Critical paths and floatEdexcel A-Level Further Maths: Revision notes

In an activity network, the earliest event time e(j) is the earliest time an event can occur. Find it with a forward pass: start with e = 0 at the start event and, at each event, take the largest value of e(i) + duration over all the activities ending there. The latest event time l(i) is the latest time an event can occur without delaying the project. Find it with a backward pass: set l at the finish event equal to its earliest time and, at each event, take the smallest value of l(j) − duration over all the activities leaving it. The value of e at the finish event is the minimum project time.

For an activity from event i to event j with duration d:

• earliest start = e(i)
• earliest finish = e(i) + d
• latest finish = l(j)
• latest start = l(j) − d. When the project is given as a precedence table, the earliest start of an activity is the largest earliest finish of its immediate predecessors, and the latest finish is the smallest latest start of the activities that follow it. Example: with A 4, B 6, C 5 (after A), D 3 (after A, B), F 4 (after C, D): D starts at 6, as B finishes at 6 and A at 4.

An activity is critical if any delay to it delays the whole project. A critical activity has no spare time: its earliest start equals its latest start and its earliest finish equals its latest finish. The critical path is a path of critical activities from start to finish. It is the longest path, and its length is the minimum project time. There can be more than one critical path. Example: B–E–G–H = 6 + 7 + 2 + 3 = 18 days is the longest path through a project, so B, E, G and H are the critical activities.

The total float of an activity is the amount by which it can be delayed, or its duration extended, without delaying the project. For an activity from i to j: F(i, j) = l(j) − e(i) − duration(i, j). Critical activities have a total float of zero. An activity with float 2 can be delayed by up to 2 days without affecting the project time, but a delay greater than 2 delays the whole project. Example: activity C has e = 4, l = 11 and duration 5, so its float is 11 − 4 − 5 = 2 days. If an activity's delay exceeds its float, the project is delayed by the excess, and the critical path may change.

A Gantt (cascade) chart shows each activity as a bar on a time axis. Each activity requires only one worker. Critical activities are drawn first, as a continuous line of bars from time 0 to the project time. Other activities are drawn at their earliest start, with the spare time shown as the float. The lower bound for the number of workers needed to complete the project in the minimum time is the sum of all durations divided by the minimum project time, rounded up. Example: durations total 42 days and the project takes 24 days, so 42 ÷ 24 = 1.75 and at least 2 workers are required. This is only a lower bound. A schedule may need more workers.