An Introduction To Management Science: Quantitative Approach, 15th Edition Solution Manual

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1-1Chapter 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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Chapter 11-21.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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Introduction1-3P 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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Chapter 11-4at $70d=800-10(70) = 100b.at $26d= 800-10(26) = 540at $27d= 800-10(27) = 530If the firm increases the per unit price from $26 to $27, the number of units the firm can sell falls by 10.at $42d= 800-10(42) = 380at $43d= 800-10(43) = 370If the firm increases the per unit price from $426 to $43, the number of units the firm can sell falls by 10.Thissuggests that the relationshipbetween the per-unit price and annual demand for the product in unitsislinear between $20 and $70 and that annual demand for the product decreases by 10 units when the price isincreased by $1.c.TR=dp= (800-10p)p= 800p-10p2d.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.e.d= 800-10(40) = 400 unitsTR= $16,00012.a.TC=2000 +60xb.P=80x-(2000 +60x) =20x-2000c.Breakeven point is the value ofxwhenP= 0Thus20x-2000 = 020x=2000x= 10013.a.Total cost =9600 +(2*60)x= 9600 + 120xb.Total profit= total revenue-total cost=600x-(9600 +120x)=480x-9600c.Total profit=480(30)-9600 =4800d.480x-9600= 0x=9600/480 =20The breakeven point is20students.14.a.Profit= Revenue-Cost=46x-(160,000 +6x)=40x-160,00040x-160,000= 040x=160,000x= 4000Breakeven point = 4000

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Introduction1-5b.Profit =40(3800)-16,000 =-8000Thus, a loss of $8000 is anticipated.c.Profit=px-(160,000 + 6x)= 3800p-(160,000 + 6(3800))= 03800p= 182,800p= 48.105 or $48.11d.Profit= $50.95 (3800)-(160,000 +6(3800))= $10,810Probably go ahead with the project although the $10,810 is only a5.98% return on the total cost of$182,800.15.a.Profit=300,000x-(4,500,000 +150,000x)= 0150,000x=4,500,000x= 30b.Build the luxury boxes.Profit= 300,000 (50)-(4,500,000 + 150,000 (50))= $3,000,00016.a.Max6x+ 4yb.50x+ 30y800,00050x500,00030y450,000x,y017.a.sj=sj-1+xj-djorsj-sj-1-xj+dj= 0b.xjcjc.sjIj18.a. maximize (3.10–0.30)x1+ (3.10–0.40)x2+ (3.10–0.48)x3+ (3.20–0.30)y1+ (3.20–0.40)y2+(3.20-0.48)y3= maximize 2.80x1+ 2,70x2+ 2.62x3+2.90y1+ 2.80y2+2.72y3b.(1)x1+y1≤12,000(2)x2+y2≤20,000(3)x3+y3≤24,000c.x1≥.35(x1+x2+x3) or .65x1-.35x2-.35x3≥0(Blend A must be composed of at least 35% of oilfrom the Texas well)x2≤.50(x1+x2+x3) or-.50x1+ .50x2-.50x3≤0(BlendAmust be composed of no more than 50%of oil from the Oklahoma well)x3≥.15(x1+x2+x3) or-.15x1-.15x2+ .85x3≥0(BlendAmust be composed of at least15% of oilfrom the California well)y1≥.20(y1+y2+y3) or .80y1-.20y2-.20y3≥0(BlendBmust be composed of at least20% of oilfrom the Texas well)

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Chapter 11-6y2≥.30(y1+y2+y3) or-.30y1+ .70y2-.30y3≥0(BlendBmust be composed of at least 30% of oilfrom the Oklahoma well)y3≤.40(y1+y2+y3) or-.40y1-.40y2+ .60y3≤0(BlendBmust be composed of no more than 40%of oil from the California well)x1+x2+x3≥20,000(Long-term contracts require at least 20,000gallons ofBlend Ato be produced)y1+y2+y3≥20,000(Long-term contracts require at least 20,000gallons ofBlend Bto be produced)19.a.Profit per contemporary cabinet is $90–[2.0($15) +1.5($12) +1.3($18)] = $18.60Profit perfarmhousecabinet is $85–[2.5($15) +1.0($12) +1.2($18)] = $13.90So the objective function is maximize 18.6x+ 13.9yb.2.0x1+ 2.5y1≤3000hours available in carpentry1.5x2+ 1.0y2≤1500hours available in carpentry1.3x3+ 1.2y3≤1500hours available in carpentryc.x≥500y≥65020.a.maximize 7000x+ 4000yb.500x+ 250y≤100,000c.x≤20d.y≥50e., or2x–y≥0f.If the constraints forthe maximum number oftelevision adsto be used from part(c) and theminimum number ofinternet adsto be used from part (d) are both satisfied, then television ads canbe at most. That is,the constraint for thestipulated ratio of television adsto Internet ads cannot be satisfied. Therefore, the problem as stated is infeasible.

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2-1Chapter2An 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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Chapter22-2Solutions: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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An Introduction to Linear Programming2-33.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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Chapter22-4b.BA(0,12)(-10,0)c.BA0(10,25)Note: Point shown wasused to locate position ofthe constraint line5.BA0100200300100200300abc

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An Introduction to Linear Programming2-56.7A+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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Chapter22-68.A0200-200-1001002001331/3(100,200)B9.BA0200100-100-200200300(150,225)(150,100)100

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An Introduction to Linear Programming2-710.012345B123456Optimal 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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Chapter22-811.0100B100200Optimal 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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An Introduction to Linear Programming2-9b.(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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