project management techniques

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Project Management

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Project

Management Techniques

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Project Management

Questions:

1. Why do we need to study Project Management?

2. How does a project management technique work?

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Objective

• The main purpose is to govern the operations of a project such that all activities involved are well administrated and that we can also control its completion time

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Project management technique

Steps to solve a project management

problem:

1. to represent a ‘project problem” graphically

2. to determine its completion time

3. to carry out sensitivity analysis, if any

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1. Represent a ‘project problem” graphically

Steps:

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

2. Draw a semantic network to represent them

Special case!

Event Processing Time

Precedent constraints

A

B

C

20

30

10

--

A

B

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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 precedentconstraint

Example (to p8)

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Example

1 2 3

Path Event Proc

Time

Pred

Const

1-2

2-3

3-4

A

B

C

20

30

10

--

A

B

A

4

CB

20 30 10

Rule1: All nodes must starts from oneNode and ends with one node

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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!

• How to draw it?

Event Processing Time

Precedent constraints

A

B

C

3

5

7

--

A

A

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Case 1

1 2 3A B

C

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Wrong! Rule2: no node can havetwo outcomes and end with the same note

Solution (to p11)

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Solutions for Rule 2

Three ways to draw it:

1 2

3

4

5A

B

C

Dummy 1=0

Dummy 2 = 0

1 2

3

4A B

CDummy = 0

1 2

3

4A

B

C

Dummy = 0

Solution 1:

Solution 2:

Solution 3:

What oneis better?

A dummy activity showsa precedence relationshipReflects no processing time

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2. Determine its completion time

Consider the project network as shown in next slide

Question: Is it an easy way to find out the solution?

Answer: YES, it knows as Critical Path Method (CPM)

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The Project Network All Possible Paths for Obtaining a Solution

Figure 8.3Expanded network

for building a house showing

concurrent activities.

Table 8.1Possible Paths to

complete the House-Building

Network

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

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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 8.5Alternative paths in the

network

This is theSolution!

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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 LF– For 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

(Do you know the reason why?)

How it works? (to p16)

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How CPM works?

Steps:1. Prepare the project

network

2. Construct a table as follows:

3. Compute ES and EF

4. Compute LS and LF

5. Compute LS-ES or LF-EF

Branch ES EF LS LF

ESij = max (EFi) EFij = ESi + tij

with EF1=0Critical path when LS-ES=0

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Compute ES and EF

Note: When computing these values, the pattern is like moving zic-zac format by firstly computer ES12 and then adding it to EF12

and move to next branch by copying the max values of the branch 1-2 to say, 2-3

We compute them from top to bottom!Their relationship : Example 1:

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The starting point of ES and EF

Consider:

Then

EF1 = 0

ES12 = max (EF1) EF12 = ES12 + t12

= 0 = 0 + t12

1 2t12

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Branches ESij = max(EFi) EFij=ESij+tij

1-2

2-3

2-4

3-4

4-5

4-6

5-6

6-7

ES12= 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=The overall computation is shown in next slide(to p20)

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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

- 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

(note:you can compute these values and show in the network diagram as well)

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

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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

Figure 8.6Earliest activity start and finish times

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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

Example: computing Figure 8.3 (to p23)

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Branches LSij = LFij-tij LFij=min(LSj)1-2

2-3

2-4

3-4

4-5

4-6

5-6

6-7

LS12 = 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)=

The overall computational is shown in next slide(to p24)

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- 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)

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

Start with the end node first Same as EF67from 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

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The Project Network Activity Scheduling - Latest Times

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

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

Figure 8.7Latest activity start and finish times

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Compute LS-ES or LF-EF

Two ways you can achieve it:

1. by compiling slack, Sij

2. by showing branches

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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 8.9Activity Slack

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What does it mean?

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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 8.8Earliest activity start and finish times

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Sensitivity Analysis

• Today, we only consider one case –

“Probabilistic Activity Times”

• Refer to activity time estimates usually can not be made with certainty

• PERT is known as the solution method(to p30)

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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)

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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?

6

b 4m a t

2

6

a - b

v

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PERT

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

• ES

• EF

• LS

• LF

• S

and branches with slack = 0 still consider as critical paths

• Example. (to p33)

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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 implicationApplications

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PERT Example

• Step 1: computer t and v values

• Step 2: determine the CPM

• Step 3: determine v value

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Step 1: computer t and v values

Figure 8.11Network with mean activity times and variances

Table 8.3Activity Time Estimates for

Figure 8.10

6

b 4m a t 2

6

a - b

v

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Step 2: determine the CPM

Figure 8.12Earliest and latest activity times

Table 8.4Activity Earliest and

Latest Times and Slack

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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

Critical Path Activity Variance

1 3 3 5 5 7 7 9

1 1/916/9 4

total 62/9

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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 8.13Normal distribution of network duration Critical value

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Consider when

x = 30

x = 22

Tutorial Assignment

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Probability Analysis of the Project NetworkExample 12 = 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 8.14Probability the network will be completed in 30 weeks or less

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Probability Analysis of the Project NetworkExample 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 8.15Probability the network will be completed in 22 weeks or less

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Tutorial Assignment

• Try to use QM to solve CPM/PERT problems (see slide 19)

• Exercises (Chapter 8)– Old: 8, 10, 17– New: 4, 6, 11

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Probability Analysis of the Project NetworkCPM/PERT Analysis with QM for Windows

Exhibit 8.1

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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 8.8Earliest activity start and finish times

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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 8.9Activity Slack

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