《运营管理》课后习题答案.docx

1、运营管理课后习题答案运营管理课后习题答案LT Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.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

2、600 min. = 150,000 min. per year operating time. Total processing time by machineProductABC148,00064,00032,000248,00048,00036,000330,00036,00024,000460,00060,00030,000Total186,000208,000122,000 You would have to buy two “A” machines at a total cost of $80,000, or two “B” machines at a total cost of

3、$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) : 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 B

4、sthese have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3. Desired output = 4 Operating time = 56 minutes Task# of Following tasksPositional WeightA423B320C218D325E218F429G324H114I05a. First rule: most followers. Second rule: largest positional weight. Assembly Line Balan

5、cing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GA36B,GG6IID77B, EB25CC41IIIE410HH91IVI59b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GD72IIG68A, EA35B

6、,EB23IIIC410EE46IVH95II5c. 4. a. l. 2. Minimum Ct = 1.3 minutesTaskFollowing tasksa4b3c3d2e3f2g1h0Work StationEligibleAssignTime RemainingIdle TimeIaA1.1b,c,e, (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

7、Assign f, g, and h to station 2: 2.3 minutes 3. 4. 7.15438762Chapter 06 - Work Design and Measurement3.ElementPROTNTAFjobST1.90.46.4141.15.4762.851.5051.2801.151.47231.10.83.9131.151.05041.001.161.1601.151.334 Total 4.3328.A = 24 + 10 + 14 = 48 minutes per 4 hours 9.a.ElementPROTNTAST11.101.191.3091

8、.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 warehousePublic school1.Convenience for customersHLMHMH2.Attractiveness of buildingHLMMH3.Nearness to raw materialsLHLM4.Large amounts of powerLHLL5.Pol

9、lution controlsLHLL6.Labor cost and availabilityLMLL7.Transportation costsLMHMHM8.Construction costsMHMMHLocation (a)Location (b) 4.FactorABCWeightABC1.Business 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

10、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 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

11、 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 4Totals2520=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 Qual

12、ity1.ChecksheetWork TypeFrequencyLube and Oil12Brakes7Tires6Battery4Transmission1Total30Pareto2.The run charts seems to show a pattern of errors possibly linked 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

13、 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 shift, and those periods should also be given managements attention.4Chapter 9 - Quality Control4.SampleMeanRange179.482.6Mean Chart: A2= 7

14、9.96 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) = 0680.381.4Both 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

15、 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 too large. 7. Control limits: UCL is 16.266, LCL becomes 0. All values are within the limits.14. Let USL = Upper Specification Limit, LSL = L

16、ower Specification Limit, = Process mean, = Process standard deviationFor process H: For process K:Assuming the minimum acceptable is 1.33, since 1.0 1.33, the 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

17、 backlogs are allowedPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888883851Subcontract2031220019Output - Forecast0440033InventoryBeginning0040003Ending0400030Average022001.51.57Backlog00000000Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,

18、400Overtime 9609609609609603609606,120Subcontract 28004201,680280002,660Inventory 0202000151570Total4,4404,1804,6005,8404,4403,5754,17531,250b. Level strategyPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888888856Subcontract222222214Output - For

19、ecast065100101InventoryBeginning0061001Ending0610010Average033.5.50.5.58Backlog000990018Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,400Overtime 9609609609609609609606,720Subcontract 2802802802802802802801,960Inventory 3035505580Backlog180180360Total4,4404,4704,4754,6254,6204,4454,44531,5208.

20、 Period123456TotalForecast160150160180170140960OutputRegular150150150150160160920Overtime1010010101050Subcontract0010100020Output- Forecast01001000InventoryBeginning00101000Ending01010000Average051050020Backlog0000000Costs:Regular 7,5007,5007,5007,5008,0008,00046,000Overtime 75075007507507503,750Sub

21、contract 00800800001,600Inventory 20402080Backlog000000Total8,2508,2708,3409,0709,0508,75051,430Chapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1 J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4b. 4.Master ScheduleDayBeg. Inv.1234567Quantity100150200Table Beg. Inv.1234567Gross re