A scheme developed by Du Pont to figure out

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Rodger Fox
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project schedule given task durations and their precedence in a

1 CPM Project Management scheme. A scheme developed by Du Pont to figure out Length of a normal project schedule given task durations and their precedence in a network type layout (or Gantt chart) Two examples

2 In a project schedule: The longest path in the project layout identifying critical activities Crash costing the project if crash-cost figures are available: decreasing the completion time by pouring more resources to buy time Observation: A project layout in terms of nodes (milestones) and directed edges (activities) is a directed graph. C(1) F(3) A(3) start D(2) finish B(4) E(4) The project graph suggests the following information: This format shows activity on node network design Activity times are deterministic An arrow leads from tail to head directionally indicate ACTIVITY, a time consuming effort that is required to perform a part of the work.

3 A node is represented by a indicating an EVENT, a point in time where one or more activities start and/or finish. An alternative representation would be activity on edge (or an arrow) while a node is the completion of a task C(1) A(3) D(2) start B(4) Given these we observe the following Tasks duration Predecessors A 3 none B 4 none C 1 A D 2 A,B E 4 D F 2 C Finish D,E,F Observations: If a task has no task before it, it can begin at a 0 time.

4 All tasks without any preceding tasks could be done concurrently if resources are there. Earliest start time ES of a task K is the time it can begin without any delay. We call this t ES (K). Earliest finishing time EF of a task K is the time it can be completed without incurring any other delay. We call this t EF (K). tef ( K) = tes ( K) + d( K), where d (K) = duration of K Suppose, L 1, L 2,..., L m are tasks which must finish before task K can start. Task K has a m fold dependency as L 1 L 2 L 3 K Then, t ( K) = max{ t L( j), j [1. m] } ES L m EF Using these rules, create the following table. task durn pred ES EF A 3 none 0 3 B 4 none 0 4 C 1 A 3 4

5 D 2 A,B 4 6 E 4 D 6 10 F 2 C 4 6 Finish D,E,F It says the project would finish at time 10. Given this information in addition to what we know, we compute the rest of the table over the given set of tasks. For each task K, we compute the latest finish ( LF ) and the latest start time ( LS ) considering them in reverse order starting from finish. For all tasks connected to the finish node, the latest finish time ( LF ) would be the t ( finish) t ( finish) EF = ES The latest start time ( LS ) of a task K is t ( K) = t ( k) d( K) LS LF For any task K if we have successor tasks shown below 1, 2 L L,..., L m t LF { t L( j), j [1. ]} ( K) = min m LS

6 L 1 K L 2 L 3 L m Using these rules, proceed to compute LS and LF times of tasks from finish to start. We get the following table with ES, EF, LS and LF time estimates. tasks durn succ ES EF LS LF slack A 3 C,D B 4 D C 1 F D 2 Finish,E E 4 finish F 2 finish Finish none If a particular task shows s units of slack, that specific task could be delayed by s units of time. The critical path on the graph is the path with nodes which offer no slack.

7 None of the activities on this path could be delayed. The critical path is the one that takes longest amount of time. It is the minimum time a project needs for completion given the resources C(1) F(3) A(3) start D(2) finish B(4) E(4) Rationale for Critical Path Method Useful for a very large project Mathematically simple to work with It gives one or more critical paths and slack times Provide project documentation Useful to monitor and control the project in terms of Completion date? On schedule? Within budget? Critical activities? Can we speed up the project? At what extra cost? Dummy activities.

8 Sometimes we need to incorporate activities without any duration, but to assert a required dependency. Such activities are called dummy activities. Suppose, we require that F must wait until both C and D complete in the above diagram. We could do this by adding dummy activity as below. C(1) F(3) A(3) G(0) start D(2) finish B(4) E(4) Another example. A small project.

9 Suppose we can speed up the process up to a point if we are willing to pay extra on critical activities. This is called project crashing. From Du Pont s perspective, our project activities could be shown in terms of time duration (normal) and cost (normal) activity predecessors normal time normal cost A none 3 $4000 B none 4 $3500 C A 1 $5200 D A,B 2 $4200 E D 4 $2500 F C 2 $3000 Finish D,E,F Total cost: $22400 What if we could introduce crash time and cost as shown below in the following table? act pred Time (norm) Cost Time Cost Crash cost (norm) (crash) (crash) Per period A - 3 $ $5000 $500 B - 4 $ $6500 $1500 C A 1 $5200 * * D A,B 2 $ $5500 $1300 E D 4 $ $6000 $1750 F C 2 $ $3500 $500 fin D,E,F -

10 For instance, the table shows we can reduce the duration of B from 4 to 2, and it would cost us extra $3000. Perhaps we don t need to spend so much. If we reduce it by 1 unit, it d cost us $1500 per unit. Therefore, we have a variety of choices. Case I: Reduce CP (Critical Path) by one day. Since the current CP involves activities B, D, and E, we need to reduce one of them. The cheapest way to do this would be reduce D from 2 to 1 (extra cost: $1300) CP now: start-b-d-e-fin (time: 9 units) Total cost: $22400+$1300 = $23700 Case II: Reduce CP by two days The cheapest way to do this would be reduce D from 2 to 1 (extra cost: $1300) reduce B from 4 to 3 (extra cost: $1500) CP now: start-b-d-e-fin (time: 8 units) Total cost: $ $1500 = $25200 Case III: Reduce CP by three days The cheapest way to do this would be

11 reduce D from 2 to 1 (extra cost: $1300) reduce B from 4 to 2 (extra cost: $1500) 2 CPs now: start-b-d-e-fin (time: 7 units) and Start-A-C-F- fin Total cost: $ $1500 = $26700 In this we assumed the crash cost increases at a constant rate. In general, it may be non-linear such as shown below: Cost, $ normal cost time The total cost is usually perceived as Total cost = Overhead cost (indirect cost) + Direct cost and are like one below: overhead cost direct cost minimum

PROGRAMMING CPM. Network Analysis. Goal:

PROGRAMMING CPM. Network Analysis. Goal: PROGRAMMING CPM 5/21/13 Network Analysis Goal: Calculate completion time Calculate start and finish date for each activity Identify critical activities Identify requirements and flows of resources (materials,