*

Project management technique

Steps to solve a project management

problem:

to represent a project problem graphically

to determine its completion time

to carry out sensitivity analysis, if any

(to p6)

(to p12)

(to p29)

*

1. Represent a project problem graphically

Steps:

Gather all information and organize them in a table format that consists of: event, processing time, and precedent constraints as follows:

Draw a semantic network to represent them

EventProcessing TimePrecedent constraintsABC203010--AB

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Semantic network to represent them

Here, we use three symbols:

node to represent stage

line/branch to represent event

arrow to represent precedent

constraint

*

Example

1

2

3

A

4

C

B

20

30

10

Rule1: All nodes must starts from one

Node and ends with one node

(to p7)

PathEventProcTimePredConst1-22-33-4ABC203010--AB

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Special case!

When two or events taken places in the same time interval(known an concurrent events)Consider the following example!EventProcessing TimePrecedent constraintsABC357--AA

*

Case 1

1

2

3

A

B

C

3

5

7

Wrong!

Rule2: no node can have

two outcomes and end

with the same note

*

Solutions for Rule 2

Three ways to draw it:

1

2

3

4

5

A

B

C

Dummy 1=0

Dummy 2 = 0

1

2

3

4

A

B

C

Dummy = 0

1

2

3

4

A

B

C

Dummy = 0

Solution 1:

Solution 2:

Solution 3:

A dummy activity shows

a precedence relationship

Reflects no processing time

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The Project Network
All Possible Paths for Obtaining a Solution

Figure 8.3

Expanded network for building a house showing

concurrent activities.

Table 8.1

Possible Paths to complete the

House-Building Network

Then the completion time for paths A, B, C and D can be computed as

*

The Project Network

Completion time for:

path A: 12 3 4 6 7, 3 + 2 + 0 + 3 + 1 = 9 months (Critical Path)

path B: 1 2 3 4 5 6 7, 3 + 2 + 0 + 1 + 1 + 1 = 8 months

path C: 1 2 4 6 7, 3 + 1 + 3 + 1 = 8 months

path D: 1 2 4 5 6 7, 3 + 1 + 1 + 1 + 1 = 7 months

The critical path is the longest path through the network; the minimum time the network can be completed.

Figure 8.5

Alternative paths in the

network

This is the

Solution!

(to p12)

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Critical Path Method (CPM)

General concepts:For each branch of the project network, we firstly determine four values of ES, EF, LS and LFFor each branch, we compute their slack time,Slack time = (LS-ES) or (LF-EF)The critical path is located at branch that has

slack time = 0

*

How CPM works?

Steps:

Prepare the project network

Construct a table as follows:

Compute ES and EF

Compute LS and LF

Compute LS-ES or LF-EF

ESij = max (EFi) EFij = ESi + tij

with EF1=0

Critical path when LS-ES=0

BranchESEFLSLF

*

The starting point of ES and EF

Consider:

Then

EF1 = 0

ES12 = max (EF1)EF12 = ES12 + t12

= 0= 0 + t12

1

2

t12

*

The overall computation is shown in next slide

(to p20)

BranchesESij = max(EFi)EFij=ESij+tij1-22-32-43-44-54-65-66-7ES12= max(EF1)=ES23=max(EF2)=ES24=max(EF2)=ES34=max(EF3)=ES45=max(EF4)=ES46=max(EF4)=ES56=max(EF5)=ES67=max(EF6)=EF12=ES12+t12=EF23=ES23+t23=EF24=EF34=EF45=EF46=EF56=EF67=

*

- ES is the earliest time an activity can start.ESij = Maximum (EFi)

- EF is the earliest start time plus the activity time.EFij = ESij + tij

Add all t to note 4 and take the longest time

Max (node 3+t34, node2+t24)

max (5+0, 3+1)

=max(5,4)=5

add all ti for note 2

Max(node4+t46,node5+t56

=max(5+3,5+1)=8

Complete solution

Branch

ESij = max (EFi )

EFij = ESij + tij

1 -2

2-3

ES12 = max (EF1) = 0

ES23 = max (EF2) = 3

EF12 = ES12 + t12

= 0 + 3 =3

EF23=ES23+t23

= 3 + 2 = 5

2-4

3-4

4 -5

ES24 = max(EF2) = 3

ES34= max (EF3) = 5

ES45= max (ES4) = 5

EF24=ES24+t24

=3 + 1 = 4

EF34=ES34 + t34

= 5 + 0 = 5

EF45 = ES45 + t45

= 5 + 1 = 6

4 -6

5-6

6-7

ES46=max(EF4) = 5

ES56=max(EF5) = 6

ES67=max(EF6) =8

EF46=ES46+t46

=5 + 3 = 8

EF56=ES56 +t56

=6 + 1 = 7

EF67=ES67+t67

= 8+ 1 = 9

*

The Project Network
Activity Scheduling- Earliest Times

- ES is the earliest time an activity can start.ESij = Maximum (EFi)

- EF is the earliest start time plus the activity time.EFij = ESij + tij

Earliest activity start and finish times

*

Compute LS and LF

Note: We compute these values from the bottom to top, with assigning:

LSij = LFi -tij LFij = min LSj

with

the end of LFij = EFij

*

The overall computational is shown in next slide

BranchesLSij = LFij-tijLFij=min(LSj)1-22-32-43-44-54-65-66-7LS12 = Li12-t12 =LS23 = LF23-t23 =LS24 = LF24-t24 =LS34 = LF34-t34 =LS45 = LF45-t45 =LS46 = LF46-i46 =LS56 = LF56-t56 =LS67 = LF67-t67 =LF12=min(LS2)=LF23=min(LS3)=LF24=min(LS4)=LF34=min(LS4)=LF45=min(LS5)=LF46=min(LS6)=LF56=min(LS6)=LF67=min(LS7)=

*

- LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij

- LF is the latest finish time LFij = Minimum (LSj)

Start with the end node first

Same as EF67

from the previous slide

Again, you can place these values onto the branches

Min(node 6-t46,node5-t45)

=Min(8-3,7-1)

=Min(5,6)=5

Min(node3-t23,node4-t24)

=Min(5-2,5-1)=Min(3,4)=3

Min(node 7-t67)

=Min(9-1)=8

Branches

LSij=LFij-tij

LFij=min LSj

1-2

2-3

2-4

LS12=LF12-t12 = 3-3 =0

LS23=LF23-t23=5-2=3

LS24=LF24-t24=5-1=4

LF12 = Min(LS2) =3

LF23=Min(LS3) = 5

LF24=Min(LS4)=5

3-4

4-5

4-6

LS34=LF34-t34=5-0 = 5

LS45=LF45-t45 = 7-1=6

LS46=LF46-t46=8-3=5

LF34=Min(LS4) = 5

LF45=Min(LS5)=7

LF46=Min(LS6)=8

5-6

6-7

LS56=LF56-t56=8-1=7

LS67=LF67-t67=9-1=8

LF56=Min(LS6)=8

LF67=Min(LS67)=9

*

The Project Network
Activity Scheduling - Latest Times

- LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij

- LF is the latest finish time LFij = Minimum (LSj)

Figure 8.7

Latest activity start and finish times

*

Compute LS-ES or LF-EF

Two ways you can achieve it:

by compiling slack, Sij

by showing branches

*

The Project Network
Calculating Activity Slack Time

- Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij

Table 8.2

Activity Slack

Figure 8.9

Activity Slack

*

What does it mean?

*

The Project Network
Activity Slack

Slack is the amount of time an activity can be delayed without delaying the project. Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. Shared slack is slack available for a sequence of activities.

Figure 8.8

Earliest activity start and finish times

*

Sensitivity Analysis

Today, we only consider one case

Probabilistic Activity Times

Refer to activity time estimates usually can not be made with certaintyPERT is known as the solution method

(to p30)

*

PERT

In PERT, three different time estimations are applied:

most likely time (m),

the optimistic time (a) , and

the pessimistic time (b).

How do we make use of these three values?

(to p31)

*

Probabilistic Activity Times

We used these values to estimate the mean and variance of a beta distribution:

mean (expected time):

variance:

How to use these values to solve a project network problem?

(to p32)

*

PERT

We simply apply t values in CPM and determine the values of:ESEFLSLFS

and branches with slack = 0 still consider as critical paths

Example.

(to p33)

*

Procedures for PERT

Step 1: based on the values of a, b and m, determine the t and v values for each path

Step 2: determine the critical path by using t values in the CPM

Step 3: compute its corresponding means and standard deviations according.

Example

Result implication

Applications

(to p34)

(to p38)

(to p39)

*

PERT Example

Step 1: computer t and v valuesStep 2: determine the CPMStep 3: determine v value

(to p35)

(to p36)

(to p37)

(to p33)

*

Step 1: computer t and v values

Figure 8.11

Network with mean activity times and variances

Table 8.3

Activity Time Estimates for

Figure 8.10

(to p34)

*

Step 2: determine the CPM

Figure 8.12

Earliest and latest activity times

Table 8.4

Activity Earliest and Latest Times and Slack

(to p34)

*

Step 3: determine v value

The expected project time is the sum of the expected times of the critical path activities. The project variance is the sum of the variances of the critical path activities. The expected project time is assumed to be normally distributed (based on central limit theorum).

In example, expected project time (tp) and variance (vp) interpreted as the mean () and variance (2) of a normal distribution:

= 25 weeks

2 = 6.9 weeks

(to p34)

Critical Path ActivityVariance

1 ( 3

3 ( 5

5 ( 7

7( 9

1

1/9

16/9

4

total 62/9

*

Probability Analysis of the Project Network

- Using normal distribution, probabilities are determined by computing number of standard deviations (Z) a value is from the mean.

- Value is used to find corresponding probability in Table A.1, App. A.

Figure 8.13

Normal distribution of network duration

Critical value

(to p33)

*

Consider when

x = 30

x = 22

Tutorial Assignment

(to p40)

(to p41)

(to p42)

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Probability Analysis of the Project Network
Example 1

2 = 6.9 = 2.63

Z = (x-)/ = (30 -25)/2.63 = 1.90

Z value of 1.90 corresponds to probability of .4713 in Appendix A of p715. Probability of completing project in 30 weeks or less : (.5000 + .4713) = .9713,

or 97.13% (Why so high a probability rate?)

Figure 8.14

Probability the network will be completed in 30 weeks or less

(to p39)

*

Probability Analysis of the Project Network
Example 2

Z = (22 - 25)/2.63 = -1.14

Z value of 1.14 (ignore negative) corresponds to probability of .3729 in Table A.1, appendix A.

Probability that customer will be retained is .1271 (= 0.5- 0.3729) , or 12.71%

(Again, why so low probability rate?)

Figure 8.15

Probability the network will be completed in 22 weeks or less

(to p39)

*

Tutorial Assignment

Try to use QM to solve CPM/PERT problems (see slide 19)Exercises (Chapter 8)Old: 8, 10, 17New: 4, 6, 11

(to p43)

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Probability Analysis of the Project Network
CPM/PERT Analysis with QM for Windows

Exhibit 8.1

(to p16)

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The Project Network
Activity Slack

Slack is the amount of time an activity can be delayed without delaying the project. Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. Shared slack is slack available for a sequence of activities.

Figure 8.8

Earliest activity start and finish times

*

The Project Network
Calculating Activity Slack Time

- Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij

Table 8.2

Activity Slack

Figure 8.9

Activity Slack

*

Branch ES

ij

= max (EF

i

)

EF

ij

= ES

ij

+ t

ij

1 -2

2-3

ES12 = max (EF1) = 0

ES23 = max (EF2) = 3

EF12 = ES12 + t12

= 0 + 3 =3

EF23=ES23+t23

= 3 + 2 = 5

2-4

3-4

4 -5

ES24 = max(EF2) = 3

ES34= max (EF3) = 5

ES45= max (ES4) = 5

EF24=ES24+t24

=3 + 1 = 4

EF34=ES34 + t34

= 5 + 0 = 5

EF45 = ES45 + t45

= 5 + 1 = 6

4 -6

5-6

6-7

ES46=max(EF4) = 5

ES56=max(EF5) = 6

ES67=max(EF6) =8

EF46=ES46+t46

=5 + 3 = 8

EF56=ES56 +t56

=6 + 1 = 7

EF67=ES67+t67

= 8+ 1 = 9

Branches

LSij=LFij-tij

LFij=min LSj

1-2

2-3

2-4

LS12=LF12-t12 = 3-3 =0

LS23=LF23-t23=5-2=3

LS24=LF24-t24=5-1=4

LF12 = Min(LS2) =3

LF23=Min(LS3) = 5

LF24=Min(LS4)=5

3-4

4-5

4-6

LS34=LF34-t34=5-0 = 5

LS45=LF45-t45 = 7-1=6

LS46=LF46-t46=8-3=5

LF34=Min(LS4) = 5

LF45=Min(LS5)=7

LF46=Min(LS6)=8

5-6

6-7

LS56=LF56-t56=8-1=7

LS67=LF67-t67=9-1=8

LF56=Min(LS6)=8

LF67=Min(LS67)=9

6

b

4m

a

t

+

+

=

2

6

a

-

b

=

v

2

6

a

-

b

=

v

Critical Path Activity

Variance

1

3

3

5

5

7

7

9

1

1/9

16/9

4

total 62/9