Forward pass:
E1 = 0
E2 = 0 + 4 = 4
E3 = 0 + 1 = 1
E4 = (E2 + 1) (or) (E3 + 1) = (4 + 1) (or) (1 + 1) = 5 (or) 2
[whichever is maximum we must select forward pass]
∴ E4 = 5
E5 = 1 + 6 = 7
E6 = 7 + 4 = 11
E7 = 8 + 7 = 15
E8 = (E7 + 2) (or) (E6 + 1)
= (15 + 2) or (11 + 1)
= 17 or 12 [whichever is maximum]
∴ E8 = 17
E9 = 5 + 5 = 10
E10 = (E9 + 7) (or) (E8 + 5)
= (10 + 7) (or) (17 + 5)
= 17 (or) 22 [Which is maximum]
E10 = 22
Backward pass:
L10 = 22
L9 = 22 – 7 = 15
L8 = 22 – 5 = 17
L7 = 17 – 2 = 15
L6 = 17 – 1 = 16
L5 = (16 – 4) (or) (15 – 8) [whichever is minimum]
L5 = 7
L4 = 15 – 5 = 10
L3 = (10 – 1) (or) (7 – 6) [whichever is minimum]
L3 = 1
L2 = 10 – 1 = 9
L1 = 0
We can also find the critical path by this method, which also helps us to counter check the solution obtained by the table method.
Path |
Time |
1 - 2 - 4 - 9 - 10 |
4 + 1 + 5 + 7 = 17 |
1 - 3 - 4 - 9 - 10 |
1 + 1 + 5 + 7 = 14 |
1 - 3 - 5 - 6 - 8 - 10 |
1 + 6 + 4 + 1 + 5 = 17 |
1 - 3 - 5 - 7 - 8 - 10 |
1 + 6 + 8 + 2 + 5 = 22 |
So critical path is 1-3-5-7-8-10 as it takes 22 units to complete the project.