运营管理课后习题答案供参考文档格式.docx

1、*refer to solved problem #2Multifactor productivity dropped steadily from a high of to about .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; hence 8

2、4 $90 = .93 carts/$1.c. Labor productivity increased by % （21-16）/16）. Multifactor productivity increased by % （./.89）.*Machine ProductivityBefore: 40 = 2 carts/$1.After: 50 = carts/$1.Productivity increased by -16% （/2）Chapter 03 - Product and Service Design6. Steps for Making Cash Withdrawal from

3、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 of Transaction 5. Deposit/Withdrawal: 1） D

4、epositplace 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 RequirementsIngredientsHandlingPreparationCust

5、omer 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 output = （.4）（Design capacity） Actual output =

6、 8 jobs Utilization = .410. 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 machineProductABC48,00064,00032,00036,00024,00060,000Total186,000208,000122,000You would have to buy two “A” machines at a total co

7、st 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） : 208,000 60 = 3, hrs. x $11 = $38,133 + $60,000 = $98,133 C（1）: 122,000 60 = 2, hrs. x

8、$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 minutesTask# of Following tasksPositional Weight232018D25EF29G24H14I5a. First rule: most followers. Second rule: largest positional

9、weight. Assembly Line Balancing Table （CT = 14）Work StationTask TimeTime RemainingFeasible tasks Remaining9A,D,G6B,GII7B, EIII10IVb. First rule: Largest positional weight.Assembly Line Balancing Table （CT = 14）8A, EB,Ec. 4. a. l. 2. Minimum Ct = minutesFollowing tasksabcdefghEligibleAssignIdle Timeb

10、,c,e, （tie）f,g 3. 4. b. 1. 2. Assign a, b, c, d, and e to station 1: minutes no idle time Assign f, g, and h to station 2: minutes 3. 4. 7.Chapter 06 - Work Design and MeasurementElementPROTNTAFjobST.90.46.414.476.85.83.913 Total A = 24 + 10 + 14 = 48 minutes per 4 hours9.a.955.56.588.676 b. c. e =

11、.01 minutes Chapter 07- Location Planning and Analysis 1.FactorLocal bankSteel millFood warehousePublic school1.Convenience for customersLMH2.Attractiveness of buildingMNearness to raw materials4.Large amounts of power5.Pollution controls6.Labor cost and availabilityTransportation costsConstruction

12、costsLocation （a）Location （b） 4.WeightBusiness Services2/918/910/9Community Services1/97/96/9Real Estate Cost3/98/9Construction Costs12/9Cost of Living4/9Taxes5/9Transportation 6 7 8 6/9 7/9 8/939444553/955/954/9Each factor has a weight of 1/7.Composite Scores B or C is the best and A is least desir

13、able.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.B is the best followed by C and then A.Locationxy 4Totals=xi= yin Hence, the center of gravity is at （5,4） and therefore the optimal location.Chapter 08

14、 - Management of QualityChecksheetWork TypeFrequencyLube and Oil12BrakesTiresBatteryTransmission30Pareto.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

15、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 shift, and those periods should also be given managements attention.Chapter 9 - Quality ControlSampleMeanRangeMean Chart: A2= UCL

16、= , LCL = Range Chart: UCL = D4= = LCL = D3= 0 = 0Both 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. ., 1 defect

17、ive = .005, 2 defectives = .01, etc.Sample 10 is too large. 7. Control limits: UCL is , LCL becomes 0. All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, = Process mean, = Process standard deviationFor process H:For process K:Assuming the minim

18、um acceptable is , since , the process is not capable.For process T:Since = , the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodMar.Apr.MayJun.JulyAug.Sep.Forecast5055604051350Regular280OvertimeSubcontract19Output - Forecast43InventoryBeginningEndingAverageBacklogCosts:Regular 3,20022,400Overtime 9603606,120Subcontract 4201,6802,660Inventory 15704,4404,1804,6005,8403,5754,17531,250b. Level strategy565101.5Ba