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TCOM 5143 Lecture 11. Project Management. 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.
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TCOM 5143Lecture 11 Project Management
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.
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?
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.
Example 1.Consider the example project of buying a small business. The project activities are listed below.
A D 1 3 4 C B 2
A D 1 3 4 C B 2 • 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.
A 1 3 D C B 4 2 Example 2Consider developing the network for a project having the following activities and immediate predecessors
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.
3 D A G 4 6 E 1 F 5 B C 2 3 A D 1 E 5 B C 4 2 Completion of the network for the example project could be shown as follows:
3 D A G 4 6 E 1 5 F B C 7 2 • 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. • 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.
Project Scheduling with PERT/CPM • LAN example:Consider the project of designing a telecommunication network LAN with the following activities.
D 2 5 E A G F C I 1 4 6 7 B H 3 LAN example (cont.) Can we complete the project in less than 51 weeks? Why?
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.
3 D A 9 2 E 1 4 5 B 1 3 C 5 2 • 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.
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)
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}.
Earliest start time Earliest finish time D[5,8] 2 5 3 E[5,6] G[10,24] A [0,5] 1 F[6,10] 5 14 4 4 C[5,9] 1 4 7 I[24,26] B[0,6] 6 2 6 H[9,21] 3 12 Activity [0,5] A 2 1 5 Activity time 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.
Finish times Let • 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)
3 C 4 B E 2 4 2 1 5 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
LS LF 3 D[2,11] A[0,2] 9 2 E[11,12] 1 4 5 B[3,6] 1 3 C[6,11] 5 2 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.
D[7,10] 2 5 3 E[5,6] A [0,5] 1 F[6,10] G[10,24] 5 14 4 4 C[8,12] 1 4 7 I[24,26] B[6,12] 6 2 6 H[12,24] 3 12 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. Note in the LAN example, LF(A)=Min{LS(C), LS(D), LS(E)} = Min{8, 7, 5}=5.
Slack of an activity Definition: • 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.
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.
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.
C 2 5 1 F A 2 2 D 1 3 7 B G 1 3 E 6 3 2 Lab Upgrade Example.
LS LF ES EF C[2,3] 2 5 1[5,6] F[6,8] A [0,2] 2[6,8] 2[3,5] D[3,6] 1 3[3,6] 7 B[0,3] G[5,6] 1[7,8] 3[0,3] E[3,5] 6 3 2[5,7]
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.
The costs for the activities for the lab Upgrade example are shown in the table below.
If the activities start at their earliest starting times, the month-by-month cost forecast is shown in the table below.
ES LS • 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).
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.
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.
Cost overrun and cost underrun Let • 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.
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.
Cost overrun and cost underrun (cont.) Lab upgrade example