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3.8a · Critical Path Analysis · Sub-skill

Activity networks

Turn a list of activities and their dependencies into an activity network — and follow algorithms step by step.

See it first

Watch it happen

Scroll to the graphic and it plays by itself — press Replay to watch it again.

Build it up, step by step

Activity networks — the key ideas

Click each step below to reveal it — work through them in order the first time round.

Step 1 · Following an algorithm

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An algorithm is a finite list of clear, unambiguous steps that always produces an answer. To follow one, do the steps in order and write down the result of each step. For example: put seven scores in order, delete the highest and lowest, add up the rest, divide by how many are left. This algorithm gives a trimmed mean, which is less affected by one extreme score.

Step 2 · Activities, durations and predecessors

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A project is broken into activities. Each has a duration and a list of immediate predecessors — the activities that must be finished before it can start. A precedence table records all of this. An activity with a dash (–) has no predecessors, so it can start straight away.

Step 3 · Drawing the network

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The network algorithm: (1) draw a column of boxes for the activities with no predecessors; (2) to the right, draw boxes for activities whose predecessors are all drawn; (3) repeat until every activity is drawn; (4) draw an arrow from each immediate predecessor to the activity that follows it. Each box shows the activity and its duration, with room for an early time and a late time.

Step 4 · Reading a network

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To read a network back into a table, look at the arrows into each box: those boxes are its immediate predecessors. A box with several arrows in must wait for all of them. Activities that are not linked by a chain of arrows can happen at the same time.

Worked example

Draw an activity network for: A (3 days, –), B (5 days, –), C (4 days, after A), D (3 days, after B), E (3 days, after C and D).

Column 1: A and B (no predecessors). Column 2: C (needs A) and D (needs B). Column 3: E (needs both C and D). Arrows: A→C, B→D, C→E, D→E.

A and B can run in parallel, as can C and D. E cannot start until both C and D are finished.

Test yourself

Past-paper style question

A kitchen is being refitted. A: Remove old units (1 day, –); B: Fit electrics (2 days, after A); C: Fit plumbing (2 days, after A); D: Fit new units (3 days, after B and C); E: Lay floor (2 days, after D); F: Decorate (2 days, after D).

(a) Draw an activity network for the project.
(b) Write down one activity that can be done at the same time as C. [5 marks]

Show the answer

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(a) Box A on the left; boxes B and C to its right, each with an arrow from A; box D with arrows from both B and C; boxes E and F to the right of D, each with an arrow from D. Each box shows the letter and duration. (4 marks: all 6 boxes, correct order, arrows from every immediate predecessor, no extra arrows)

(b) B (B and C both follow A only, so they can run at the same time). (1 mark)

Practice

Activity networks worksheet

Five short questions. Work through them, then reveal the mark scheme to check.

  1. Follow this algorithm. Step 1: write the numbers 8, 3, 9, 4, 6. Step 2: cross out the largest and the smallest. Step 3: add up those left. Step 4: divide the total by how many numbers are left. What is the answer?
  2. A table has P (–), Q (after P), R (after P) and S (after Q and R). Which activities have no immediate predecessors, and what are the immediate predecessors of S?
  3. For the table in question 2, how many arrows are needed in the activity network?
  4. Activity T has immediate predecessors U and V. Can T start as soon as U finishes? Explain.
  5. A (–), B (after A), C (after A), D (after B and C). Which two activities can be done at the same time?

Mark scheme

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  1. Cross out 9 and 3 → 8, 4, 6. Total 18. 18 ÷ 3 = 6.
  2. Only P has no immediate predecessors. The immediate predecessors of S are Q and R.
  3. P→Q, P→R, Q→S, R→S: 4 arrows.
  4. No. T must wait for all of its immediate predecessors, so it cannot start until both U and V have finished.
  5. B and C — both depend only on A, and neither depends on the other.
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