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Critical path analysis questions
Some jobs can wait and some cannot. A forward pass finds the earliest anything can happen, a backward pass finds the latest it may, and the jobs where those agree decide how long the whole project takes.
6 original questions · 22 marks · the critical path analysis notes · Decision Mathematics 1
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Explain what a dummy activity is and give one reason it is needed.
Worked answer
An arrow of zero duration that carries a dependency without representing any work. One reason is that two activities share some but not all of their predecessors, so a dummy is needed to keep the dependencies exact. A second is that two activities would otherwise run between the same pair of events, and every activity must have a unique pair. B1 for the zero-duration arrow carrying a dependency, B1 for either reason.A project has A (3 days, no predecessors), B (5, none), C (4, after A), D (2, after A), E (6, after B and C), F (3, after D) and G (2, after E and F). Carry out the forward pass.
Worked answer
A finishes at 3 and B at 5. C runs 3 to 7, D runs 3 to 5. E waits for the later of B and C, so it runs 7 to 13; F runs 5 to 8. G waits for the later of E and F, so it runs 13 to 15: the project takes 15 days. M1 for taking the later of the incoming times, A1 for the early times through the network, A1 for 15 days.Carry out the backward pass for the same project and state the critical path.
Worked answer
Work back from 15. G must start by 13. E and F must both finish by 13, so E starts by 7 and F by 10. C must finish by 7, starting by 3; D must finish by 10, starting by 8; B must finish by 7, starting by 2. A must finish by 3. Early and late starts agree for A, C, E and G, so that is the critical path, and 3 + 4 + 6 + 2 = 15 confirms it. M1 for working back from 15, A1 A1 for the late times, A1 for the critical path. Put both numbers in every event box. The backward pass takes the smallest late start of the following activities, and taking the largest is the commonest way to lose all four marks.Explain why the critical path is not simply the route with the most activities on it, using this project as an example.
Worked answer
What matters is the total duration along a route, not how many activities it contains. Here A, C, E, G has four activities and takes 15 days, while A, D, F, G also has four and takes only 10. A route with more but shorter activities can easily be shorter overall. The critical path is defined by zero float, which is the same as being the longest route in time. B1 for duration mattering rather than the count, B1 for the comparison of the two four-activity routes, B1 for zero float or the longest route.An activity has earliest start 6, duration 4 and latest finish 13. Is it critical?
Worked answer
Its latest start is 13 − 4 = 9, later than its earliest start of 6. The total float is 3 days, so the activity is not critical. M1 for the latest start 13 − 4, A1 for a float of 3 days, A1 for the conclusion. It could be delayed by up to three days without extending the project.A project has A (3 days, no predecessors), B (5, none), C (4, after A), D (2, after A), E (6, after B and C), F (3, after D) and G (2, after E and F); it takes 15 days along the critical path A, C, E, G. Activity B is found to take 8 days rather than 5. Find the new project duration and the new critical path. Hence find the greatest duration B could have without displacing the original critical path.
Worked answer
B had a total float of 2 days, so 3 extra days uses it up and pushes past it. B now finishes at day 8, so E cannot start until 8 and finishes at 14, and G finishes at 16. The project has lengthened by one day.
The new critical path is B, E, G, with 8 + 6 + 2 = 16. A and C now carry a day of float, since C could finish at 8 rather than 7.
E cannot begin until both B and C are done, and C finishes at 7. So B governs the start of E only once its duration exceeds 7. For B of duration 7 days or less the path A, C, E, G still sets the duration, and at exactly 7 both paths are critical together. M1 for a fresh forward pass, A1 for 16 days, A1 for the new critical path, B1 for the float now carried by A and C, M1 for comparing B with the time C needs, A1 for 7 days, A1 for both paths being critical at exactly 7.
A change to a non-critical activity does nothing until it exceeds the float, and then it changes the critical path rather than simply lengthening the old one.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise critical path analysis one question at a time
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