(a) 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 earliest start time (EST), earliest finish time (EFT), latest start time (LST) and latest finish time (LFT) of each activity, and determine the critical path of the project and duration to complete the project.
OR
(b) Differentiate the function with respect to .
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Start your 14-day free trial to unlock the full solution →(a) Forward/backward pass gives project duration 36 days and critical path 1–2–3–5–8. (b) Logarithmic differentiation gives .
Part (a): CPM network. Activities (with durations): 1-2:7, 1-6:6, 2-3:14, 2-4:5, 3-5:11, 4-5:7, 6-7:11, 5-8:4, 7-8:18.
Forward pass (earliest event times ):
Project duration days.
Backward pass (latest event times ), :
Activity times (EST , EFT EST, LFT , LST LFT):
| Activity | Dur | EST | EFT | LST | LFT | Float |
|---|---|---|---|---|---|---|
| 1-2 | 7 | 0 | 7 | 0 | 7 | 0* |
| 1-6 | 6 | 0 | 6 | 1 | 7 | 1 |
| 2-3 | 14 | 7 | 21 | 7 | 21 | 0* |
| 2-4 | 5 | 7 | 12 | 20 | 25 | 13 |
| 3-5 | 11 | 21 | 32 | 21 | 32 | 0* |
| 4-5 | 7 | 12 | 19 | 25 | 32 | 13 |
| 6-7 | 11 | 6 | 17 | 7 | 18 | 1 |
| 5-8 | 4 | 32 | 36 | 32 | 36 | 0* |
| 7-8 | 18 | 17 | 35 | 18 | 36 | 1 |
Zero-float (critical) activities: 1-2, 2-3, 3-5, 5-8.
Critical path: 1–2–3–5–8, duration days.
Part (b): Differentiate . Use logarithmic differentiation.
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