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Dr. Honghui Deng

MIS 746 IS Project Management. Dr. Honghui Deng. Associate Professor MIS Department UNLV. Q&A. What are the Three Constraints of Project Management? What are the Time Killers for Project? Is time a resource for project management?. Discussion.

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Dr. Honghui Deng

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  1. MIS 746 IS Project Management Dr. Honghui Deng Associate Professor MIS Department UNLV

  2. Q&A • What are the Three Constraints of Project Management? • What are the Time Killers for Project? • Is time a resource for project management?

  3. Discussion • Consider the idea that time is a resource but also a constraint for IT project management, • Is this a contradiction?

  4. Review of PERT/CPM Video 5 PERT/CPM

  5. PERT/CPM PERT (Program Evaluation Review Technique)/ CPM (Critical Path Method) is a method of scheduling tasks. It shows sequence by detailed activities and time (hours, days, weeks). It helps determine which task may become a bottleneck and delay the entire project. PERT is a planning and control tool that helps accomplish project objectives on time. It graphically illustrates the interrelationships of events and activities required to bring a project to its successful conclusion.

  6. A management tool As a management tool, PERT helps the analysts to • Coordinate the various project tasks. • See the relative importance of each activity. • Determine the time to complete each activity. • Identify tasks that will delay the entire project if they are not completed as scheduled. • Reschedule activities in order to reduce the total time. • Controlproject progress.

  7. A network of activities PERT is defined in terms of network. The network consists of events and activities. All of the required events are connected by arrows that indicate the preceding and succeeding events. An event (A, B, C, D, or E) is the beginning or ending of an activity. An event can be considered as a milestone. Events have no time dimension and usually are represented by a circle. An activity(presented by an arrow)links two successive events together and represents the work required between these two events. An activity must be accomplished before the following event can occur. Let’s look at an example.

  8. A simple example Consider the list of four activities for buying a software for your office: ActivityDescriptionImmediate predecessors A Identify needs and a vendor - B Identify the source for funding - C Develop a proposal for funding B D Submit the proposal A,C The immediate predecessors for any activity are those activities that, when completed, enable the start of that activity.

  9. Sequence of activities • We can start work on activities A and B anytime, since neither of these activities depends upon the completion of prior activities. • C activity cannot start until activity B has been completed, and activity D cannot start until both activities A and C have been completed. • The graphical representation (next slide) is referred to as the PERT/CPM network for project.

  10. Network of four activities Arrows indicate project activities A D 1 3 4 C B 2 Nodes indicate start and finish of activities

  11. Another example Develop the network for a project with following activities and immediate predecessors: ActivityImmediate predecessors A - B - C B D A, C E C F C G D,E,F First, attempt for the first five (A,B,C,D,E) activities

  12. Network of first five activities A D 1 3 4 E B C 5 2 We need to introduce a dummy activity

  13. Network of seven activities A D 1 3 4 G 7 E B C 5 F 2 6 • Note how the network correctly identifies D, E, and F as the immediate predecessors for activity G. • Dummy activities can be used to identify precedence relationships correctly as well as to eliminate the possible confusion of two or more activities having the same starting and ending nodes.

  14. Scheduling with activity time ActivityImmediate Completion predecessorsTime (week) A - 5 B - 6 C A 4 D A 3 E A 1 F E 4 G D,F 14 H B,C 12 I G,H 2 Total …… 51 This information indicates that the total time required to complete activities is 51 weeks. However, we can see from the network that several of the activities can be carried out simultaneously (A and B, for example).

  15. Network with activity time D 3 5 2 G 14 E 1 F 4 A 5 7 4 I 2 C 4 1 6 B 6 H 12 3 Each activity letter is written above and each activity time is written bellow the arc To complete the total project completion time we will have to analyze the network and identify what is called critical path. A path is a sequence of connected activities that leads from the starting node (1) to the completion node (7).

  16. Earliest start & earliest finish time • We are interested in the longest path through the network, i.e., the critical path. • Starting at the network’s origin (node 1) and using a starting time of 0, we compute an earliest start (ES) and earliest finish (EF) time for each activity in the network. • The expression EF = ES + t can be used to find the earliest finish time for a given activity. For example, for activity A, ES = 0 and t = 5; thus the earliest finish time for activity A is EF = 0 + 5 = 5

  17. Arc with ES & EF time ES = earliest finish time ES = earliest start time Activity 2 A [0,5] 5 1 t = expected activity time

  18. Network with ES & EF time D[5,8] 3 5 2 G[10,24] 14 F[6,10] 4 E[5,6] 1 A[0,5] 5 7 I[24,26] 2 4 C[5,9] 4 1 6 H[9,21] 12 B[0,6] 6 3 Earliest start time 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.

  19. Latest start & latest finish time • To find the critical path we need a backward pass calculation. • Starting at the completion point (node 7) and using a latest finish time (LF) of 26 for activity I, we trace back through the network computing a latest start (LS) and latest finish time for each activity. • The expression LS = LF – t can be used to calculate latest start time for each activity. For example, for activity I, LF = 26 and t = 2, thus the latest start time for activity I is LS = 26 – 2 = 24

  20. Network with LS & LF time D[5,8] 3[7,10] 5 2 G[10,24] 14[10,24] F[6,10] 4[6,10] E[5,6] 1[5,6] A[0,5] 5[0,5] 7 I[24,26] 2[24,26] 4 C[5,9] 4[8,12] 1 6 H[9,21] 12[12,24] B[0,6] 6[6,12] 3 Latest finish time 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.

  21. Activity, duration, ES, EF, LS, LF EF = earliest finish time ES = earliest start time Activity 3 C [5,9] 4 [8,12] 2 LF = latest finish time LS = latest start time

  22. Slack or free time Slack = LS – ES = LF – EF Slack is the length of time an activity can be delayed without affecting the completion date for the entire project. For example, slack for C = LS – ES = 8 – 5 = 3 weeks or LF – EF = 12 – 9 = 3 weeks. This means activity C can be delayed up to 3 weeks (start anywhere between weeks 5 and 8).

  23. Activity schedule for our example

  24. Important questions • What is the total time to complete the project? • 26 weeks if all individual activities are completed on schedule. • What are the scheduled start and completion times for each activity? • ES, EF, LS, LF are given for each activity. • What activities are critical and must be completed as scheduled in order to keep the project on time? • Critical path activities: A, E, F, G, and I. • How long can non-critical activities be delayed before they cause a delay in the project’s completion time • Slack time available for all activities are given.

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