运营管理课件全 10习题答案

《运营管理》课程习题及标准答案-修订版(1)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题标准答案

《运营管理》课后习题答案————————————————————————————————作者：————————————————————————————————日期：2Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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; 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 Design6. Steps for Making Cash Withdrawal from an ATM1. Insert Card: Magnetic Strip Should be Facing Down2. Watch Screen for Instructions3. Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4. Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5. Deposit/Withdrawal: 1) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.TechnicalRequirements IngredientsHandlingPreparationCustomer RequirementsTaste √√ Appearance√ √√Texture/consistency√√Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==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 = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===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 machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $111,000B (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 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.3 adf752 b4 c4 e9 h5 i6 gDesired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasksRemainingIF 5 9 A,D,G A 3 6 B,G G6 – – II D7 7 B, E B 2 5 C C4 1 – III E 4 10 H H9 1 – IV I59–b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasks RemainingIF 5 9 A,D,G D7 2 – II G 6 8 A, E A 3 5 B,E B2 3 – III C 4 10 E E4 6 – IV H 95 I I5–c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1habd cfeghWork StationEligible Assign Time RemainingIdle TimeIa A 1.1 b,c,e, (tie)B 0.7C 0.4E 0.3 0.3 II d D 0.0 0.0 IIIf,g F 0.5G 0.2 0.2 IVh H 0.1 0.10.63. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 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 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7. 1 5 4 3 8 762Chapter 06 - Work Design and Measurement3. Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = 5.0 -y =∑y i= 20 = 4.0 n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30Pareto127641 Lube & Oil Brakes Tires Battery Trans.2 .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 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 management’s attention.4Power Per LamMissDidn’Not OutletDefectBurn LoosLampOtheCordbreak lunch3 2•• •• •• • ••• • ••• •••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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 too large.7. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriod Mar. Apr. May Jun. July Aug. Sep. TotalForecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - Forecast 0 4 –4 0 0 3 –3 Inventory Beginning 0 0 4 0 0 0 3 Ending 0 4 0 0 0 3 0 Average 0 2 2 0 0 1.5 1.5 7 Backlog 0 0 0 0 0 0 0 0 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 360 960 6,120 Subcontract 280 0 420 1,680280 0 0 2,660 Inventory 0 20 20 0 0 15 15 70 Total4,4404,1804,6005,8404,4403,575 4,17531,250b. Level strategyPeriod Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 OutputRegular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 8 8 56 Subcontract 2 2 2 2 2 2 2 14 Output - Forecast 0 6 –5 –10 0 10 –1InventoryBeginning 0 0 6 1 0 0 1Ending 0 6 1 0 0 1 0Average 0 3 3.5 .5 0 .5 .5 8 Backlog 0 0 0 9 9 0 0 18 Costs:Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 960 960 6,720 Subcontract 280 280 280 280 280 280 280 1,960 Inventory 30 35 5 0 5 5 80 Backlog 180 180 360 Total 4,440 4,470 4,475 4,625 4,620 4,445 4,445 31,520 8.Period 1 2 3 4 5 6 TotalForecast 160 150 160 180 170 140 960OutputRegular 150 150 150 150 160 160 920Overtime 10 10 0 10 10 10 50Subcontract 0 0 10 10 0 0 20Output- Forecast 0 10 0 –10 0 0InventoryBeginning 0 0 10 10 0 0Ending 0 10 10 0 0 0Average 0 5 10 5 0 0 20Backlog 0 0 0 0 0 0 0Costs:Regular 7,500 7,500 7,500 7,500 8,000 8,000 46,000Overtime 750 750 0 750 750 750 3,750Subcontract 0 0 800 800 0 0 1,600Inventory 20 40 20 80Backlog 0 0 0 0 0 0Total 8,250 8,270 8,340 9,070 9,050 8,750 51,430Chapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4D: 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 Schedule Day Beg. Inv. 1 2 3 4 5 6 7 Quantity100 150 200 TableBeg. Inv. 1 2 3 4 5 6 7 Gross requirements 100 150 200 Scheduled receipts Projected on hand Net requirements 100 150 200 Planned-order receipts 100 150 200 Planned-order releases 100 150 200Wood Sections Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 200300 400 Scheduled receipts 100 Projected on hand 100100 Net requirements 100 300 400 Planned-order receipts 100 300 400 Planned-order releases400 400Braces Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 300 450 600 Scheduled receipts Projected on hand 60 60 60 60 Net requirements 240 450 600 Planned-order receipts 240 450 600Planned-order releases 240 450 600StaplerTopBaseCoveSpri SlideBase Strik RubberSlidSpriLegs Beg.Inv.1 2 3 4 5 6 7Gross requirements 400 600 800Scheduled receiptsProjected on hand 120 120 120 120 88 88 71 Net requirements 280 600 800Planned-order receipts 308 660 880Planned-order releases 968 88010. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. Shifting to an earlier week would result in added carrying costs;shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase due toovertime premium, a probable decrease in productivity, and possible increase inaccidents.Chapter 12 - Inventory Management2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items. DollarItemUnit Cost UsageUsageCategoryK34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr. S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 20015. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+4.851.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $5.00(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal.Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35• •• •495 497 500 5031,0004,000 6,000QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date Days Job (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD: Job time Flow time Due date DaysJob (days) (days) (days) tardyC 7 7 15 0B 10 17 16 1D 6 23 17 6A 14 37 20 1784 24Critical RatioJob Processing Time(Days) Due Date Critical Ratio CalculationA 14 20 (20 – 0) / 14 = 1.43B 10 16 (16 – 0) /10 = 1.60C 7 15 (15 – 0) / 7 = 2.14D 6 17 (17 – 0) / 6 = 2.83Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 14) /10 = 0.20C 7 15 (15 – 14) / 7 = 0.14D 6 17 (17 – 14) / 6 = 0.50Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 21) /10 = –0.50C –––D 6 17 (17 – 21) / 6 = –0.67Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is A –C –D –B and the makespan is 37 days. Critical Ratio sequenceProcessing Time(Days)Flow time Due Date TardinessA 14 14 20 0 C 7 21 15 6 D 6 27 17 10 B1037 16 21 ∑9937b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.86 2.142.272.67c. SPT is superior.9.Time (hr.) Sequence of assignment:Order Step 1 Step 2A 1.20 1.40 .80 [C] last (or 7th)B 0.90 1.30 .90 [B] firstC 2.00 0.80 1.20 [A] 2ndD 1.70 1.50 1.30 [G] 3rdE 1.60 1.80 1.60 [E] 4thF 2.20 1.75 1.50 [D] 6th G1.301.401.75[F]5thThus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

运营管理课后习题答案

Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40OverheadCost @MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,9002 33,600 3,360 5,040 2,820 11,2203 32,200 3,360 5,040 2,760 11,1604 35,400 3,840 5,760 2,880 12,480*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 84 ÷ $90 = .93 carts/$1.c. Labor productivity increased by % ((21-16)/16).Multifactor productivity increased by % ((./.89).*Machine ProductivityBefore: 80 ÷ 40 = 2 carts/$1.After: 84 ÷ 50 = carts/$1.Productivity increased by -16% (/2)Chapter 03 - Product and Service Design6. Steps for Making Cash Withdrawal from an ATM1. Insert Card: Magnetic Strip Should be Facing Down2. Watch Screen for Instructions3. Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4. Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5. Deposit/Withdrawal: 1) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==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 = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===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 machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $111,000B (2) : 208,000 ÷ 60 = 3, hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 ÷ 60 = 2, hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1h3. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: minutes [no idle time]Assign f, g, and h to station 2: minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Chapter 06 - Work Design and Measurement3. Element PR OT NT AF jobST1 .90 .46 .414 .4762 .853 .83 .913 4Total8. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST12 .83 .9553.56.588.676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = -y =∑y i= 20 = n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30ParetoLube & Oil Brakes Tires Battery Trans.2 .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 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 management’s attention.4break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range1Mean Chart: =X ± A 2-R = ± 2 = ±3UCL = , LCL =4 Range Chart: UCL = D 4-R = = 5LCL = D 3-R = 0 = 06[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fraction represented by each data point is half the amount shown. ., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large.7. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is , LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is , since < , the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since = , the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Totalb.Level strategyPeriodForecastOutputRegularOvertimeSubcontractOutput - ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractOutput- ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotalChapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4D: 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.MasterSchedule10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor % % 80% %Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. Shifting to an earlier week would result in added carrying costs;shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase due toovertime premium, a probable decrease in productivity, and possible increase inaccidents.Chapter 12 - Inventory Management2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1, – $1,350] = $4.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6, (only $ higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr.S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q ' =TC =S QD H 2I max + TC orig. = $1, TC rev. = $Savings would be $D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.15. Range PHQ D = 4,900 seats/yr. 0–999 $ $ 495 H = .4P 1,000–3,999 497 NF S = $50 4,000–5,999 500 NF 6,000+503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($ + 4,900($50) + $(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($ + 4,900($50) + $(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($ + 4,900($50) + $(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 gal Risk = 9% Z = Solving, σd LT = 3% Z = , ss= x = gal.Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = hr. = 80 = 3C = 45 45X = .35QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date Days Job (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD:Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow time Average flow time Number of jobs Days tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CRc. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题及答案修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题标准答案

《运营管理》课后习题答案————————————————————————————————作者：————————————————————————————————日期：2Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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; 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 Design6. Steps for Making Cash Withdrawal from an ATM1. Insert Card: Magnetic Strip Should be Facing Down2. Watch Screen for Instructions3. Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4. Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5. Deposit/Withdrawal: 1) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.TechnicalRequirements IngredientsHandlingPreparationCustomer RequirementsTaste √√ Appearance√ √√Texture/consistency√√Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==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 = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===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 machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $111,000B (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 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.3 adf752 b4 c4 e9 h5 i6 gDesired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasksRemainingIF 5 9 A,D,G A 3 6 B,G G6 – – II D7 7 B, E B 2 5 C C4 1 – III E 4 10 H H9 1 – IV I59–b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasks RemainingIF 5 9 A,D,G D7 2 – II G 6 8 A, E A 3 5 B,E B2 3 – III C 4 10 E E4 6 – IV H 95 I I5–c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1habd cfeghWork StationEligible Assign Time RemainingIdle TimeIa A 1.1 b,c,e, (tie)B 0.7C 0.4E 0.3 0.3 II d D 0.0 0.0 IIIf,g F 0.5G 0.2 0.2 IVh H 0.1 0.10.63. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 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 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7. 1 5 4 3 8 762Chapter 06 - Work Design and Measurement3. Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = 5.0 -y =∑y i= 20 = 4.0 n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30Pareto127641 Lube & Oil Brakes Tires Battery Trans.2 .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 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 management’s attention.4Power Per LamMissDidn’Not OutletDefectBurn LoosLampOtheCordbreak lunch3 2•• •• •• • ••• • ••• •••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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 too large.7. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriod Mar. Apr. May Jun. July Aug. Sep. TotalForecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - Forecast 0 4 –4 0 0 3 –3 Inventory Beginning 0 0 4 0 0 0 3 Ending 0 4 0 0 0 3 0 Average 0 2 2 0 0 1.5 1.5 7 Backlog 0 0 0 0 0 0 0 0 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 360 960 6,120 Subcontract 280 0 420 1,680280 0 0 2,660 Inventory 0 20 20 0 0 15 15 70 Total4,4404,1804,6005,8404,4403,575 4,17531,250b. Level strategyPeriod Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 OutputRegular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 8 8 56 Subcontract 2 2 2 2 2 2 2 14 Output - Forecast 0 6 –5 –10 0 10 –1InventoryBeginning 0 0 6 1 0 0 1Ending 0 6 1 0 0 1 0Average 0 3 3.5 .5 0 .5 .5 8 Backlog 0 0 0 9 9 0 0 18 Costs:Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 960 960 6,720 Subcontract 280 280 280 280 280 280 280 1,960 Inventory 30 35 5 0 5 5 80 Backlog 180 180 360 Total 4,440 4,470 4,475 4,625 4,620 4,445 4,445 31,520 8.Period 1 2 3 4 5 6 TotalForecast 160 150 160 180 170 140 960OutputRegular 150 150 150 150 160 160 920Overtime 10 10 0 10 10 10 50Subcontract 0 0 10 10 0 0 20Output- Forecast 0 10 0 –10 0 0InventoryBeginning 0 0 10 10 0 0Ending 0 10 10 0 0 0Average 0 5 10 5 0 0 20Backlog 0 0 0 0 0 0 0Costs:Regular 7,500 7,500 7,500 7,500 8,000 8,000 46,000Overtime 750 750 0 750 750 750 3,750Subcontract 0 0 800 800 0 0 1,600Inventory 20 40 20 80Backlog 0 0 0 0 0 0Total 8,250 8,270 8,340 9,070 9,050 8,750 51,430Chapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4D: 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 Schedule Day Beg. Inv. 1 2 3 4 5 6 7 Quantity100 150 200 TableBeg. Inv. 1 2 3 4 5 6 7 Gross requirements 100 150 200 Scheduled receipts Projected on hand Net requirements 100 150 200 Planned-order receipts 100 150 200 Planned-order releases 100 150 200Wood Sections Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 200300 400 Scheduled receipts 100 Projected on hand 100100 Net requirements 100 300 400 Planned-order receipts 100 300 400 Planned-order releases400 400Braces Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 300 450 600 Scheduled receipts Projected on hand 60 60 60 60 Net requirements 240 450 600 Planned-order receipts 240 450 600Planned-order releases 240 450 600StaplerTopBaseCoveSpri SlideBase Strik RubberSlidSpriLegs Beg.Inv.1 2 3 4 5 6 7Gross requirements 400 600 800Scheduled receiptsProjected on hand 120 120 120 120 88 88 71 Net requirements 280 600 800Planned-order receipts 308 660 880Planned-order releases 968 88010. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. Shifting to an earlier week would result in added carrying costs;shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase due toovertime premium, a probable decrease in productivity, and possible increase inaccidents.Chapter 12 - Inventory Management2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items. DollarItemUnit Cost UsageUsageCategoryK34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr. S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 20015. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+4.851.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $5.00(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal.Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35• •• •495 497 500 5031,0004,000 6,000QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date Days Job (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD: Job time Flow time Due date DaysJob (days) (days) (days) tardyC 7 7 15 0B 10 17 16 1D 6 23 17 6A 14 37 20 1784 24Critical RatioJob Processing Time(Days) Due Date Critical Ratio CalculationA 14 20 (20 – 0) / 14 = 1.43B 10 16 (16 – 0) /10 = 1.60C 7 15 (15 – 0) / 7 = 2.14D 6 17 (17 – 0) / 6 = 2.83Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 14) /10 = 0.20C 7 15 (15 – 14) / 7 = 0.14D 6 17 (17 – 14) / 6 = 0.50Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 21) /10 = –0.50C –––D 6 17 (17 – 21) / 6 = –0.67Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is A –C –D –B and the makespan is 37 days. Critical Ratio sequenceProcessing Time(Days)Flow time Due Date TardinessA 14 14 20 0 C 7 21 15 6 D 6 27 17 10 B1037 16 21 ∑9937b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.86 2.142.272.67c. SPT is superior.9.Time (hr.) Sequence of assignment:Order Step 1 Step 2A 1.20 1.40 .80 [C] last (or 7th)B 0.90 1.30 .90 [B] firstC 2.00 0.80 1.20 [A] 2ndD 1.70 1.50 1.30 [G] 3rdE 1.60 1.80 1.60 [E] 4thF 2.20 1.75 1.50 [D] 6th G1.301.401.75[F]5thThus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题和答案解析_修订版

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第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中,处于核心地位的是:（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（ )。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置,加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（)A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（ )A.服务过程B.生产过程 C。

计划过程 D。

管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程 C。

服务过程 D。

生产过程6、制造业企业与服务业企业最主要的一个区别是（ )A.产出的物理性质B.与顾客的接触程度 C。

产出质量的度量 D。

对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发 B。

产品设计 C.生产运营活动 D。

生产系统的选择8、下列哪项不是生产运作管理的目标（ )A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在( ）A。

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题答案

Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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; 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 Design6. Steps for Making Cash Withdrawal from an ATM1. Insert Card: Magnetic Strip Should be Facing Down2. Watch Screen for Instructions3. Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4. Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5. Deposit/Withdrawal:1) Deposit —place in an envelope (which you’ll fin d near or in theATM) and insert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash 6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==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 = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===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 machineProductA B C 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,00060,00030,000Total 186,000 208,000 122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $111,000B (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 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasks PositionalWeight A 4 23 B 3 20 C 2 18 D 3 25 E 2 18 F 4 29 G 3 24 H 1 14 I5a. First rule: most followers. Second rule: largest positionalweight.Assembly Line Balancing Table (CT = 14)b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d2e 3f 2g 1h3. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 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 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Chapter 06 - Work Design and Measurement3.Element PR OT NT AF job ST1 .90 .46 .414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10 .83 .913 1.15 1.050 41.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05 .56 .588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭ c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractivenessof building H L M M–H3. Nearness to rawmaterials L H L M4. Large amountsof power L H L L5. Pollutioncontrols L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C WeightA B C1.Business Services 9 5 5 2/9 18/9 10/9 10/92.Community Services 7 6 7 1/9 7/9 6/9 7/93.Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94.Construction Costs 5 6 5 2/9 10/9 12/9 10/95.Cost of Living 4 7 8 1/9 4/9 7/9 8/96.Taxes 5 5 5 1/9 5/9 5/9 4/97.Transportation 6 7 8 1/9 6/9 7/9 8/9 Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a . Composite Scores39 44 457 7 7B orC 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 ScoresA B C53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = 5.0 -y =∑y i= 20 = 4.0 n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimallocation.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30ParetoLube & Oil Brakes Tires Battery Trans.2 .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 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 management’s attention.4break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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 too large.7. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL XFor process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T: 33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Inventory Beginning Ending Average Backlog Costs:RegularOvertimeSubcontractInventoryTotalb. Level strategyPeriodForecastOutputRegularOvertimeSubcontractInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotalChapter 11 - MRP and ERP1. a. F: 2G: 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 = 44. Master Schedule10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. Capacity utilization exceeds 100% for both labor and machine in week2, and for machine alone in week 4.Production could be shifted to earlier or later weeks in which capacityis underutilized. Shifting to an earlier week would result in addedcarrying costs; shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase dueto overtime premium, a probable decrease in productivity, and possibleincrease in accidents.Chapter 12 - Inventory Management2.The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, suchas cost of a stockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TC Increase by [$1,428.71 – $1,350] = $78.714. D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 isacceptable.)7. H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 = $181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr.S = $100H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD ==e. Q ' = 258.2D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.TC =S QDH 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be$774.5015.RangePHQD = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,9994.95 1.98 497 NFS = $50 4,000–5,9994.90 1.96 500 NF6,000+ 4.85 1.94 503 NFCompare TC 495 with TC for all lower price breaks:TC 495 = 495 ($2) + 4,900($50) + $5.00(4,900) = $25,4902 495TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units22. d = 30 gal./day ROP = 170 gal.QuantityTCLT = 4 days,ss = Zσd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.313% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal. Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hour CT = 75 min. = 1.25 hr.= 80(1.25)(1.35) = 3C = 45 45X = .354. The smallest daily quantity evenly divisible into all four quantities is3. Therefore, use three cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 –75 = 405. Takt time = 405/300 units per day = 1.35 minutes.Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date DaysJob (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD:completed on day 14. After the completion of Job A, the revised critical ratios are:completed on day 21. After the completion of Job C, the revised criticalJob D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.86 2.14 2.27 2.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

《运营管理》教材作业参考答案

《生产与运作管理》教材习题参考答案第七章：P215 练习题1：1）网络图如下图所示：2）最早开始时间和最迟结束时间，如图中标示所示。

工期为：8+4+3=153）关键路线为：1-3-4-7（红色粗线标示）练习2：解：1)2 关于赶工问题参照教材P202，例7.2。

略第五章：P140答案参见教材：P118，例5.6，略第四章：P95 练习题1：12）车间布置：根据关系分数排序，题目条件或约束，确定好车间7、2、4、1（注意车间1不能位于厂区中间位置，故列于最左下）；车间3与车间2的关系为A ，故安排于其左或右；车间8与车间4关系为A ，可安排于其上或其右；车间3、5、8的安排根据其相邻的所有车间的关系密切程度之和可以计算（略），可知车间5位于车间4上，车间8位于车间4右，车间3位于车间1上。

车间6与车间8的关系为A ，故车间6位于车间8上。

最终结果如图示：第三章：P61练习题1：解：Y=0.5X40+0.35X30+0.20X10=32.5 城市2综合得分高，因此是最佳选择。

练习题3．解：设采用营口冷库，则如下表所示：其运输总费用=210*10+60*12+70*17+10*20+150*11+160*7=6980 由此可知，采用营口冷库更佳。

练习题5：解：首先，建立地图坐标系 9 8 7 65 4 3 2 11 23 4 5 6 7 8 9 10 11 设处理中心（X ，Y ），则利用重心法公式计算如下：∑∑=ii i Q Q x X /)403025926/()408302254942610(++++⨯+⨯+⨯+⨯+⨯==6∑∑=i i i Q Q y Y /)403025926/()40730625791265(++++⨯+⨯+⨯+⨯+⨯==6处理中心坐标为（6，6），位置如图中红色点标示。

第九章：P302 练习题1：解：这是一个存在折扣的库存模型问题。

1）无论订购多少，订购单价为10元时，库存维持费为：H=10*10%=1 ，订货费用S=225 年需求量为：D=3600，则：EOQ （10）=12731225360022=⨯⨯=H DS单位2）当订购量为大于等于500时，库存维持费为：H=9*10%=0.9，订货费用S=225，年需求量为：D=3600，则： EOQ （9）=13429.0225360022=⨯⨯=H DS单位3）当订购量为大于等于1000时，库存维持费为：H=8*10%=0.8，订货费用S=225，年需求量为：D=3600，则： EOQ （8）=14238.0225360022=⨯⨯=H DS单位以上三种经济订货量单位都大于1000的订量单位，显然都是可行的订货量。

《运营管理》课程习题及答案修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

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