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Question
The following table gives the characteristics of the project
| Activity | 1 - 2 | 1 - 3 | 2 - 3 | 3 - 4 | 3 - 5 | 4 - 6 | 5 - 6 | 6 - 7 |
| Duration (in days) |
5 | 10 | 3 | 4 | 6 | 6 | 5 | 5 |
Draw the network for the project, calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and find the critical path. Compute the project duration.
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Solution
E1 = 0
E2 = 0 + 5 = 5
E3 = (5 + 3) or (0 + 10)
Whichever is maximum
= 10
E4 = 10 + 4 = 14
E5 = 10 + 6 = 16
E6 = (14 + 6) or (16 + 5)
Whichever is maximum
= 21
E7 = 21 + 5 = 26
L7 = 26
L6 = 26 − 5 = 21
L5 = 21 − 5 = 16
L4 = 21 − 6 = 15
L3 = (15 − 4) or (16 − 6)
Whichever is minimum
= 10
L2 = 10 − 3 = 7
L1 = (7 − 5) or (10 − 10)
Whichever is minimum
L1 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 5 | 0 | 5 | 7 − 5 = 2 | 7 |
| 1 - 3 | 10 | 0 | 10 | 10 − 10 = 0 | 10 |
| 2 - 3 | 3 | 5 | 8 | 10 − 3 = 7 | 10 |
| 3 - 4 | 4 | 10 | 14 | 15 − 4 = 11 | 15 |
| 3 - 5 | 6 | 10 | 16 | 16 − 6 = 10 | 16 |
| 4 - 6 | 6 | 14 | 20 | 21 − 6 = 15 | 21 |
| 5 - 6 | 5 | 16 | 21 | 21 − 5 = 16 | 21 |
| 6 - 7 | 5 | 21 | 26 | 26 − 5 = 21 | 26 |
Since EFT and LFT values are same in 1 - 3, 3 - 5, 5 - 6 and 6 - 7.
Hence the critical path is 1 - 3 - 5 - 6 - 7 and the duration of time taken is 26 days.
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