​https://www.youtube.com/watch?v=rxGcV0tuxRU&t=251s&ab_channel=OnlinePMCourses-MikeClayton

WBS

Start by examining the WBS, which should detail all the tasks required to complete the project. Each task should have a defined duration and dependencies with other tasks, if applicable.

Review the Work Breakdown Structure (WBS):

Step 1

Determine the dependencies between tasks. Tasks can be either sequential (where one task must be completed before another can start) or parallel (where tasks can be executed simultaneously).

In our WBS, Tasks AB & D do not have any predecessor activities

Identify Task Dependencies:

Step 2

​Place a start symbol (this is optional).Tasks A, B and D can start at the same time since they do not depend on any preceding task. Align all 3 tasks vertically.

Since tasks A, B and D have commenced, the next activity is 'C' which is dependent on A and B. Place activity 'C' where it is easy to draw links from both tasks without possibility of crossing any lines

Identify and Position Next Task:

Step 3

Move alphabetically and place any activity that has already been inserted in the chart. Make sure to position activities strategically so its easy to draw arrows to predecessors without intersecting any other arrow.

Step 4

Link with arrows. Take note of predecessor activities.

Step 5

Insert estimated durations or days for each activity.

Step 6

C is dependent on A and B to be completed before it can start. Therefore, both activities would be completed on day 4 so the earliest possible time C can start is 4. The earliest finish time for C will be the sum of the earliest start (4) and number of days required to complete activity C (5) [4+5=9]. Apply same steps to F. ES for F is 1 and EF will be 1+2=3

Step 7

E is dependent on C so the ES is 9. EF will therefore be 9+5=14. G is dependent on E & F to be completed before it can start. Therefore, ES is 14 and EF will be 14+5=19. H is dependent on G to be completed before it can start. Hence, ES will be 19 and EF will be 19+6=25

Step 8

Overall duration of this project is 25days

Starting from the end of the network diagram with activity H, coming from G, this is a converging task coming at the end so the values for LS and LF will be the same as ES and EF i.e 19 and 25. For activity G, the latest possible finish is 19 because activity H starts on day 19. LS for G will be 19 – 5 (duration of activity G) = 14. Day 14 is the latest time that G can start for the project to remain on track

Step 9

Now we will start the second pass to find latest start time and latest finish time

LF for activity E is 14. LS will be 14-5=9. For activity F, LF is 14 because activity E has a longer duration of 14 which G also depends on to be completed before it starts. LS for ‘F’ will be 14-2=12. For activity D, LF will be 12 and LS will be 12-1=11. Activity C will have a latest finish of 9 and LS will be 9-5=4. For activity B, LF will be 4 and LS will be 4-3=1. Finally for A, LF will be 4 and LS will be 4-4 = 0.

Step 10

Apply the same approach to determine E to A

Use the Total slack formular to calculate total slack for each activity. H = 25-25=0, G=19-19=0, E=14-14=0, C=9-9=0, A = 4-4=0. F=14-3=11, D=12-1=11, B=4-3=1

Step 11

Calculate total slack or slippage. This will be used to determine the critical path.

From the network diagram, take note of the path that has 0(zeros) as all its total slack time. The path A, C, E, G and H is our critical path. This implies that all the activities along this path are crucial in ensuring that the project is completed on time. Having zero slack or slippage means we have no leeway with regards the schedule of the project. On the other hand, the path B, D and F has a lot of leeway. E.g, Task D and F has up to 11days slack to be completed even though they only require 1 and 2 days respectively to be completed.

Step 12

Determining the critical path