Solution Manual For Introduction To Management Science, 11th Edition

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1-1Chapter One: Management SciencePROBLEM SUMMARY1.Total cost, revenue, profit, andbreak-even2.Total cost, revenue, profit, andbreak-even3.Total cost, revenue, profit, andbreak-even4.Break-even volume5.Graphical analysis (1−2)6.Graphical analysis (1−4)7.Break-even sales volume8.Break-even volume as a percentageof capacity (1−2)9.Break-even volume as a percentageof capacity (1−3)10.Break-even volume as a percentageof capacity (1−4)11.Effect of price change (1−2)12.Effect of price change (1−4)13.Effect of variable cost change (1−12)14.Effect of fixed cost change (1−13)15.Break-even analysis16.Effect of fixed cost change (1−7)17.Effect of variable cost change (1−7)18.Break-even analysis19.Break-even analysis20.Break-even analysis21.Break-even analysis; volume andprice analysis22.Break-even analysis; profit analysis23.Break-even analysis24.Break-even analysis; profit analysis25.Break-even analysis; price and volume analysis26.Break-even analysis; profit analysis27.Break-even analysis; profit analysis28.Break-even analysis; profitanalysis29.Linear programming30.Linear programming31.Linear programming32.Linear programming33.Forecasting/statistics34.Linear programming35.Waiting lines36.Shortest routePROBLEM SOLUTIONS1.a)=====+=+=====−=fvfv300,$8,000,$65 per table,$180;TC$8,000(300)(65)$27,500;TR(300)(180)$54,000;$54,00027,500$26,500 per monthvccpcvcvpZb)fv8,00069.56 tables per month18065cvpc===−−2.a)=====+=+=====−=fvfv12,000,$60,000,$9,$25; TC60,000(12,000)(9)$168,000;TR(12,000)($25)$300,000;$300,000168,000$132,000 per yearvccpcvcvpZb)===−−fv60,0003,750 tires per year259cvpc3.a)=====+=+=====−= −fvfv18,000,$21,000,$.45,$1.30;TC$21,000(18,000)(.45)$29,100;TR(18,000)(1.30)$23, 400;$23, 40029,100$5,700 (loss)vccpcvcvpZb)4.======−−fvfv$25,000,$.40,$.15,25,000100,000 lb per month.40.15cpccvpcfv21,00024,705.88 yd per month1.30.45cvpc===−−

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1-25.6.7.−−fv$25,000=== 1,250 dolls3010cvpc8.====Break-even volume as percentage of capacity3,750.46946.9%8,000vk9.====Break-even volume as percentage of capacity24,750.88.98898.8%25,000vk10.====Break-even volume as percentage of100,000capacity.83383.3%120,000vk11.fv60,0002,727.3tires319per year; it reduces the break-evenvolume from 3,750 tires to 2,727.3tires per year.cvpc===−−12.===−−fv25,00055,555.55 lb.60.15per month; it reduces the break-evenvolume from 100,000 lb per monthto 55,555.55 lb.cvpc13.fv25,00065,789.47 lb.60.22per month; it increases the break-evenvolume from 55,555.55 lb per monthto 65,789.47 lb per month.cvpc===−−14.===−−fv39,000102,613.57 lb.60.22per month; it increases the break-evenvolume from 65,789.47 lb per monthto 102,631.57 lb per month.cvpc15.fvfvInitial profit:(9,000)(.75)4,000(9,000)(.21)6,7504,0001,890$860 per month; increase in price:(5,700)(.95)4,000(5,700)(.21)5, 4154,0001,197$218 per month; the dairZvpcvcZvpcvc=−−=−−=−−==−−=−−=−−=y should notraise its price.16.−fv35,000=== 1,75030–10cvpcThe increase in fixed cost from $25,000 to$35,000 will increase the break-even point from1,250 to 1,750 or 500 dolls;thus, he should notspend the extra $10,000 for advertising.17.Original break-even point (from problem 7) = 1,250New break-even point:===−−fv17,0001,062.53014cvpcReduces BE point by 187.5 dolls.18.a)===−−fv$27,0005,192.30 pizzas8.953.75cvpcb)=5,192.3259.6 days20c)Revenue for the first 30 days = 30(pv−vcv)= 30[(8.95)(20)−(20)(3.75)]= $3,120$27,000−3,120 = $23,880, portion of fixed costnot recouped after 30 days.===−−fv$23,880New5,685.7 pizzas7.953.75cvpcTotal break-even volume = 600 + 5,685.7 =6,285.7 pizzas

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1-35,685.7Total time to break-even3020314.3 days=+=19.a)Cost of Regular plan = $55 + (.33)(260 minutes)= $140.80Cost of Executive plan = $100 + (.25)(60 minutes)= $115Select Executive plan.b)55 + (x−1,000)(.33) = 100 + (x−1,200)(.25)−275 + .33x= .25x−200x= 937.50 minutes permonthor 15.63 hrs.20.a)=−7,50014,000.35pp= $0.89 to break evenb)If the team did not perform as well as expectedthe crowds could be smaller; bad weather couldreduce crowds and/or affect what fans eat at thegame; the price she charges could affect demand.c)This will be a subjective answer, but $1.25 seemsto be a reasonable price.Z=vp−cf−vcvZ= (14,000)(1.25)−7,500−(14,000)(0.35)= 17,500−12,400= $5,10021.a)cf= $1,700cv= $12 per pupilp= $75=−1,7007512v= 26.98 or 27 pupilsb)Z=vp−cf−vcv$5,000 =v(75)−$1,700−v(12)63v= 6,700v= 106.3 pupilsc)Z=vp−cf−vcv$5,000 = 60p−$1,700−60(12)60p= 7,420p= $123.6722.a)cf= $350,000cv= $12,000p= $18,000=−fvcvpc=−350,00018,00012,000= 58.33 or 59 studentsb)Z= (75)(18,000)−350,000−(75)(12,000)= $100,000c)Z= (35)(25,000)−350,000−(35)(12,000)= 105,000This is approximately the same as the profit for75 students and a lower tuition in part (b).23.p= $400cf= $8,000cv= $75Z= $60,000+=−fvZcvpc+=−60,0008,00040075vv= 209.23 teams24.Fixed cost (cf) = 875,000Variable cost (cv) = $200Price (p) = (225)(12) = $2,700v=cf/(p–cv) = 875,000/(2,700–200)= 350With volume doubled to 700:Profit (Z) = (2,700)(700)–875,000–(700)(200)= $875,00025.Fixed cost (cf)= 100,000Variable cost (cv) =$(.50)(.35) + (.35)(.50) + (.15)(2.30)= $0.695Price (p) = $6Profit (Z) = (6)(45,000)–100,000–(45,000)(0.695)= $138,725This is not the financial profit goal of $150,000.The price to achieve the goal of $150,000 is,p= (Z+cf+vcv)/v= (150,000 + 100,000 + (45,000)(.695))/45,000= $6.25

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1-4The volume to achieve the goal of $150,000 is,v= (Z+cf)/(p−cv)= (150,000 + 100,000)/(6−.695)= 47,12526.a)Monthly fixed cost (cf) = cost of van/60 months+ labor (driver)/month= (21,500/60) + (30.42days/month)($8/hr)(5 hr/day)= 358.33 + 1,216.80= $1,575.13Variable cost (cv) = $1.35 + 15.00= $16.35Price (p) = $34v=cf/(p−vc)= (1,575.13)/(34−16.35)v= 89.24 orders/monthb)89.24/30.42 = 2.93 orders/day−MondaythroughThursdayDouble for weekend = 5.86 orders/day−FridaythroughSundayOrders per month = approximately (18 days)(2.93 orders) + (12.4 days)(5.86 orders)= 125.4 delivery orders per monthProfit = total revenue−total cost=vp–(cf+vcv)= (125.4)(34)−1,575.13–(125.4)(16.35)= 638.1827.a)==−−fv5003014cvpcv= 31.25 jobsb)(8 weeks)(6 days/week)(3 lawns/day) = 144lawnsZ= (144)(30)−500−(144)(14)Z= $1,804c)(8 weeks)(6 days/week)(4 lawns/day) = 192lawnsZ= (192)(25)−500−(192)(14)Z= $1,612No, she would make less money than (b)28.a)==−−fv700353cvpcv= 21.88 jobsb)(6 snows)(2 days/snow)(10 jobs/day) = 120 jobsZ= (120)(35)−700−(120)(3)Z= $3,140c)(6 snows)(2 days/snow)(4 jobs/day) = 48 jobsZ= (48)(150)−1800−(48)(28)Z= $4,056Yes, better than (b)d)Z= (120)(35)−700−(120)(18)Z= $1,340Yes,still a profit with one more person29.There are two possible answers, or solution points:x= 25,y= 0 orx= 0,y= 50Substituting these values in the objective function:Z= 15(25) + 10(0) = 375Z= 15(0) + 10(50) = 500Thus, the solution isx= 0 andy= 50This is a simple linear programming model, thesubject of the next several chapters. The studentshould recognize that there are only two possiblesolutions, which are the corner points of thefeasible solution space, only one of which isoptimal.30.The solution is computed by solvingsimultaneous equations,x= 30,y= 10,Z= $1,400It is the only, i.e., “optimal” solution becausethere is only one set of values forxandythatsatisfy both constraints simultaneously.

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1-531.Labor usageClay usageProfitPossible# bowls# mugs12x+ 15y<=609x+5y<=30300x+250ysolution?01155250yes10129300yes112714550yes023010500yes202418600yes124219800yes213923850yes2254281100yes, best solution034515750yes303627900yes1357241050yes3151321150no2369331350no3266371400no3381421650no4048361200no0460201000yes1472291300no4163411450no2484381600no4278461700no

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1-632.MaximizeZ= $30xAN+ 70xAJ+ 40xBN+ 60xBJsubject toxAN+xAJ= 400xBN+xBJ= 400xAN+xBN= 500xAJ+xBJ= 300The solution isxAN= 400,xBN= 100,xBJ= 300, andZ= 34,000This problem can be solved by allocating as much aspossible to the lowest cost variable,xAN= 400, thenrepeating this step until all the demand has been met.This is a similar logic to the minimum cell cost method.33.This is virtually a straight linear relationship betweentime and site visits;thus, a simple linear graph wouldresult in a forecast of approximately 34,500 site visits.34.Determine logical solutions:CakesBreadTotal Sales1.02$122.12$223.31$364.40$40Each solution must be checked to see ifit violatestheconstraints for baking time and flour. Some possiblesolutions can be logically discarded because they areobviously inferior. For example, 0 cakes and 1 loaf ofbread is clearly inferior to 0 cakes and 2 loaves ofbread. 0 cakes and 3 loaves of bread is not possiblebecause there is not enough flour for 3 loaves of bread.Using this logic, there are four possible solutions asshown. The best one, 4 cakes and 0 loaves of bread,results in the highest total sales of $40.36.The shortest route problem is one of the topics ofChapter 7. At this point, the most logical “trial anderror” way that most students will probablyapproach this problem is to identify all the feasibleroutes and compute the total distance for each, asfollows:1-2-6-9 = 2281-2-5-9 = 2131-3-5-9 = 2111-3-8-9 = 2761-4-7-8-9 = 275Obviously inferior routes like 1-3-4-7-8-9 and1-2-5-8-9 that include additional segments to theroutes listed above can be logically eliminatedfrom consideration. As a result,the route 1-3-5-9is the shortest.An additional aspect to this problem could be tohave the students look at these routes on a realmap and indicate which they think might“practically” be the best route. In this case,1-2-5-9 would likely be a better route, becauseeven though it’s two miles farther it is Interstatehighway the whole way, whereas 1-3-5-9encompasses U.S. 4-lane highways and stateroads.35.This problem demonstrates the cost trade-offinherent in queuing analysis, the topic of Chapter 13.In this problem the cost of service, i.e., the cost ofstaffing registers, is added to the cost of customerswaiting, i.e., the cost of lost sales and ill will, asshown in the following table.Registersstaffed12345678Waitingtime (mins)2014941.710.50.1Cost of service ($)60120180240300360420480Cost of waiting ($)850550300500000Total cost ($)910670480290300360420480The total minimum cost of $290 occurs with 4 registers staffed

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1-7CASE SOLUTION:CLEAN CLOTHESCORNER LAUNDRYa)===−−fv1,7002,000 items per month1.10.25cvpcb)Solution depends on number of months; 36 usedhere. $16,200 ÷ 36 = $450 per month, thusmonthly fixed cost is $2,150===−−fv2,1502,529.4 items per month1.10.25cvpc529.4 additional items per monthc)Z=vp−cf−vcv= 4,300(1.10)−2,150−4,300(.25)= $1,505 per monthAfter 3 years,Z= $1,955 per monthd)===−−=−−=−−=fvfv1,7002,297.3.99.253,800(.99)1,7003,800(.25)$1,112 per monthcvpcZvpcvce)With both options:Z=vp−cf−vcv= 4,700(.99)−2,150−4,700(.25)= $1,328She should purchase the new equipmentbut notdecrease prices.CASE SOLUTION: OCOBEE RIVERRAFTING COMPANY=fAlternative 1:$3,000c=$20p=v$12c===−−f1v3,000375 rafts2012cvpc=fAlternative 2:$10,000c=$20p=v$8c===−−f2v10,000833.37208cvpcIf demand is less than 375 rafts, the students should notstart the business.If demand is less than 833 rafts, alternative 2 should notbe selected, and alternative 1 should be used if demand isexpected to be between 375 and 833.33 rafts.If demand is greater than 833.33 rafts, which alternativeis best? To determine the answer, equate the two costfunctions.3,000 + 12v= 10,000 + 8v4v= 7,000v= 1,750This is referred to as the point of indifference betweenthe two alternatives. In general, for demand lower than thispoint (1,750) the alternative should be selected with thelowest fixed cost; for demand greater than thispoint the alternative with the lowest variable cost shouldbe selected. (This general relationship can be observed bygraphing the two cost equations and seeing where theyintersect.)Thus, for the Ocobee River Rafting Company, thefollowing guidelines should be used:demand < 375, do not start business; 375 < demand< 1,750, select alternative 1; demand > 1,750, selectalternative 2Since Penny estimates demand will be approximately1,000 rafts, alternative 1 should be selected.Z=vp−cf−vcv= (1,000)(20)−3,000−(1,000)(12)Z= $5,000CASE SOLUTION: CONSTRUCTINGA DOWNTOWN PARKING LOTIN DRAPERa)The annual capital recovery payment for a capitalexpenditure of $4.5 million over 30 yearsat8% is,(4,500,000)[0.08(1 + .08)30] / (1 + .08)30−1= $399,723.45

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2-1Chapter Two: Linear Programming: Model Formulation and Graphical SolutionPROBLEMSUMMARY1.Maximization (1–28 continuation), graphicalsolution2.Maximization, graphical solution3.Minimization, graphical solution4.Sensitivity analysis (2–3)5.Minimization, graphical solution6.Maximization, graphical solution7.Slack analysis (2–6)8.Sensitivity analysis (2–6)9.Maximization, graphical solution10.Slack analysis (2–9)11.Maximization, graphical solution12.Minimization, graphical solution13.Maximization, graphical solution14.Sensitivity analysis (2–13)15.Sensitivity analysis (2–13)16.Maximization, graphical solution17.Sensitivity analysis (2–16)18.Maximization, graphical solution19.Sensitivity analysis (2–18)20.Maximization, graphical solution21.Standard form (2–20)22.Maximization, graphical solution23.Standard form (2–22)24.Maximization, graphical solution25.Constraint analysis (2–24)26.Minimization, graphical solution27.Sensitivity analysis (2–26)28.Sensitivity analysis (2–26)29.Sensitivity analysis (2–22)30.Minimization, graphical solution31.Minimization, graphical solution32.Sensitivity analysis (2–31)33.Minimization, graphical solution34.Maximization, graphical solution35.Minimization, graphical solution36.Maximization, graphical solution37.Sensitivity analysis (2–34)38.Minimization, graphical solution39.Maximization, graphical solution40.Maximization, graphical solution41.Sensitivity analysis (2–38)42.Maximization, graphical solution43.Sensitivity analysis (2–40)44.Maximization, graphical solution45.Sensitivity analysis (2–42)46.Minimization, graphical solution47.Sensitivity analysis (2–44)48.Maximization, graphical solution49.Sensitivity analysis (2–46)50.Maximization, graphical solution51.Sensitivity analysis (2–48)52.Maximization, graphical solution53.Minimization, graphical solution54.Sensitivity analysis (2–53)55.Minimization, graphical solution56.Sensitivity analysis (2–55)57.Maximization, graphical solution58.Minimization, graphical solution59.Sensitivity analysis (2–52)60.Maximization, graphical solution61.Sensitivity analysis (2–54)62.Multiple optimal solutions63.Infeasible problem64.Unbounded problemPROBLEMSOLUTIONS1.a)x1= # cakesx2= # loaves of breadmaximizeZ= $10x1+ 6x2subject to3x1+ 8x2≤20 cups of flour45x1+ 30x2≤180 minutesx1,x2≥0

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2-2b)2.a)MaximizeZ= 6x1+ 4x2(profit, $)subject to10x1+10x2≤100 (line1, hr)7x1+ 3x2≤42(line2, hr)x1,x2≥0b)3.a)MinimizeZ= .05x1+ .03x2(cost, $)subject to8x1+ 6x2≥48 (vitamin A, mg)x1+2x2≥12(vitamin B, mg)x1,x2≥0b)4.The optimal solution point would changefrom point A to point B, thus resulting in theoptimal solutionx1=12/5x2=24/5Z= .4085.a)MinimizeZ= 3x1+ 5x2(cost, $)subject to10x1+2x2≥20 (nitrogen, oz)6x1+ 6x2≥36 (phosphate, oz)x2≥2(potassium, oz)x1,x2≥0b)6.a)MaximizeZ= 400x1+100x2(profit, $)subject to8x1+10x2≤80 (labor, hr)2x1+ 6x2≤36 (wood)x1≤6 (demand, chairs)x1,x2≥0

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2-3b)7.In order to solve this problem, you mustsubstitute the optimal solution into theresource constraint for wood and theresource constraint for labor and determinehow much of each resourceis left over.Labor8x1+10x2≤80 hr8(6) +10(3.2)≤8048 + 32≤8080≤80There is no labor left unused.Wood2x1+ 6x2≤362(6) + 6(3.2)≤3612+19.2≤3631.2≤3636−31.2=4.8There is 4.8 lb of wood left unused.8.The new objective function,Z= 400x1+500x2, is parallel to the constraint for labor,which results in multiple optimal solutions.PointsB(x1= 30/7,x2= 32/7) andC(x1= 6,x2= 3.2) are the alternate optimal solutions,each with a profit of $4,000.9.a)MaximizeZ=x1+ 5x2(profit, $)subject to5x1+ 5x2≤25 (flour, lb)2x1+ 4x2≤16 (sugar, lb)x1≤5 (demand for cakes)x1,x2≥0b)10.In order to solve this problem, you mustsubstitute the optimal solution into theresource constraints for flour and sugar anddetermine how much of each resource is leftover.Flour5x1+ 5x2≤25 lb5(0) + 5(4)≤2520≤2525−20 = 5There are 5 lb of flour left unused.Sugar2x1+ 4x2≤162(0) + 4(4)≤1616≤16There is no sugar left unused.11.12.a)MinimizeZ= 80x1+ 50x2(cost, $)subject to3x1+x2≥6 (antibiotic1, units)x1+x2≥4 (antibiotic2, units)2x1+ 6x2≥12(antibiotic 3, units)x1,x2≥0b)

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2-413.a)MaximizeZ= 300x1+ 400x2(profit, $)subject to3x1+2x2≤18 (gold, oz)2x1+ 4x2≤20 (platinum, oz)x2≤4 (demand, bracelets)x1,x2≥0b)14.The new objective function,Z= 300x1+600x2,is parallel to the constraint line forplatinum, which results in multiple optimalsolutions. PointsB(x1=2,x2= 4) andC(x1= 4,x2= 3) are the alternate optimalsolutions, each with a profit of $3,000.The feasible solution space will change. Thenew constraint line, 3x1+ 4x2=20, isparallel to the existing objective function.Thus, multiple optimal solutions will also bepresent in this scenario. The alternateoptimal solutions are atx1=1.33,x2= 4 andx1=2.4,x2= 3.2, each with a profit of$2,000.15.a)Optimal solution:x1= 4 necklaces,x2= 3bracelets. The maximum demand is notachieved by the amount of one bracelet.b)The solution point on the graph whichcorresponds to no bracelets being producedmust be on thex1axis wherex2= 0. This ispointDon the graph. In order for pointDtobe optimal, the objective function “slope”must change such that it is equal to or greaterthan the slope of the constraint line, 3x1+2x2=18. Transforming this constraint into theformy=a+bxenables us to compute theslope:2x2=18−3x1x2= 9−3/2x1From this equation the slope is−3/2. Thus,the slope of the objective function must be atleast−3/2. Presently, the slope of theobjective function is−3/4:400x2=Z−300x1x2=Z/400−3/4x1The profit for a necklace would have toincrease to$600to result in a slope of−3/2:400x2=Z−600x1x2=Z/400−3/2x1However, this creates a situation where bothpointsCandDare optimal, ie., multipleoptimal solutions, as are allpoints on the line segment betweenCandD.16.a)MaximizeZ= 50x1+ 40x2(profit, $)subjectto3x1+ 5x2≤150 (wool, yd2)10x1+ 4x2≤200 (labor, hr)x1,x2≥0b)17.The feasible solution space changes from thearea0ABCto0AB'C', as shown on thefollowing graph.

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2-5The extreme points to evaluate are nowA,B', andC'.A:x1= 0x2= 30Z=1,200*B':x1=15.8x2=20.5Z=1,610C':x1=24x2= 0Z=1,200PointB'is optimal18.a)MaximizeZ=23x1+ 73x2subject tox1≤40x2≤25x1+ 4x2≤120x1,x2≥0b)19.a)No, not thiswinter, but they might after theyrecover equipment costs, which should beafter the2ndwinter.b)x1= 55x2=16.25Z=1,851No, profit will go downc)x1= 40x2=25Z=2,435Profit will increase slightlyd)x1= 55x2=27.72Z= $2,073Profit will go down from (c)20.21.MaximizeZ=1.5x1+x2+ 0s1+ 0s2+ 0s3subject tox1+s1= 4x2+s2= 6x1+x2+s3= 5x1,x2≥0A:s1= 4,s2=1,s3= 0B:s1= 0,s2= 5,s3= 0C:s1= 0,s2= 6,s3=122.23.MaximizeZ= 5x1+ 8x2+ 0s1+ 0s3+ 0s4subject to3x1+ 5x2+s1= 50

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2-62x1+ 4x2+s2= 40x1+s3= 8x2+s4=10x1,x2≥0A:s1= 0,s2= 0,s3= 8,s4= 0B:s1= 0,s2= 3.2,s3= 0,s4= 4.8C:s1=26,s2=24,s3= 0,s4=1024.25.It changes the optimal solution to pointA(x1= 8,x2= 6,Z=112), and the constraint,x1+x2≤15, is no longer part of the solutionspace boundary.26.a)MinimizeZ= 64x1+ 42x2(labor cost, $)subject to16x1+12x2≥450 (claims)x1+x2≤40 (workstations)0.5x1+1.4x2≤25 (defective claims)x1,x2≥0b)27.Changing the pay for afull-time claimsprocessor from $64 to $54 will change thesolution to pointAin the graphical solutionwherex1=28.125 andx2= 0, i.e., there willbe no part-time operators. Changing the payfor a part-time operator from $42to $36 hasno effect on the number of full-time and part-time operators hired, although the total costwill be reduced to $1,671.95.28.Eliminating the constraint for defectiveclaims would result in a new solution,x1= 0 andx2= 37.5, where only part-timeoperators would be hired.29.The solution becomes infeasible; there arenot enough workstations to handle theincrease in the volume of claims.30.31.32.The problem becomes infeasible.

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2-733.34.35.36.a)MaximizeZ= $4.15x1+ 3.60x2(profit, $)subject to1212112212115 (freezer space, gals.)0.930.7590 (budget, $)2 or20 (demand)1,0xxxxxxxxx x++−b)37.No additional profit, freezer space is not abinding constraint.38.a)MinimizeZ=200x1+160x2(cost, $)subject to6x1+2x2≥12(high-grade ore, tons)2x1+2x2≥8 (medium-grade ore, tons)4x1+12x2≥24 (low-grade ore, tons)x1,x2≥0b)

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2-839.a)MaximizeZ= 800x1+ 900x2(profit, $)subject to2x1+ 4x2≤30 (stamping, days)4x1+2x2≤30 (coating, days)x1+x2≥9 (lots)x1,x2≥0b)40.a)MaximizeZ= 30x1+ 70x2(profit, $)subjectto4x1+10x2≤80 (assembly, hr)14x1+ 8x2≤112(finishing, hr)x1+x2≤10 (inventory, units)x1,x2≥0b)41.The slope of the original objective functionis computed as follows:Z= 30x1+ 70x270x2=Z−30x1x2=Z/70−3/7x1slope =−3/7The slope of the new objective function iscomputed as follows:Z= 90x1+ 70x270x2=Z−90x1x2=Z/70−9/7x1slope=−9/7The change in the objective function notonly changes theZvalues but also results ina new solution point,C. The slope of thenew objective function is steeper and thuschanges the solution point.A:x1= 0C:x1= 5.3x2= 8x2= 4.7Z= 560Z= 806B:x1= 3.3D:x1= 8x2= 6.7x2= 0Z= 766Z= 72042.a)MaximizeZ= 9x1+12x2(profit, $1,000s)subject to4x1+ 8x2≤64 (grapes, tons)5x1+ 5x2≤50 (storage space, yd3)15x1+ 8x2≤120 (processing time, hr)x1≤7 (demand, Nectar)x2≤7 (demand, Red)x1,x2≥0b)

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2-943.a)15(4) + 8(6)≤120 hr60 + 48≤120108≤120120−108 =12hr left unusedb)PointsCandDwould be eliminated and anew optimal solution point atx1= 5.09,x2= 5.45, andZ=111.27 would result.44.a)MaximizeZ= .28x1+ .19x212211296 cans2,0xxxxx x+b)45.The model formulation would become,maximizeZ= $0.23x1+ 0.19x2subject tox1+x2≤96–1.5x1+x2≥0x1,x2≥0The solution isx1= 38.4,x2= 57.6, andZ= $19.78The discount would reduce profit.46.a)MinimizeZ= $0.46x1+ 0.35x2subject to.91x1+ .82x2= 3,500x1≥1,000x2≥1,000.03x1−.06x2≥0x1,x2≥0b)47.a)MinimizeZ= .09x1+ .18x2subject to.46x1+ .35x2≤2,000x1≥1,000x2≥1,000.91x1−.82x2= 3,500x1,x2≥0b)477−445 = 32fewer defective items48.a)MaximizeZ= $2.25x1+1.95x2subject to8x1+ 6x2≤1,9203x1+ 6x2≤1,4403x1+2x2≤720x1+x2≤288x1,x2≥0

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2-10b)49.A new constraint is added to the model in121.5xxThe solution isx1=160,x2=106.67,Z= $56850.a)MaximizeZ= 400x1+ 300x2(profit, $)subject tox1+x2≤50 (available land, acres)10x1+ 3x2≤300 (labor, hr)8x1+20x2≤800 (fertilizer, tons)x1≤26 (shipping space, acres)x2≤37 (shipping space, acres)x1,x2≥0b)51.The feasible solution space changes if thefertilizer constraint changes to20x1+20x2≤800 tons. The new solution space isA'B'C'D'. Two of the constraints now haveno effect.The new optimal solution is pointC':A':x1= 0*C':x1=25.71x2= 37x2=14.29Z=11,100Z=14,571B':x1= 3D':x1=26x2= 37x2= 0Z=12,300Z=10,40052.a)MaximizeZ= $7,600x1+22,500x2subject tox1+x2≤3,500x2/(x1+x2)≤.40.12x1+ .24x2≤600x1,x2≥0

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2-11b)53.a)MinimizeZ=$(.05)(8)x1+(.10)(.75)x2subject to5x1+x2≥8001251.5xx=8x1+ .75x2≤1,200x1,x2≥0x1=96x2=320Z=$62.40b)54.The new solution isx1=106.67x2=266.67Z=$62.67If twice as many guests prefer winetobeer,then the Robinsons would be approximately10 bottles of wine short and they would haveapproximately 53 more bottles of beer thanthey need. The waste is more difficult tocompute. The model in problem53assumesthat the Robinsons are ordering more wineand beer than they need, i.e., a buffer, andthus there logically would be some waste,i.e., 5% of the wine and10% of the beer.However, if twice as many guests preferwine,then there would logically be no wastefor wine but only for beer. This amount“logically” would be the waste from266.67bottles, or $20, and the amount from theadditional 53 bottles, $3.98, for a total of$23.98.55.a)MinimizeZ= 3700x1+ 5100x2subject tox1+x2= 45(32x1+14x2) / (x1+x2)≤21.10x1+ .04x2≤6112.25()xxx+212.25()xxx+x1,x2≥0b)

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2-1256.a)No, the solution would not changeb)No, the solution would not changec)Yes, the solution would change to China (x1)=22.5, Brazil (x2) =22.5, andZ= $198,000.57.a)x1= $ invested in stocksx2= $ invested in bondsmaximizeZ= $0.18x1+ 0.06x2(averageannual return)subject tox1+x2≤$720,000 (availablefunds)x1/(x1+x2)≤.65 (% of stocks).22x1+ .05x2≤100,000 (total possibleloss)x1,x2≥0b)58.x1= exams assigned to Bradx2= exams assigned to SarahminimizeZ= .10x1+ .06x2subject tox1+x2=120x1≤(720/7.2) or100x2≤50(600/12)x1,x2≥059.If the constraint for Sarah’s time becamex2≤55 with an additional hour then thesolution point at A would move tox1= 65,x2= 55 andZ= 9.8. If the constraintfor Brad’s time becamex1≤108.33 with anadditional hour then the solution point (A)would not change. All of Brad’s time is notbeing used anyway so assigning him moretime would not have an effect.One more hour of Sarah’s time wouldreduce the number of regraded exams from10 to 9.8, whereas increasing Brad by onehour would have no effect on the solution.This is actually the marginal (or dual) valueof one additional hour of labor, for Sarah,which is 0.20 fewer regraded exams,whereas the marginal value of Brad’s iszero.60.a)x1= # cups of Pomonax2= # cups of CoastalMaximizeZ= $2.05x1+1.85x2subject to16x1+16x2≤3,840 oz or (30 gal.×128 oz)(.20)(.0625)x1+ (.60)(.0625)x2≤6 lbs.Colombian(.35)(.0625)x1+ (.10)(.0625)x2≤6 lbs.Kenyan(.45)(.0625)x1+ (.30)(.0625)x2≤6 lbs.Indonesianx2/x1= 3/2x1,x2≥0

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2-13b)Solution:x1= 87.3 cupsx2=130.9 cupsZ= $421.0961.a)The only binding constraint is forColombian; the constraints for Kenyan andIndonesian are nonbinding and therearealready extra, or slack, pounds of thesecoffees available. Thus, only getting moreColombian would affect the solution.One more pound of Colombian wouldincrease sales from $421.09 to $463.20.Increasing the brewing capacity to 40gallons would have no effect since there isalready unused brewing capacity with theoptimal solution.b)If the shop increased the demand ratio ofPomona to Coastal from1.5 to1to2to1itwould increase daily sales to $460.00, so theshop should spend extra on advertising toachieve this result.62.Multiple optimal solutions;AandBalternateoptimal63.64.

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2-14CASE SOLUTION:METROPOLITANPOLICE PATROLThe linear programming model for this caseproblem isMinimizeZ=x/60 +y/45subject to2x+2y≥52x+2y≤12y≥1.5xx,y≥0The objective function coefficients aredetermined by dividing the distance traveled,i.e.,x/3, by the travel speed, i.e.,20 mph.Thus, thexcoefficient isx/3 ÷20, orx/60. Inthe first two constraints,2x+2yrepresents the formula for theperimeter of a rectangle.The graphical solution is displayed asfollows.The optimal solution isx=1,y=1.5, andZ= 0.05. This means that a patrol sector is1.5miles by1mile and the response time is0.05 hr, or 3 min.CASE SOLUTION:“THEPOSSIBILITY” RESTAURANTThe linear programming model formulationisMaximize =Z= $12x1+16x2subject tox1+x2≤60.25x1+ .50x2≤20x1/x2≥3/2or2x1−3x2≥0x2/(x1+x2)≥.10 or .90x2−.10x1≥0x1x2≥0The graphical solution is shown as follows.Changing the objective function toZ= $16x1+16x2would result in multipleoptimal solutions, the end points beingBandC.The profit in each case would be $960.Changing the constraint from.90x2−.10x1≥0 to .80x2−.20x1≥0has no effect on the solution.CASE SOLUTION:ANNABELLEINVESTS IN THE MARKETx1= no. of shares of indexfundx2= no. of shares of internet stock fundMaximizeZ= (.17)(175)x1+ (.28)(208)x2=29.75x1+ 58.24x2subject to12122112175208$120, 000.332,0+=xxxxxxxxx1=203x2= 406Z = $29,691.3721Eliminating the constraint.33xxwill have no effect on the solution.12Eliminating the constraint2xxwillchange the solution tox1=149,x2= 451.55,Z= $30,731.52.

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2-15Increasing the amount available to invest(i.e., $120,000 to $120,001) will increaseprofit fromZ= $29,691.37 toZ= $29,691.62or approximately $0.25.Increasing by another dollar will increaseprofit by another $0.25, and increasing theamount available by one more dollarwill again increase profit by $0.25. Thisindicates that for each extra dollar invested areturn of $0.25 might be expected with thisinvestment strategy.Thus, themarginal valueof an extra dollarto invest is $0.25, which is also referred toas the “shadow” or “dual” price as describedin Chapter 3.

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3-1Chapter Three: Linear Programming: Computer Solution and Sensitivity AnalysisPROBLEM SUMMARY1.QM for Windows2.QM for Windows and Excel3.Excel4.Excel solution5.Excel6.Graphical solution; sensitivity analysis7.Model formulation8.Graphical solution; sensitivity analysis (3–7)9.Sensitivity analysis (3–7)10.Model formulation11.Graphical solution; sensitivity analysis (3–10)12.Sensitivity analysis (3–10)13.Model formulation14.Graphical solution; sensitivity analysis (3–13)15.Computer solution; sensitivity analysis (3–13)16.Model formulation17.Graphical solution; sensitivity analysis (3–16)18.Computer solution; sensitivity analysis (3–16)19.Model formulation20.Graphical solution; sensitivity analysis (3–19)21.Computer solution; sensitivity analysis (3–19)22.Model formulation23.Graphical solution; sensitivity analysis (3–22)24.Computer solution; sensitivity analysis (3–22)25.Model formulation26.Graphical solution; sensitivity analysis (3–25)27.Computer solution; sensitivity analysis (3–25)28.Model formulation29.Graphical solution; sensitivity analysis (3–28)30.Computer solution; sensitivity analysis (3–28)31.Model formulation; graphical solution32.Computer solution; sensitivity analysis (3–31)33.Model formulation34.Graphical solution; computer solution; sensitivityanalysis (3–33).35.Model formulation36.Graphical solution; sensitivity analysis (3–35)37.Computer solution; sensitivity analysis (3–35)38.Model formulation; computer solution39.Model formulation; computer solution40.Computer solution; sensitivity analysis41.Model formulation42.Graphical solution; sensitivity analysis (3–41)43.Computer solution; sensitivity analysis (3–41)44.Model formulation; computer solution45.Sensitivity analysis (3–44)46.Model formulation; computer solution47.Sensitivity analysis (3–46)48.Model formulation49.Computer solution; sensitivity analysis (3–48)50.Model formulation51.Computer solution; sensitivity analysis (3–50)52.Model formulation53.Computer solution; sensitivity analysis (3–52)54.Model formulation55.Computer solution; sensitivity analysis (3–54)56.Computer solution57.Model formulation58.Sensitivity analysis (3–57)59.Model formulation60.Computer solution; sensitivity analysis (3–59)61.Model formulation; graphical solution62.Computer solution; sensitivity analysis (3–61)PROBLEM SOLUTIONS1.2.QM for Windows establishes a “template” forthe linear programming model based on theuser’s specification of the type of objectivefunction, the number of constraints, andnumber of variables. Then the modelparameters are input and the problem is solved.In Excel the model “template” must bedeveloped by the user.

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3-23.Set Target cell: B13Changing cells: B10:B12Profit: = B10*C4 + B11*D4 + B12*E4Constraints: B10:B12≥0G6≤F6G7≤F74.x1= 0x2= 9Z= $545.F6: = C6*B12 + D6*B13F7: = C7*B12 + D7*B13F8: = C8*B12 + D8*B13F9: = C9*B12 + D9*B13G6: = E6−F6G7: = E7−F7G8: = E8−F8G9: = E9−F9B14: = C4*B12 + D4*B13x1= 0x2= 5.2Z= 81.66.The slope of the constraint line is−70/60. Theoptimal solution is at point A wherex1= 0 andx2= 70.To change the solution to B,c1mustincrease such that the slope of the objectivefunction is at least as great as the slope oftheconstraint line,−c1/50 =−70/60c1= 58.33Alternatively,c2must decrease such that theslope of the objective function is at least asgreat as the slope of the constraint line,–30/c2=–70/60c2= 25.71Thus, ifc1increases to greaterthan 58.33 orc2decreases to less than 25.71, B will become optimal.7.a)x1= no. of basketballsx2= no. of footballsmaximizeZ= 12x1+ 16x2subject to3x1+ 2x2≤5004x1+ 5x2≤800x1,x2≥0b)maximizeZ= 12x1+ 16x2+ 0s1+ 0s2subject to3x1+ 2x2+s1= 5004x1+ 5x2+s2= 800x1,x2,s1,s2≥08.a)A:3(0) + 2(160) +s1= 500s1= 1804(0) + 5(160) +s2= 800s2= 0B:3(128.5) + 2(57.2) +s1= 500s1= 04(128.5) + 2(57.2) +s2= 800s2= 0C:2(167) + 2(0) +s1= 500s1= 04(167) + 5(0) +s2= 800s2= 132b)Z= 12x1+ 16x2and,x2=Z/16−12x1/16The slope of the objective function,−12/16,would have to become steeper (i.e., greater)than the slope of the constraint line 4x1+ 5x2=800, for the solution to change.The profit,c1, for a basketball that wouldchange the solution point is,−4/5 =−c1/165c1= 64c1= 12.8Since $13 > 12.8 the solution point wouldchange toBwherex1= 128.5,x2= 57.2.The newZvalue is $2,585.70.For a football,−4/5 =−12/c24c2= 60c2= 15Thus, if the profit for a football decreased to$15 or less, pointBwill also be optimal (i.e.,multiple optimal solutions). The solution atBisx1= 128.5,x2= 57.2, andZ= $2,400.

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3-3c)If the constraint line for rubber changes to 3x1+ 2x2= 1,000, it moves outward, eliminatingpointsBandC. However, sinceAis theoptimalpoint, it will not change and the optimal solutionremains the same,x1= 0,x2= 160, andZ= 2,560.There will be an increase in slack,s1, to 680 lbs.If the constraint line for leather changes to 4x1+5x2= 1,300, pointAwill move to a newlocation,x1= 0,x2= 250,Z= $4,000.9.a)Forc1the upper limit is computed as−4/5 =−c1/165c1= 64c1= 12.8and the lower limit is unlimited.Forc2the lower limitis,−4/5 =−12/c24c2= 60c2= 15and the upper limit is unlimited.Summarizing,∞≤c1≤12.815≤c2≤∞Forq1the upper limit is∞since no matter howmuchq1increases the optimal solution pointAwill not change.The lower limit forq1is at the point where theconstraint line(3x1+ 2x2=q1)intersects withpointAwherex1= 0,x2= 160,3x1+ 2x2=q13(0) + 2(160) =q1q1=320Forq2the upper limit is at the point where therubber constraint line (3x1+ 2x2= 500)intersects with the leather constraint line(4x1+ 5x2= 800) along thex2axis, i.e.,x1= 0,x2= 250,4x1+ 5x2=q24(0) + 5(250) =q2q2= 1,250The lower limit is 0 since that is the lowestpoint on thex2axis the constraint line candecrease to.Summarizing,320≤q1≤∞0≤q2≤1,250c)Z= 2,560.000VariableValueReducedCostx10.000.800x2160.0000.000ConstraintSlack/SurplusShadowPricec1180.000.00c20.003.20The shadow price for rubber is $0. Since thereis slack rubber left over at the optimal point,extra rubber would have no marginal value.The shadow price for leather is $3.20. For eachadditional ft.2of leather that the company canobtain profit would increase by $3.20, up tothe upper limit of the sensitivity range forleather (i.e., 1,250 ft.2).10.a)x1= no. of units ofAx2= no. of units ofBmaximizeZ= 9x1+ 7x2subject to12x1+ 4x2≤604x1+ 8x2≤40x1,x2≥0b)maximizeZ= 9x1+ 7x2+ 0s1+ 0s2subject to12x1+ 4x2+s1= 604x1+ 8x2+s2= 40x1,x2,s1,s2≥0b)Objective Coefficient RangesVariablesLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasex1No limit12.00012.8000.800No limitx215.00016.000No limitNo limit1.000Right Hand Side RangesConstraintsLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasec1320.000500.000No limitNo limit180.000c20.000800.0001,250.000450.000800.000

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3-411.a)A:12(0) + 4(5) +s1= 60s1= 404(0) + 8(5) +s2= 40s2= 0B:12(4) + 4(3) = 60s1= 04(4) + 8(3) +s2= 40s2= 0C:12(5) + 4(0) +s1= 60s1= 04(5) + 8(0) +s2= 40s2= 20b)The constraint line 12x1+ 4x2= 60 would moveinward resulting in a new location for pointBatx1= 2,x2= 4, which would still be optimal.c)In order for the optimal solution point to changefromBtoAthe slope of the objective functionmust be at least as flat as the slope of theconstraint line, 4x1+ 8x2= 40, which is−1/2.Thus, the profit for productBwould have to be,−9/c2=−1/2c2= 18If the profit for productBis increased to $15the optimal solution point will not change,althoughZwould change from $57 to $81.If the profit for productBis increased to$20the solution point will change fromBtoA,x1= 0,x2= 5,Z= $100.12.a)Forc1the upper limit is computed as,−c1/7 =−3c1= 21and the lower limit is,−c1/7 =−1/2c1= 3.50Forc2the upper limit is,−9/c2=−1/2c2= 18and the lower limit is,−9/c2=−3c2= 3Summarizing,3.50≤c1≤213≤c2≤18b)Z= 57.000VariableValueReducedCostx14.0000.000x23.0000.000ConstraintSlack/SurplusShadowPricec10.0000.550c20.0000.600Objective Coefficient RangesVariablesLowerLimitCurrentValuesUpperLimitAllowableIncreaseAllowableDecreasex13.5009.00021.00012.0005.500x23.0007.00018.00011.0004.000Right Hand Side RangesConstraintsLowerLimitCurrentValuesUpperLimitAllowableIncreaseAllowableDecreasec120.00060.000120.00060.00040.000c220.00040.000120.00080.00020.000c)The shadow price for line 1 time is $0.55 per

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3-5hour, while the shadow price for line 2 time is$0.60 per hour. The company would prefer toobtain more line 2 time since it would result inthe greater increase in profit.13.a)x1= no. of yards of denimx2= no. of yards of corduroymaximizeZ= $2.25x1+ 3.10x2subject to5.0x1+ 7.5x2≤6,5003.0x1+ 3.2x2≤3,000x2≤510x1,x2≥0b)maximizeZ= $2.25x1+ 3.10x2+ 0s1+ 0s2+ 0s3subject to5.0x1+ 7.5x2+s1= 6,5003.0x1+ 3.2x2+s2= 3,000x2+s3=510x1,x2,s1,s2,s3≥014.a)5.0(456) + 7.5(510) +s1= 6,500s1=6,500−6,105s1= 395 lbs.3.0(456) + 3.2(510) +s2= 3,000s2= 0 hrs.510 +s3= 510s3= 0Therefore demand for corduroy is met.b)In order for the optimal solution point tochangefromBtoCthe slope of the objectivefunction must be at least as great as the slopeof the constraint line, 3.0x1+ 3.2x2= 3,000,which is−3/3.2. Thus, the profit for denimwould have to be,−c1/3.0 =−3/3.2c1= 2.91If the profit for denim is increased from $2.25 to$3.00 the optimal solution would change to pointC, wherex1= 1,000,x2= 0,Z= 3,000.Profit for corduroy has no upper limit thatwould change the optimal solution point.c)The constraint line for cotton would moveinward as shown in the following graph wherepointCis optimal.15.Z= 2,607.000VariableValueReducedCostx1456.0000.000x2510.0000.000ConstraintSlack/SurplusShadowPricec1395.0000.000c20.0000.750c30.0000.700

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3-6Objective Coefficient RangesVariablesLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasex10.0002.2502.9060.6562.250x22.4003.100No limitNo limit0.700Right Hand Side RangesConstraintsLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasec16,015.0006,500.000No limitNo limit395.000c21,632.0003,000.0003,237.000237.0001,368.000c30.000510.000692.308182.308510.000a)The company should select 237 additionalhours of processing time, with a shadow priceof $0.75 per hour. Cotton has a shadow priceof $0 because there is already extra (slack)cotton available and not being used, so anymore would have no marginal value.b)0≤c1≤2.9066,105≤q1≤∞2.4≤c2≤∞1,632≤q2≤3,2370≤q3≤692.308The demand for corduroy can decrease to zeroor increase to 692.308 yds. without changingthe current solution mix of denim andcorduroy. If the demand increases beyond692.308 yds., then denim would no longer beproduced and only corduroy would beproduced.16.x1= no. of days to operate Mill 1x2= no. of days to operate Mill 2minimizeZ= 6,000x1+ 7,000x2subject to6x1+ 2x2≥122x1+ 2x2≥84x1+ 10x2≥5x1,x2≥017.a)6(4) + 2(0)−s1= 12s1= 122(4) + 2(0)−s2= 8s2= 04(4) + 10(0)−s3= 5s3= 11b)The slope of the objectivefunction,−6,000/7,000,must become flatter(i.e., less) than the slope of the constraint line,2x1+ 2x2= 8, for the solution tochange. The cost of operating Mill 1,c1, thatwould change the solution point is,−c1/7,000 =−1c1= 7,000Since $7,500 > $7,000, the solution point willchange toB, wherex1= 1,x2= 3,Z= $28,500.c)If the constraint line for high-grade aluminumchanges to 6x1+ 2x2= 10, it moves inward butdoes not change the optimal variable mix.Bremains optimal but moves to a new location,x1= 0.5,x2= 3.5,Z= $27,500.18.Z= 24,000VariableValuex14.000x20.000ConstraintSlack/SurplusShadowPricec112.0000.000c20.000−3,000.000c311.0000.000

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3-7Objective Coefficient RangesVariablesLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasex10.0006,000.0007,000.0001,000.0006,000.000x26,000.0007,000.000No limitNo limit1,000.000Right Hand Side RangesConstraintsLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasec1No limit12.00024.00012.000No limitc24.0008.000No limitNo limit4.000c3No limit5.00016.00011.000No limita)There is surplus high-grade and low-gradealuminum so the shadow price is $0 for both.The shadow price for medium-gradealuminum is $3,000, indicating that for everyton that this constraint could be reduced, costwill decrease by $3,000.b)0≤c1≤7,000∞≤q1≤246,000≤c2≤∞4≤q2≤∞∞≤q3≤16c)There will be no change.19.x1= no. of acres of cornx2= no. of acres of tobaccomaximizeZ= 300x1+ 520x2subject tox1+x2≤410105x1+ 210x2≤52,500x2≤100x1,x2≥020.a)x1= 320,x2= 90320 + 90 +s1= 410s1= 0 acres uncultivated90 +s3= 100s3=10 acres of tobaccoallotment unusedb)At pointDonly corn is planted. In order forpointDto be optimal the slope of theobjective function will have to be at least asgreat (i.e., steep) as the slope of the constraintline,x1+x2= 410, which is−1. Thus, theprofit for corn is computed as,−c1/520 =−1c1= 520The profit for corn must be greater than $520 forthe Bradleys to plant only corn.c)If the constraint line changes fromx1+x2= 410 tox1+x2= 510, it will moveoutward to a location which changes thesolution to the point where 105x1+210x2= 52,500 intersects with the axis. Thisnew point isx1= 500,x2= 0,Z= $150,000.d)If the constraint line changes fromx1+x2= 410 tox1+x2= 360, it moves inward to alocation which changes the solution point tothe intersection ofx1+x2= 360 and 105x1+210x2= 52,500. At this pointx1= 260,x2=100, andZ= $130,000.21.Z= 142,800.000VariableValuex1320.000x290.000ConstraintSlack/SurplusShadowPricec10.00080.000c20.0002.095c310.0000.000

Solution Manual For Introduction To Management Science, 11th Edition - Page 31 preview image

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3-8Objective Coefficient RangesVariablesLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasex1260.000300.000520.000220.00040.000x2300.000520.000600.00080.000220.000Right Hand Side RangesConstraintsLower LimitCurrent ValuesUpper LimitAllowable IncreaseAllowable Decreasec1400.000410.000500.00090.00010.000c243,050.00052,500.00053,550.0001,050.0009,450.000c390.000100.000No limitNo limit10.000a)No, the shadow price for land is $80 per acreindicating that profit will increase by no morethan $80 for each additional acre obtained. Themaximum price the Bradleys should pay is $80and the most they should obtain is at the upperlimit of the sensitivity range for land. This limitis 500 acres, or 90 additional acres. Beyond 90acres the shadow price would change.b)The shadow price for the budget is $2.095.Thus, for every $1 borrowed they couldexpect a profit increase of $2.095. If theyborrowed $1,000 it would change theamountof corn and tobacco they plant;x1=310.5acres of corn andx2= 99.5 acres of tobacco.22.x1= no. of sausage biscuitsx2= no. of ham biscuitsmaximizeZ= .60x1+ .50x2subject to.10x1≤30.15x2≤30.04x1+ .04x2≤16.01x1+ .024x2≤6x1,x2≥023.a)x1= 300,x2= 100,Z= $230.10(300) +s1= 30s1= 0 leftover sausage.15(100) +s2= 30s2= 15 lbs. leftover ham.01(300) + .024(100) +s4= 6s4= 0.6 hr.b)The slope of the objective function,−6/5,must become flatter (i.e.,less) than the slopeof the constraint line, .04x1+ .04x2= 16, forthe solution to change. The profit for ham,c2,that would change the solution point is,−0.6/c2=−1c2= .60Thus, an increase in profit for ham of 0.60will create a second optimal solution pointatCwherex1= 257,x2= 143, andZ= $240.(PointDwould also continue to be optimal,i.e.,multiple optimalsolutions.)c)A change in the constraint line from .04x1+.04x2= 16 to .04x1+ .04x2= 18 would move theline outward, eliminating both pointsCandD.The new solution point occurs at the intersectionof 0.01x1+ .024x2= 6 and .10x= 30. This pointisx1= 300,x2= 125, andZ= $242.50.24.Z= 230.000VariableValuex1300.000x2100.000ConstraintSlack/SurplusShadowPricec10.0001.000c215.0000.000c30.00012.500c40.6000.000

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