《运营管理》课后习题标准答案

1、运营管理课后习题答案作者: 日期:#Solutions Problems OM 11e Stevenson15 / 243.(1)(2)(3)(4)(5)(6)(7)WorkerOverheadMaterialTotalMFPWeekOut putCostCost 1.5Cost$6Cost(2)子$12x40130,0002,8804,3202,7009,9003.03233,6003,3605,0402,82011,2202.99332,2003,3605,0402,76011,1602.89435,4003,8405,7602,88012,4802.84Chapter 02 - Comp

2、 etitiveness. Strategy, and Productivity*refer to solved p roblem #2Multifactor p roductivity dropped steadily from a high of 3.03 to about 2.84.4.a.Before: 80 m 5 = 16 carts per worker per hour.b.After: 84 子 4 = 21 carts per worker per hour.Before: ($10 x 5 = $50) + $40 = $90; hence 80$90 = .89 art

3、s/$1.After: ($10 x 4 = $40) + $50 = $90; hence 84$90 = .93 carts/$1.c.Labor p roductivity in creased by 31.25%(21-16)/16).Multifactor p roductivity in creased by 4.5%(.93-.89)/.89).*Machi ne P roductivityBefore: 80 +40 = 2 carts/$1.After:84 + 50 = 1.68 carts/$1.Productivity in creased by -16%(1.68-2

4、)/2)Cha pter 03 - Product and Service Design6.Ste ps for Maki ng Cash Withdrawal from an ATM1.In sert Card: Mag netic Strip Should be Fac ing Down2.Watch Scree n for I nstructio ns3.Select Tran sact ion Op ti ons:1)Depo sit4.2)3)4)WithdrawalTran sferOtherEn ter In formatio n:1) PIN Number2) Select a

5、 Transaction and Account3) Enter Amount of Transaction5.Dep osit/Withdrawal:2) Withdrawal lift the1) Depo sit pl ace in an en vel ope (which you ll find n ear or in the ATM) and in sert it into the depo sit slot6.Withdrawal Door, ” being careful to remove all cashRemove card and rece ip t (which ser

6、ves as the tra nsact ion record)8.Tech ni calRequireme nts Customer Requireme ntsIn gredie ntsHan dli ngPrep arati onTasteVVApp eara neeVVVTexture/c on siste ncyVVChap ter 04 - Strategic Cap acity Planning for Products and Services2.Efficiency=Actual output二*。%Effective capacityActual out put = .8 (

7、Effective cap acity) Effective cap acity = .5 (Desig n cap acity) Actual out put = (.5)(.8)(Effective cap acity) Actual out put = (.4)(Desig n cap acity) Actual out put = 8 jobsUtilizati on = .4Utilizatio n 二 Actual outputDesig n cap acityr . 小 .Actual out putDesign Capacity =Effective capacityW =20

8、 jobs10.a. Give n: 10 hrs. or 600 min. of op erati ng time per day.250 days x 600 min. = 150,000 min. per year op erati ng time.Na186,000150,000= 1.24 上2machi neNb208,000150,000= 1.38 s： 2 machineNc122,000150,000= .81 止1 machi neTotal processing time by machineProductABC148,00064,00032,000248,00048,

9、00036,000330,00036,00024,000460,00060,00030,000Total186,000208,000122,000You would have to buy two“ A machi nes at a total cost of $80,000, or two“ B” machines at a total cost of $60,000, or one“ C” machine at $80,000.b. Total cost for each type of machine:A (2): 186,000 min - 60 = 3,100 hrs. x $10

10、= $31,000 + $80,000 = $111,000B ：208,000 - 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133C(1): 122,000 - 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bsthese have the lowest total cost.Cha pter 05 - Process Selection and Facility Layout3.Desired out put = 4Op erat ing time =

11、56 minu tesCT Op erati ng time 56 m inu tes per hourCT = 14mi nu tes per unitDesired out put4un its per hourTask # of Following tasks Positional Weight23 20 18 25 18 29 24 14a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing TabieCT = 14)Work Statio nTaskTa

12、sk TimeTime Remai ningFeasible tasksRemai ningIF59A,D,GA36B,GG6一一IID77B, EB25CC41一IIIE410HH91一IVI59一b. First rule: Largest po siti onal weight.Assembly Line Balancing Table (CT = 14)Work Statio nTaskTask TimeTime Remai ningFeasible tasksRemai ningIF59A,D,GD72一IIG68A, EA35B,EB23一IIIC410EE46一IVH95II5一

13、c.Efficiency =Total time=兰=80.36%CT x no. of stations 564.a.l.2. Min imumCt=1.3mi nutesTaskFollow ing tasksWork Statio n 1 EligibleAssig nTime Remai ningIdle TimeIaA1.1b,c,e, (tie)B0.7C0.4E0.30.3IIdD0.00.0IIIf,gF0.5G0.20.2IVhH3.丄 Z(idle time) .6Idle percent= NX CT 二帀 W.54 percent7.4.b.OT 42

14、0 min /dayOutput =323.1 (rounds to 323)copiers /dayCT 1.3 min. / cycle.Total time =4.6, CT = Total time =仝=2.3 minutesN2Assig n a, b, c, d, and e to stati on 1: 2.3 minu tes no idle timeAssig n f, g, and h to stati on 2: 2.3 minu tesOut put =二420-=182.6 copi ers /dayCT 2.3Maximum Ct is 4.6. O

15、ut put二 420 min./day = 91.30 copiers / day4.6 min. / cycle154387623.Eleme ntPROTNTAF jobST141.15.4762.851.5051.2801.151.47131.151.05041.001.161.1601.151.334Total4.332Cha pter 06 - Work Design and MeasurementA = 24 + 10 + 14 = 48 mi nutes per 4 hours空=.20240NT =6(.95) =5.70mi n.=7

16、.125min .9.a. Eleme ntPROTNTAST091.151.505551.151.09831.05.56.5881.15.676b.= .83s =.034/zsz =2.00ax2 zv)2(.034) 丿1.01(.83)丿= 67.12 68 observationsA =.01c.e = .01 minu tesfzsY (2(.034) 丫I = Ivej = 46.24, round to 47I .01 丿Chap ter 07- Location Planning and Analysis1.FactorLocal

17、bankSteel millFood warehouse Public scl1.ConvenienceforcustomersHLM -HM H2.Attractive nessofbuildi ngHLMM -H3.Nearness torawmaterialsLHLM4.4.po werLHLLP olluti on con trolsLHLLLabor cost and availabilityLMLLTransp ortati on costsLM -HM -HMCon struct ion costsMHMM HLocati on (a)Locati on (b)FactorABC

18、WeightABCBusin ess Services9552/918/910/910/9Commu nity Services7671/97/96/97/9Real Estate Cost3871/93/98/97/9Con structio n Costs5652/910/912/910/9Cost of Livi ng4781/94/97/98/9Taxes5551/95/95/94/9Transp ortati on6781/96/97/98/9Total3944451.053/955/954/9Large amounts of..Each factor h

19、as a weight of 1/7.5.6.a.394445Compo site ScoresB or C is the best and A is least desirable.b.Bus in ess Services and Con structi on Costs both have a weight of 2/9; the have a weight of 1/9.other factors eachc.5 x + 2 x + 2 x =仁 x = 19ABCCompo site Scores 53/955/954/9B is the best followed by C and

20、 then A.5.Locati on X yTotals2520Zxjx =25= 5.020=4.0Hence, the cen ter of gravity is at (5,4) and therefore the op timal locati on.1.Cha pter 08 - Management of QualityChecksheetWork TypeFreque ncyLube and Oil12Brakes7Tires6Battery4Tran smissi on1Total30P areto12Solutions Problems OM 11e Stevenson17

21、/24breaklunchThe run charts seems to show a p atter n of errors p ossibly lin ked to break times or the end of the shift. Perha ps workers are beco ming fatigued. If so, p erha ps two 10 minute breaks in the morning and aga in in the after noon in stead of one 20 minute break could reduce some error

22、s. Also, errors are occurri ng duri ng the last few minu tes before noon and the end of the shift, and those p eriods should also be give n man ageme nts attention.DdnPerLamMiss / / Bur n LoosLam pPower /NorOutlet /Defect/OtheCordSolutions Problems OM 11e Stevenson23 / 244.Samp leMea nRan ge179.482.

23、6Mean Chart: X A2R = 79.96 0.58(1.87)280.142.3=79.96 1.08380.141.2UCL = 81.04, LCL = 78.88479.601.7Range Chart: UCL = D 4R = 2.11(1.87) = 3.95580.022.0LCL = D 3R = 0(1.87) = 0680.381.4Both charts suggest the p rocess is in con trol: Neither has anyCha pter 9 - Quality Controlpoints outside the limit

24、s.6.n = 200Control Limits =p 2jp(17.14.2513(200)=.0096巴=7.85714=.00962咤空= .0096 .0138200Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fract ion rep rese nted by each data point is half the amount show n. Eg, 1 defective = .005, 2 defectives =.01, etc.Sam pie 10 is too large.Control limits:

25、 c 3乙=7.857 8.409UCL is 16.266, LCL becomes 0.All values are within the limits.Let USL = Upper Sp ecificati on Limit, LSL = Lower Sp ecificati on Limit,X = P rocess mea n, c = P rocess sta ndard deviati onFor p rocess H:X -LSL 15-14.1cc 933b (3)(.32) -USLX1615=1.043b (3)(.32)Cpk =mi n938,1.04=93.93

26、1.0, not capableFor process K:X _LSL3b=注=1.0(1)USL X=36口 =1.17(3)(1)Cp k =mi n 1.0,1.17 =1.0Assuming the minimum acceptable Cpk is 1.33, since 1.0 22.ROPLTssRisk = 9%3%=30 gal./day=170 gal.=4 days,=Zod LT = 50 gal Z = 1.34 Z = 1.88,Solving,航 lt = 37.31 ss=1.88 x 37.31 = 70.14 gal.Cha pter 13 - JIT a

27、nd Lean Op erations1. NDT(1 + X)=80 pi eces per hour=75 min. = 1.25 hr.80(1.25) (1.35)=3=4545X = .35Solutions Problems OM 11e Stevenson4.The smallest daily qua ntity eve niy divisible into all four qua ntities is 3. Therefore, use three cycles.35/24ProductDaily quantityUn its per cycle2121/3 = 71212/3 = 43/3 = 11515/3 = 55.a. Cycle 1234b. Cycle 121111c. 4 cy