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प्रश्न
A project schedule has the following characteristics
| 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 |
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 + 4 = 4
E3 = 0 + 1 = 1
E4 = 4 + 1 = 5
E5 = 1 + 6 = 7
E6 = 7 + 4 = 11
E7 = 8 + 7 = 15
E8 = 15 + 2 = 7
E9 = 5 + 5 = 10
E10 = 17 + 5 = 22
L7 = 31
L10 = 22
L9 = 22 – 7 = 15
L8 = 22 – 5 = 17
L7 = 17 – 2 = 15
L6 = 17 – 1 = 16
L5 = (16 – 4) or (15 – 8)
whichever is minimum = 7
L4 = 15 – 5 = 10
L3 = (10 – 1) or (7 – 6)
whichever is minimum = 1
L2 = 10 – 1 = 9
L1 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 4 | 0 | 4 | 9 – 4 = 5 | 9 |
| 1 - 3 | 1 | 0 | 1 | 1 – 1 = 0 | 1 |
| 2 - 4 | 1 | 4 | 5 | 10 – 1 = 9 | 10 |
| 3 - 4 | 1 | 1 | 2 | 10 – 1 = 9 | 10 |
| 3 - 5 | 6 | 1 | 7 | 7 – 6 =1 | 7 |
| 4 - 9 | 5 | 5 | 10 | 15 – 5 =10 | 15 |
| 5 - 6 | 4 | 7 | 11 | 16 4 = 12 | 16 |
| 5 - 7 | 8 | 7 | 15 | 15 – 8 = 7 | 15 |
| 6 - 8 | 1 | 11 | 12 | 17 – 1 = 16 | 17 |
| 7 - 8 | 2 | 15 | 17 | 17 – 2 = 15 | 17 |
| 8 - 10 | 5 | 17 | 22 | 22 – 5 = 17 | 22 |
| 9 - 10 | 7 | 10 | 17 | 22 – 7 = 15 | 22 |
Since EFT and LFT is same on 1 - 3, 3 - 5, 5 - 7 and 7 - 8 and 8 - 10 the critical path is 1 - 3 - 5 - 7 - 8 - 10 and the duration is 22 time units.
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संबंधित प्रश्न
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| Activity code | A | B | C | D | E | F | G | H | I | J | K |
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The following table gives the characteristics of the project
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| Duration (in days) |
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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.
