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TCOM 5143Lecture 11

Project Management

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What’s Project Management?

• Project management involves planning and scheduling the different activities of a project.

• A project is made of a number of activities that all need to be completed in order to finish the project.

• Activities are usually interdependent on each other; i.e., some activities cannot start unless other activities are finished.

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Project management techniques such PERT/CPM can be used to answer questions such as the following:

•What is the total time to complete (finish) the project?

•What are the scheduled start and finish dates for each specific activity?

•Which activities are “critical” and must be completed exactly as scheduled in order to keep the project in schedule?

•How long can “noncritical” activities be delayed before they cause a delay in the total project?

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PERT/CPM networks

• PERT stands for Program Evaluation and Review Technique

• CPM stands for Critical Path Method

• The main step in project scheduling is to identify all activities or jobs of the project and the immediate predecessors of every activity.

• The immediate predecessors of a particular activity are the activities that should be completed first before the activity in question can start.

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Example 1.Consider the example project of buying a small business. The project activities are listed below.

Activity Description ImmediatePredecessor

A Develop a list of sources offinancing

__

B Analyze the financial recordsof the business

__

C Develop a business plan(e.g., sales and cash flowprojections)

B

D Submit a proposal to alending institution

A,C

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1

2

43

B

A

C

D

Activity Description ImmediatePredecessor

A Develop a list of sources offinancing

__

B Analyze the financial recordsof the business

__

C Develop a business plan(e.g., sales and cash flowprojections)

B

D Submit a proposal to alending institution

A,C

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• The activities are shown on the branches or arcs of the network.

• The nodes of the network correspond to the beginning and ending of the activities.

• The completion of all the activities that lead into a node is called an event. For example, node 3 corresponds to the event that both activities A and C have been completed.

1

2

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B

A

C

D

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Example 2Consider developing the network for a project having the following activities and immediate predecessors

1

2

4

3

B

A

CD

Activity Immediatepredecessor

A __

B __

C BD A,CE CF CG D,E,F

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Example 2 (cont.)

• Activity E starts after the completion of activity C. To represent this, it is incorrect to show activity E beginning at node 3 because if E starts at node 3 this implies that both A and C precede E, which is incorrect.

• To avoid the above problem, we insert in the network a dummy activity. A dummy variable is a fictitious variable used to ensure that the proper precedence relationships among the activities are shown in the network. A dummy activity is represented by a dashed line.

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Completion of the network for the example project could be shown as follows:

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D

1

2

4

3

B

A

C

E

5

D

1

2

4

3

B

A

C

E

F

6G

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•Rule: It is not acceptable to have two activities that start at the same node and finish at the same node.•To solve the above problem, we add a dummy activity.

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D

1

2

4

3

B

A

C

E

F

6G

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•Note: Dummy activities can be used to show precedence relationships correctly and to eliminate the possible confusion of two or more activities having the same starting and ending nodes.

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Project Scheduling with PERT/CPM• LAN example:Consider the project of

designing a telecommunication network LAN with the following activities.

Activity Immediateprodecessor

Completion time(weeks)

A -- 5B -- 6C A 4D A 3E A 1F E 4G D, F 14H B, C 12I G,H 2

Total time=51

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LAN example (cont.)

4

D

1

3

52

B

A

C

E

F

6

G

7

H

I

Can we complete the project in less than 51 weeks? Why?

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The critical path

• A path is a sequence of connected activities that leads from the starting node of the project to the completion node of the project.

• A path does not need to pass through all the nodes in the network.

• The longest path is called the critical path.

• The activities on the critical path are called the critical activities of the project.

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• Is the critical path useful?• Rule: The critical path determines the total time

required to finish the project.• Consider the following small example

• The longest path is the path 1-3-4-5 with total time equal to 12. A, D and E are critical activities

• Rule: The only way for the project manager to reduce the total project time is to reduce the length of the critical path by shortening the durations of the critical activities.

3

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5

1

2

4

3

B

A

C

51

E

D

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Start times• We assume that the project starts at time 0.

• Let

• ES(j)=Earliest start time for activity j

• EF(j)=Earliest finish time for activity j

• T(j)=duration time for activity j

• Rule: EF(j) = ES(j) + T(j)

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Start times (cont.)

• Rule: The earliest start time for an activity leaving a particular node is equal to the largest of the earliest finish times for all activities entering the node.

Equivalently,

• Let activity j leave node k, then

• ES(j)=Max{EF(i) for all activities i such that i enters node k}.

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1

2A [0,5]

Activity

Earliest start time

Earliest finish time

5Activity time

4

D[5,8]

1

3

52

B[0,6]

A [0,5]

C[5,9]

E[5,6]

F[6,10]

6

G[10,24]

7

H[9,21]

I[24,26]

5

3

14

6

4

12

14

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Rule: The total time required to finish the project is equal to the earliest finish time for the last activity of the project.

For the LAN example, it is 26 weeks.

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Finish timesLet

• LS(j)=Latest start time for activity j

• LF(j)=Latest finish time for activity j

• T(j)=duration time for activity j

Rule: LS(j) = LF(j) - T(j)

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Finish times (cont.)Rule: The latest finish time for an activity entering a particular node is equal to the smallest of the latest start times for all activities leaving the node.

Equivalently,• If we let activity j enter node k. Then• LF(j)=Min{LS(i) for all activities i such that i leaves

node k}.

In the example above, if LS(C)=12 and LS(E)=9, what is LF(B)?LF(B)=Min{LS(C),LS(E)}=Min12,9}=9

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3

B

C

51

E

4

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Finish times (cont.)

• We compute the latest start times and latest finish times by making a backward pass calculation. That is, using the total duration of the project as the latest finish time for the last activity of the project, we trace back trough the network, computing a latest start and latest time for each activity.

• In the small example, the duration of the project is 12.

3

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5

1

2

4

3

B[3,6]

A[0,2]

C[6,11]

51

E[11,12]

D[2,11]

LFLS

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Finish times (cont.)

• Let’s compute the latest finish times and latest start times for the activities of the LAN project.

• The latest finish time for activity I is 26 because it is the last activity in the project.

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D[7,10]

1

3

52

B[6,12]

A [0,5]

C[8,12]

E[5,6]

F[6,10]

6

G[10,24]

7

H[12,24]

I[24,26]

5

3

14

6

4

12

14

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Note in the LAN example, LF(A)=Min{LS(C), LS(D), LS(E)} = Min{8, 7, 5}=5.

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Slack of an activityDefinition:

• The slack of an activity is defined as the length of time an activity can be delayed without affecting the total time required to finish the project.

• Slack(j)=LS(j)-ES(j) = LF(j)-EF(j)

• The activities with zero slack correspond to the critical activities of the project.

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Activity schedule for the LAN example

Activity Duration ES LS EF LF Slack Critical?A 5 0 0 5 5 0 YesB 6 0 6 6 12 6 NoC 4 5 8 9 12 3 NoD 3 5 7 8 10 2 NoE 1 5 5 6 6 0 YesF 4 6 6 10 10 0 YesG 14 10 10 24 24 0 YesH 12 9 12 21 24 3 NoI 2 24 24 26 26 0 Yes

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Let’s answer the following questions regarding scheduling the activities of the LAN project.

1.What is the total time to complete the project?

Answer: The project can be completed in 26 weeks.

2. What are the scheduled start and completion times for each activity?

Answer: The activity schedule (see previous table) shows the earliest start, latest start, earliest finish, and latest finish times for each activity.

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3. Which activities are critical and must be completed exactly as scheduled in order the keep the project on schedule?

Answer: The five activities, A, E, F, G, and I, are the critical activities.

4. How long can “noncritical” activities be

delayed before they cause a delay in the

completion time for the project?

Answer: The activity schedule (see previous table) shows the slack time associated with each activity.

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Lab Upgrade Example.

C

1

3

52

B

A

D

6

F

7

E

G

2

1

2

3

3

2

1

28

C[2,3]

1

3

52

B[0,3]

A [0,2]

D[3,6]

6

F[6,8]

7

E[3,5]

G[5,6]

2[3,5]

1[5,6]

2[6,8]

3[0,3]

3[3,6]

2[5,7]

1[7,8]

ES EF

LS LF

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Activity ES LS EF LF Slack Critical?A 0 3 2 5 3 NoB 0 0 3 3 0 YesC 2 5 3 6 3 NoD 3 3 6 6 0 YesE 3 5 5 7 2 NoF 6 6 8 8 0 YesG 5 7 6 8 2 No

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Planning and scheduling project costs: PERT/Cost technique

• The budgeting process for project involves the following: Identify the costs for all the activities of the project. Develop a schedule or forecast of when the costs are

expected to incur. During various stages of project completion, compare

actual (incured) costs to scheduled (budgeted) costs, and take corrective actions if necessary.

In using the PERT/Cost technique, it is assumed that the cost of each activity occurs at a constant rate over the duration of the activity.

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• The costs for the activities for the lab Upgrade example are shown in the table below.

Activity Duration Budgeted cost Budgeted costper month

A 2 10000 5000B 3 30000 10000C 1 3000 3000D 3 6000 2000E 2 20000 10000F 2 10000 5000G 1 80000 8000 Total project budget = $ 87 000

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If the activities start at their earliest starting times, the month-by-month cost forecast is shown in the table below.

Budgeted costs for an earliest startingtime schedule (in $1000s) MonthActivity1 2 3 4 5 6 7 8

A 5 5B 10 10 10C 3D 2 2 2E 10 10F 5 5G 8Monthlycost

15 15 13 12 12 10 5 5

Totalproject cost

15 30 43 55 67 77 82 87

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• The figure below shows the forecasted total project costs for both the earliest and the latest starting time schedules.

• The area between the two curves shows the possible budgets for the project.

• For example, by month 3 total project costs should be between $30 000 (latest starting time schedule) and $43 000 (earliest starting time schedule).

Feasible Budgets for total project costs

0100002000030000400005000060000700008000090000

100000

1 2 3 4 5 6 7 8 9

Months

Bud

gete

d to

tal p

roje

ct c

ost

ES

LS

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Controlling project costs• Periodically (for example biweekly or monthly)

throughout the project's duration, the project manager should do the following:• Collect up-to-date information on the cost status of all

completed and in-process activities.• Identify the sources of cost overruns and cost

underruns and take corrective actions where necessary.

• There is a cost overrun for an activity when actual cost exceeds budgeted cost.

• There is a cost underrun when actual cost is less than budgeted cost.

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Lab upgrade example• At the end of the fourth month of the project, the

project manager has collected the actual cost and percent completion data for the project activities as shown in the table below.

Activity Duration Budgeted cost Actual cost Percentcompletion

A 2 10000 12000 100B 3 30000 30000 100C 1 3000 1000 50D 3 6000 2000 33.33E 2 20000 10000 25F 2 10000 0 0G 1 80000 0 0

Total = $55000

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Cost overrun and cost underrunLet

• ACi = actual cost to date for activity i

• Bi = budget for activity i

• pi = percent completion for activity i

• Vi = value of work completed for activity i

• Vi = (pi/100) Bi.

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Cost overrun and cost underrun (cont.)

• Cost overrun or cost underrun (Di) for an activity is computed as the difference between the actual cost and the value for work completed (that is, the appropriate budgeted cost).

Di = ACi - Vi

• If Di is positive, there is a cost overrun.

• If Di is negative, there is a cost underrun.

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Cost overrun and cost underrun (cont.)

Lab upgrade example

Activity Percentcompletion

Budgeted cost Value of workcompleted (Vi)

Actual cost(ACi)

Overrun/underrun(Di)

A 100 10000 10000 12000 2000B 100 30000 30000 30000 0C 50 3000 1500 1000 -500D 33.33 6000 2000 2000 0E 25 20000 5000 10000 5000F 0 10000 0 0 0G 0 80000 0 0 0

Total = $48500 Total =$55000

Total = $6500