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प्रश्न
The following table use the activities in a construction projects and relevant information
| Activity | 1 - 2 | 1 - 3 | 2 - 3 | 2 - 4 | 3 - 4 | 4 - 5 |
| Duration (in days) |
22 | 27 | 12 | 14 | 6 | 12 |
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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उत्तर
E1 = 0
E2 = 22 + 0 = 22
E3 = (0 + 27) or (22 + 12)
Whichever is maximum
E3 = 34
E4 = (22 + 14) or (34 + 6)
Whichever is maximum
E4 = 40
E5 = 40 + 12 = 52
L5 = 32
L4 = 52 – 12 = 40
L3 = 40 – 6 = 34
L2 = (40 – 14) or (34 – 12)
whichever is minimum
= 22
L1 = 22 – 22 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 22 | 0 | 22 | 22 – 22 = 0 | 22 |
| 1 - 3 | 27 | 0 | 27 | 34 – 27 = 7 | 34 |
| 2 - 3 | 12 | 22 | 34 | 34 – 12 = 26 | 34 |
| 2 - 4 | 14 | 22 | 36 | 40 – 14 = 26 | 40 |
| 3 - 4 | 6 | 34 | 40 | 40 – 6 = 34 | 40 |
| 4 - 5 | 12 | 40 | 52 | 52 – 12 = 40 | 52 |
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 of time taken is 52 days.
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संबंधित प्रश्न
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| Activity | 1 - 2 | 1 - 3 | 2 - 4 | 3 - 4 | 3 - 5 | 4 - 9 | 5 - 6 | 5 - 7 | 6 - 8 | 7 - 8 | 8 - 10 | 9 - 10 |
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| Duration | 5 | 8 | 6 | 7 | 5 | 4 | 8 |
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| Activity code | A | B | C | D | E | F | G | H | I | J | K |
| Predecessor activity | - | A | A | A | B | C | C | C, D | E, F | G, H | I, J |
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| Activity code | A | B | C | D | E | F | G |
| Predecessor activity | - | - | A | A | B | C | D, E |
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| 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.
