Critical path illustrates the activities and tasks that will delay the entire project if they are not completed on time. CPM helps managers to replan, reschedule, and reallocate labor and resources to complete critical tasks by identifying non-critical activities.
Process of Finding Critical Path:
1. Construct a project network:
The first step is to divide the entire project into significant activities in accordance with the work breakdown structure. Activity on Node (AON) is the most commonly used approach for drawing project networks, where nodes designate activities.
2. Perform forward passes and backward passes: Forward pass is a process that identifies all the early times. Backward pass is an activity that finds all the late start and late finish times. Here, we calculate two distinct starting and ending times for each activity. These are defined as follows:
Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed
Earliest finish (EF) = earliest time at which an activity can be finished
Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project
Latest finish (LF) = latest time by which an activity has to finish so as to not delay the completion time of the entire project
3. Determine project completion time
4. Calculate Slack values
5. State the Critical path
Example: Find the
critical path
A Nod inlcudes: A (Activity), t (Activity Time), ES (Early Start), EF (Early Finish), LS (Late Start), and LF (Late Finish)
Perform 'Forward Pass'
Activity A:
ES: Activity A has no predecessor, so A's earliest start time (ES) will be 0.
EF = ES + t . Since A has 7 weeks to be completed (t), its earliest finish time (EF) will be 0+7 = 7.
Activity B:
ES:Activity B has no predecessor, so its earliest start time (ES) will be 0.
EF = ES + t. Since it has 9 weeks to be completed (t), its earliest finish time (EF) will be 0+9 = 9.
Activity C:
ES: Activity C has a predecessor; 'A' (C needs A to be completed before it can start), Since the Earliest Finish (EF) Time for A is 7, then the Earliest Time (ET) C can start is 7.
EF = ES + t. Since C has 12 weeks to be completed (t), its earliest finish time (EF) will be 7+12 = 19.
Activity D:
ES: Activity D has a two predecessors; A& B (D needs A & B to be completed before it can start), Since the Earliest Finish (EF) Time for A and B are 7 and 9 respectively, and D needs both of them to finish in order to start, then the Earliest Time (ET) D can start is 9 which is the highest time of both predecessors.
EF = ES + t.Since D has 8 weeks to be completed (t), its earliest finish time (EF) will be 9+8 = 17.
Activity E:
ES: Activity E has only one predecessor; D, so its earliest start time (ES) will be 17.
EF = ES + t. Since E has 9 weeks to be completed (t), its earliest finish time (EF) will be 17+9 = 26.
Activity F:
ES: Activity F has a two predecessors; C&E (F needs C & E to be completed before it can start), Since the Earliest Finish (EF) Time for C and E are 19 and 26 respectively, and F needs both of them to finish in order to start, then the Earliest Time (ET) F can start is 26 which is the highest time of both predecessors.
EF = ES + t. Since F has 6 weeks to be completed (t), its earliest finish time (EF) will be 26+6 = 32.
Activity G:
ES: Activity G has only one predecessor; E, so G's earliest start time (ES) will be 26.
EF = ES + t. Since E has 5 weeks to be completed (t), its earliest finish time (EF) will be 26+5 = 31
Finish Time: The project's completion time is higher earliest finish time (EF). (Here, even though G is the last activity, F has highest earliest finish (EF) time.) Hence, Finish time of the project is 32.
Perform Backward Pass:
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