第九章网络计划技术(MBA)

1、Chapter 12 PERT/CPM Models for Project Management1 Project Networks 计划网络图计划网络图2 Scheduling a Project with PERT/CPM 网络时间与关键路线网络时间与关键路线3 Dealing with Uncertain Activity Durations 4 网络优化网络优化1项目管理项目管理 Project Managementn项目特征Characteristics of ProjectsqUnique, one-time operationsqInvolve a large number o

2、f activities that must be planned and coordinatedqLong time-horizonqGoals of meeting completion deadlines and budgetsnExamplesqBuilding a houseqPlanning a meetingqIntroducing a new productnPERTProject Evaluation and Review Technique计划评审技术CPMCritical Path Method 关键路线法qA graphical or network approach

3、for planning and coordinating large-scale projects.可信建筑公司项目可信建筑公司项目 Reliable Construction Company Project P285nThe Reliable Construction Company has just made the winning bid of $5.4 million to construct a new plant for a major manufacturer.可信建筑公司中标540万美元的项目nThe contract includes the following provi

4、sions:qA penalty of $300,000 if Reliable has not completed construction within 47 weeks. 如果不能在47周内完成项目，将被罚款30万美元。q如果能在40周内完成，将获得15万美元的 奖励A bonus of $150,000 if Reliable has completed the plant within 40 weeks.Questions: 286How can the project be displayed graphically to better visualize the activiti

5、es?如何用图来表示项目中活动流程What is the total time required to complete the project if no delays occur?When do the individual activities need to start and finish?What are the critical bottleneck activities?对于非关键的活动For other activities, how much delay can be tolerated承受?What is the probability the project can b

6、e completed in 47 weeks?What is the least expensive way to complete the project within 40 weeks?1.How should ongoing costs be monitored to try to keep the project within budget?如何对成本进行实时监控，使项目成本控制在预算内活动明细表活动明细表 Activity List for Reliable ConstructionActivityActivity Description紧前紧前ImmediatePredecess

7、ors估计工期估计工期 (Weeks)AExcavate挖掘2BLay the foundation打地基A4CPut up the rough wall承重墙施工B10DPut up the roof 封顶C6EInstall the exterior plumbing安装外部管道C4FInstall the interior plumbing安装内部管道E5GPut up the exterior siding外墙施工D7HDo the exterior painting外部粉刷E, G9IDo the electrical work电路铺设C7JPut up the wallboard竖

8、墙板F, I8KInstall the flooring铺地板J4LDo the interior painting内部粉刷J5MInstall the exterior fixtures设施H2NInstall the interior fixtures 安装内部设备K, L6项目网络项目网络 Project Networksn表示整个项目的网络称为项目网络A network used to represent a project is called a project network.nA project network consists of a number of 节点nodes an

9、d a number of arcs.nTwo types of project networks:qActivity-on-arc (AOA弧表示活动): each activity is represented by an arc. A node is used to separate an activity from its predecessors. The sequencing of the arcs shows the precedence relationships.qActivity-on-node (AON节点表示活动): each activity is represent

10、ed by a node. The arcs are used to show the precedence 优先relationships.nAdvantages of AON (used in this textbook): P288qconsiderably easier to construct 易于构建qeasier to understand 易于理解qeasier to revise when there are changes更便于修改Reliable Construction Project Network ASTARTGHMFJKLNActivity Code A. Exc

11、avate B. Foundation C. Rough wall D. Roof E. Exterior plumbing F. Interior plumbing G. Exterior siding H. Exterior painting I. Electrical work J. Wallboard K. Flooring L. Interior painting M. Exterior fixtures N. Interior fixtures24107467958456200FINISHDIECBMicrosoft Project Gantt Chart 甘特图W1/1表示第一周

12、第一天； W2/5 表示第二周第五天；Microsoft ProjectProject Network活动名称编号工期最早开始时间最早结束时间P290W1/1第一周第一天开始W2/5 第二周第五天完成二周AOA Project Network1235ABCD6GEI4871011139H12FJKLNM 关于关于AOA计划网络图计划网络图 统筹方法的第一步工作就是绘制计划网络图，也就是将工序（或称为统筹方法的第一步工作就是绘制计划网络图，也就是将工序（或称为活动）进度表转换为统筹方法的网络图。下面用一个例子来说明活动）进度表转换为统筹方法的网络图。下面用一个例子来说明 例例3、某公司研制新产品

13、的部分工序与所需时间以及它们之间的相互、某公司研制新产品的部分工序与所需时间以及它们之间的相互关系都显示在其工序进度表如表关系都显示在其工序进度表如表12-8所示，请画出其统筹方法网络图所示，请画出其统筹方法网络图。 工序代号工序代号工序内容工序内容所需时间（天所需时间（天)紧前工序紧前工序abcde产品设计与工艺设产品设计与工艺设计计外购配套零件外购配套零件外购生产原料外购生产原料自制主件自制主件主配可靠性试验主配可靠性试验601513388-aacb,d2表表12-8 网络图中的网络图中的节点节点表示一个事件，是一个或若干个工序的开始或结束，是相邻工序在时间表示一个事件，是一个或若干个工序

14、的开始或结束，是相邻工序在时间上的分界点，点圆圈表示，数字表示点的编号。上的分界点，点圆圈表示，数字表示点的编号。弧弧表示一个工序（或活动），弧的表示一个工序（或活动），弧的方向是从工序开始指向工序的结束，弧上是各工序的代号，下面标以完成此工序所方向是从工序开始指向工序的结束，弧上是各工序的代号，下面标以完成此工序所需的时间（或资源）等数据，即为对此弧所赋的权数需的时间（或资源）等数据，即为对此弧所赋的权数虚工序：虚工序：只表示相邻工序的只表示相邻工序的前后逻辑关系。需要人力、物力等资源与时间。前后逻辑关系。需要人力、物力等资源与时间。画网络图的规则：画网络图的规则：两节点之间只能有一条弧（箭

15、线）。两事件一工序两节点之间只能有一条弧（箭线）。两事件一工序网络图只能有一个开始节点（事项）和一个终点事项网络图只能有一个开始节点（事项）和一个终点事项不允许有回路，不能有缺口不允许有回路，不能有缺口 abcb, c的紧前工序为aabcdc , d的紧前工序为a, babcdc 的紧前工序为a。d的紧前工序为a和b3绘制网络图分绘制网络图分3步步：(1) 任务分解任务分解 WBS Work breakdown Structure 工作项目明细工作项目明细表：工序名称表：工序名称 ，代号代号；前后逻辑关系，消耗资源。前后逻辑关系，消耗资源。(2)按表绘制网络图。（按表绘制网络图。（3）节点统一

16、编号。采用平行和交叉作业技术节点统一编号。采用平行和交叉作业技术 例、把例的工序进度表做一些扩充，如表例、把例的工序进度表做一些扩充，如表12-9，请画出其统筹方法的网络图。，请画出其统筹方法的网络图。 表表12-9工序代号工序代号所需时间（天）所需时间（天）紧前工序紧前工序工序代号工序代号所需时间（天）所需时间（天）紧前工序紧前工序abcd60151338aacefgh810165b，dde，abcde601383815图图12-44 解：我们把工序扩充到图解：我们把工序扩充到图12-412-4发生了问题，由于的紧前工序是，发生了问题，由于的紧前工序是，故的结束应该是的开始，所以代表的弧的起

17、点应该是故的结束应该是的开始，所以代表的弧的起点应该是，由，由于工序的结束也是于工序的结束也是，所以工序也成了工序的紧前工序，与，所以工序也成了工序的紧前工序，与题意不符。题意不符。为此我们利用虚工序。由于为此我们利用虚工序。由于e e的紧前工序是的紧前工序是b b和，在和，在和和之间加入之间加入虚工序虚工序 8e15243a60b151013dc38f图图12-55 在网络图上添加、工序得网络图在网络图上添加、工序得网络图12-612-6。 在统筹方法的网络图中不允许两个点之间多于一条弧，因此增加了一个点在统筹方法的网络图中不允许两个点之间多于一条弧，因此增加了一个点和虚工序如图和虚工序如图

18、12-7。615243a60b158e1013dc38fg16h515243a60b158e1013dc38fg16h57681ActivityTime (Weeks)ImmediatePredecessorsActivityTime (Weeks)ImmediatePredecessorsABCDE241064ABCCHIJKL97845E, GCF, IJJF5EM2HG7DN6K, L2A23B4C4105D67EI46G87F7510J89H9KLM542111312N6关键路线The Critical PathnA path through a network is one of t

19、he routes following the arrows (arcs) from the start node to the finish node.路是指沿着箭头从开始点路是指沿着箭头从开始点到终点的一条路线到终点的一条路线nThe length of a path is the sum of the (estimated) durations of the activities on the path.路长度为路上所有活动时间总和路长度为路上所有活动时间总和nThe (estimated) project duration equals the length of the longes

20、t path through the project network.项目工期等于项目网络中最长路的长项目工期等于项目网络中最长路的长度度nThis longest path is called the critical path. (If more than one path tie for the longest, they all are critical paths.)项目网络中最长路称为关项目网络中最长路称为关键路键路The Paths for Reliables Project NetworkPathLength (Weeks)StartA B C D G H M Finish2

21、+ 4 + 10 + 6 + 7 + 9 + 2 = 40Start A B C E H M Finish2 + 4 + 10 + 4 + 9 + 2 = 31Start A B C E F J K N Finish2 + 4 + 10 + 4 + 5 + 8 + 4 + 6 = 43Start A B C E F J L N Finish2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 = 44Start A B C I J K N Finish2 + 4 + 10 + 7 + 8 + 4 + 6 = 41Start A B C I J L N Finish2 + 4 + 10

22、+ 7 + 8 + 5 + 6 = 42关键路线关键路线critical pathP292最早开始时间和最早结束时间nThe starting and finishing times of each activity if no delays occur anywhere in the project are called the earliest start time and the earliest finish time.n用用ES表示一个活动最早开始时间；表示一个活动最早开始时间；EF表示一个活动最早结束时间。表示一个活动最早结束时间。最早可能开始时间和结束时间。最早可能开始时间和结束

23、时间。 nEarliest Start Time Rule: ES = Largest EF of the immediate predecessors. 最早开始时间最早开始时间=所有紧前工序所有紧前工序EF的最大者的最大者ABCiES=Max20,23,19=23EF=20EF=23EF=19HCBAHEF=20EF=23EF=19ES=23n计算所有活动的计算所有活动的ES时间时间:For each activity that starts the project (including the start node), set its ES = 0.项目的最早开始时间为项目的最早开始时间

24、为0 0最早结束时间EF = ES + 活动工期For each activity whose ES has just been obtained, calculate EF = ES + duration对于每个新活动，用最早开始时间法则求ES For each new activity whose immediate predecessors now have EF values, obtain its ES by applying the earliest start time rule. Apply step 2 to calculate EF.1.重复上述3个步骤Repeat ste

25、p 3 until ES and EF have been obtained for all activities.最早结束时间EF = 最早开始时间ES + 活动工期从开始(左)向后计算ES只有单个只有单个紧前工序的工序紧前工序的工序ES与与EF的计算的计算 ASTA RTGHMFJF IN IS HKLNDIECB241074679584562ES = 0 EF = 2ES = 2 EF = 6ES = 16 EF = 22ES = 16 EF = 20ES = 16 EF = 23ES = 20 EF = 25ES = 22 EF = 29ES = 6 EF = 1600ES and E

26、F Times for Reliable Construction ASTARTGHMFJFINISHKLNDIECB241074679584562ES = 0 EF = 2ES = 2 EF = 6ES = 16 EF = 22ES = 16 EF = 20ES = 16 EF = 23ES = 20 EF = 25ES = 22 EF = 29ES = 6 EF = 16ES = 0 EF = 0ES = 25 EF = 33ES = 33 EF = 38ES = 38 EF = 44ES = 33 EF = 37ES = 29 EF = 38ES = 38 EF = 40ES = 44

27、EF = 4400多个紧前工序的多个紧前工序的ES与与EF的计算的计算ES=Max25,23=25ES=Max29,20=29最迟开始时间和最迟结束时间最迟开始时间和最迟结束时间nThe latest start time for an activity is the latest possible time that it can start without delaying the completion of the project (so the finish node still is reached at its earliest finish time).在不影响项目完成总时间的前

28、在不影响项目完成总时间的前提下，一个活动提下，一个活动最晚必须开始的时间最晚必须开始的时间。对于终点，最晚时间仍然是最对于终点，最晚时间仍然是最早结束时间早结束时间, LF=EFqLS = Latest start time for a particular activity 最晚开始时间qLF = Latest finish time for a particular activity 最晚结束时间nLatest Finish Time Rule: LF = Smallest LS of the immediate successors.最晚时间等于其所有紧后活动最晚开始时间的最最晚时间等于

29、其所有紧后活动最晚开始时间的最小者小者ABCLF=Min30,28,27=27LS=30LS=28LS=27HCBAHLS=30LS=28LS=27LF=27LS and LF Times for Reliables Project ASTARTGHMFJFINISHKLNDIECB241074679584562LS = 0 LF = 2LS = 2 LF = 6LS = 20 LF = 26LS = 16 LF = 20LS = 18 LF = 25LS = 20 LF = 25LS = 26 LF = 33LS = 6 LF = 16LS = 0 LF = 0LS = 25 LF = 33

30、LS = 33 LF = 38LS = 38 LF = 44LS = 34 LF = 38LS = 33 LF = 42LS = 42 LF = 44LS = 44 LF = 4400Min 20,33=20F 20,25 20,25EFLSLFESH 29,38 33,42EFLSLFESThe Complete Project Network ASTARTGHMFJFINISHKLNDIECB241074679584562S = (0, 0) F = (2, 2)S = (2, 2) F = (6, 6)S = (16, 20) F = (22, 26)S = (16, 16) F = (

31、20, 20)S = (16, 18) F = (23, 25)S = (20, 20) F = (25, 25)S = (22, 26) F = (29, 33)S = (6, 6) F = (16, 16)S = (0, 0) F = (0, 0)S = (25, 25) F = (33, 33)S = (33, 33) F = (38, 38)S = (38, 38) F = (44, 44)S = (33, 34) F = (37, 38)S = (29, 33) F = (38, 42)S = (38, 42) F = (40, 44)S = (44, 44) F = (44, 44

32、)00Critical path:Start A B C E F J L N FinishES，LSEF，LFTimes for Reliables Project AOA1235ABCD6GEI4871011139H12FJKLNM0, 20, 222, 62, 646, 166, 1610716,2316, 20416, 226722, 2929, 38938, 402520, 2525, 338533,3833,37438, 44638, 4442, 4434,3833,3831, 4225, 3320, 2518, 2516, 2024, 3118, 24Slack for Relia

33、bles Activities活动（工序）的时差活动（工序）的时差活动活动LSESLFEF总时差总时差 (LSES)是否关键是否关键工序工序A00220YesB22440YesC4416160YesD201626224NoE161620200YesF202025250YesG262233294NoH332942384NoI181625232NoJ252533330YesK343338371NoL333338380YesM423844404NoN383844440Yes总时差总时差=最迟开始时最迟开始时间最早开始时间间最早开始时间=最迟结束时间最最迟结束时间最早结束时间早结束时间TS=LS ES

34、 =LFEFThe slack of an activity 例、例、某公司装配一条新的生产线，具体过程如表某公司装配一条新的生产线，具体过程如表12-10,12-10,求：完成此工求：完成此工程的最少时间，关键路线及相应的关键工序，各工序的最早开始时间程的最少时间，关键路线及相应的关键工序，各工序的最早开始时间和非关键工序在不影响工程完成时间的前提下，其开始时间与结束时和非关键工序在不影响工程完成时间的前提下，其开始时间与结束时间可以推迟多久。间可以推迟多久。7工序代号工序代号工序内容工序内容所需时间（天）所需时间（天）紧前工序紧前工序abcdefghij生产线设计生产线设计外购零配件外购零

35、配件下料、锻件下料、锻件工装制造工装制造1木模、铸件木模、铸件机械加工机械加工1工装制造工装制造2机械加工机械加工2机械加工机械加工3装配调试装配调试60451020401830152535/aaaacdd,egb,i,f,h解：据表解：据表12-10,绘制网络图如图绘制网络图如图12-8。 图图12-8关键路线：如图关键路线：如图12-8 ,-就是一条关键路线，我们要干完所有的工就是一条关键路线，我们要干完所有的工序就必须走完所有这样的路线，由于很多工序可以同时进行，所以网络中序就必须走完所有这样的路线，由于很多工序可以同时进行，所以网络中最长的路线就决定了完成整个工程所需的最少时间，这条路

36、线称为关键路最长的路线就决定了完成整个工程所需的最少时间，这条路线称为关键路线。线。12346785a60b45echj35ig1030d204025f18159 图图12-10 其次其次,从网络的收点开始计算出在不影响整个工程最早结束时间的情从网络的收点开始计算出在不影响整个工程最早结束时间的情况下各个工序的最晚开始时间况下各个工序的最晚开始时间(缩写为缩写为LS)和最晚结束时间（缩写为和最晚结束时间（缩写为LF),显然对同一工序有显然对同一工序有 LS=LF-t1236785a0,6060b60,10545e60.100c60,70h100,115j135,17035i110.135g80

37、,11030d60.80204025f70,88184101511 运用此法则，可以从首点开始计算出每个工序的运用此法则，可以从首点开始计算出每个工序的LF与与LS，如图，如图12-11所示。所示。 接着，可以计算出每一个工序的时差，把在不影响工程最早结束时间接着，可以计算出每一个工序的时差，把在不影响工程最早结束时间的条件下，工序最早开始（或结束）的时间可以推迟的时间，成为该工序的条件下，工序最早开始（或结束）的时间可以推迟的时间，成为该工序的时差，对每个工序来说其时差记为的时差，对每个工序来说其时差记为Ts有有 Ts=LS-ES=LF-EF1236785a0,60600,60b60,105

38、4590,135e60.100c 60,70h 100,115j135,17035135,170i110.135g80,1103080,110d60.802060,804080,12025110,135f70,8818117,135410107,11715 120,13512最后将各工序的时差，以及其他信息构成工序时间表如表最后将各工序的时差，以及其他信息构成工序时间表如表12-11所示。所示。 这样就找到了一条由关键工序这样就找到了一条由关键工序a,d,g,i和和j依次连接成的从发点到收点的关依次连接成的从发点到收点的关键路线。键路线。13The PERT 三种估计方式Most likely

39、 estimate (m) = Estimate of most likely value of the duration 最大可能估计 Optimistic estimate (o) = Estimate of duration under most favorable conditions. 乐观估计时间Pessimistic estimate (p) = Estimate of duration under most unfavorable conditions. 悲观估计时间 Elapsed timep0Beta distributionmo处理不确定活动工期分布分布时间时间均值和标准

40、差Mean and Standard DeviationAn approximate formula for the variance 活动时间方差 (2) of an activity is 2po62An approximate formula for the mean 活动时间的均值(m) of an activity is mo 4mp6乐观估计乐观估计P和悲观和悲观估计时间估计时间O的差的差大致为大致为6 6个标准差个标准差可信公司活动的均值和方差ActivityompMeanVarianceA12321/9B23.5841C6918104D45.51061E14.5544/9F44

41、1051G56.51171H581794I37.5971J39981K44440L15.5751M12321/9N55.5964/9P306P306Pessimistic Path Lengths for Reliables Project悲观估计时间的路径与路径长度PathPessimistic Length (Weeks)采用悲观估计时间采用悲观估计时间StartA B C D G H M Finish3 + 8 + 18 + 10 + 11 + 17 + 3 = 70Start A B C E H M Finish3 + 8 + 18 + 5 + 17 + 3 = 54Start A B

42、 C E F J K N Finish3 + 8 + 18 + 5 + 10 + 9 + 4 + 9 = 66Start A B C E F J L N Finish3 + 8 + 18 + 5 + 10 + 9 + 7 + 9 = 69Start A B C I J K N Finish3 + 8 + 18 + 9 + 9 + 4 + 9 = 60Start A B C I J L N Finish3 + 8 + 18 + 9 + 9 + 7 + 9 = 63三个简化的近似 PERT/CPM P306 The mean critical path will turn out to be th

43、e longest path through the project network.假设均值关键路是项目中最长的一条路假设均值关键路是项目中最长的一条路统计独立性统计独立性The durations of the activities on the mean critical path are statistically independent. Thus, the three estimates of the duration of an activity would never change after learning the durations of some of the othe

44、r activities.正太分布假设正太分布假设The form of the probability distribution of project duration is the normal distribution. By using simplifying approximations 1 and 2, there is some statistical theory (one version of the central limit theorem) that justifies this as being a reasonable approximation if the nu

45、mber of activities on the mean critical path is not too small（活动数至少（活动数至少5个）个）.计算项目的均值和方差Activities on Mean Critical PathMeanVarianceA21/9B41C104E44/9F51J81L51N64/9Project durationmp = 442p = 9Probability of Meeting Deadline 44 (Mean)47 (Deadline)Project duration (in weeks)2 = 9pd - m = 47 - 44 = 1p

46、3pPzdPzpp)()(mz 为为d超过超过p的标准差个数，的标准差个数，查统计表可得概率值查统计表可得概率值Probability of Meeting a DeadlineP(T d)P(T d)3.00.001400.502.50.00620.250.602.00.0230.50.691.750.0400.750.771.50.0671.00.850.891.0330.750.231.750.9600.50.312.00.9770.250.402.50.993800.503.00.9986 dmpp dmppSpreadsheet for P

47、ERT Three-Estimate Approach3456789101112131415161718BCDEFGHIJKTime EstimatesOn MeanActivityompCritical PathmA123*20.1111Mean CriticalB23.58*41PathC6918*104m44D45.510619E14.55*40.4444F4410*51P(T=d) = 0.8413G56.51171whereH581794d = 47I37.5971J399*81K44440L15.57*51M12320.1111N55.59*60.4444时间和成本权衡Consid

48、ering Time-Cost Trade-OffsQuestion: If extra money is spent to expedite the project, what is the least expensive way of attempting to meet the target completion time (40 weeks)?CPM Method of Time-Cost Trade-Offs:qCrashing an activity 赶工完成一项活动refers to taking special costly measures to reduce the dur

49、ation of an activity below its normal value. Special measures might include overtime, hiring additional temporary help, using special time-saving materials, obtaining special equipment, etc.qCrashing the project赶工完成项目refers to crashing a number of activities to reduce the duration of the project bel

50、ow its normal value.Time-Cost Graph for an Activity Activity durationActivity costCrash costNormal costNormalCrashCrash timeNormal time赶工常规赶工成本正常成本活动成本常规常规时间赶工时间活动时间Time-Cost Trade-Off Data for Reliables Project Time (weeks) 成本成本Cost MaximumReductionin Time (weeks)Crash Costper WeekSavedActivityNorm

51、alCrashNormalCrashA21$180,000$280,0001$100,000B42320,000420,000250,000C107620,000860,000380,000D64260,000340,000240,000E43410,000570,0001160,000F53180,000260,000240,000G74900,0001,020,000340,000H96200,000380,000360,000I75210,000270,000230,000J86430,000490,000230,000K43160,000200,000140,000L53250,000

52、350,000250,000M21100,000200,0001100,000N63330,000510,000360,000Sum of normal cost= 4550,000; Sum of crash cost = 6150,000P400Marginal Cost Analysis for Reliables ProjectInitial TableLength of PathActivitytoCrashCrashCostABCDGHMABCEHMABCEFJKNABCEFJLNABCIJKNABCIJLN403143444142最长路最长路Marginal Cost Analy

53、sis for Reliables ProjectTable After Crashing One WeekLength of PathActivitytoCrashCrashCostABCDGHMABCEHMABCEFJKNABCEFJLNABCIJKNABCIJLN403143444142J$30,000403142434041Marginal Cost Analysis for Reliables ProjectTable After Crashing Two WeeksLength of PathActivitytoCrashCrashCostABCDGHMABCEHMABCEFJKN

54、ABCEFJLNABCIJKNABCIJLN403143444142J$30,000403142434041J$30,000403141423940J只大只能赶工2天Marginal Cost Analysis for Reliables ProjectTable After Crashing Three WeeksLength of PathActivitytoCrashCrashCostABCDGHMABCEHMABCEFJKNABCEFJLNABCIJKNABCIJLN403143444142J$30,000403142434041J$30,000403141423940F$40,000

55、403140413940Marginal Cost Analysis for Reliables ProjectFinal Table After Crashing Four WeeksLength of PathActivitytoCrashCrashCostABCDGHMABCEHMABCEFJKNABCEFJLNABCIJKNABCIJLN403143444142J$30,000403142434041J$30,000403141423940F$40,000403140413940F$40,000403139403940Project Network After Crashing AST

56、ARTGHMFJFINISHKLNDIECB241074679364562S = (0, 0) F = (2, 2)S = (2, 2) F = (6, 6)S = (16, 16) F = (22, 22)S = (16, 16) F = (20, 20)S = (16, 16) F = (23, 23)S = (20, 20) F = (23, 23)S = (22, 22) F = (29, 29)S = (6, 6) F = (16, 16)S = (0, 0) F = (0, 0)S = (23, 23) F = (29, 29)S = (29, 29) F = (34, 34)S

57、= (34, 34) F = (40, 40)S = (29, 30) F = (33, 34)S = (29, 29) F = (38, 38)S = (38, 38) F = (40, 40)S = (40, 40) F = (40, 40)00Using LP to Make Crashing DecisionsnRestatment of the problem: Consider the total cost of the project, including the extra cost of crashing activities. The problem then is to

58、minimize this total cost, subject to the constraint that project duration must be less than or equal to the time desired by the project manager.nThe decisions to be made are the following:The start time of each activity.每一个活动的开始时间The reduction in the duration of each activity due to crashing.每个活动工期的

59、减少量The finish time of the project (must not exceed 40 weeks).nThe constraints are:Time Reduction Max Reduction (for each activity).Project Finish Time Desired Finish Time.Activity Start Time Activity Finish Time of all predecessors (for each activity).1.Project Finish Time Finish Time of all immedia

60、te predecessors of finish node.Spreadsheet Model3456789101112131415161718192021222324BCDEFGHIJKMaximumCrash CostTimeCostTimeper WeekStartTimeFinishActivityNormalCrashNormalCrashReductionsavedTimeReductionTimeA21$180,000$280,0001$100,000002B42$320,000$420,0002$50,000206C107$620,000$860,0003$80,000601