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EnglandFurther MathsSyllabus dot point

How do you schedule a project, find its shortest completion time and identify the activities that cannot be delayed?

Activity networks, forward and backward passes to find earliest and latest times, the critical path, float, and resource scheduling with Gantt charts.

A focused answer to the AQA A-Level Further Mathematics critical path analysis content, covering activity networks, forward and backward passes to find earliest and latest times, the critical path, float, and resource scheduling with Gantt charts.

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  1. What this dot point is asking
  2. Activity networks and the two passes
  3. The critical path and float
  4. Resource scheduling

What this dot point is asking

AQA wants you to draw an activity network from a precedence table, carry out a forward pass for earliest event times and a backward pass for latest event times, identify the critical path and the minimum project duration, calculate float, and schedule activities with a Gantt chart and a given number of workers.

Activity networks and the two passes

You draw the network from a precedence (dependence) table, in which each activity lists its immediate predecessors. AQA uses an activity-on-arc convention: arcs are activities and nodes are events, with each node split into the earliest and latest event times. Sometimes a dummy activity (a dotted arc of zero duration) is needed to preserve the correct logic when two activities share a predecessor but not all dependencies, or to give two activities distinct start and end events. Once the network is drawn, you sweep through it twice.

The critical path and float

Resource scheduling

A Gantt chart (cascade chart) plots each activity against time as a horizontal bar, with critical activities fixed in place and non-critical activities drawn at their earliest start with a trailing block showing their float. Stacking the bars by time lets you read off the number of workers required at each instant, the resource histogram. If the peak demand exceeds the workforce available, you slide non-critical activities later within their float to flatten the peaks, a process called resource levelling. Because critical activities have no float, they cannot be moved, so levelling only ever shifts the non-critical work, and the project duration is unchanged provided no float is exceeded.

Besides total float, two finer measures sometimes appear. Independent float is the spare time available even if every other activity is as late as possible, and interfering float is the difference between total and independent float, the part shared with neighbouring activities. The key exam discipline is to apply the maximum rule on the forward pass and the minimum rule on the backward pass without mixing them, to read the critical path as the longest route through the network, and to remember that using an activity's float can consume the float of others further along the same chain.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20197 marksA project has activities with durations (days): A(4), B(3), C(6), D(5), E(2). A and B can start immediately; C follows A; D follows B; E follows both C and D. Carry out a forward and backward pass to find the project duration and identify the critical path.
Show worked answer →

Forward pass (earliest event times). Start at time 00. A finishes at 44 and B finishes at 33. C follows A, finishing at 4+6=104 + 6 = 10. D follows B, finishing at 3+5=83 + 5 = 8. E follows both C and D, so it cannot start until max(10,8)=10\max(10, 8) = 10, finishing at 10+2=1210 + 2 = 12.

So the project duration is 1212 days (the earliest time of the final event).

Backward pass (latest event times). The end is fixed at 1212. E must start by 122=1012 - 2 = 10. Therefore C must finish by 1010 (latest finish 1010, no float) and D must finish by 1010, so D has latest start 55 against earliest start 33, giving 22 days of float. A must finish by 44 (no float); B must finish by 55, giving B float of 22.

Critical path: the zero-float chain A, C, E, of length 4+6+2=124 + 6 + 2 = 12.

Markers reward a correct forward pass (max rule at merge events), backward pass (min rule), the duration, and the critical path with reasons.

AQA 20224 marksAn activity has earliest start time 66, latest finish time 1515 and duration 55. Calculate its total float and explain what a float of zero would indicate.
Show worked answer →

Total float =latest finishearliest startduration= \text{latest finish} - \text{earliest start} - \text{duration}.

Substitute: total float =1565=4= 15 - 6 - 5 = 4 days.

So the activity can be delayed by up to 44 days without delaying the whole project.

A float of zero would indicate the activity is critical: any delay to it delays the entire project, so it lies on the critical path.

Markers reward the float formula, the correct value of 44, and the interpretation of zero float as a critical activity.

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