会议运营与管理课后答案
会议运行管理测试题2
会议运行管理（专，2019春）题目1正确获得5.00分中的5.00分标记题目题干如果某个项目的成本发生偏差，虽然在预算范围但也需要改进。
选择一项：对错反馈正确的答案是“错”。
题目2正确获得5.00分中的5.00分标记题目题干举办商务会议之前，由主办方组成会议策划委员会来作出会议预算计划。
选择一项：对错反馈正确的答案是“错”。
题目3正确获得5.00分中的5.00分标记题目题干收人预算是收入预测的一种特定类型，是规划未来会议收人额的预算。
选择一项：对错反馈正确的答案是“对”。
题目4正确获得5.00分中的5.00分标记题目题干会议的成本控制监视各项活动以保证它们按计划进行并纠正各种重要偏差的过程。
选择一项：对错反馈正确的答案是“对”。
题目5正确获得5.00分中的5.00分标记题目题干固定费用预算随着会议人数而变动，当实际收人少于预期收人时，固定费用随之变动选择一项：对错反馈正确的答案是“错”。
题目6正确获得5.00分中的5.00分标记题目题干在筹备商务会议的过程中，用货币形式表示会议的各项活动是必不可少的一个环节，掌握了会议预算就掌握了整个会议。
选择一项：对错反馈正确的答案是“对”。
题目7正确获得5.00分中的5.00分标记题目题干在收人既定的情况下，较低的费用意味着较高的收益率。
当竞争激烈或收人来源紧张时，首先削减费用预算的做法是比较得当的。
选择一项：对错反馈正确的答案是“对”。
题目8正确获得5.00分中的5.00分标记题目题干预算既可以自下而上制定，也可以自上而下制定。
选择一项：对错反馈正确的答案是“对”。
题目9正确获得5.00分中的5.00分标记题目题干会议的收人主要由与会人员或公司的交费以及广告费、赞助费等构成。
选择一项：对错反馈正确的答案是“对”。
题目10正确获得5.00分中的5.00分标记题目题干商务会议的策划委员会通常是从主办公司的各个部门抽调人员组成，需要为其支付一定报酬，委员会的支出由他们花费的差旅费、餐饮费和一些设备材料费用组成。
酒店运营与管理范文及答案
酒店运营与管理范文及答案一、引言随着旅游业的快速发展，酒店行业作为旅游服务的重要组成部分，也迅速发展壮大。
然而，酒店运营与管理面临着诸多挑战和机遇。
本文将重点探讨酒店运营和管理的相关范文及答案，帮助读者更好地理解和应对酒店行业的变化。
二、酒店运营范文1. 酒店运营管理的重要性酒店运营管理是酒店的核心业务，涉及到酒店的各个方面，包括客房管理、餐饮服务、市场推广、人员管理等。
良好的酒店运营管理可以提高酒店的竞争力，增加利润，提升客户满意度，打造品牌效应。
因此，酒店运营管理的重要性不言而喻。
2. 酒店运营管理的主要内容酒店运营管理的主要内容包括： - 客户关系管理：建立良好的客户关系，提供个性化的服务，增加客户忠诚度。
- 营销策划与推广：制定有效的营销策略，利用各种营销手段提高酒店的知名度和市场份额。
- 服务质量管理：培训员工，提升服务质量，确保客户满意度。
- 成本控制与利润管理：合理控制成本，提高利润，实现经济效益最大化。
- 设备设施维护和更新：定期维护和更新酒店的设备设施，保持酒店的良好状态。
3. 酒店运营管理的挑战与应对酒店运营管理面临着一系列的挑战，包括： - 日益激烈的竞争：酒店行业竞争激烈，酒店需要找到差异化竞争的策略，提供独特的服务和体验。
- 人员管理：酒店需要建立一支高素质、专业化的员工队伍，提供优质的人员服务。
- 技术创新：随着科技的发展，酒店需要跟上时代的步伐，引入新技术，提升酒店的服务质量和效率。
三、酒店管理范文1. 酒店管理的层级结构酒店管理的层级结构通常包括总经理、副总经理、部门经理等，每个层级都有相应的权责和职责。
总经理负责全面领导和管理酒店，副总经理协助总经理履行职责，部门经理负责具体的部门管理工作，如客房部经理、餐饮部经理等。
2. 酒店管理的主要职责酒店管理的主要职责包括： - 制定酒店的发展战略和目标，规划酒店的发展方向。
- 确定酒店的管理体系和工作流程，确保各项工作有序进行。
会议运营管理选择题题库
会议运营管理选择题题库1. 会议运营管理概述1.1 会议运营管理是指什么？ - A. 策划和组织会议的过程 - B. 议程和日程的安排- C. 会议参与人员的招募工作 - D. 会议场地的预订和布置1.2 以下哪项不是会议运营管理的重要内容？ - A. 参会人员的注册和签到 - B. 演讲嘉宾的邀请与安排 - C. 会议主题的确定与发布 - D. 酒店住宿及交通安排1.3 有效的会议运营管理能够带来哪些好处？ - A. 提高会议参与度和满意度 - B. 降低会议成本 - C. 加强会议的专业性 - D. 优化会务人员的工作效率2. 会议策划与组织2.1 会议策划的关键步骤包括哪些？ - A. 制定目标和目的 - B. 确定参会人员 - C. 策划议程安排 - D. 确定预算和资源2.2 在制定会议目标时，应考虑哪些因素？ - A. 主题和目标受众 - B. 参会人员的规模和背景 - C. 会议的时间和地点 - D. 赞助商和合作伙伴的需求2.3 会议组织的关键工作包括哪些？ - A. 寻找合适的会议场地 - B. 确定会议宣传策略 - C. 邀请演讲嘉宾和专家 - D. 安排会议的技术支持和设备3. 会议日程与议程安排3.1 会议日程的编排原则是什么？ - A. 从重要到次要依次安排 - B. 根据参会人员的意见和需求安排 - C. 尽量减少不同议题之间的时间间隔 - D. 关注会议的时长和参会人员的疲劳程度3.2 以下哪项是设置议程时应注意的事项？ - A. 考虑各议题的时长和顺序 - B. 确定每个议题的发言人和主持人 - C. 留出适当的时间进行讨论和提问 - D. 给与会人员提供足够的休息时间3.3 在编写会议议程时，应注意哪些方面？ - A. 突出会议主题与目标 - B. 分配时间合理并充分考虑到细节 - C. 设定明确的议题和发言顺序 - D. 考虑调整议程以适应实际情况4. 参会人员管理4.1 参会人员注册的目的是什么？ - A. 统计参会人数和背景信息 - B. 安排会议的餐饮和住宿需求 - C. 确定参会人员的签到流程 - D. 主办方与参会人员的沟通和联系4.2 参会人员签到的作用是什么？ - A. 确认参会人员的到场情况 - B. 发放会议资料和礼品 - C. 确保会场的安全和秩序 - D. 进行参会人员信息的核实4.3 以下哪项属于参会人员管理的工作内容？ - A. 发送参会邀请函和会议通知 -B. 配发参会人员的胸卡和会议资料 -C. 安排参会人员的餐饮和住宿 -D. 指导参会人员进行交流和合作5. 会议评估与改进5.1 会议评估的目的是什么？ - A. 确认会议的成功与否 - B. 判定会议运营管理的效果 - C. 收集参会人员的反馈和意见 - D. 提供数据支持以便改进下一次会议5.2 在会议评估过程中，应收集哪些方面的数据和反馈？ - A. 参会人员的满意度和参与度 - B. 各项议程和活动的效果和反响 - C. 演讲嘉宾和主持人的表现 - D. 会议的成本和收益情况5.3 改进会议的方式和方法有哪些？ - A. 根据参会人员的意见和建议调整议程 -B. 加强宣传和推广以吸引更多参与者 -C. 定期组织培训和交流会议运营管理经验 -D. 引入新的技术和工具提高会议效率以上是关于会议运营管理的选择题题库。
2016年 会议运营管理复习资料
会议运营管理复习资料单项选择题：1*18=18分填空题：1*10=10分名词解释：3*4=12分简答题：6*5=30分论述题：15*2=30分一、选择题1、congress：指普通大会、政治性的代表大会或专业人员代表大会，Panel 专题研讨会。
要求有两位甚至更多的发言人讲述其观点，发言人和与会者一起进行充分的讨论,convention：大型会议。
conference：专业性会议，forum：论坛,s ession：一届会议。
2、注册台：设置在大厅，用于与会者进行注册（报到）和缴费。
存储区：存放会议用品、会议资料等3、会议室的布置：中空圆（正方，多边）形：规格较高、与的身份都重要的讨论会。
中空椭圆（长方）形：但体现出主次之分。
宴会形：用于与酒会、饮食结合在一起的会议。
4、会议专业活动：大会报告、分组会报告、张贴报告、专业参观5、话筒：主持人使用台式固定话筒，演讲人使用小型无线话筒，提问话筒采用无线话筒和立式固定话筒。
视听设备：6、最终程序和会议指南：会议日程安排、会议内容摘要、当地交通安排表、参加会议者名单（住房安排表）7、会议名卡：胸卡、挂卡、IC卡8、会议午餐：统一安排，自助餐和非酒精饮料9、附设展览对会议主办者的好处：增加会议收入；对与会者的好处：看到新产品，了解专业、行业的产品发展情况；对参展商的好处：从而获得的商机。
10、PCO的收费方式：包价式，基本费+参会者人头费，参会者人头费，按小时收费11、会议局可以是以政府机构，也可以是一个非官方机构或半官方机构。
会议局的特点：非营利性、中立性、以会议客户为中心。
12、国际社团会议审批时限：省部级，提前8-12个月申请报批；国务院，提前12个月13、会议开幕式安排：在会议正式开始前一天的晚上，也可以安排在会议正式开始的第一天上午；地点取决于会议代表的人数14、欢迎招待会（宴会）的安排：正式前一天晚上或正式开始的第一天晚上，时间可以稍晚一点。
欢送（闭幕）宴会的安排：正式结束的前一天晚上或会议结束的当天晚上15、茶歇：每天安排2次16、餐饮费是会议可变支出中比例最大的一笔费用17、注册费是会议最主要的收入，是决定会议能否做到收支平衡的绝对因素。
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.。
会议组织与活动管理课后测试
会议组织与活动管理课后测试第一部分：会议组织会议组织是一项重要的管理技能，对于确保会议的顺利进行至关重要。
以下是一些有关会议组织的问题。
1.请解释会议组织的目标和目的。
为什么会议组织对于组织的成功非常重要？2.在会议筹备过程中，你将会面临什么挑战？请列举一些常见的挑战，并提供如何应对的建议。
3.请描述会议议程的重要性。
议程应包含哪些内容？如何制定一个具有吸引力的议程？4.解释什么是会议纪要？为什么它对于会议组织和参与者都很重要？请提供一个简要的会议纪要模板。
第二部分：活动管理活动管理是一项复杂的任务，要求组织者具备良好的计划和执行能力。
以下是一些有关活动管理的问题。
1.请解释活动管理的重要性。
为什么活动管理是一个关键的组织任务？2.列举一些活动管理过程中可能遇到的挑战。
如何应对这些挑战并确保活动的成功？3.解释什么是活动预算？为什么预算对于活动管理非常重要？请提供一个简单的活动预算表格模板。
4.请阐述活动营销的重要性。
活动营销策略中应包含哪些关键元素？我们可以使用哪些渠道来推广活动？第三部分：附加题1.请列举一些促进有效沟通的技巧，特别是在组织会议和管理活动时。
2.在会议组织和活动管理中，团队合作是至关重要的。
请描述一些可以提高团队合作的策略和方法。
3.列举一些评估会议成功的指标。
我们如何衡量会议的成功程度？4.请分享一次您组织或参与的成功会议或活动的经验，并从中总结出一些教训和启示。
以上是关于会议组织与活动管理的测试题。
请准备好您的答卷，并在规定的时间内完成测试。
祝您成功！。
《运营管理》课后习题答案
《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 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 Measurement 3.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 school 1. 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 0Chapter 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 strategyPeriodForecastOutputRegularOvertimeSubcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs:Regular Overtime Subcontract Inventory BacklogTotal8.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 pH DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I m ax ==-=Average is92.154248.309:2I m ax =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) = $25,4902 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 units22. 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 Operations 1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35) = 3C = 45 45X = .35QuantityTC。
会议运营与管理判断题
会议运营与管理判断题1.会议运营和管理的目标是什么？会议运营和管理的目标是确保会议的顺利进行，达到预期效果。
这包括确定会议的目的和议程、选择合适的会场和设备、组织和协调与会人员、确保会议的顺利进行、及时处理突发情况等。
2.会议运营和管理可以忽略会议的目的和议程，只需要提供好会场和设备即可。
错误。
会议的目的和议程是会议的核心，它们决定了会议的内容和方向。
会议运营和管理不仅需要提供好会场和设备，还需要对会议的目的和议程进行合理规划和组织，确保会议的顺利进行。
3.在会议运营和管理中，会场选择是可以忽略的，只需要确保设备的正常运作即可。
错误。
会场选择是很重要的一环，合适的会场可以提供舒适的会议环境，有利于与会人员的参与和交流。
会场的大小、布局、设施设备等都需要考虑，以满足会议的需求。
4.与会人员的组织和协调不是会议运营和管理的职责，可以由与会人员自行协调。
错误。
与会人员的组织和协调是会议运营和管理的重要职责之一。
这包括确定与会人员名单、发送邀请函、跟进与会人员的参会意愿、协调与会人员的行程等。
确保与会人员的顺利参会是会议运营和管理的重要任务之一。
5.在会议运营和管理中，突发情况是不可预测的，无需事先做好应急准备。
错误。
尽管突发情况无法预测，但会议运营和管理需要在事前做好应急准备，以应对可能出现的问题。
这包括制定应急预案、分配应急责任人、准备必要的应急设备等，以保障会议的顺利进行。
6.会议运营和管理不需要评估会议的效果和总结经验教训，只需要负责会议的现场运营。
错误。
会议运营和管理的工作不仅仅是现场运营，还需要对会议的效果进行评估和总结经验教训。
通过评估会议效果，可以了解到会议的优点和改进的空间，为将来的会议提供参考。
同时，总结经验教训可以帮助提升会议运营和管理的水平。
7.会议运营和管理只关注会议的运行和管理，不需要考虑与会人员的需求和体验。
错误。
与会人员的需求和体验是会议运营和管理的重要考虑因素之一。
会议运营和管理需要关注与会人员的参与度和满意度，提供良好的服务和体验，以增强会议的效果和影响力。
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会议运营与管理课后答案
一、会议运营的重要性
会议是组织内外部沟通和协作的重要方式，对于企业的发展和运营具有重要的
意义。
合理高效的会议运营能够提升沟通效率，促进团队合作，加强信息共享和交流，进而提升组织的综合竞争力。
1.沟通有效性：会议是团队成员集体讨论和决策的场所，能够促进信
息的传递和共享，提高沟通的质量和效率，防止信息的断层和错漏。
2.决策效率：会议能够汇集各方意见和专业能力，协调利益关系，加
快决策的速度，提高决策的质量，保证决策的科学性和合理性。
3.团队协作：会议是团队成员交流和协作的重要场所，通过会议可以
增进团队成员之间的理解和信任，促进团队成员的合作和协作，提高团队的凝聚力和战斗力。
4.信息共享：会议是组织内外信息共享的重要渠道，可以及时传递重
要信息，促进组织内外部的交流和合作，提高组织运转效率。
综上所述，会议运营对于组织的发展和运营至关重要，必须高度重视和合理运营。
二、会议运营的注意事项
1.确定会议目标和议程：会议前应明确会议的目标和议程，确保会议
的目的明确，议程合理，以提高会议的效率和结果。
2.邀请合适的参会人员：会议参与者的选择应根据会议的内容和目标
来确定，确保参会人员的能力和权威，提高会议的决策质量和可行性。
3.合理安排会议时间和地点：会议时间和地点的选择应适当考虑参会
人员的日程安排和交通便利性，避免对参会人员产生不必要的困扰和阻碍。
4.提前准备会议材料：主持人和参会人员应提前准备会议所需的材料，
包括议程、资料、报告等，以便参会人员充分了解会议背景和内容，提高会议讨论的效果和品质。
5.会议纪要和行动计划：会议结束后应及时整理会议纪要和行动计划，
明确会议的决策结果和具体行动，以方便参会人员和相关人员的跟进和执行。
6.评估会议效果：会议结束后应对会议效果进行评估，了解会议的优
点和不足，及时改进会议的运营方式和组织方式，以提高会议的效率和成效。
三、会议运营的常见问题及解决方法
1.会议无效率：会议过长、议程不清晰、参会人员不专注等原因导致
会议效率低下。
解决方法是在会议前制定清晰的议程，控制会议时间，提醒参会人员专注并及时纠正偏离议题的讨论。
2.会议决策不达预期：会议决策往往需要综合各方意见和专业能力，
但可能出现意见冲突、权威缺乏等问题。
解决方法是提前与参会人员沟通，整合意见，重视专业能力，由决策权威进行决策。
3.会议交流不畅：会议中可能存在语言障碍、参会人员不愿发言等问
题，导致交流不畅。
解决方法是提前预留时间给参会人员准备发言，营造良好的交流氛围，重视沟通技巧培训。
4.会议纪要不准确：会议纪要是记录会议讨论和决策的重要文件，如
有缺失或不准确会影响后续工作。
解决方法是由专人负责记录会议讨论内容，并及时与参会人员核对纪要的准确性。
5.会议执行不到位：会议的决策与具体实施可能存在偏差或延迟，影
响会议运营效果。
解决方法是在会议结束后及时制定行动计划，并明确责任人和时间节点，定期跟进执行情况。
四、会议运营的优化建议
1.合理利用技术手段：借助远程会议、在线会议等现代技术手段，提
高会议的开展效率和参与度。
2.注重参会人员的反馈与建议：定期邀请参会人员对会议的运营方式
和效果进行评价，并据此进行改进。
3.持续提升主持人和参会人员的沟通能力：加强主持人和参会人员的
相关培训，提高沟通和协作效果。
4.定期组织会议运营经验分享：通过组织经验分享会、案例学习等方
式，促进会议运营经验的交流和共享。
综上所述，会议运营与管理对于组织的发展和运营具有重要的意义。
通过合理运营、有效管理会议，能够提高沟通效率、决策效率，促进团队协作，加强信息共享和交流，进而提升组织的综合竞争力。