1

7.1 Introduction

• A project is a collection of tasks that must be completed in minimum time or at minimal cost.

• Tasks are called “activities.” – Estimated completion time (and sometimes costs) are

associated with each activity.– Activity completion time is related to the amount of

resources committed to it. – The degree of activity details depends on the

application and the level of specificity of data.

2

• Determining precedence relations among the activities is crucial to developing optimal schedules.

• Objective of Project Scheduling.

– Project Scheduling is used to plan and control a project

efficiently and economically.

– With Project Scheduling we can:• characterize the activities• monitor the project progress.

3

– The minimal expected completion time for a project.– The critical activities.– The earliest and latest time to start and finish an activity.– The amount of slack time for each activity.– The most cost effective completion alternatives.– The activities on which extra resources should be spent.– Whether or not a project is on schedule or within budget.– A schedule that offers a constant level of resources

utilization.– A schedule that completes a project in minimum time give

limited resources.

• Project Scheduling provides us with the following information:

4

7.2 Identifying the Activities of a Project

KLONE COMPUTERS, INC.

• KLONE Computers manufactures personal computers.

• It is about to design, manufacture, and market the Klonepalm 2000 palmbook computer.

5

• There are three major tasks to perform:– Manufacture the new computer.– Train staff and vendor representatives.– Advertise the new computer.

• KLONE needs to develop a precedence relations chart.

• The chart gives a concise set of tasks and their immediate predecessors.

6

Activity Description Activity Description

A Prototype model designB Purchase of materials

Manufacturing C Manufacture of prototype model activities D Revision of design

E Initial production run

F Staff trainingTraining activities G Staff input on prototype models

H Sales training

Advertising activities I Preproduction advertising campaign J Production advertising campaign

7

From the activity description chart, we can determine immediate predecessors for each activity.

8

Immediate EstimatedActivity Predecessor Completion Time

A None 90B A 15C B 5D G 20E D 21F A 25G C,F 14H D 28I A 30J D,I 45

Precedence Relationships Chart

9

7.4 Building PERT/CPM Networks• The PERT/CPM approach to project

scheduling – A network representation of the project that reflects

activity precedence relationships.

– It is designed to perform a thorough analysis of

possible project schedules.

– Primary objectives of PERT/CPM• To determine the minimal possible completion time for the

project.• To determine a range of start and finish time for each

activity, so that the project can be completed in minimal time.

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• PERT is a scheduling method in which activity completion times are treated as a random variable.

• CPM is a scheduling method which assumes the completion time for an activity is deterministic, dependent only on the amount of money spent to complete it.

• Both methods require one to identify the activities and the precedence relationships between activities.

11

7.5 The PERT/CPM Approach for Project Scheduling

KLONE COMPUTERS, INC. - Continued

• Management at KLONE would like to schedule

the activities so that the project is completed in

minimal time.

12

• Management wishes to know:

– The earliest completion date for the project.

– The earliest and latest start times for each activity which will not alter

this date.

– The earliest finish times for each activity which will not alter this date.

– Activities with rigid schedule and activities that have slack in their

schedules.

13

An Activity On Node (AON) Network Representationof the Klonepalm 2000 Computer Project

A90

J45

H28

E21

D20

I30

B15

G14

F25

C5

FINIS

H

Immediate EstimatedActivity Predecessor Completion Time

A None 90B A 15C B 5D G 20E D 21F A 25G C,F 14H D 28I A 30J D,I 45

NoneA

A

A

B

14

• Earliest start time / Earliest finish time

– Make a forward pass through the network as follows:

• Evaluate all the activities which have no immediate predecessors.

– The earliest start for such an activity is zero ES = 0.– The earliest finish is the activity duration EF = Activity duration.

• Evaluate the ES of all the nodes for which EF of all the immediate predecessor has been determined.

– ES = Max EF of all its immediate predecessors. – EF = ES + Activity duration.

• Repeat this process until all nodes have been evaluated– EF of the finish node is the earliest finish time of the project.

15

Earliest Start / Earliest finish

A90

B15

C5

F25

I30

G14

D20

E21

H28

J45

FIN

ISH

Evaluate all the activities which have no immediate predecessor

0,90

A

Evaluate the ES of all the nodes for which EF has been determined

B90,105

F

90,115

I

90,120

C105,110

110,124

G

115,129

D

129,149

E

149,170

H

149,177

120,165

J

149,194

170

177

194

FINISH

194

EARLIEST FINISH TIME

16

• Latest start time / Latest finish time

– Make a backward pass through the network as follows:

• Evaluate all the activities that immediately precede the finish node.

– The latest finish for such an activity is LF = minimal project completion time.

– The latest start for such an activity is LS = LF - activity duration.

• Evaluate the LF of all the nodes for which LS of all the immediate successors has been determined.

– LF = Min LS of all its immediate successors. – LS = LF - Activity duration.

• Repeat this process backward until all nodes have been evaluated.

17

B

F

C

A

I

Latest Start / Latest finish

E

DG

FINISHFINISHHH28

166,194

JJ45

149,194

E21

173,19490,105

90,115

90,120

105,110

115,129 129,149

149,170

149,177

149,194

153,173146,166

194129,149

0,90

129,149

D20

129,149129,149129,149129,149129,149129,149129,149

G14

115,129

I30

119,149

29,119

C5

110,115B1595,110

5,95F25

90, 115

0,90A90

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• The Critical Path and Slack Times

– Activity start time and completion time may be delayed by planned reasons as well as by unforeseen reasons.

– Some of these delays may affect the overall completion date.

– To learn about the effects of these delays, we calculate the slack time, and form the critical path.

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– Slack time is the amount of time an activity can be delayed without delaying the project completion date, assuming no other delays are taking place in the project.

Slack Time = LS - ES = LF - EF

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Critical activities

must be rigidly

scheduled

Critical activities

must be rigidly

scheduled

Activity LS - ES SlackA 0 -0 0B 95 - 90 5C 110 - 105 5C 119 - 119 0D 173 - 149 24E 90 - 90 0F 115 - 115 0G 166 - 149 17H 119 - 90 29I 149 - 149 0

Slack time in the Klonepalm 2000 Project

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• The critical path is a set of activities that have no slack,connecting the START node with the FINISH node.

• The critical activities (activities with 0 slack) form at least one critical path in the network.

• A critical path is the longest path in the network.

• The sum of the completion times for the activities on the critical path is the minimal completion time of the project.

- The Critical Path - The Critical Path

22

B

F

C

A

I

The forward and backward passes provided the earliest and latest times to start and finish each activity.

E

DG

FINISHFINISHHH28

166,194

JJ45

149,194

E21

173,19490,105

90,115

90,120

105,110

115,129 129,149

149,170

149,177

149,194

D20

194

0,90129,149

G14

115,129

I30

119,149

A90

C5

110,115B1595,110

F25

90, 115

0,90

THE CRITICAL PATHTHE CRITICAL PATH

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• Analysis of possible delays – We observe two different types of delays:

• Single delays.

• Multiple delays.

– In some cases the overall project completion time will be delayed.

– The conditions that specify each case are presented next.

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– Single delays

• A delay of a certain amount in a critical activity, causes the entire project to be delayed by the same amount.

• A delay of a certain amount in a noncritical activity will delay the project by the amount the delay exceeds the slack time.

• When the delay is less than the slack, the entire project is not delayed.

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– Multiple delays

• Observe three cases of delays in two noncritical activities.

– Case 1: There is no path linking the two non-critical activities.– Case 2: The non-critical activities are on the same path, separated

by a critical activity.– Case 3: The two non-critical activities are on the same path, with no

critical activity separating them.

• In each one of the above cases, the amount specified will cause no delay in the entire project.

26

A90

J45

H28

E21

D20

I30

G14

F25

C5

B15

ES=149

LS=173 DELAYED START=149+15=164

ES=90

LS =119 DELAYED START=90+15 =105

Case 1: Activity E and I are each delayed 15 days.THE PROJECT COMPLETION TIME THE PROJECT COMPLETION TIME IS NOT DELAYEDIS NOT DELAYED

FINISH

27

A90

B15

C5

F25

I30

G14

D20

E21

H28

J45

FINISHES=149

LS =173 DELAYED START=149+15 =164

ES=90DELAYED START =94

LS =95

Case 2: Activity B is delayed 4 days; Activity E is delayed 15 days.

Activity B and E are separatedby the critical activities G and D.

THE PROJECT COMPLETION TIME IS NOT DELAYEDTHE PROJECT COMPLETION TIME IS NOT DELAYED

28

A90

B15

C5

F25

I30

G14

D20

E21

H28

J45

FINISH

DELAYED START= 109 + 4 =113;

ES= 90

DELAYED START =94DELAYED FINISH =94+15=109

LS =110

• Case 3: Activity B is delayed 4 days;

Activity C is delayed 4 days.3 DAYS DELAYIN THE ENTIRE PROJECT

THE PROJECT COMPLETION TIME IS DELAYED 3 DAYSTHE PROJECT COMPLETION TIME IS DELAYED 3 DAYS

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7.7 The Probability Approach to Project Scheduling

• Activity completion times are seldom known with 100% accuracy.

• PERT is a technique that treats activity completion times as random variables.

• Completion time estimates are obtained by theThree Time Estimate approach

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• The three time estimate approach provides completion time estimate for each activity.

• We use the notation:a = an optimistic time to perform the activity.m = the most likely time to perform the activity.b = a pessimistic time to perform the activity.

• Approximations for the mean and the standard deviation of activity completion time are based on the Beta distribution.

= the mean completion time =a + 4m + b

6= the standard deviation =

b - a6

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To calculate the mean and standard deviation of the project completion time we make some simplifying assumptions.

– Assumption 1• A critical path can be determined by using the mean

completion times for the activities. • The project mean completion time is determined solely by the

completion time of the activities on the critical path.

– Assumption 2• The time to complete one activity is independent of the time

to complete any other activity.

– Assumption 3• There are enough activities on the critical path so that the

distribution of the overall project completion time can be approximated by the normal distribution.

32

The three assumptions imply that the overall project completion time is normally distributed,

with

Mean = Sum of mean completion times along

the critical path.

Variance = Sum of completion time variances

along the critical path.

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Activity Optimistic Most Likely PessimisticA 76 86 120B 12 15 18C 4 5 6D 15 18 33E 18 21 24F 16 26 30G 10 13 22H 24 18 32I 22 27 50J 38 43 60

KLONE COMPUTERS - revised

• Three time estimates were given for each activity

34

• Management at KLONE is interested in

– The probability that the project will be completed within

194 days.

– An interval estimate of the project completion time.

– The probability that the project will be completed within 180 days.

– The probability that the project will take longer than 210 days.

– An upper limit for the number of days within which it can virtually be sure the project will be completed.

35

• Calculation of the means and the variancesA = [76+4(86)+120]/6 = 90

= (120 - 76)/6 = 7.33 A2 = (7.33)2 = 53.78

– For the rest of the activities we have

Activity A 90 7.33 53.78B 15 1.00 1.00C 5 0.33 0.11D 20 3.00 9.00E 21 1.00 1.00F 25 2.33 5.44G 14 2.00 4.00H 28 1.33 1.78I 30 4.67 21.78J 45 3.67 13.44

Activity A 90 7.33 53.78B 15 1.00 1.00C 5 0.33 0.11D 20 3.00 9.00E 21 1.00 1.00F 25 2.33 5.44G 14 2.00 4.00H 28 1.33 1.78I 30 4.67 21.78J 45 3.67 13.44

36

• The mean times are the same as in the CPM problem, previously solved for KLONE.

• Thus, the critical path is A - F- G - D - J.

– Expected completion time = A + F + G + D + J = 194

– The project variance =A2 +F

2 +G2 +D

2 +J2 = 85.66

– The standard deviation = = 9.255

Under the assumptions stated above the completion time of

the project is normally distributed with= 194 days and

= 9.255 days.

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• Now the quantities of concern for management can be calculated. – The probability of completion in 194 days =

• An interval in which we are reasonably sure the completion date lies is

– The interval is = 194 1.96(9.255) {175, 213} days.• There is 95% confidence that the real completion time lies within 175

days and 213 days.

P(X 194) = P(Z194 -194

9.255 ) ( ) .P Z 0 0 5

z 0.025

38

XZ

1940

The probability of completion in 180 days = P(X 180) = P(Z -1.51) = 0.5 - 0.4345 = 0.0655

180-1.51

0.0655

.4345

– The probability that the completion time is longer

than 210 days = 0.0418=0.458-0.5=1.73)P(Z=210)P(X

2101.73

.45820.0418

x2.33

0.01

– Assume a 99% certain completion date is acceptable P(Z 2.33) = 0.99; x=+ z =194 + 2.33(9.255) = 215.56 days.

0.99

39

7.8 Computer Solution of PERT/CPM

• A linear programming approach

– Variables• Xi = The start time of the activities for i=A, B, C, …,J• X(FIN) = Finish time of the project

– Constraints• Each immediate predecessor relationship has one constraint • The constraints state that the start time of an activity must not

occur before the finish time of the immediate predecessor.

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– The Model

X(FIN) XE + 21

X(FIN) XH + 28

X(FIN) XJ + 45 XD XG + 14

XE XD + 20 XG XC + 5

XH XD + 20 XG XF + 25

XJ XD + 20 XI XD + 90

XJ XI + 30 XF XA + 90

XC XB + 15

XD XG + 14

XB XA + 90

C5

F25

G

All X’s are nonnegative

Minimize X(FIN)ST

41

• A schedule of (not necessarily the earliest) start time that will complete the project in minimal time.

• The earliest starts, returned when using the following objective function

Minimize XA + XB + XC +…+ XJ + X(FIN)

• The latest starts that can be obtained (after solving for the project earliest finish) as follows

– Add the constraint X(FIN) = 194– Change the objective function to

Maximize XA + XB + XC +…+ XJ + X(FIN).

– An Excel solution provides:

42

• A Network Approach– WINQSB provides

• Activity schedule chart (ES, EF, LS, LF).• Mean and standard deviation of the critical path.• Optional: Earliest and latest time Gantt chart.• Optional: The probability of completing the project within

a specified time.

– No programming is needed. – Input data consists of

• Immediate predecessors for each activity.• Completion time (for a, m, and b).

43

44

Critical Activities

WINQSB Scheduling Solution

45

7.9 Cost Analysis Using the Expected Value Approach

• Spending extra money, in general should decrease project duration.

• Is this operation cost effective?

• The expected value criterion is used to answer this question.

46

KLONE COMPUTERS -

Cost Analysis Using Probabilities

• Analysis indicated:

– Completion time within 180 days yields an additional profit of $1 million.

– Completion time between 180 days and 200 days, yields an additional profit of $400,000.

47

KLONE COMPUTERS -

Cost Analysis Using Probabilities

• Completion time reduction can be achieved by additional training.

• Two possible activities are considered for training.– Sales personnel training:

• Cost $200,000; • New time estimates are a = 19, m= 21, and b = 23 days.

– Technical staff training: • Cost $250,000; • New time estimates are a = 12, m = 14, and b = 16.

Which option should be pursued?

48

• Evaluation of spending on sales personnel training.– This activity (H) is not critical.– Under the assumption that the project completion time is

determined solely by critical activities, this option should not be considered further.

• Evaluation of spending on technical staff training.– This activity (F) is critical.– This option is further studied as follows:

• Calculate expected profit when not spending $250,000.• Calculate expected profit when spending $250,000.• Select the decision with a higher expected profit.

49

– Case 1: Do not spend $250,000 on training.

This case represents the current situation.Expected gross additional profit =P(completion time is less than 180)($1 million) + P(180<Completion time<200)($400,000) +P(Completion time> 200)(0) = P(X<180)($1 million) + P(180<X<200)($400,000) + P(X>200)(0)

[P(Z< )=P(Z< -1.51) = .0655; P(-1.51<Z<.65) = .6767]

Expected gross additional profit = .06555(1M)+.6767(400K)= = $366,180.

180-1949.255

50

– Case 2: Spend $250,000 on training. • The new time estimates change the expected time and standard deviation of a critical activity.

• The new estimates are= (12 + 4 (14) + 16)/6 = 14= (16 -12)/6 =0.67= 0(.67)2 =0.44

• The project has a new critical path (A-B-C-G-D-J), with a mean time = 189 days, and a standard deviation of = 9.0185 days.

180-1899.0185

P(Z< ) = P(Z< -.99) = 0.1611; P( - 0.99<Z< 1.22) = 0.7277

Expected Gross Additional Profit = P( completion time<180)(1M)+

P(180<completion time<200)(400K)+

P(completion time>200)(0) = 0.1611(1M)+0.7277(400K) = $452,180

Expected Gross Additional Profit = P( completion time<180)(1M)+

P(180<completion time<200)(400K)+

P(completion time>200)(0) = 0.1611(1M)+0.7277(400K) = $452,180

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– The expected net additional profit =

452,180-250,000 = $202,180 < $336,180

Expected additional net profit whenspending $250,000 on training

Expected profit without spending $250,000 on training

Conclusion: Management should not spend money on additional training of technical personnel

Conclusion: Management should not spend money on additional training of technical personnel

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7.10 The Critical Path Method (CPM)

• The critical path method (CPM) is a deterministic approach to project planning.

• Completion time depends only on the amount of money allocated to the activity.

• Reducing an activity’s completion time is called “crashing.”

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• There are two crucial completion times to consider for each activity.

– Normal completion time (NT). – Crash completion time (CT). – NT is achieved when a usual or normal cost (NC) is spent to complete the activity. – CT is achieved when a maximum crash cost (CC) is spent to complete the activity.

54

The Linearity Assumption

[Normal Time - Crash Time] [Normal Time]

=

[Crash Cost - Normal Cost]

[Normal Cost]

55

Time

Cost ($100)

20

18

1614

1210

8

642

5 10 15 20 25 30 35 40 45

NormalNC = $2000NT = 20 days

A demonstration of the

Linearity assumption

Add to the normal cost...

…and save oncompletion time

Add more to the normal cost...

CrashingCC = $4400CT = 12 days

…and save more on completion time Add 25% of the extra

cost to the normal cost

Save 25% of the max. time reduction

Total Cost = $2600Job time = 18 days

Marginal Cost = Additional Cost to get Max. Time Reduction

Maximum Time reduction= (4400 - 2000)/(20 - 12) = $300 per day

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• Meetings a Deadline at Minimum Cost

– Let D be the deadline date to complete a project.

– If D cannot be met using normal times, additional resources must be spent on crashing activities.

– The objective is to meet the deadline D at minimal additional cost.

– Tom Larkin’s political campaign problem illustrates the concept.

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TOM LARKIN’S POLITICAL CAMPAIGN• Tom Larkin has to plan 26 weeks of his mayoral campaign activities.• The campaign consists of the following activities

Immediate Normal Schedule Normal Schedule Reduced ScheduleActivity Predecessor Time Cost Time Cost

A. Hire campain staff None 4 2.0K 2 5.0KB. Prepare position paper None 6 3.0 3 9C. Recruit volunteers A 4 4.5 2 10D. Raise funds A,B 6 2.5 4 10E. File candidacy papers D 2 0.5 1 1F. Prepare campaign material E 13 13.0 8 25G. Locate/staff headquarters E 1 1.5 1 1.5H. Run personal campaign C,G 20 6.0 10 23.5I. Run media campaign F 9 7.0 5 16

58

A

D

C

BE

F

G

I

H

FINISH

Project completion (normal) time = 36 weeks

NETWORK PRESENTATIONNETWORK PRESENTATION

To meet the deadline date of 26 weeks some activities must be crashed.

60

Activity NT NC($) CT CC T M($)A 4 2000 2 5000 2 $1,500B 6 3000 3 9000 3 2000C 4 4500 2 10000 2 2750D 6 2500 4 10000 2 3750E 2 500 1 1000 1 500F 13 13000 8 25000 5 2400G 1 1500 1 1500 *** ***H 20 6000 10 23500 10 1750I 9 7000 5 16000 4 2250

Mayoral Campaign Crash Schedule

61

• Linear Programming Approach– Variables

Xj = start time for activity j.

Yj = the amount of crash in activity j.

– Objective FunctionMinimize the total additional funds spent on crashing activities.

– Constraints• The project must be completed by the deadline date D.• No activity can be reduced more than its Max. time reductioN.• Start time of an activity takes place not before the Finish time of an

immediate predecessor.

SOLUTION

62

JHFEDCBA Y2250Y17500Y2400Y500Y37502750Y2000YMin1500Y Minimize total crashing costs

26)FIN(X

STMeet the deadline

10

5

1

2

2

3

2

H

F

E

D

C

B

A

Y

Y

Y

Y

Y

Y

Y

Maximum time-reductionconstraints

AAC

AAD

BBD

DDE

EEF

EEG

CCH

GH

FFI

HH

II

YXX

YXX

YXX

YXX

YXX

YXX

YXX

XX

YXX

YXFINX

YXFINX

4

4

6

6

2

2

4

1

13

20)(

9)(

Activity canstart only afterall the predecessorsare completed.

A

D

C

BE F

G

I

H

FINISH

63

• Other cases of project crashing

– Operating optimally within a given budget.• When a budget is given, minimizing crashing costs is a constraint,

not an objective.• In this case the objective is to minimize the completion time.

64

TOM LARKIN - Continued• The budget is $75,000.

The objective function becomes a constraint Minimize

1500 YA+ 2000 YB + 2750 YC + 3750 YD + 500 YE + 2400 YF

+1750 YH + 2250 YJ

1500 YA+ 2000 YB + 2750 YC + 3750 YD + 500 YE + 2400 YF

+1750 YH + 2250 YJ 75,000 - 40,000 = 35,000

This constraint becomes the objective functionX(FIN) 26

X(FIN)

The rest of other crashing model constraints remain the same.