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प्रश्न
A Project has the following time schedule
| Activity | 1 - 2 | 1 - 6 | 2 - 3 | 2 - 4 | 3 - 5 | 4 - 5 | 6 - 7 | 5 - 8 | 7 - 8 |
| Duration (in days) | 7 | 6 | 14 | 5 | 11 | 7 | 11 | 4 | 18 |
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.
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उत्तर
E1 = 0
E2 = 0 + 7 = 7
E3 = 7 + 14 = 21
E4 = 7 + 5 = 12
E5 = 21 + 11 or (12 + 7)
Whichever is maximum
= 32
E6 = 0 + 6 = 6
E7 = 6 + 11 = 17
E8 = 17 + 18 = 35
L8 = 35
L7 = 35 – 18 = 17
L6 = 17 – 11 = 6
L5 = 35 – 4 = 31
L4 = 32 – 7 = 25
L3 = 32 – 11 = 21
L2 = (21 – 14) or (25 – 5)
whichever is minimum
= 7
L1 = 7 – 7 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 7 | 0 | 7 | 7 – 7 = 0 | 7 |
| 1 - 6 | 6 | 0 | 6 | 7 – 6 = 1 | 7 |
| 2 - 3 | 14 | 7 | 21 | 21 – 14 = 7 | 21 |
| 2 - 4 | 5 | 7 | 12 | 25 – 5 = 20 | 25 |
| 3 - 5 | 11 | 21 | 32 | 31 – 11 = 21 | 32 |
| 4 - 5 | 7 | 12 | 19 | 32 – 25 = 7 | 14 |
| 6 - 7 | 11 | 6 | 17 | 18 – 11 = 7 | 18 |
| 5 - 8 | 4 | 32 | 36 | 36 – 4 = 32 | 36 |
| 7 - 8 | 18 | 17 | 36 | 36 – 18 = 18 | 36 |
Since EFT and LFT are same in 1 - 2, 2 - 3, 3 - 5 and 5 - 8.
Hence the critical path is 1 - 2 - 3 - 5 - 8 and the duration of time taken is 36 days.
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संबंधित प्रश्न
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Activities A, D, E can start simultaneously; B, C > A; G, F > D, C; H > E, F.
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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 |
| Time | 4 | 1 | 1 | 1 | 6 | 5 | 4 | 8 | 1 | 2 | 5 | 7 |
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| Duration | 6 | 5 | 10 | 3 | 4 | 6 | 2 | 9 |
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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 |
