Project Management

Project Management Techniques

Project Management Questions: Why do we need to study Project Management? How does a project management technique work? ( to p4) ( to p5)

Questions:

Why do we need to study Project Management?

How does a project management technique work?

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 ( to p3)

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

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)

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

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 Special case! ( to p7) ( to p9) ( to p4) -- A B 20 30 10 A B C Precedent constraints Processing Time Event

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

Special case!

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 ( to p8) ( to p6)

Here, we use three symbols:

node to represent stage

line/branch to represent event

arrow to represent precedent

constraint

Example

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) -- A B 20 30 10 A B C 1-2 2-3 3-4 Pred Const Proc Time Event Path

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? ( to p10) -- A A 3 5 7 A B C Precedent constraints Processing Time Event

When two or events taken places in the same time interval

(known an concurrent events)

Consider the following example!

How to draw it?

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 Solution ( to p11)

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: What one is better? A dummy activity shows a precedence relationship Reflects no processing time ( to p6)

Three ways to draw it:

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) ( to p15) ( to p13)

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)

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 ( to p14)

The Project Network Completion time for: path A: 1  2  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)

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)

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?

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 ES ij = max (EF i ) EF ij = ES i + t ij with EF 1 =0 Critical path when LS-ES=0 ( to p4) ( to p17) ( to p26) ( to p22) LF LS EF ES Branch

Steps:

Prepare the project network

Construct a table as follows:

Compute ES and EF

Compute LS and LF

Compute LS-ES or LF-EF

Compute ES and EF Note: When computing these values, the pattern is like moving zic-zac format by firstly computer ES 12 and then adding it to EF 12 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: ( to p18) ( to p22) ( to p19)

Note:

When computing these values, the pattern is like moving zic-zac format by firstly computer ES 12 and then adding it to EF 12 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:

The starting point of ES and EF Consider: Then EF 1 = 0 ES 12 = max (EF 1 ) EF 12 = ES 12 + t 12 = 0 = 0 + t 12 1 2 t 12 ( to p17)

Consider:

Then

EF 1 = 0

ES 12 = max (EF 1 ) EF 12 = ES 12 + t 12

= 0 = 0 + t 12

The overall computation is shown in next slide ( to p20) EF 12 =ES 12 +t 12 = EF 23 =ES 23 +t 23 = EF 24 = EF 34 = EF 45 = EF 46 = EF 56 = EF 67 = ES 12 = max(EF 1 )= ES 23 =max(EF 2 )= ES 24 =max(EF 2 )= ES 34 =max(EF 3 )= ES 45 =max(EF 4 )= ES 46 =max(EF 4 )= ES 56 =max(EF 5 )= ES 67 =max(EF 6 )= 1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 EF ij =ES ij +t ij ES ij = max(EF i ) Branches

- ES is the earliest time an activity can start. ES ij = Maximum (EF i ) - EF is the earliest start time plus the activity time. EF ij = ES ij + t ij (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 t i for note 2 Max(node4+t46,node5+t56 =max(5+3,5+1)=8 Complete solution ( to p4) ( to p21)

The Project Network Activity Scheduling- Earliest Times - ES is the earliest time an activity can start. ES ij = Maximum (EF i ) - EF is the earliest start time plus the activity time. EF ij = ES ij + t ij Figure 8.6 Earliest activity start and finish times ( to p20)

Compute LS and LF Note: We compute these values from the bottom to top, with assigning: LS ij = LF i -t ij LF ij = min LS j with the end of LF ij = EF ij Example: computing Figure 8.3 ( to p23)

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

LS ij = LF i -t ij LF ij = min LS j

with

the end of LF ij = EF ij

Example: computing Figure 8.3

The overall computational is shown in next slide ( to p24) LF 12 =min(LS 2 )= LF 23 =min(LS 3 )= LF 24 =min(LS 4 )= LF 34 =min(LS 4 )= LF 45 =min(LS 5 )= LF 46 =min(LS 6 )= LF 56 =min(LS 6 )= LF 67 =min(LS 7 )= LS 12 = L i12 -t 12 = LS 23 = LF 23 -t 23 = LS 24 = LF 24 -t 24 = LS 34 = LF 34 -t 34 = LS 45 = LF 45 -t 45 = LS 46 = LF 46 - i46 = LS 56 = LF 56 -t 56 = LS 67 = LF 67 -t 67 = 1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 LF ij =min(LS j ) LS ij = LF ij -t ij Branches

- LS is the latest time an activity can start without delaying critical path time. LS ij = LF ij - t ij - LF is the latest finish time LF ij = Minimum (LS j ) 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 ( to p25) ( to p22)

The Project Network Activity Scheduling - Latest Times - LS is the latest time an activity can start without delaying critical path time. LS ij = LF ij - t ij - LF is the latest finish time LF ij = Minimum (LS j ) Figure 8.7 Latest activity start and finish times ( to p24)

Compute LS-ES or LF-EF Two ways you can achieve it: by compiling slack, S ij by showing branches ( to p27) ( to p28) ( to p16)

Two ways you can achieve it:

by compiling slack, S ij

by showing branches

The Project Network Calculating Activity Slack Time - Slack, S ij , computed as follows: S ij = LS ij - ES ij or S ij = LF ij - EF ij Table 8.2 Activity Slack Figure 8.9 Activity Slack * What does it mean? ( to p26)

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 ( to p26)

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.

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)

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

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)

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?

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)

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?

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)

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.

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)

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

PERT Example Step 1: computer t and v values Step 2: determine the CPM Step 3: determine v value ( to p35) ( to p36) ( to p37) ( to p33)

Step 1: computer t and v values

Step 2: determine the CPM

Step 3: determine v value

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 (t p ) and variance (v p ) interpreted as the mean (  ) and variance (  2 ) of a normal distribution:  = 25 weeks  2 = 6.9 weeks ( to p34)

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 (t p ) and variance (v p ) interpreted as the mean (  ) and variance (  2 ) of a normal distribution:

 = 25 weeks

 2 = 6.9 weeks

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)

Consider when

x = 30

x = 22

Tutorial Assignment

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)

 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?)

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, 17 New: 4, 6, 11 (to p43)

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

Exercises (Chapter 8)

Old: 8, 10, 17

New: 4, 6, 11

Probability Analysis of the Project Network CPM/PERT Analysis with QM for Windows Exhibit 8.1 (to p16)

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

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.

The Project Network Calculating Activity Slack Time - Slack, S ij , computed as follows: S ij = LS ij - ES ij or S ij = LF ij - EF ij Table 8.2 Activity Slack Figure 8.9 Activity Slack *