12(2)2A+1B+1S3=16(3)A,B,S1,S2,S30

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b.02468B2468Optimal SolutionAA= 20/3,B= 8/3Value = 30 2/31012141012(2)(1)(3)c.S1= 8 +A–2B= 8 + 20/3-16/3 = 28/3S2= 12-A–2B= 12-20/3-16/3 = 0S3= 16–2A-B= 16-40/3-8/3 = 020.a.Max3A+2Bs.t.A+B-S1=43A+4B+S2=24A-S3=2A-B-S4=0A,B,S1,S2,S3,S40

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b.c.S1= (3.43 + 3.43)-4 = 2.86S2= 24-[3(3.43) + 4(3.43)] = 0S3= 3.43-2 = 1.43S4= 0-(3.43-3.43) = 0

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21.a. and b.010203040B10203040Optimal SolutionA2A+ 3B= 6050607050608070809090100Constraint 3Constraint 2Constraint 1Feasible Regionc.Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2,B= 10 +ASubstituting forBin constraint 1 we obtain5A+ 5(10 +A)= 4005A+ 50 + 5A= 40010A= 350A= 35B= 10 +A= 10 + 35 = 45Optimal solution isA= 35,B= 45d.Because the optimal solution occurs at the intersection of constraints 1 and 2,these are bindingconstraints.

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e.Constraint 3 is the nonbinding constraint. At the optimal solution 1A+ 3B= 1(35) + 3(45) = 170.Because 170 exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associatedwith this constraint.22.a.0500100015002000C500100015002000Feasible RegionA5A+ 4C= 400025003000350025003000Inspection andPackagingCutting andDyeingSewingNumber of All-Pro Footballs54321b.Extreme PointCoordinatesProfit1(0, 0)5(0) + 4(0) = 02(1700, 0)5(1700) + 4(0) = 85003(1400, 600)5(1400) + 4(600) = 94004(800, 1200)5(800) + 4(1200) = 88005(0, 1680)5(0) + 4(1680) = 6720Extreme point 3 generates the highest profit.c.Optimal solution isA= 1400,C= 600d.The optimal solution occurs at the intersection of the cutting and dyeing constraint and theinspectionand packaging constraint. Therefore these two constraints are the binding constraints.e.New optimal solution isA= 800,C= 1200Profit = 4(800) + 5(1200) = 9200

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23.a.LetE=number of units of the EZ-Rider producedL=number of units of the Lady-Sport producedMax2400E+1800Ls.t.6E+3L2100Engine timeL280Lady-Sport maximum2E+2.5L1000Assembly and testingE, L0b.c.The binding constraints are the manufacturing time and the assembly and testing time.24.a.LetR= number of units of regular model.C= number of units of catcher’s model.Max5R+8Cs.t.1R+3/2C900Cutting and sewing1/2R+1/3C300Finishing1/8R+1/4C100Packing and ShippingR,C00LProfit = $960,000Optimal Solution100200300400500600700100200300400500EEngineManufacturing TimeFrames for Lady-SportAssembly and TestingE= 250,L= 200Number of Lady-Sport ProducedNumber of EZ-Rider Produced

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b..800400400800CR200600100020060001000Optimal Solution(500,150)FC & SP & SRegular ModelCatcher's Modelc.5(500) + 8(150) = $3,700d.C & S1(500) +3/2(150) = 725F1/2(500) +1/3(150) = 300P & S1/8(500)+1/4(150) = 100e.DepartmentCapacityUsageSlackC & S900725175 hoursF3003000 hoursP & S1001000 hours25.a.LetB= percentage of funds invested in the bond fundS= percentage of funds invested in the stock fundMax0.06B+0.10Ss.t.B0.3Bond fund minimum0.06B+0.10S0.075Minimum returnB+S=1Percentage requirementb.Optimal solution:B= 0.3,S= 0.7Value of optimal solution is 0.088 or 8.8%

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26.a.a.LetN= amount spent on newspaper advertisingR= amount spent on radio advertisingMax50N+80Rs.t.N+R=1000BudgetN250Newspaper min.R250Radio min.N-2R0News2 RadioN,R0b.500500RN010001000Optimal SolutionN= 666.67,R= 333.33Value = 60,000Newspaper MinBudgetRadio MinFeasible regionis this line segmentN= 2R27.LetI= Internet fund investment in thousandsB= Blue Chip fund investment in thousandsMax0.12I+0.09Bs.t.1I+1B50Available investment funds1I35Maximum investment in the internet fund6I+4B240Maximum risk for a moderate investorI, B0

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Internet fund$20,000Blue Chip fund$30,000Annual return$ 5,100b.The third constraint for the aggressive investor becomes6I+ 4B320This constraint is redundant; the available funds and the maximum Internet fund investmentconstraints define the feasible region. The optimal solution is:Internet fund$35,000Blue Chip fund$15,000Annual return$ 5,550The aggressive investor places as much funds as possible in the high return but high risk Internetfund.c.The third constraint for the conservative investor becomes6I+ 4B160This constraint becomes a binding constraint. The optimal solution isInternet fund$0Blue Chip fund$40,000Annual return$ 3,6000BObjectiveFunction0.12I+ 0.09BOptimal Solution102030405060601020304050IRisk ConstraintMaximumInternet FundsAvailable Funds$50,000I= 20,B= 30$5,100Internet Fund (000s)Blue Chip Fund (000s)

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The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fundis still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. Theremaining $10,000 could be invested in low-risk bonds or certificates of deposit.28.a.LetW= number of jars of Western Foods Salsa producedM= number of jars of Mexico City Salsa producedMax1W+1.25Ms.t.5W7M4480Whole tomatoes3W+1M2080Tomato sauce2W+2M1600Tomato pasteW,M0Note: units for constraints are ouncesb.Optimal solution:W= 560,M= 240Value of optimal solution is 86029.a.LetB= proportion of Buffalo's time used to produce component 1D= proportion of Dayton's time used to produce component 1Maximum Daily ProductionComponent 1Component 2Buffalo20001000Dayton6001400Number of units of component 1 produced: 2000B+ 600DNumber of units of component 2 produced: 1000(1-B) + 600(1-D)For assembly of the ignition systems, the number of units of component 1 produced must equal thenumber of units of component 2 produced.Therefore,2000B+ 600D= 1000(1-B) + 1400(1-D)2000B+ 600D= 1000-1000B+ 1400-1400D3000B+ 2000D= 2400Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we canmaximize the number of electronic ignition systems produced by maximizing the number of units ofsubassembly 1 produced.Max 2000B+ 600DIn addition,B1 andD1.

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The linear programming model is:Max2000B+600Ds.t.3000B+ 2000D= 2400B1D1B,D0The graphical solution is shown below.Optimal Solution:B= .8,D= 0Optimal Production PlanBuffalo-Component 1.8(2000) = 1600Buffalo-Component 2.2(1000) = 200Dayton-Component 10(600) = 0Dayton-Component 21(1400) = 1400Total units of electronic ignition system = 1600 per day.0DB.2.4.6.81.01.2.2.4.6.81.01.23000B+ 2000D= 24002000B+ 600D= 300OptimalSolution

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30.a.LetE=number of shares of Eastern CableC=number of shares of ComSwitchMax15E+18Cs.t.40E+25C50,000Maximum Investment40E15,000Eastern Cable Minimum25C10,000ComSwitch Minimum25C25,000ComSwitch MaximumE, C0b.c.There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000)d.Optimal solution isE= 625,C= 1000Total return = $27,375050010005001000CE150015002000Minimum Eastern CableMaximum ComswitchMinimum ConswitchMaximum InvestmentNumber of Shares of Eastern CableNumber of Shares of ComSw itch

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