数据、模型与决策(运筹学)课后习题和案例答案008.pdf

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1、CHAPTER 8PERT/CPM MODELS FOR PROJECT MANAGEMENTReview Questions8.1-1 The bid is for$5.4 million with a penalty of$300,000 if the deadline of 47 weeks is notmet.In addition,a bonus of$150,000 will be paid if the plant is completed within 40weeks.8.1-2 H e has decided to fbcus on meeting the deadline

2、of 47 weeks.8.1-3 An immediate predecessor is an activity that must be completed just prior to starting thegiven activity.Given the immediate predecessors of an activity,this activity then becomesthe immediate successor of each of these immediate predecessors.8.1-4(1)the activities of the project;(2

3、)the immediate predecessors of the activities;and(3)theestimated duration of the activities8.2-1 (1)activity infonnation;(2)precedence relationships;and(3)time information(duration)8.2-2 In an AOA network,each activity is represented by an arc,while in an AON network,eachactivity is represented by a

4、 node.AON networks are being used here.8.2-3 The bars in a Gantt chart show the scheduled start and finish times fbr activities in a project.8.3-1 (a)A path through a project network is one of the routes following the arrows(arcs)fromthe start node to the finish node;(b)the length of a path is the s

5、um of the estimateddurations of the activities on the path;(c)the longest path is called the critical path.8.3-2(1)The actual duration of each activity must be the same as its estimated duration;and(2)each activity must begin as soon as all its immediate predecessors are finished.8.3-3 The earliest

6、start time of an activity is equal to the largest of the earliest finish times of itsimmediate predecessors.8.3-4 A forward pass is the process of starting with the initial activities and working forward intime toward the final activities.8.3-5 It is a last chance schedule because anything later wil

7、l delay the completion of the project.8.3-6 The latest finish time of an activity is equal to the smallest of the latest start times of itsimmediate successors.8.3-7 A backward pass starts with the final activities and works backward in time toward theinitial activities instead of starting with the

8、initial activities.8.3-8 Any delay along the critical path will delay project completion.8.3-9(1)Identify the longest path through the project network;and(2)identify the activities withzero slack-they are on the critical path.8-18.4-1 The three estimates are the most likely estimate,optimistic estim

9、ate,and pessimisticestimate.8.4-2 The optimistic and pessimistic estimates are meant to lie at the extremes of what is possible,whereas the most likely estimate provides the highest point of the probability distribution.8.4-3 It is assumed that the mean critical path will turn out to be the longest

10、path through theproject network.8.4-4 It is assumed that the durations of the activities on the mean critical path are statisticallyindependent.8.4-5|ip=sum of the means of the durations for the activities on the mean critical path.8.4-6 Op-=sum of the variances of the durations fbr the activities o

11、n the mean critical path.8.4-7 It is assumed that the form of the probability distribution of project duration is the normaldistribution.8.4-8 It is usually higher than the true probability.8.5-1 Using overtime,hiring additional labor,and using special materials or equipment are allways of crashing

12、an activity.8.5-2 The two key points are labeled normal and crash.The normal point shows the time andcost of the activity when it is performed in the normal way.The crash point shows the timeand cost when the activity is fully crashed.8.5-3 No,only crashing activities on the critical path will reduc

13、e the duration of the project.8.5-4 Crash costs per week saved are being examined.8.5-5 The decisions to be made are the start time of each activity,the reduction in the duration ofeach activity due to crashing,and the finish time of the project.8.5-6 An activity cannot start until its immediate pre

14、decessor starts and then completes itsduration.8.5-7 Because of uncertainty,the plan fbr crashing the project only provides a 50%chance ofactually finishing within 40 weeks,so the extra cost of the plan is not justified.8.6-1 PERT/Cost is a systematic procedure to help the manager plan,schedule,and

15、controlproject costs.8.6-2 It begins by developing an estimate of the cost of each activity when it is performed in theplanned way.8.6-3 A common assumption is that the costs of performing an activity are incurred at a constantrate throughout its duration.8.6-4 A work package is a group of related a

16、ctivities.8.6-5 PERT/Cost uses earliest start time and latest start time schedules as a basis fbr developingcost schedules.8-28.6-6 A PERT/Cost schedule of costs shows the weekly project cost and cumulative project costfor each time period.8.6-7 A PERT/Cost report shows the budgeted value of the wor

17、k completed of each activity andthe cost overruns to date.8.6-8 Since deviations from the planned work schedule may occur,a PERT/Cost report is neededto evaluate the cost perfbnnance of individual activities.8.7-1 Planning,scheduling,dealing with uncertainty,time-cost trade-offs,and controlling cost

18、sare addressed by PERT/CPM.8.7-2 Computer implementation has allowed for application to larger projects,faster revisions inproject plans and effortless updates and changes in schedules.8.7-3 The accuracy and reliability of end-point estimates are not as good for points that are not atthe extremes of

19、 the probability distribution.8.7-4 The technique of computer simulation to approximate the probability that the project willmeet its deadline is an alternative for improving on PERT/CPM.8.7-5 The Precedence Diagramming Method has been developed as an extension of PERT/CPMto deal with overlapping ac

20、tivities.8.7-6 PERT/CPM assumes that each activity has available all the resources needed to perform theactivity in a normal way.8.7-7 It encourages effective interaction between the project manager and subordinates that leadsto setting mutual goals fbr the project.8.7-8 New improvements and extensi

21、ons are still being developed but have not been incorporatedmuch into practice yet.8-3Problems8.1 a)Start 0Finish 00b)Start f A C f FinishLength=4 weeksStart A D E FinishLength=7 weeksStart B C FinishLength=5 weeksStart B f D E f FinishLength=8 weeks*critical pathActivityESEFLSLFSlackCritical PathSt

22、art00000YesA02131NoB03030Yesc35683NoD35350YesE58580YesFinish88880YesCritical Path:Start f B f D f E Finishd)No,this will not shorten the length of the project because the activity is not on thecritical path.8-48.2 a)b)Start f A D FinishLength=4 weeksStart f A f E FinishLength=5 weeksStart f A F K Fi

23、nisLength=8 weeks*critical pathStart A、G-H I FinishLength=8 weeks*critical pathStart f B D f FinishLength=3 weeksStart B C f E f FinishLength=6 weeksStart B C H I J FinishLength=8 weeks*critical pathStart f B C f K FinishLength=7 weeks8-5c)ActivityESEFLSLFSlackCritical PathStart00000YesA02020YesB010

24、10Yesc13130YesD24684NoE36582NoF24240YesG23230YesH34340YesI45450YesJ58580YesK48480YesFinish88880YesCritical Paths:Start A f F f K FinishStart A G H I J FinishStart B C H ,I J Finishd)No,this will not shorten the length of the project because A is not on all of the criticalpaths.8-68.3 a)8-7b,c,&d)Act

25、ivityESEFLSLFSlackCritical PathStart00000YesA01010YesB13130Yesc37370YesD36360YesE35461NoF7107100YesG6116110YesH101110110YesI111511150YesJ151715170YesK151816191NoL172017200YesM182319241NoN202420240YesFinish242424240YesCritical Paths:Start A B-C-H -I-J L f N f FinishStart A B D G I J L N Finish8-88.4

26、a)Start 0b)Start A B J L FinishStart f C f D J-L f FinishStart f E f F J-L FinishStart-,G-H I J L FinishStart f K f L FinishLength=75 minutes*critical pathLength=45 minutesLength=72 minutesLength=67 minutesLength=45 minutes8-9Critical Path:Start J L Finishd&e)ActivityESEFLSLFSlackCritical PathStart0

27、0000YesA0300300YesB303530350Yesc02303230NoD25323530NoE073103NoF73210353NoG0158238NoH152523338NoI252733358NoJ354535450YesK015304530NoL457545750YesFinish757575750Yesf)Dinner will be delayed 3 minutes because of the phone call.If the food processor isused then dinner will not be delayed because there w

28、as 3 minutes of slack and 5minutes of cutting time saved and the call only used 6 minutes of the 8 total.8.5 a)Start A D H M FinishStart f B f E-J f M t FinishStart f C f F f K f N t FinishStart f A f I f M FinishStart C G L N FinishLength=19 weeksLength=20 weeksLength=16 weeksLength=17 weeksLength=

29、20 weeks*critical path*critical path8-10b)ActivityESEFLSLFSlackCritical PathStart00000YesA06171NoB03030Yesc04040YesD6107111NoE3103100YesF488124NoG4104100YesH101311141NoI6119143NoJ101410140YesK81112154NoL101510150YesM142014200YesN152015200YesFinish202020200YesKen will be able to meet his deadline if

30、no delays occur.c)Critical Paths:Start-B E J M FinishStart f C f G f L f N f FinishFocus attention on activities with 0 slack(those in the critical paths).d)If activity I takes 2 extra weeks there will be no delay because its slack is 3.If activityH takes 2 extra weeks then there will be a delay of

31、1 week because its slack is 1.Ifactivity J takes 2 extra weeks there will be a delay of 2 weeks because it has no slack.8-118.6 a)b)ActivityESEFLSLFSlackCritical PathStart00000YesA05050YesB01111211Noc13121411NoD597112NoE5115110YesF111411140YesFinish141414140YesCritical Path:Start f A E f F f FinishL

32、ength=14 monthsc)6 months8-128.7Total duration=70 weeksActivityESEFLSLFSlackCritical PathStart00000YesA03030YesB3113110Yesc112911290YesD293929390YesE293430351NoF344435451NoG395039500YesH506750670YesI293836457NoJ445345541NoK535757614NoL536054611NoM677067700YesN606961701NoFinish707070700YesCritical Pa

33、th:Start A B C D G H M Finish8.8ActivityESEFLSLFSlackCritical PathStart00000YesA01010YesB13130Yesc39390YesD91311152NoE9109100YesF101410140YesG131815202NoH182320252NoI91211142NoJ141714170YesK172117210YesL171820213NoM232425262NoN212621260YesFinish262626260YesCritical Path:Start A f B C E F J K N Finis

34、hTotal duration=26 weeks8-138.9ActivityESEFLSLFSlackCritical PathStart00000YesA01010YesB13130Yesc3103100YesD101413173NoE101310130YesF131613160YesG141817213NoH182421273NoI101511161NoJ162216220YesK222522250YesL222522250YesM242527283NoN252825280YesFinish282828280YesCritical Paths:Start f A-B f C E F J

35、K N f FinishStart A B ,C E F J L ,N-FinishTotal duration=28 weeks8.108.11pi=(o+4/?2+p)/6=30+(4)(36)+48/6=37C T2=(p-o)/62=(48-30)/62=9a)Start A-E-I f FinishStart A C-F-1 FinishStart B f D G J FinishStart B f H J FinishLength=17 monthsLength=17 monthsLength=17 monthsLength=18 months*critical pathb)By

36、Table 8.7,P(T A-E f I Finishd/22-171By Table 8.7,P(T A C F 1-Finish=096By Table 8.7,P(T D G J Finishd-/up _ 22-17=0.95By Table 8.7,P(T A E I f FinishBC|D|EFGHlJK3Time EstimatesOn Mean4Activityo m pCritical Patho25A 1.5 2 15*B 2 3.5 21C 1 1.5 18D 0.5 1 15E 3 5 24*F 1 2 16G 0.5 1 14H 2.5 3.5 25I 1 3 1

37、8*J 2 3 184.08335.0625Mean Critical66.166710.027778Path74.16678.027777817.0833325.3402883.255.8402778a2=97.833312.2500104.16676.25P(T A C F I FinishBc I D I EFGHIJK3Time EstimatesOn Mean4Activityo m pCritical PathMC T25A 1.5 2 15*B 2 3.5 21C 1 1.5 18*D 0.5 1 15E 3 5 24F 1 2 16*G 0.5 1 14H 2.5 3.5 25

38、I 1 3 18*J 2 3 184.08335.0625Mean Critical66.166710.027778Path74.16678.027777817.5833327.3680683.255.8402778(r=97.833312.2500104.16676.25P(T B D G-J FinishBCDEFGHIJK3Time EstimatesOn Mean4Activity0mPCritical PathM(r5A1.52154.08335.063Mean Critical6B23.5216.166710.028Path7C11.5184.16678.02817.833338D

39、0.5115ltr3.255.840a2=28.041679E35247.833312.25010F12164.16676.250P(T B H J FinishBCDEFGHIJK3Time EstimatesOn Mean4Activity0mPCritical Path025A1.52154.08335.063Mean Critical6B23.5216.166710.028Path7C11.5184.16678.02818.416678D0.51153.255.840G2=31.201399E35247.833312.25010F2164.16676.250P(T A E-I-Fini

40、shStart A C F I FinishStart B D G J FinishStart B-H J f FinishLength=17.08 monthsLength=17.58 monthsLength=17.83 monthsLength=18.42 months*critical pathb)P(Start-B H J-Finish 22 months)=0.7394c)P(Start f A E f I f Finish 22)=0.8356P(Start f Af Cf Ff l f Finish 22)=0.8007P(Start f f Finish A B f C Fi

41、nishStart-A B-E FinishStart A D E FinishLength=10.83 weeksLength=9.17 weeksLength=10.17 weeks*critical pathd)A/0.361Then,by Appendix A,P(T 11 weeks)=0.512,or by Table 8.7,P(TA C E F Finish Length=51 daysStart f B f D Finish Length=50 days*critical pathc)d-N p 57-51 n而“By Appendix A,P(T 57 days)=0.97

42、72.By Table 8.7,P(T 57 days)=0.997.d)=1.4.By Appendix A,P(T 57 days)=0.9192.By Table 8.7,P(T C f J f FinishStart f B F J f FinishStart B E f H f FinishStart f B E I f FinishStart f B f D-G f H f FinishStart B D G-I FinishLength=85.7 weeksLength=99.1 weeks*critical pathLength=80 weeksLength=93.7 week

43、sLength=80.7 weeksLength=94.4 weeks8.16、d-M 100-99.1.c)；=/-=i=0.136.By Appendix A,P(T 100 weeks)=0.5557.By Table 8.7,P(T 100 weeks)is between 0.5 and 0.6.d)Highera)False.The optimistic estimate is the duration under the most favorable conditions.Therefore activity durations are assumed to be no smal

44、ler than the optimistic estimate.Similarly,activity durations are assumed to be no larger than the pessimistic estimate.(Pg.340)b)False.PERT also assumes that the form of the probability distribution is a betadistribution.(Pg.340)c)False.It is assumed that the mean critical path will turn out to be

45、the longest paththrough the project network.(Pg.343)8.17Activityto CrashCrashCostLength of PathA-CB-D1416B$5,0001415B$5,0001414C,D$4,000+$6,0001313C,D$4,000+$6,00012128-198.18 a)Let A=reduction in A due to crashingC=reduction in C due to crashingMinimize Crashing Cost=$5,000A+$4,000Csubject to A 3 m

46、onthsC 2 monthsand A 0,C 0.Optimal solution:(xp 检)=Q,Q =(0,2)and Crashing Cost=$8,000.8-20b)Let B=reduction in B due to crashingD=reduction in D due to crashingMinimize Crashing Cost=$5,0003+$6,000。subject to B 2 monthsD 4 monthsand B 0,)0.Optimal solution:(x.必)=(B,D)=(2,2)and Crashing Cost=$22,000.

47、c)Let A=reduction in A due to crashingB=reduction in B due to crashingC=reduction in C due to crashingD=reduction in D due to crashingMinimize Crashing Cost=$5,000A+$5,0003+$4,000C+$6,0000subject to A 3 monthsB 2 monthsC 2 monthsD2 monthsS+4 monthsand A 0,B 0,C 0,D 0.Optimal solution(A,B,C,D)=(0,2,2

48、,2)and Crashing Cost=$30,000.8-21d)ABcDEFGH1J1NormalCrashMaximum TimeCrash Cost2TimeTimeNormalCrashReductionper MonthStartTimeFinish3Activity(months)(months)CostCost(months)SavedTimeReductionTime4A85$25,000$40,0003$5,0000o85B97$20,000$30,0002$5,0000276C64$16,000$24,0002$4,00082127D74$27,000$45,0003$

49、6,000721289Max Time10Project Completion Time(months)l 12=121112Total Cost_Deadline of 11 months:ABcDEFGHIJ1NormalCrashMaximum TimeCrash Cost2TimeTimeNormalCrashReductionper MonthStartTimeFinish3Activity(months)(months)CostCost(months)SavedTimeReductionTime4A85$25.000S40.0003$5.0000175B97$20,000$30,0

50、002$5,0000276C64$16,000$24,0002$4,00072117D74$27.000S45.0003S6.000731189Max Time10Project Completion Time(months)11=111112Total CostDeadline of 13 months:ABcDEFGHIJ1NormalCrashMaximum TimeCrash Cost2TimeTimeNormalCrashReductionper MonthStartTimeFinish3Activity(months)(months)CostCost(months)SavedTim