运营管理课后习题答案供参考.docx

1、运营管理课后习题答案供参考Chapter 02 - Competitiveness, Strategy, and Productivity3.(1)(2)(3)(4)(5)(6)(7) WeekOutputWorker Cost$12x40Overhead Cost 1.5Material Cost$6Total CostMFP (2) (6)130,0002,8804,3202,7009,9003.03233,6003,3605,0402,82011,2202.99332,2003,3605,0402,76011,1602.89435,4003,8405,7602,88012,4802.84

2、*refer to solved problem #2Multifactor productivity dropped steadily from a high of 3.03 to about 2.84.4. a. Before: 80 5 = 16 carts per worker per hour. After: 84 4 = 21 carts per worker per hour.b. Before: ($10 x 5 = $50) + $40 = $90; hence 80 $90 = .89 carts/$1. After: ($10 x 4 = $40) + $50 = $90

3、; hence 84 $90 = .93 carts/$1.c. Labor productivity increased by 31.25% (21-16)/16). Multifactor productivity increased by 4.5% (.93-.89)/.89).*Machine ProductivityBefore: 80 40 = 2 carts/$1.After: 84 50 = 1.68 carts/$1.Productivity increased by -16% (1.68-2)/2)Chapter 03 - Product and Service Desig

4、n6. Steps for Making Cash Withdrawal from an ATM 1. Insert Card: Magnetic Strip Should be Facing Down 2. Watch Screen for Instructions 3. Select Transaction Options: 1) Deposit 2) Withdrawal 3) Transfer 4) Other 4. Enter Information: 1) PIN Number 2) Select a Transaction and Account 3) Enter Amount

5、of Transaction 5. Deposit/Withdrawal: 1) Depositplace in an envelope (which youll find near or in the ATM) and insert it into the deposit slot 2) Withdrawallift the “Withdrawal Door,” being careful to remove all cash 6. Remove card and receipt (which serves as the transaction record) 8.Technical Req

6、uirementsIngredientsHandlingPreparationCustomer RequirementsTasteAppearanceTexture/consistencyChapter 04 - Strategic Capacity Planning for Products and Services2. Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual out

7、put = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4 10. a. Given: 10 hrs. or 600 min. of operating time per day. 250 days x 600 min. = 150,000 min. per year operating time. Total processing time by machineProductABC148,00064,00032,000248,00048,00036,000330,00036,00024,000460,00060,00

8、030,000Total186,000208,000122,000 You would have to buy two “A” machines 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 = $31,000 + $80,000 = $111,000 B (2) :

9、208,000 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400 Buy 2 Bsthese have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3. Desired output = 4 Operating time = 56 minutes Task# of Following tasksPo

10、sitional WeightA423B320C218D325E218F429G324H114I05a. First rule: most followers. Second rule: largest positional weight. Assembly Line Balancing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GA36B,GG6IID77B, EB25CC41IIIE410HH91IVI59b. First rule: Largest posit

11、ional weight.Assembly Line Balancing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GD72IIG68A, EA35B,EB23IIIC410EE46IVH95II5c. 4. a. l. 2. Minimum Ct = 1.3 minutesTaskFollowing tasksa4b3c3d2e3f2g1h0Work StationEligibleAssignTime RemainingIdle TimeIaA1.1b,c,e,

12、(tie)B0.7C0.4E0.30.3IIdD0.00.0IIIf,gF0.5G0.20.2IVhH0.10.10.6 3. 4. b. 1. 2. Assign a, b, c, d, and e to station 1: 2.3 minutes no idle time Assign f, g, and h to station 2: 2.3 minutes 3. 4. 7.15438762Chapter 06 - Work Design and Measurement3.ElementPROTNTAFjobST1.90.46.4141.15.4762.851.5051.2801.15

13、1.47231.10.83.9131.151.05041.001.161.1601.151.334 Total 4.3328.A = 24 + 10 + 14 = 48 minutes per 4 hours 9.a.ElementPROTNTAST11.101.191.3091.151.50521.15.83.9551.151.09831.05.56.5881.15.676 b. c. e = .01 minutes Chapter 07- Location Planning and Analysis 1.FactorLocal bankSteel millFood warehousePub

14、lic school1.Convenience for customersHLMHMH2.Attractiveness of buildingHLMMH3.Nearness to raw materialsLHLM4.Large amounts of powerLHLL5.Pollution controlsLHLL6.Labor cost and availabilityLMLL7.Transportation costsLMHMHM8.Construction costsMHMMHLocation (a)Location (b) 4.FactorABCWeightABC1.Business

15、 Services9552/918/910/910/92.Community Services7671/97/96/97/93.Real Estate Cost3871/93/98/97/94.Construction Costs5652/910/912/910/95.Cost of Living4781/94/97/98/96.Taxes5551/95/95/94/97.Transportation 6 7 81/9 6/9 7/9 8/9Total3944451.053/955/954/9Each factor has a weight of 1/7.a.Composite Scores

16、394445777B or C is the best and A is least desirable.b.Business Services and Construction Costs both have a weight of 2/9; the other factors each have a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c.Composite Scores ABC53/955/954/9B is the best followed by C and then A.5.LocationxyA37B82C46D41E 6 4Tota

17、ls2520=xi=25= 5.0=yi=20= 4.0n5n5 Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork TypeFrequencyLube and Oil12Brakes7Tires6Battery4Transmission1Total30Pareto2.The run charts seems to show a pattern of errors possibly linked

18、 to break times or the end of the shift. Perhaps workers are becoming fatigued. If so, perhaps two 10 minute breaks in the morning and again in the afternoon instead of one 20 minute break could reduce some errors. Also, errors are occurring during the last few minutes before noon and the end of the

19、 shift, and those periods should also be given managements attention.4Chapter 9 - Quality Control4.SampleMeanRange179.482.6Mean Chart: A2= 79.96 0.58(1.87)280.142.3= 79.96 1.08380.141.2UCL = 81.04, LCL = 78.88479.601.7Range Chart: UCL = D4= 2.11(1.87) = 3.95580.022.0LCL = D3= 0(1.87) = 0680.381.4Bot

20、h charts suggest the process is in control: Neither has any points outside the limits.6. n = 200 Control Limits = Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is t

21、oo large. 7. Control limits: UCL is 16.266, LCL becomes 0. All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, = Process mean, s = Process standard deviationFor process H: For process K:Assuming the minimum acceptable is 1.33, since 1.0 1.33, th

22、e process is not capable.For process T:Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888883851Subcontract2031220019Output

23、 - Forecast0440033InventoryBeginning0040003Ending0400030Average022001.51.57Backlog00000000Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,400Overtime 9609609609609603609606,120Subcontract 28004201,680280002,660Inventory 0202000151570Total4,4404,1804,6005,8404,4403,5754,17531,250b. Level strategyPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecas