Early and late times — Critical Path Analysis

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Early and late times — the key ideas

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Step 1 · Forward pass — early times

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The early time of an activity is the earliest it can start. Activities with no predecessors have early time 0. For any other activity, work out (early time + duration) for each immediate predecessor and take the largest — it has to wait for all of them.

Step 2 · Minimum completion time

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The project cannot finish before the last activity has finished. The minimum completion time is the largest value of (early time + duration) over the final activities — the length of the longest route through the network.

Step 3 · Backward pass — late times

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The late time of an activity is the latest time it can finish without delaying the project. The final activities have late time = minimum completion time. For any other activity, work out (late time − duration) for each of its successors and take the smallest — it must be done in time for all of them.

Step 4 · Recording and checking

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Write the early time in the left cell and the late time in the right cell of each box. Check: activities with no predecessors have early time 0, and the final activity's late time equals the minimum completion time. Early time + duration can never be more than the late time.

Worked example

Using A (3), B (5), C (4, after A), D (3, after B), E (3, after C and D), find every early and late time.

Forward: A: 0, B: 0, C: 0+3 = 3, D: 0+5 = 5, E: max(3+4, 5+3) = max(7, 8) = 8. Minimum completion time = 8 + 3 = 11.

Backward: E: 11. C: 11−3 = 8. D: 8. A: 8−4 = 4. B: 8−3 = 5.

Test yourself

Past-paper style question

A project has activities A (4 days, –), B (6 days, –), C (5 days, after A), D (2 days, after B), E (7 days, after A) and F (3 days, after C and D).

(a) Draw the network and complete a forward pass to find each early time.
(b) Complete a backward pass to find each late time.
(c) State the minimum completion time. [6 marks]

Show the answer

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(a) Early times: A 0, B 0, C 4, D 6, E 4, F = max(4+5, 6+2) = max(9, 8) = 9. (3 marks)

(b) Minimum time is 12 (see c), so F's late time is 12. E: 12. C: 12−3 = 9. D: 9. A: min(9−5, 12−7) = min(4, 5) = 4. B: 9−2 = 7. (2 marks)

(c) F finishes at 9+3 = 12; E finishes at 4+7 = 11. Minimum completion time = 12 days. (1 mark)

Practice

Early and late times worksheet

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

Activity X follows U and V. U has early time 2 and duration 4; V has early time 3 and duration 5. Find the early time of X.
Activity P is followed by Q and R. Q has late time 15 and duration 4; R has late time 12 and duration 3. Find the late time of P.
A (3 days) is followed by B (4 days), which is followed by C (2 days). Find each early time and the minimum completion time.
Why do we take the largest value on the forward pass?
The final activity has early time 14 and duration 5. What is its late time?

Mark scheme

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U: 2+4 = 6. V: 3+5 = 8. Early time of X = max(6, 8) = 8.
Q: 15−4 = 11. R: 12−3 = 9. Late time of P = min(11, 9) = 9.
A: 0, B: 3, C: 7. Minimum completion time = 7 + 2 = 9 days.
An activity cannot start until all its predecessors are finished, so it must wait for the one that finishes last — the largest value.
The late time of the final activity equals the minimum completion time = 14 + 5 = 19.

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