Critical path analysis, float and scheduling

What you’ll learn

• Translate immediate predecessors into an activity network.
• Perform forward and backward passes.
• Identify critical activities, total float and scheduling constraints.

Understand the idea

A project network expresses precedence as well as duration. In an activity-on-arc network, arrows represent activities and nodes represent events. A dummy arc has zero duration and can preserve a required dependency without adding work. A predecessor table lists immediate predecessors; an activity cannot start until all have finished.

The forward pass takes the maximum of incoming finish times to find each earliest event time. Starting with the project completion time, the backward pass takes the minimum of outgoing latest-start times. An activity's total float is the latest event time at its finish minus the earliest event time at its start minus its duration.

Zero-float activities are critical: delaying one delays the earliest possible project completion unless other conditions change. A Gantt chart shows start and finish intervals and can use float to schedule staff or equipment. The critical-path completion time assumes sufficient resources; a resource limit can force a longer feasible schedule.

Key formulae

$$E_j=\max_{i\to j}(E_i+d_{ij})$$

$$L_i=\min_{i\to j}(L_j-d_{ij})$$

$$F_{ij}=L_j-E_i-d_{ij}$$

$$\text{worker lower bound}=\left\lceil\frac{\text{total worker-time}}{\text{project duration}}\right\rceil$$

Worked example

Activity durations and immediate predecessors are: A(3), none; B(4), none; C(5), A; D(2), A; E(3), B and C; F(2), D and E. Find the minimum project duration, a critical path, and the total float of D. Assume unlimited resources.

1. Earliest intervals are A: 0–3, B: 0–4, C: 3–8 and D: 3–5.
2. E waits for both B and C, so E runs 8–11. F waits for D and E, so F runs 11–13.
3. The critical chain is A–C–E–F, with duration 3 + 5 + 3 + 2 = 13.
4. D can finish as late as 11 without delaying F. Its latest start is 9; its earliest start is 3, giving total float 6.

Answer: Project duration 13; critical path A–C–E–F; total float of D is 6.