最新《运营管理》课后习题答案
《运营管理》课程习题和答案解析_修订版【范本模板】
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。
《运营管理》课程习题及答案-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 envelop e (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 = 3,466.67 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 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 = 1.3 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: 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 1.05041.00 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 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:}{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 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 = 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. 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 [$1,428.71 – $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 200 days/yr. = 4,000 tons/yr.15. 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+ 4.85 1.94503 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 = .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 = 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 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 2.142.272.67c. SPT is superior.9.Thus, 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。
智慧树运营管理课后答案
智慧树运营管理课后答案1. 简介本文是智慧树课程《运营管理》的课后答案。
运营管理是管理学的重要分支，旨在帮助企业有效组织、规划和控制生产过程，以提高生产效率并提供高质量的产品和服务。
2. 课后答案以下是智慧树课程《运营管理》中各章节的课后答案：第一章：运营管理导论1.运营管理是指企业对生产系统和分销系统进行规划、组织、实施和控制的管理活动。
2.运营管理的目标是提高生产效率、提供高质量的产品和服务、降低成本并满足顾客需求。
3.运营管理的主要职能包括生产计划和控制、库存管理、供应链管理、质量管理等。
第二章：产品和服务设计1.产品设计是指确定产品的功能、特性和性能，并为其提供合适的外观和包装。
2.服务设计是指确定服务的特性、流程和交付方式，并为其提供合适的服务环境和设施。
3.产品和服务设计的目标是满足顾客需求、提高产品价值和降低生产成本。
第三章：过程设计和分析1.过程设计是指确定产品或服务的生产流程和操作步骤。
2.过程分析是指研究和改进生产过程，以提高生产效率和质量。
3.过程设计和分析的工具和技术包括流程图、价值流图、工艺分析和工作测量等。
第四章：物料管理和供应链管理1.物料管理是指对物料的采购、储存和使用进行有效管理。
2.供应链管理是指对供应商、生产商和分销商之间的物流、信息流和资金流进行协调和管理。
3.物料管理和供应链管理的目标是降低成本、提高响应速度和提供优质的产品和服务。
第五章：质量管理和绩效评价1.质量管理是指通过一系列控制和改进措施，确保产品和服务的质量符合顾客的要求和期望。
2.绩效评价是指对企业运营绩效进行量化和评估，以判断是否达到预期目标。
3.质量管理和绩效评价的工具和技术包括质量控制图、质量成本、平衡计分卡等。
3. 总结本文总结了智慧树课程《运营管理》中各章节的课后答案。
通过学习运营管理，我们可以了解企业如何通过有效组织、规划和控制生产过程，提高生产效率、降低成本并提供高质量的产品和服务。
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.。
运营管理答案(最新版本)
运营管理（11.04）第一章绪论一、单选、填空1、运营是企业（创造价值）活动的主要环节。
2、企业之间的竞争最终体现在（运营过程的结果上）3、有形产品的变换过程通常也称为（生产过程）P24、无形产品的变换过程有时称为（服务过程）5、企业运营管理有两大对象：（运营过程）和（运营系统）。
P56、运营管理的目标可以用一句话来概括：“在需要的时候，以（适宜的价格），向顾客提供具有适当质量的（产品和服务）。
”。
P9二、多项选择题1、服务运营管理的特殊性体现在（）ACDE P6-8A.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、资源要素的管理包括的内容主要有（）BCD P10-11A.环境要素管理B.信息管理C.人员管理D.物料管理E.成本管理3、运营管理中的决策内容包括（）ABD P13A.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.营销决策三、名词解释1、运营活动：是一个“投入→变换→产出”的过程，即投入一定的资源，经过一系列、多种形式的变换，使其价值增值，最后以某种形式的产出提供给社会的过程。
2、运营系统：是指使“投入→变换→产出”的运营过程得以实现的手段的总称。
四、简答题1、运营管理的目标和基本问题是什么？P9-11（1）目标：在需要的时候，以适宜的价格，向顾客提供具有适当质量的产品和服务。
基本问题：①产出要素管理；②资源要素管理；③环境要素管理。
五、论述题1、试述运营管理的基本问题。
P9-11（1）产出要素管理：（1）质量；（2）时间；（3）成本；（4）服务。
（2）资源要素管理：（1）设施设备管理；（2）物料管理；（3）人员管理；（4）信息管理。
（3）环境要素管理：（1）从“产出”的角度来说，企业有必要在产品设计和运营过程中考虑如何保护环境；（2）从“投入”的角度来说，在资源获取和利用上尽量节约、合理使用自然资源，并考虑各种资源的再生利用问题。
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4,320
2,700
Total Cost
9,900
MFP (2) (6)
3.03
2
33,600
3,360
5,040
2,820
10
5,040
2,760
11,160
2.89
4
35,400
3,840
5,760
2,880
12,480
2.84
*refer to solved problem #2
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Total processing time by machine
Product 1
A 48,000
B 64,000
C 32,000
2
48,000
48,000
36,000
3
30,000
36,000
24,000
4
60,000
60,000
30,000
Total
186,000 208,000 122,000
186,000
NA
1.24 2 machine
150,000
208, 000
NB
1.38 2 machine 150, 000
122,000 NC 150,000 .81 1machine
You would have to buy two
“ A” machines at a total cost of $80,000, or two
8.
Technical Requirements Customer Requirements
Ingredients
Taste
√
Appearance
√
Texture/consistency
√
Handling √
Preparation
√ √ √
Chapter 04 - Strategic Capacity Planning for Products and Services
Utilizatio n Actual output Design capacity
Actual output 8
Design Capacity
20 jobs
Effective capacity .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.
2) Select a Transaction and Account
3) Enter Amount of Transaction
5. Deposit/Withdrawal:
1) Deposit — place in an envelope (which you insert it into the deposit slot
b. Before: ($10 x 5 = $50) + $40 = $90; hence 80 $90 = .89÷carts/$1.
After: ($10 x 4 = $40) + $50 = $90; hence 84 $90 = .9÷3 carts/$1.
c. Labor productivity increased by 31.25%
“ B” machines at a total cost of $60,000, or one
Productivity increased by -16%
((1.68-2)/2)
Chapter 03 - Product and Service Design
6.
Steps for Making Cash Withdrawal from an ATM
1. Insert Card: Magnetic Strip Should be Facing Down
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Chapter 02 - Competitiveness, Strategy, and Productivity
3.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Week 1
Output 30,000
Worker Cost@ $12x40
2,880
Overhead Material Cost @1.5 Cost@$6
’ ll find near or in the ATM) and
2) Withdrawal — lift the “ Withdrawal Door, ” being careful to remove all cash
6. Remove card and receipt (which serves as the transaction record)
((21-16)/16).
Multifactor productivity increased by 4.5% ((.93-.89)/.89).
*Machine Productivity
Before: 80 ÷ 40 = 2 carts/$1. After: 84 ÷ 50 = 1.68 carts/$1.
Multifactor 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.
2. Watch Screen for Instructions
3. Select Transaction Options:
1) Deposit
2) Withdrawal
3) Transfer
4) Other
4. Enter Information:
1) PIN Number
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Actual output
2.
Efficiency
80%
Effective capacity
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