ch 9ch 9

pert and cpmpert and cpm

Learning objectivesLearning objectives::

11 - -Give a general description of/techniquesGive a general description of/techniques..

22 - -Construct simple net work diagramsConstruct simple net work diagrams..

33 - -Analyze net works that have deterministic timeAnalyze net works that have deterministic time..

44 - -Describe activity crashing “and solve simple problemDescribe activity crashing “and solve simple problem..

SummarySummary

Projects are comprised f a unique set of activities Projects are comprised f a unique set of activities established to achieve a given set of objectives during a established to achieve a given set of objectives during a

limited life spanlimited life span. .

PERI/CPM are two commonly used techniques for PERI/CPM are two commonly used techniques for developing and monitoring projectsdeveloping and monitoring projects..

Glossary Activity A taskActivity A task

CPM (critical Path method )CPM (critical Path method ) Pert/CPM were developedPert/CPM were developed

Separately at about the Separately at about the same time in the late same time in the late 1950s. Both techniques 1950s. Both techniques are concerned with are concerned with integration of a number of integration of a number of different tasksdifferent tasks..

PERT program evaluation and Review Technique) See CPMPERT program evaluation and Review Technique) See CPM

CrashingCrashing

Critical PathCritical Path

Critical ActivityCritical Activity

Accelerating a project by Accelerating a project by speeding up those critical- path speeding up those critical- path activities that have lowest activities that have lowest ratio of incremental cost to ratio of incremental cost to

incremental time savedincremental time saved. .

The path in a network diagram The path in a network diagram that takes the longest time for that takes the longest time for

completioncompletion. .

An activity on the critical PathAn activity on the critical Path. .

Pert / CPMPert / CPM

Program Evolution Critical path MethodProgram Evolution Critical path Method and Review Techniqueand Review Technique

Most projects have Certain elements in commonMost projects have Certain elements in common 11 . .The involve Considerable CostThe involve Considerable Cost 22 . .Usually have a long time horizonUsually have a long time horizon

33 . .Involve a large number of activities that must carefully, planned Involve a large number of activities that must carefully, planned and performance guidelinesand performance guidelines . .

Planning and Scheduling with Gantt chartsPlanning and Scheduling with Gantt charts: :

Grantt chart is a popular tool for planning and Scheduling simple projectsGrantt chart is a popular tool for planning and Scheduling simple projects..

This tool enable a manager to initially Schedule project activities and This tool enable a manager to initially Schedule project activities and than monitor progress over time by comparing planned progress to than monitor progress over time by comparing planned progress to actual progressactual progress..

For exampleFor example-: -:

Consider the case of bank that planned to great a new direct marketing Consider the case of bank that planned to great a new direct marketing departmentdepartment . .

He developed a list of the major activities that would be requiresHe developed a list of the major activities that would be requires

Activity time to accomplish week afterActivity time to accomplish week after in week project start in week project start

11 . .Locate new facility 8 immediateLocate new facility 8 immediate

22 . .Interview staff 4 immediateInterview staff 4 immediate

33..Hire + train staff 9 4Hire + train staff 9 4

44 . .Select and order furniture 6 8Select and order furniture 6 8

55 . .Remodel and intall phones 11 8Remodel and intall phones 11 8

66 . .Receive Furniture and set up 3 14Receive Furniture and set up 3 14

77 . .Move in/ start up 1 19Move in/ start up 1 19

prepare a Gantt chart for this projectprepare a Gantt chart for this project. .

Gantt chart for bank caseGantt chart for bank case: :

Activity Start 2 4 6 8 10 12 14 16 18 20Activity Start 2 4 6 8 10 12 14 16 18 20

3

4

5

6

7

1

2

start 2 6 8 10 12 14 16 18 204

PERT and CPMPERT and CPM

By using pert or cpm , managers are able to obtainBy using pert or cpm , managers are able to obtain: : 11 . .A graphical display of project activitiesA graphical display of project activities. .

22 . .An estimate of how long the project will takeAn estimate of how long the project will take. . 33..An indication of which activities are the most critical to An indication of which activities are the most critical to

timely completion of the projecttimely completion of the project . . 44 . .An indication of how long any activity can be delayed An indication of how long any activity can be delayed

without lengthening the projectwithout lengthening the project..Note : pert : stressed probabilistic activity time estimates Note : pert : stressed probabilistic activity time estimates because the environment in which it was developed was because the environment in which it was developed was

typified by high uncertaintytypified by high uncertainty. .

Cpm: originally made no provision for variable time estimatesCpm: originally made no provision for variable time estimates

The net work diagramThe net work diagram

Recall the bank caseRecall the bank case::

Nodes: represent both beginning and endings of activitiesNodes: represent both beginning and endings of activities.. arrows : represent the project activities that lead from arrows : represent the project activities that lead from

the starting node to the finishing nodethe starting node to the finishing node..

1

2

4

3

5

6

Path: is a sequence of activities that lead from the startingPath: is a sequence of activities that lead from the starting

node to the finishing nodenode to the finishing node. .

from bank 1 - 2 – 4 - 5 – 6from bank 1 - 2 – 4 - 5 – 6

Case digram 1 – 2- 5 – 5Case digram 1 – 2- 5 – 5

11 – – 33 – – 55 – – 66

Critical path : The longest path which referred to critical Critical path : The longest path which referred to critical activitiesactivities. .

Deterministic time EstimatesDeterministic time Estimates

Deterministic time estimatesDeterministic time estimates::

ExampleExample-: -:

Given the information provided in the a companying net Given the information provided in the a companying net work diagram , determine each of the followingwork diagram , determine each of the following

11 . .The length of each pathThe length of each path

22..The critical pathThe critical path

33 . .The expected length of the projectThe expected length of the project

44 . .The amount of slack time for each pathThe amount of slack time for each path. .

1

..

1

2 4

35

6

Solution :-Solution :-(The start and end “activities” have no time they merely (The start and end “activities” have no time they merely serve as reference points )serve as reference points )

11 . .The path length areThe path length are

A . 1- 2- 4- 5- 6A . 1- 2- 4- 5- 6

88 + + 66 + + 33+ + 11 = = 1818 weeksweeks..

B . 1 -2 – 5- 6B . 1 -2 – 5- 6

88 + + 1111 + + 11 = = 2020 weeksweeks. .

C . 1 - 3 – 5 - 6C . 1 - 3 – 5 - 6

44 + + 99 + + 11 = = 1414 weeksweeks . .

22 . . The longest 20 week (1 -2 -5 -6 )The longest 20 week (1 -2 -5 -6 )

is the critical pathis the critical path . .

33 . .The expected length of the project is equal to the length of The expected length of the project is equal to the length of the critical path 20 weeksthe critical path 20 weeks . .

44 . .The amount of slack time for each pathThe amount of slack time for each path

path length ( week) slackpath length ( week) slack

1-2-4-5-61-2-4-5-6 8+6+3+18+6+3+1 = =1818 20-1820-18 = = 22

11 - -22 - -55 – – 66 88 + +1111 + +11 = = 2020 20-2020-20 = = 00

11 - -33 – – 55 – – 66 44 + + 99+ + 11 = = 1414 20-1420-14 = = 66

NoteNote -: -:

slack time = length of critical path – path lengthslack time = length of critical path – path length. .

A Computing ALogrithmA Computing ALogrithm

Algorithm is used to develop four pieces of info. About the Algorithm is used to develop four pieces of info. About the net worknet work

ActivitiesActivities -: -:

Es, earliest time the activity can start , assuming all preceding Es, earliest time the activity can start , assuming all preceding activities start as early as possible activities start as early as possible

EF , earliest time the activity can finishEF , earliest time the activity can finish

..

LS , latest time the activity can start and not delay the LS , latest time the activity can start and not delay the projectproject. .

Once these values have been determined , they can be used Once these values have been determined , they can be used to find to find

11 . .Expected duration of the projectExpected duration of the project

22 . .Slack timeSlack time

33 . .Which activities are on the critical pathWhich activities are on the critical path. .

Computing Es and EF TimesComputing Es and EF Times

1

24

3 5 6

8

199

Computation of earliest starting and finished time is aidedComputation of earliest starting and finished time is aidedby tow simple rulesby tow simple rules: :

11 . .THE EF is equal to its Es plus it expected duration ,tTHE EF is equal to its Es plus it expected duration ,t::

?? ??EF=Es + tEF=Es + t

22 . .For nodes with one entering arrowFor nodes with one entering arrow: :

Es for activities at such nodes is equal to EF of entering Es for activities at such nodes is equal to EF of entering arrow. For nodes with Multiple entering arrows : equals arrow. For nodes with Multiple entering arrows : equals activities leaving such nodes equals largest EF of activities leaving such nodes equals largest EF of

entering arrowsentering arrows. .

Compute Es and EF for our exampleCompute Es and EF for our example: :

SolutionSolution: : Assume an Es of o for activities without predecessorsAssume an Es of o for activities without predecessors..

EF EF 1-2 = 0 +8=8 and EF 1-3=0+4=4

Likewise for the other activities. These results are Summarized in the table below:

Activity Duration ,t Es EF1-2 8 0 81-3 4 0 4

2-4 6 8 14 2-5 11 8 193-5 9 4 134-6 3 14 17

5-6 1 19 20

T+Es

Note : last EF is the project duration , thus ,the expected Note : last EF is the project duration , thus ,the expected length of the projectlength of the project

is 20 weekis 20 week..

ES, EF forward passES, EF forward pass????

Computing LS and LF timesComputing LS and LF times Computation of the Ls +LF times is aided by the use of tow Computation of the Ls +LF times is aided by the use of tow

rulesrules: : 11..The L is equal to its LF minus its expected durationThe L is equal to its LF minus its expected duration: :

LS= LF -1LS= LF -122 . .For nodes with one arrow :LF for arrows entering that node For nodes with one arrow :LF for arrows entering that node

equals the LS of the leaving arrowsequals the LS of the leaving arrows For nodes with multiple leaving arrowsFor nodes with multiple leaving arrows..LS, LF back passLS, LF back pass??Compute LF, LS in our examplesCompute LF, LS in our examples

Set LF of the last activity equal to the EF of that activity . ThusSet LF of the last activity equal to the EF of that activity . Thus LF 5-6 =EF 5-6 = 20 weeksLF 5-6 =EF 5-6 = 20 weeks

LS 5-6 =LF 5-6 -1LS 5-6 =LF 5-6 -1 = = 20-120-1- - 1919

For LF 4-5 =LF 2-5 = LF 3 -5 =19For LF 4-5 =LF 2-5 = LF 3 -5 =19

For the restFor the rest LS 4-5 = 19-3 =16LS 4-5 = 19-3 =16

LS 2-5 = 19 – 11= 18LS 2-5 = 19 – 11= 18 LS 2-5 = 19 – 9 = 10LS 2-5 = 19 – 9 = 10

There are two arrow leaving Node 2There are two arrow leaving Node 2 , , 2-42-4 with LS = 10 , 2-5 Ls = 8with LS = 10 , 2-5 Ls = 8

The latest Finish for activity 2-1 thus become 8 which is the smallest Ls The latest Finish for activity 2-1 thus become 8 which is the smallest Ls for a leaving arrow the LF for 1-3 is equal to the LS for 3-5for a leaving arrow the LF for 1-3 is equal to the LS for 3-5 . .

LF 1-3 LS 3-5 =10LF 1-3 LS 3-5 =10 The LS for activity 1-3 isThe LS for activity 1-3 is LS 1-3 = 10 - 4 =6LS 1-3 = 10 - 4 =6The LS for activity 1-2 is LS 1-2 isThe LS for activity 1-2 is LS 1-2 is

LS 1-2 =LFLS 1-2 =LF1-21-2 –t –t

= = 88 - - 88 = = 00

The LS , LF computation are Summarized The LS , LF computation are Summarized in the table below in the table below : :

Activity Duration LF LSActivity Duration LF LS

5-65-6 11 2020 1919

4-54-5 33 1919 1616

2-52-5 1111 1919 88

3-53-5 99 1919 1010

2-42-4 66 1616 1010

1-21-2 88 88 00

1-31-3 44 1010 66

Computing Activity slack timesComputing Activity slack times

Activity slack = LS –ES or LF- ESActivity slack = LS –ES or LF- ES

Activity LS ES slackActivity LS ES slack

1-21-2 00 00 00

1-31-3 66 00 66

2-42-4 1010 88 22

2-52-5 88 88 00

3-53-5 1010 44 66

4-54-5 1616 1414 22

5-65-6 1919 1919 00

LS -ES

Are all Critical

Activities

Probabilistic time estimatesProbabilistic time estimates

In the preceding example we assumed that activity times In the preceding example we assumed that activity times were known and not subject to variationwere known and not subject to variation . .

For other situations we need to use probabilistic approachFor other situations we need to use probabilistic approach

Probabilistic time estimates use three time estimates for Probabilistic time estimates use three time estimates for each activity Instead of oneeach activity Instead of one: :

11 - -Optimistic time (a) length of time required under optimum Optimistic time (a) length of time required under optimum conditioncondition. .

22 - -Pessimistic time (b) time under worse conditionsPessimistic time (b) time under worse conditions

33 - -Most – likely time (m) most probable amount timeMost – likely time (m) most probable amount time. .

Beta distribution is commonly , used to describe inherentBeta distribution is commonly , used to describe inherent

Variability in time estimatesVariability in time estimates. .

A Beta DistributionA Beta Distribution

Of special interest in network analysisOf special interest in network analysis

Are the Average or expected timeAre the Average or expected time

For each activity te and the variance of each activity timeFor each activity te and the variance of each activity time..

The expected time is computed as a weighted averageThe expected time is computed as a weighted average

Of three time estimatesOf three time estimates: :

te = a + 4m + bte = a + 4m + b

Standard deviation of each activityStandard deviation of each activity

Time is estimated 1/6 of difference betweenTime is estimated 1/6 of difference between

the pessimistic and optimistic time estimatethe pessimistic and optimistic time estimate. .

The variance is found by squiring the standard deviation , thusThe variance is found by squiring the standard deviation , thus: :

6622 = =22

oror

Size of the variance reflects the degree of uncertainty associated with an Size of the variance reflects the degree of uncertainty associated with an activity timeactivity time

The larger the variance , the greater the uncertaintyThe larger the variance , the greater the uncertainty

It is also desirable to compute the standard deviation of the expectedIt is also desirable to compute the standard deviation of the expected

Time for each pathTime for each path. .

6path6path = =

Go to exampleGo to example

pathtivitiesoniancesofac(var4

6

ab 36

)( 2ab

ExampleExample::

The net work diagram for a shown in the accompanying figure The net work diagram for a shown in the accompanying figure , with three time estimates for each activity, with three time estimates for each activity , ,

Activity times are in month , do the followingActivity times are in month , do the following: :

.1.1Compute the expected time for each activity and the Compute the expected time for each activity and the expected duration for each pathexpected duration for each path

22 . .Identify the Critical PathIdentify the Critical Path. .

33 . .Compute the variance for each activity and the variance Compute the variance for each activity and the variance for each pathfor each path. . a m b

Solution: times te =a+4m=b pathSolution: times te =a+4m=b path1. path activity a m b 6 total 1. path activity a m b 6 total a -b - c a 1 3 4 2.83 a -b - c a 1 3 4 2.83 b 2 4 6 4.00 =10.00 b 2 4 6 4.00 =10.00 c 2 3 5 3.17 c 2 3 5 3.17

d –e -f d 3 4 5 4.00d –e -f d 3 4 5 4.00 e 3 5 7 5.00 =16.00 e 3 5 7 5.00 =16.00 f 5 7 9 7.00 largest critical path f 5 7 9 7.00 largest critical path

g- h –i g 2 3 6 3.33g- h –i g 2 3 6 3.33 h 4 6 8 6.00 13.50 h 4 6 8 6.00 13.50 i 3 4 6 4.17 i 3 4 6 4.17

2.2.* * critical path is d- h -i which has the largest path total critical path is d- h -i which has the largest path total

16.00 weeks .16.00 weeks .Which mean the expected duration of the projectWhich mean the expected duration of the project. .

33..VarianceVariance: :

Path activity a m b 5Path activity a m b 52 2 act= (b-a)act= (b-a)22 5 52 2 path pathpath path 3636

a-b – c a 1 3 4 (4-1)a-b – c a 1 3 4 (4-1)22/36=9/36/36=9/36 b 2 4 6 (6-2)b 2 4 6 (6-2)22/36=16/36 /36=16/36 34/63=.944 .9734/63=.944 .97

c 2 3 5 (5-2)c 2 3 5 (5-2)22/36= 9/36/36= 9/36

D- e-f d 3 4 5 (5-3)D- e-f d 3 4 5 (5-3)22/36= 4/36/36= 4/36 e 3 5 7 (7-3)e 3 5 7 (7-3)22/36= 16/36 /36= 16/36 36/36=1.00 1.0036/36=1.00 1.00

f 5 7 9 (9-5)f 5 7 9 (9-5)22/36= 16/36/36= 16/36

g- h- i g 2 3 6 (6-2)g- h- i g 2 3 6 (6-2)22/36=16/36/36=16/36 h 4 6 8 (8-4)h 4 6 8 (8-4)22/36=16/36 /36=16/36 41/36=1.39 41/36=1.39

1.071.07

i 3 4 6 (6-3)i 3 4 6 (6-3)22/36= 9/36/36= 9/36

The previous Solution enable a manager to make The previous Solution enable a manager to make probabilistic estimates of the project completion time , probabilistic estimates of the project completion time , such assuch as

11 . .The probability that the project will completed with The probability that the project will completed with months of startmonths of start? ?

22 . .The probability that the project will take longer than The probability that the project will take longer than specific months more than it’s completionspecific months more than it’s completion

Go to exampleGo to example

The next example illustrates the use of normal The next example illustrates the use of normal distribution to determine the probabilities for distribution to determine the probabilities for

various completion timevarious completion time. .

Before we look at that example , it is important Before we look at that example , it is important to make note of two pointsto make note of two points

11 . .Related to independence, it is assumed that Related to independence, it is assumed that path duration times are independent of each path duration times are independent of each otherother

22..A project is not completed until all of it’s A project is not completed until all of it’s activities have been completed (included those activities have been completed (included those on the critical path)on the critical path)..

ExampleExample-: -:

using the information from the preceding exampleusing the information from the preceding example , ,

Answer the following questionAnswer the following question: :

11 . .Can the paths be considered independent ? WhyCan the paths be considered independent ? Why? ?

22 . .Determine the probability that the project will be Determine the probability that the project will be completed within it months of it’s startcompleted within it months of it’s start? ?

33..Determine the probability that the project will be Determine the probability that the project will be completed with 15 months of it’s startcompleted with 15 months of it’s start..

44 . .What is the probability that the project will not be What is the probability that the project will not be completed within 15 months of it’s startcompleted within 15 months of it’s start? ?

SolutionSolution -: -:

11 . .Yes , the paths can be considered Independent because Yes , the paths can be considered Independent because no activity is on information than on path , and we have no activity is on information than on path , and we have information that would suggest that any activity times information that would suggest that any activity times

are interrelatedare interrelated. .

22 . .In order to answer Q’s of this nature , we must take into In order to answer Q’s of this nature , we must take into account the degree to witch the path distribution the account the degree to witch the path distribution the specified completion time. See the figuresspecified completion time. See the figures

path 17path 17

monthsmonths

100%100%

A – b – cA – b – c

10.010.0 monthsmonths

D – e – fD – e – f

16.016.0 monthsmonths

99.9599.95

q – h – iq – h – i

13.513.5 monthsmonths

The shaded portion of each distribution corresponds to the probability The shaded portion of each distribution corresponds to the probability the part will be completed within the specified time , observe that the part will be completed within the specified time , observe that pathspaths::

A -b - c and g -h -i are far enough to the left of the A -b - c and g -h -i are far enough to the left of the specified time that is highly likely that both will finished specified time that is highly likely that both will finished by month 17 but that critical path overlaps the specified by month 17 but that critical path overlaps the specified completion timecompletion time..

Hence , we need only consider the distribution of pathHence , we need only consider the distribution of path

d- e - f in assessing the probability of completion by month d- e - f in assessing the probability of completion by month 17.to do so, we must first compute the value of Z using 17.to do so, we must first compute the value of Z using

the relationshipthe relationship: :

Z = specified time – Expected timeZ = specified time – Expected time

path standard deviationpath standard deviation

In our exampleIn our example-: -:

expected time for path d – e – fexpected time for path d – e – f

is 16.0is 16.0, ,

Z = 17-16 =+1.00Z = 17-16 =+1.00

1.001.00

From table B with Z = +1.00From table B with Z = +1.00

Normal distributionNormal distribution

. . 84138413

Probability 84% to finish the project within 17 monthsProbability 84% to finish the project within 17 months

33..The question illustrates how to handle a problem in which more The question illustrates how to handle a problem in which more than one of the distribution over laps the specified timethan one of the distribution over laps the specified time

See belowSee below??

1515 monthsmonths

100%100%

A – b – cA – b – c

10.010.0 monthsmonths

. . 15871587

D – e – fD – e – f

16.016.0 monthsmonths

1515 monthsmonths

. . 91929192

Q – h – iQ – h – i

13.513.5 monthsmonths

Note that d –e – f and g – h –i paths overlaps month 15Note that d –e – f and g – h –i paths overlaps month 15..This means that both paths have the potential for delaying the This means that both paths have the potential for delaying the project beyond 15 monthproject beyond 15 month..

Back to Z equationBack to Z equationPath Z= 15-expected path duration from normalPath Z= 15-expected path duration from normal

path standard deviation distribution tablepath standard deviation distribution table , , B1B1,, P(Z)P(Z)

A –b - c 15-10 = +5.15 1.00A –b - c 15-10 = +5.15 1.00. . 9797

D –e - f 15-16 =-1.00 .1587D –e - f 15-16 =-1.00 .1587 1.001.00

G –h - i 15-13.5 =+1.40 .9192G –h - i 15-13.5 =+1.40 .9192 1.071.07

NoteNote-: -:

Any path with a Z of more than +2.5 is assigned a Any path with a Z of more than +2.5 is assigned a probability of 1.00probability of 1.00. .

The joint probability of finishing before month 15 is the The joint probability of finishing before month 15 is the product of these probabilitiesproduct of these probabilities

) ) 1.001.00. ) (. ) (15871587.) (.) (91929192. = (. = (14591459?? ??

Time – Cost Trade – OFFs – CrashingTime – Cost Trade – OFFs – Crashing

The desire to shorten the length of a project merely reflects o The desire to shorten the length of a project merely reflects o attempt to reduce the indirect cost associated with running attempt to reduce the indirect cost associated with running

the projectthe project. .

Such as : Facilities of equipment cost , supervision and labor Such as : Facilities of equipment cost , supervision and labor and personal costsand personal costs..

To avoid late penaltiesTo avoid late penalties..

To crashing desirable , a manager needs the following infoTo crashing desirable , a manager needs the following info

11 . .Regular time and crash time estimates for each activityRegular time and crash time estimates for each activity. .

22 . .Regular cost and crash cost estimates for each activityRegular cost and crash cost estimates for each activity

33 . .A list of activities that are on the critical pathA list of activities that are on the critical path. .

The general procedure for crashing isThe general procedure for crashing is::

11 . .Obtain estimates of regular and crash time and costs Obtain estimates of regular and crash time and costs for each activityfor each activity. .

22..Determine the lengths of all paths and path slack timesDetermine the lengths of all paths and path slack times

33 . .Determine which activities are on the critical pathDetermine which activities are on the critical path. .

44 . .Crash critical activities , in order of increasing costs do Crash critical activities , in order of increasing costs do not exceed benefitsnot exceed benefits. .

ExampleExample-: -:

Using the info . Below , develop an optimum time- cost Using the info . Below , develop an optimum time- cost solution , Assume that indirect project costs are $1000per solution , Assume that indirect project costs are $1000per

dayday . .

Cost per dayCost per day

ActivityActivity Normal time crash time to crash Normal time crash time to crash

1-21-2 66 66

2-52-5 1010 88 $500$500

1-31-3 55 44 300300

3-43-4 44 11 700700

4-54-5 99 77 600600

5-65-6 22 11 800800

SolutionSolution-:-:

11..Determine which activities are on the critical path it’s Determine which activities are on the critical path it’s length , and the length of the other pathlength , and the length of the other path: :

path lengthpath length

1-2-5-61-2-5-6 1818

1-3-4-5-61-3-4-5-6 2020 critical pathcritical path

1

2

5

3 4

6

6 10

5 49

2Start End

22 . .Rank the critical path activities in order of lost wet Rank the critical path activities in order of lost wet crashing cost and determine the umber of days each can crashing cost and determine the umber of days each can be crashedbe crashed

cost per day availablecost per day available

activity to crash daysactivity to crash days

1-31-3 $300$300 11

4-54-5 600600 22

3-43-4 700700 33

5-65-6 800800 11

33..Begin shorting the project , one day at a time , and chek Begin shorting the project , one day at a time , and chek after each reducing to see which path is critical .( After after each reducing to see which path is critical .( After a certain point, another path may equal the length of the a certain point, another path may equal the length of the shortened critical path ) thusshortened critical path ) thus..

44..A shorten Activity 1-3 one day at cost of $ 300, the length A shorten Activity 1-3 one day at cost of $ 300, the length of the critical path now become 19 daysof the critical path now become 19 days..

b. Activity 1-3 cannot be shortened any more shorten b. Activity 1-3 cannot be shortened any more shorten activity 4-5 one day of a cost of $600the length of pathactivity 4-5 one day of a cost of $600the length of path

1-3-4-5-61-3-4-5-6 now become 18 days which is the same as the now become 18 days which is the same as the length of path 1-2-5-6length of path 1-2-5-6

c. Since the paths are now both critical , further c. Since the paths are now both critical , further improvements will necessitate shorting one activity on improvements will necessitate shorting one activity on eacheach..

The remaining points for crashing and their costs oneThe remaining points for crashing and their costs one::

PathPath activityactivity crash cost per daycrash cost per day

1-2-5-61-2-5-6 1-21-2 no reduction possibleno reduction possible

2-52-5 $500$500 5-65-6 800800

1-3-4-5-61-3-4-5-6 1-31-3 no further reduction no further reduction possiblepossible

3-43-4 $700$700

4-54-5 600600

5-65-6 800800

Activity 5-6 would not be advantageous because it has highest Activity 5-6 would not be advantageous because it has highest crashing cost . How ever activity 5-6 is on paths and he projectcrashing cost . How ever activity 5-6 is on paths and he project? ?

* *The option of shortening the least expensive activity on each path The option of shortening the least expensive activity on each path would cost $500 for 2-5 and $600 for 4-5would cost $500 for 2-5 and $600 for 4-5* *

Thus , shorten Activity 5-6 by one day the project duration is now 17 Thus , shorten Activity 5-6 by one day the project duration is now 17 daysdays. .

d. At this point , no additional improvement is feasible . The cost to crash d. At this point , no additional improvement is feasible . The cost to crash Activity 4-5 is $600 for a total of $1.100 and that would exceed the Activity 4-5 is $600 for a total of $1.100 and that would exceed the project costs of $1000 per dayproject costs of $1000 per day. .

E. The crashing sequence is Summarized belowE. The crashing sequence is Summarized below

length afterlength after

Path crashing n daysPath crashing n days

n=0 1 2 3n=0 1 2 3

1-2-5-61-2-5-6 1818 1818 1818 1717

1-3-4-5-61-3-4-5-6 2020 1919 1818 1717

Activity crashed 1-3 4-5 5-6Activity crashed 1-3 4-5 5-6

Cost $300 $600 $800Cost $300 $600 $800