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प्रश्न
The following table gives the characteristics of the project
| 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.
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उत्तर
E1 = 0
E2 = 0 + 5 = 5
E3 = (5 + 3) or (0 + 10)
Whichever is maximum
= 10
E4 = 10 + 4 = 14
E5 = 10 + 6 = 16
E6 = (14 + 6) or (16 + 5)
Whichever is maximum
= 21
E7 = 21 + 5 = 26
L7 = 26
L6 = 26 − 5 = 21
L5 = 21 − 5 = 16
L4 = 21 − 6 = 15
L3 = (15 − 4) or (16 − 6)
Whichever is minimum
= 10
L2 = 10 − 3 = 7
L1 = (7 − 5) or (10 − 10)
Whichever is minimum
L1 = 0
| Activity | Duration tij |
EST | EFT = EST + tij | LST = LFT – tij | LFT |
| 1 - 2 | 5 | 0 | 5 | 7 − 5 = 2 | 7 |
| 1 - 3 | 10 | 0 | 10 | 10 − 10 = 0 | 10 |
| 2 - 3 | 3 | 5 | 8 | 10 − 3 = 7 | 10 |
| 3 - 4 | 4 | 10 | 14 | 15 − 4 = 11 | 15 |
| 3 - 5 | 6 | 10 | 16 | 16 − 6 = 10 | 16 |
| 4 - 6 | 6 | 14 | 20 | 21 − 6 = 15 | 21 |
| 5 - 6 | 5 | 16 | 21 | 21 − 5 = 16 | 21 |
| 6 - 7 | 5 | 21 | 26 | 26 − 5 = 21 | 26 |
Since EFT and LFT values are same in 1 - 3, 3 - 5, 5 - 6 and 6 - 7.
Hence the critical path is 1 - 3 - 5 - 6 - 7 and the duration of time taken is 26 days.
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संबंधित प्रश्न
Draw the event oriented network for the following data:
| Events | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Immediate Predecessors | - | 1 | 1 | 2, 3 | 3 | 4, 5 | 5, 6 |
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.
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 |
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.
Which of the following is not correct?
In the context of network, which of the following is not correct
In critical path analysis, the word CPM mean
Draw a network diagram for the following activities.
| 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 |
Draw the network diagram for the following activities.
| Activity code | A | B | C | D | E | F | G |
| Predecessor activity | - | - | A | A | B | C | D, E |
A Project has the following time schedule
| Activity | 1 - 2 | 2 - 3 | 2 - 4 | 3 - 5 | 4 - 6 | 5 - 6 |
| Duration (in days) |
6 | 8 | 4 | 9 | 2 | 7 |
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.
