Critical Path
Critical path consists of a specific sequence of tasks that must be completed on time in order for the project to meet its deadline. It is the longest path in a chain of interdependent activities from start to finish in a project.
Learning Objective
By completing this tutorial you will be able to
- Identify critical path in the project
- Adopt a systematic approach to project planning and execution with critical path analysis.
- Learn how to reduce the project duration by optimizing the critical path.
- Avoid the risk of over schedule by identifying the tasks that are most critical.
Business Scenario
Leester group is working on a construction project for Donnelley apartments. This will involve offsite and onsite employees and supervisors collaborating on the different tasks and transactions. Let us consider a task list and determine the critical path for that task list. It contains the following tasks:
| Task | Predecessor | Duration(in days) |
| A: Basement tiles | -- | 5 |
| B: Tank material testing | -- | 7 |
| C: Overhead tank installation | A | 9 |
| D: Check leakages | A,B | 6 |
| E: Force test | D | 7 |
| F: Batt insulation | C,E | 4 |
| G: Cauk and air seal | E | 5 |
Let's now determine the critical path. There are four steps in the critical path analysis.
Build the project network.
Calculate the early start, early finish, late start, and late finish dates in the project.
Determine the project completion time.
- Identify the critical path and and calculate slack.
Critical Path Analysis
Step 1: Derive the network model
Let's draw the network path based on the given inputs. Activities A and B have no predecessors so they can begin at the start. C needs A to be completed before it can start. D should complete before E starts and C and E should finish before F. G depends on E. Since F and G have no successors they go to finish.
Before proceeding to the next step click here to understand the terms used in them.
Step 2: Forward Pass
Let's do the forward pass and calculate the early start, early finish, and late start, late finish dates.
Forward pass formula
EF= ES + Duration
ES of successor= Maximum(EF of all predecessors)
Backward pass formula
LS = LF – Duration.
LF of predecessor = Minimum (LS of all the successors)
Slack= LS - ES or LF - EF
Slack= LS - ES or LF - EF
Forward Pass
A has no predecessor so its earliest start time will be 0. Since it has 5 days to finish , it's earliest finish time will be 0 + 5 which is 5 days.
B also can start at 0 and duration is 7 days, it will have an earliest finish time of 7.
Since the earliest finish time for A is 5, the earliest time C can start is 5 and with a task duration of 9 days, C will have the earliest finish time of 9 + 5= 14 days.
Since the earliest finish time for A and B are 5 and 7 respectively, and D needs both of them to finish in order to start, then the earliest time D can start is 7 i.e. the highest of the earliest finish times preceding a task will be the task’s earliest start time.
E here has only one predecessor D and can start on day 13 and finish earliest on day 20.
F has predecessors, C and E. Since the higher earliest finish time is 24 days, F can start earliest on day 20 and finish on day 24.
G can also start earliest on days 20 since it has only one predecessor and G can finish earliest days at 20 + 5 which gives 25 days.
- Note that G gives the highest earliest finish of 25 days. In essence, the project’s completion time is the highest earliest finish time at the finish node.
Backward Pass
At the finish node F and G has to be 25 i.e. F and G cannot take more than 25 days. Latest start times are obtained by subtracting the duration from the latest finish.
For G, the latest start time will be 25 - 5= 20 days. For F, the latest start will be 25-4= 21.
E has two successors F and G. As a result, the latest time E has to finish is 20 days in order for G to start. In essence, when doing a backward pass, the latest finish time of a task must be the minimum of the latest start times of its successors. Thus the latest start time for E will be 20 - 7 = 13 days.
D has only one successor E, so the latest finish time for D will be the latest start time for E which is 13 days. And, LS will be 13 - 6 = 7 days.
C also has one successor F. Therefore the latest finish will be 21 days for C and latest start will be 12 days.
A has two successors, C and D. The minimum of their latest start is 7 days. So the latest finish for A will be 7 days and its latest start will be 2 days.
B has one successor D with the latest start of 7 days. So the latest finish for B will be 7 days and its latest start will be 0.
Next step is to calculate the slack value. It can be done in two ways. Either by calculating the difference between the latest start and earliest start time ( LS - ES) or difference between the latest finish and earliest finish time ( LF - EF). Slack for A is 2 - 0 or 7 - 5 = 2 days. For B, slack is 0. For C the slack is 7 days. The slack values for D, E, and G are 0.
- In our scenario, the activity C can begin anytime between 5 and 12 days and it can finish anytime between 14 and 21 days. Thus C can be delayed for up to 7 days and the project will still be completed in 25 days. On the other hand tasks B, D, E and G cannot be delayed without extending the project’s completion time. For instance, if D is delayed by 2 days then the project completion time will be extended by 2 days, from 25 to 27 days. Thus, these sequence of tasks with zero slack are called critical tasks. The critical tasks must be completed on time in order for the project to meet its scheduled deadline. Critical path forms the longest path in the network. The critical path is B-D-E-G.