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Question
The following table gives the activities of a project and their duration in days
Activity | 1 - 2 | 1 - 3 | 2 - 3 | 2 - 4 | 3 - 4 | 3 - 5 | 4 - 5 |
Duration | 5 | 8 | 6 | 7 | 5 | 4 | 8 |
Construct the network and calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.
Solution
E1 = 0
E2 = 0 + 5 = 5
E3 = (0 + 8) or (5 + 6)
Whichever is maximum
= 11
E4 = (11 + 5) or (5 + 7)
Whichever is maximum
= 16
E5 = (11 + 4) or (16 + 8)
Whichever is maximum
= 24
L5 = 24
L4 = 24 – 8 = 16
L3 = (24 – 4) or (16 – 5)
whichever is minimum
= 11
L2 = (16 – 6) or (16 – 7)
whichever is minimum
= 5
L1 = (5 – 5) or (11 – 8)
whichever is minimum
L1 = 0
Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
1 - 2 | 5 | 0 | 5 | 5 – 5 = 0 | 5 |
1 - 3 | 8 | 0 | 8 | 11 – 8 = 3 | 11 |
2 - 3 | 6 | 5 | 11 | 11 – 6 = 5 | 11 |
2 - 4 | 7 | 5 | 12 | 16 – 7 = 9 | 16 |
3 - 4 | 5 | 11 | 16 | 16 – 5 = 11 | 16 |
3 - 5 | 4 | 11 | 15 | 24 – 4 = 20 | 24 |
4 - 5 | 8 | 16 | 24 | 24 – 8 = 16 | 24 |
Since EFT and LFT are same in 1 - 2, 2 - 3, 3 - 4 and 4 - 5.
Hence the critical path is 1 - 2 - 3 - 4 - 5 and the duration time taken is 24 days.
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