How do you model a project as a network and find the shortest time to complete it?
Modelling a project with an activity network, finding the critical path and minimum completion time, and calculating the float (slack) of non-critical activities.
A focused answer to the SQA Higher Applications of Mathematics planning content, covering activity networks, precedence, the forward and backward pass, the critical path and minimum completion time, and the float of non-critical activities.
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What this dot point is asking
The SQA wants you to model a project as an activity network, work out the order activities must follow, find the critical path and the minimum time to complete the project, and calculate how much spare time (float) each non-critical activity has. This is the core of the planning area, used to schedule real projects.
Activities, durations and precedence
A project breaks into activities, each with a duration and a set of precedence rules saying which activities must finish before it can start. For example "lay foundations" must finish before "build walls", which must finish before "fit roof". Listing each activity, its duration and its immediate predecessors is the first step.
Finding the minimum completion time
Because every path must be completed, the project takes as long as its longest path. That longest path is the critical path, and its length is the minimum completion time.
You find it by listing every path from start to finish, adding the durations along each, and taking the largest total.
The forward and backward pass
For larger networks, two passes find the timings at each stage. The forward pass computes the earliest time each activity can start, working from the start: an activity starts as soon as all its predecessors finish. The backward pass computes the latest time each activity can start without delaying the project, working back from the finish.
An activity is critical when its earliest and latest start times are equal, because it has no spare time. The critical path is the chain of critical activities from start to finish.
Float (slack)
The float of an activity is the spare time it has: how long it could be delayed without delaying the whole project.
Float tells a project manager where there is flexibility (high-float activities can absorb delay or share resources) and where there is none (critical activities must be watched closely).
Why this matters
Critical-path analysis tells a manager the shortest possible schedule, which activities must not slip, and where there is slack to move people or money. It turns a list of tasks into a workable plan, which is exactly what the planning area examines.
Try this
Q1. A project has X (), Y (, after X), Z (, after X), then W (, after Y and Z). Find the minimum completion time. [3 marks]
- Cue. Route X-Y-W ; route X-Z-W ; minimum time days.
Q2. State the critical path for the project in Q1. [1 mark]
- Cue. X-Y-W, the longest route at days.
Q3. Find the float of activity Z in the Q1 project. [2 marks]
- Cue. The route through Z takes days against a project length of , so Z has days of float.
Exam-style practice questions
Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
SQA Higher Apps style5 marksA project has activities A ( days), B ( days, after A), C ( days, after A) and D ( days, after B and C). Find the minimum completion time and state the critical path.Show worked answer →
A must finish before B and C, and both B and C must finish before D. The two routes through the project are A then B then D, and A then C then D (1 mark).
Route A-B-D takes days; route A-C-D takes days (2 marks).
The minimum completion time is the longer route, days, and the critical path is A-C-D (2 marks). Markers reward computing both route lengths, taking the longest as the minimum time, and naming the critical path.
SQA Higher Apps style4 marksIn the project above, activity B lies on a non-critical route. Calculate the float of activity B and explain what it means.Show worked answer →
The critical route through B (A-B-D) takes days, but the project lasts days, so there are spare days along that route (2 marks).
The float of activity B is days: B could start up to days late, or take up to extra days, without delaying the project (1 mark).
Critical activities (A, C, D) have zero float, so any delay to them delays the whole project; B's float of days gives flexibility (1 mark). Markers reward identifying the spare time on the non-critical route and interpreting float as the allowable delay.
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