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प्रश्न
Draw 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.
| Jobs | 1 - 2 | 1 - 3 | 2 - 4 | 3 - 4 | 3 - 5 | 4 - 5 | 4 - 6 | 5 - 6 |
| Duration | 6 | 5 | 10 | 3 | 4 | 6 | 2 | 9 |
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उत्तर
E1 = 0
E2 = 0 + 6 = 6
E3 = 0 + 5 = 5
E4 = 6 + 10 = 16
E5 = (5 + 4) or (16 + 6)
Whichever is maximum
= 22
E6 = (16 + 2) or (22 + 9)
Whichever is maximum
= 31
L6 = 31
L5 = 31 – 8 = 22
L4 = 22 – 6 = 16 or (31 – 2)
whichever is minimum
L3 = 22 – 4 = 18
L2 = 16 – 10 = 6
L1 = 6 – 6 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 6 | 0 | 6 | 6 – 6 = 0 | 6 |
| 1 - 3 | 5 | 0 | 5 | 18 – 5 = 13 | 18 |
| 2 - 4 | 10 | 6 | 16 | 16 – 10 = 6 | 16 |
| 3 - 4 | 3 | 5 | 8 | 16 – 3 = 13 | 16 |
| 3 - 5 | 4 | 5 | 9 | 22 – 4 = 18 | 22 |
| 4 - 5 | 6 | 16 | 22 | 22 – 6 = 16 | 22 |
| 4 - 6 | 2 | 16 | 18 | 31 – 2 = 29 | 31 |
| 5 - 6 | 9 | 22 | 31 | 31 – 9 = 22 | 31 |
Since EFT and LFT is same on 1 - 2, 2 - 4, 4 - 5 and 5 - 6, the critical path is 1 - 2 - 4 - 5 - 6 and duration time taken is 31 days.
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| Activity | 0 - 1 | 1 - 2 | 1 - 3 | 2 - 4 | 2 - 5 | 3 - 4 | 3 - 6 | 4 - 7 | 5 - 7 | 6 - 7 |
| Duration (in week) | 3 | 8 | 12 | 6 | 3 | 3 | 8 | 5 | 3 | 8 |
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| Time | 4 | 1 | 1 | 1 | 6 | 5 | 4 | 8 | 1 | 2 | 5 | 7 |
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| Activity code | A | B | C | D | E | F | G |
| Predecessor activity | - | - | A | A | B | C | D, E |
