Quick Recap

Monitoring and Controlling

Difference and connection

In put

Transformation processes

Out put

Feedforward control

Feedback control

Concurrent control

Scope Control

• Scope control involves controlling changes to the project scope

• Goals of scope control are to:– Influence the factors that cause scope changes– Assure changes are processed according to

procedures developed as part of integrated change control

– Manage changes when they occur• Variance is the difference between planned and

actual performance. Control variances

3

Lesson 12: Monitoring and Controlling Project Schedule and CostsTopic 12A: Control the Project ScheduleTopic 12B: Control Project Costs

5

Control Project Scheduling

6

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

• Objectives of Project Scheduling– Completing the project as early as possible by

determining the earliest start and finish of each activity.

– Calculating the likelihood a project will be completed within a certain time period.

– Finding the minimum cost schedule needed to complete the project by a certain date.

5.1 Introduction

7

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

• Objectives of Project Scheduling

– Investigating the results of possible delays in activity’s completion time.

– Progress control.– Smoothing out resource allocation over the

duration of the project.

5.1 Introduction

8

• 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.

Task Designate

9

Identifying the Activities of a Project

• To determine optimal schedules we need to– Identify all the project’s activities.– Determine the precedence relations among activities.

• Based on this information we can develop managerial tools for project control.

10

Identifying Activities, Example

XYZ. COMPUTERS, INC.

• XYZ. Computers manufactures personal computers.

• It is about to design, manufacture, and market the New Computer.

11

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

• XYZ. needs to develop a precedence relations chart.• The chart gives a concise set of tasks and their immediate predecessors.

XYZ. COMPUTERS, INC

12

Activity DescriptionA 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 Pre-production advertising campaign

J Post-redesign advertising campaign

XYZ. COMPUTERS, INC

13

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

Activity A is an immediate predecessorof activity B, because it must be competedjust prior to the commencement of B.

A B

XYZ. COMPUTERS, INC

14

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

XYZ. COMPUTERS, INC

15

The PERT/CPM Approach for Project Scheduling

• The PERT/CPM approach to project scheduling uses network presentation of the project to– Reflect activity precedence relations– Activity completion time

• PERT/CPM is used for scheduling activities such that the project’s completion time is minimized.

16

XYZ. COMPUTERS, INC. - Continued

• Management at XYZ. would like to schedule the activities so that the project is completed in minimal time.

• Management wishes to know:– The earliest and latest start times for each activity

which will not alter the earliest completion time of the project.

– 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.

17

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.

Earliest Start / Earliest Finish – Forward Pass

A90

B15

C5

F25

I30

G14

D20

E21

H28

J45

90,105

90,115

90,120

105,110

110,124115,129 129,149

149,170

149,177

120,165149,194

170

194

A

0,90

B

I

F

C

G D

E

H

J

177

194

EARLIEST FINISH

19

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.

20

B

F

C

A

I

E

DG HH28

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

194

129,149

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

Latest Start / Latest Finish – Backward Pass

21

• 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.

Slack Times

22

– 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

Slack Times

23

Critical activities

must be rigidly

scheduled

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

Slack time in the XYZ. Project

• 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

25

B

F

C

A

I

E

DG HH28

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

0,90129,149

G14

115,129

I30

119,149

A90

C5

110,115B15

95,110

F25

90, 1150,90

The Critical Path

26

• We observe two different types of delays:

– Single delays.– Multiple delays.

• Under certain conditions the overall project completion time will be delayed.

• The conditions that specify each case are presented next.

Possible Delays

27

• 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 non-critical 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.

Single delays

28LS =119

A90

J45

H28

E21

D20

I30

G14

F25

C5

B15

ES=149

LS=173 DELAYED START=149+15=164

ES=90DELAYED START=90+15 =105

Activity E and I are each delayed 15 days.

THE PROJECT COMPLETION TIME IS NOT DELAYED

FI NI SHMultiple delays of non critical activities:

Case 1: Activities on different paths

29

A90

90

B15

Gantt chart demonstration of the (no) effects on the project completion time when delayingactivity “I” and “E” by 15 days.

Activity I

F25

I30

105

C5

115

G14

129

D

20149

E21

H28

J

45

194

194Activity E

30

A90

B15

C5

F25

I30

G14

D20

E21

H28

J45

FINISH

ES=149

LS =173 DELAYED START=149+15 =164

ES=90DELAYED START =94

LS =95

THE PROJECT COMPLETION TIME IS NOT DELAYED

Multiple delays of non critical activities:Case 2: Activities are on the same path,

separated by critical activities.

Activity B is delayed 4 days, activity E is delayed 15 days

31

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

3 DAYS DELAYIN THE ENTIRE

PROJECT

Activity B is delayed 4 days; Activity C is delayed 4 days.

THE PROJECT COMPLETION TIME IS DELAYED 3 DAYS

Multiple delays of non critical activities:Case 2: Activities are on the same path, no

critical activities separating them.

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A Linear Programming Approach to PERT/CPM

• Variables– Xi = The start time of the activities for i=A, B, C,

…,J– X(FIN) = Finish time of the project

• Objective function– Complete the project in minimum time.

• Constraints– For each arc a constraint states that

the start time of M must not occur before the finish time of its immediate predecessor, L.

ML

33

A Linear Programming Approach

Define X(FIN) to be the finish time of the project. The objective then is

Minimize X(FIN)

While this objective function is intuitiveother objective functions provide moreinformation, and are presented later.

34

X(FIN) ³ XE + 21 X(FIN) ³ XH + 28X(FIN) ³ XJ + 45 XD ³ XG + 14 XE ³ XD + 20

XG ³ XC+ 5 XH ³ XD + 20 XG ³ XF+ 25 XJ ³ XD + 20 XI ³ XD+ 90XJ ³ XI + 30 XF ³ XA+ 90

XC ³ XB+ 15 XD ³ XG+ 14 XB ³ XA+ 90

G

C5

F25

All X s are nonnegative

Minimize X(FIN) ST

A Linear Programming Approach

35

Minimize XA+XB+…+XJ

This objective function ensures that the optimal X values are the earliest start times of all the activities. The project completion time is minimized.

Maximize XA+XB+…+XJ

S.T. X(FIN) = 194and all the other constraints as before.

This objective function and the additional constraint ensure that the optimal X values arethe latest start times of all the activities.

A Linear Programming Approach

36

Obtaining Results Using Excel

MEAN 194STANDARD DEVIATION* 0 * Assumes all critical activities are on one critical path

VARIANCE* 0 If not, enter in gold box, the variance on one critical path of interest.

PROBABILITY COMPLETE BEFORE =

Acitivty Node Critical ES EF LS LF SlackDesign A * 90 0 90 0 90 0Materials B 15 90 105 95 110 5Manufacture C 5 105 110 110 115 5Design Revision D * 20 129 149 129 149 0Production Run E 21 149 170 173 194 24Staff Training F * 25 90 115 90 115 0Staff Input G * 14 115 129 115 129 0Sales Training H 28 149 177 166 194 17Preprod. Advertise I 30 90 120 119 149 29Post. Advertise J * 45 149 194 149 194 0

CRITICAL PATH ANALYSIS

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Gantt Charts

• Gantt charts are used as a tool to monitor and control the project progress.

• A Gantt Chart is a graphical presentation that displays activities as follows:– Time is measured on the horizontal axis. A horizontal

bar is drawn proportionately to an activity’ s expected completion time.

– Each activity is listed on the vertical axis.• In an earliest time Gantt chart each bar begins and ends

at the earliest start/finish the activity can take place.

Here‘s how we build an Earliest Time Gantt Chart

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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

A90

90

B15

F25

I30

105

C5

115

G14

129

D

20

149

E21

H28

J

45

194

194

40

• Gantt chart can be used as a visual aid for tracking the progress of project activities.

• Appropriate percentage of a bar is shaded to document the completed work.

• The manager can easily see if the project is progressing on schedule (with respect to the earliest possible completion times).

Gantt Charts- Monitoring Project Progress

41

A90

B15

F25

I30

C5

G14

D

20

E21

H28

J

45

194

194

135

Monitoring Project Progress

The shaded bars representcompleted work BY DAY 135.

Do not conclude that the project is behind schedule.

Activity “I” has a slack and therefore can be delayed!!!

42

• Advantages. – Easy to construct– Gives earliest completion date.– Provides a schedule of earliest possible start and finish

times of activities.

• Disadvantages– Gives only one possible schedule (earliest).– Does not show whether the project is behind schedule.– Does not demonstrate the effects of delays in any one

activity on thestart of another activity, thus on the project completion time.

Gantt Charts – Advantages and Disadvantages

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Resource Leveling and Resource Allocation

• It is desired that resources are evenly spread out throughout the life of the project.

• Resource leveling methods (usually heuristics) are designed to: – Control resource requirements– Generate relatively similar usage of resources over

time.

44

• A heuristic approach to “level” expenditures– Assumptions

• Once an activity has started it is worked on continuously until it is completed.

• Costs can be allocated equally throughout an activity duration.

Step 1: Consider the schedule that begins each activity at its ES. Step 2: Determine which activity has slack at periods of peak spending.Step 3: Attempt to reschedule the non-critical activities performed during these peak periods to periods of less spending, but within the time period between their ES and LF.

Resource Leveling – A Heuristic

45

• Management wishes to schedule the project such that – Completion time is 194 days.– Daily expenditures are kept as constant as possible.

• To perform this analysis cost estimates for each activity will be needed.

Resource Leveling –XYZ. COMPUTERS. - continued

46

Total Total CostCost Time per

Activity Description (x10000) (days) DayA Prototype model design 2250 90 25B Purchase of materials 180 15 12C Manufacture of prototype 90 5 18D Revision of design 300 20 15E Initial production run 231 21 11F Staff training 250 25 10G Staff input on prototype 70 14 5H Sales traini ng 392 28 14I Pre-production advertisement 510 30 17J Post-production advertisement 1350 45 30

Total cost = 5,623

Resource Leveling –XYZ. COMPUTERS,– cost estimates

47

Cumulative Daily Expenditure – Earliest Times vs. Latest Times55

50

45

40

35

30

25

201510 5

20 40 60 80 100 120 140 160 180 200

Level Budget

Earliest Start-Earliest FinishBudget

Latest Start-Latest FinishBudget

FeasibleBudgets

Time

48

55

50

45

40

35

30

25

201510 5

20 40 60 80 100 120 140 160 180 200

E

H

JJ

Cost Leveling

E

J

E

J

H

J

H

C

F A

B

F

I

I

H

27

15

44

30

Daily Expenditure of the ES Schedule55

I I I I I

25I

I39

45I

ES = 90I

LS = 110

22

32

I

I

I

F

G

D

49

G

55

50

45

40

35

30

25

201510 5

20 40 60 80 100 120 140 160 180 200

E

H

JJ

IE

J

E

J

H

JC

F A

B

F

I

I

H

25 27

22

15

44

3032

55

H

H

Cost Leveling

ID

I

F

50

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

51

• 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.

The Probability Approach – Three Time Estimates

52

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

6= the standard deviation =

b - a6

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

The Distribution, Mean, and Standard Deviation of an Activity

53

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

The Project Completion Time Distribution - Assumptions

54

• 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.

The Project Completion Time Distribution - 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.

55

Mean = Sum of mean completion times along

the critical path.

The three assumptions imply that the overall project

completion time is normally distributed, the following

parameters:

The Project Completion Time Distribution

Variance = Sum of completion time variances

along the critical path.

Standard deviation = ÖVariance

56

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

The Probability Approach – XYZ. COMPUTERS

57

• Management at XYZ. is interested in information regarding the completion time of the project.

• The probabilistic nature of the completion time must be considered.

The Probability Approach – XYZ. COMPUTERS

58

A = [76+4(86)+120]/6 = 90A = (120 - 76)/6 = 7.33 A

2 = (7.33)2 = 53.78

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

XYZ. COMPUTERS – Finding activities’ mean and variance

59

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

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

– Expected completion time = mA +mF +mG +mD

+mJ=194.

– The project variance =sA2 +sF

2 +sG2 +sD

2 +sJ2 = 85.66

– The standard deviation = = 9.255

XYZ. COMPUTERS – Finding mean and variance for the critical

path

60

• The probability of completion in 194 days =

P(X 194) = P(Z194 - 194

9.255 ) ( ) .P Z 0 0 5

The Probability Approach – Probabilistic analysis

194

61

• An interval in which we are reasonably sure the completion date lies is ± z 0.025

The Probability Approach – Probabilistic analysis

.95

• The interval is = 194 ± 1.96(9.255) @ [175, 213] days.• The probability that the completion time lies in the interval

[175,213] is 0.95.

62

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

The Probability Approach – Probabilistic analysis

63

• The probability that the completion time is longer than 210 days =

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

XZ

1940

.4582

2101.73

?0.0418

The Probability Approach – Probabilistic analysis

The Probability Approach – Critical path spreadsheet

MEAN 194STANDARD DEVIATION* 9.255629 * Assumes all critical activities are on one critical path

VARIANCE* 85.66667 If not, enter in gold box, the variance on one critical path of interest.

PROBABILITY COMPLETE BEFORE 180 = 0.065192

Acitivty Node Critical ES EF LS LF SlackDesign A * 90 7.333333 53.77778 0 90 0 90 0Materials B 15 1 1 90 105 95 110 5Manufacture C 5 0.333333 0.111111 105 110 110 115 5Design Revision D * 20 3 9 129 149 129 149 0Production Run E 21 1 1 149 170 173 194 24Staff Training F * 25 2.333333 5.444444 90 115 90 115 0Staff Input G * 14 2 4 115 129 115 129 0Sales Training H 28 1.333333 1.777778 149 177 166 194 17Preprod. Advertise I 30 4.666667 21.77778 90 120 119 149 29Post. Advertise J * 45 3.666667 13.44444 149 194 149 194 0

CRITICAL PATH ANALYSIS

The Probability Approach – critical path spreadsheet

MEAN 189STANDARD DEVIATION* 9.0185 * Assumes all critical activities are on one critical path

VARIANCE* 81.33333 If not, enter in gold box, the variance on one critical path of interest.

PROBABILITY COMPLETE BEFORE 180 = 0.159152

Acitivty Node Critical ES EF LS LF SlackDesign A * 90 7.333333 53.77778 0 90 0 90 0Materials B * 15 1 1 90 105 90 105 0Manufacture C * 5 0.333333 0.111111 105 110 105 110 0Design Revision D * 20 3 9 124 144 124 144 0Production Run E 21 1 1 144 165 168 189 24Staff Training F 14 0.666667 0.444444 90 104 96 110 6Staff Input G * 14 2 4 110 124 110 124 0Sales Training H 28 1.333333 1.777778 144 172 161 189 17Preprod. Advertise I 30 4.666667 21.77778 90 120 114 144 24Post. Advertise J * 45 3.666667 13.44444 144 189 144 189 0

CRITICAL PATH ANALYSISA comment – multiple critical paths • In the case of multiple critical paths (a not unusual situation), determine the probabilities for each critical path separately using its standard deviation. • However, the probabilities of interest (for example, P(X ³ x)) cannot be determined by each path alone. To find these probabilities, check whether the paths are independent.• If the paths are independent (no common activities among

the paths), multiply the probabilities of all the paths: [Pr(Completion time³x) = Pr(Path 1³x)P(Path 2³x)…Path k³x)]• If the paths are dependent, the calculations might become

very cumbersome, in which case running a computer simulation seems to be more practical.