Questions on critical path analysis

Question 1

Explain what a dummy activity is and give one reason it is needed.

Accepted 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.

Question 2

State the six numbers an activity's box carries in an activity-on-node network, and state how its float is read from them.

Accepted Answer

The box carries the activity's name and its duration, the earliest start and earliest finish, and the latest start and latest finish. B1 for the six entries. The total float is the latest start minus the earliest start, equivalently the latest finish minus the earliest finish, and an activity with a float of zero is critical. B1 for the float.
Both readings give the same number, because the duration is subtracted from both ends, so either one serves as a check on the other.

Question 3

A project has P (4 days, no predecessors), Q (6, none), R (5, after P), S (3, after P and Q) and T (2, after R and S). Find the project duration and the critical activities.

Accepted Answer

Forward pass: P finishes at 4 and Q at 6. R runs 4 to 9. S waits for the later of P and Q, so it runs 6 to 9. T waits for the later of R and S, so it runs 9 to 11. M1 for taking the later value at each merge, A1 for 11 days.
P, R, T take 4 + 5 + 2 = 11 and Q, S, T take 6 + 3 + 2 = 11, so both routes are as long as the project and every one of the five activities is critical. A1.
Two critical paths through one project is an ordinary situation, not a sign of an error, and shortening one of them on its own does not shorten the project.

Question 4

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.

Accepted 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.

Question 5

Carry out the backward pass for the same project and state the critical path.

Accepted 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; taking the largest gives a late time that the following activities cannot meet.

Question 6

Explain why the critical path is not simply the route with the most activities on it, using this project as an example.

Accepted 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.

Question 7

An activity has earliest start 6, duration 4 and latest finish 13. Is it critical?

Accepted 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.

Question 8

In an activity-on-node network, activity K has duration 5. The activities immediately before it have earliest finishes 4 and 6; the activities immediately after it have latest starts 14 and 12. Find the six numbers in K's box, and determine whether K is critical.

Accepted Answer

The earliest start is the larger of the incoming earliest finishes, max(4, 6) = 6, so the earliest finish is 6 + 5 = 11. M1 for taking the larger, A1 for the pair. The latest finish is the smaller of the outgoing latest starts, min(14, 12) = 12, so the latest start is 12 − 5 = 7. M1 for taking the smaller. The float is 7 − 6 = 1, which is not zero, so K is not critical. A1 for the float and the conclusion.

Question 9

A project has P (4 days, no predecessors), Q (6, none), R (5, after P), S (3, after P and Q) and T (2, after R and S). Draw the activity-on-arc network for it, using a dummy where one is needed, and state the dependency your dummy carries.

Accepted Answer

P and Q both leave the start event. R needs P alone, so R leaves the event at the end of P. S needs both P and Q, so S leaves the event where Q finishes, and a dummy of zero duration runs from the end of P into that event. R and S then both arrive at the event T leaves from, and T runs on to the finish. B1 for P and Q leaving the start event, M1 for R and S leaving different events, A1 for the dummy from the end of P to the start of S, A1 for the completed network with T closing it.
The dummy carries the dependency S cannot start until P has finished without representing any work. B1 for that dependency stated. Drawing R and S from the same event would make R wait for Q as well, which the table does not say; drawing them from separate events with no dummy would free S from P, which it also does not say.

Question 10

At one event of an activity-on-arc network, three activities arrive with earliest finishing times 12, 9 and 15, and two leave with latest starting times 20 and 17. Write down the earliest and the latest event times, saying which pass produces each, and find the total float carried by the activity that arrives with earliest finish 9.

Accepted Answer

Three arrivals make this a merge on the way forward, and the forward pass takes the largest arriving time: the earliest event time is max(12, 9, 15) = 15. B1.
Two departures make it a burst on the way back, and the backward pass takes the smallest departing time: the latest event time is min(20, 17) = 17. B1.
The activity arriving with earliest finish 9 may finish as late as the latest event time, so its total float is 17 − 9 = 8 days. M1 for the subtraction, A1 for 8.
The two rules only ever bite at a merge going forward and at a burst coming back; everywhere else there is one incoming value and one outgoing value and nothing to choose between.

Question 11

A project has U (3 days, no predecessors), V (5, none), W (4, after U), X (6, after U and V) and Y (2, after W and X). Find the project duration and the critical activities, and state what the activity-on-arc drawing of this project needs that the activity-on-node drawing does not.

Accepted Answer

Forward pass: U finishes at 3 and V at 5. W runs 3 to 7. X waits for the later of U and V, so it runs 5 to 11. Y waits for the later of W and X, so it runs 11 to 13. M1 for taking the later value at each merge, A1 for the project duration of 13 days.
Backward pass: Y must start by 11, so W must finish by 11 and start by 7, and X must finish by 11 and start by 5. U must finish by 5 and start by 2; V must finish by 5 and start by 0. Early and late starts agree for V, X and Y, and 5 + 6 + 2 = 13 confirms it. A1 for the critical activities.
U carries 2 days of float and W carries 4. B1.
The arc drawing needs a dummy: W depends on U alone while X depends on U and V, so the two cannot leave the same event, and a zero-duration arrow has to carry U's dependency into X. The node drawing needs nothing extra, because there an arrow says only that one box precedes another. B1.

Question 12

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.

Accepted 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.

Question 13

Explain why an activity-on-arc network sometimes needs dummy activities while an activity-on-node network never does.

Accepted Answer

On an arc network an arrow does two jobs at once: it is the activity itself and, through the events at its ends, it carries the dependencies. When two activities share some but not all of their predecessors, or would otherwise run between the same pair of events, no honest arrangement of ordinary arrows exists, and a zero-duration dummy carries the missing dependency without adding work. B1 for the arrow doing two jobs, B1 for either situation that forces a dummy. On a node network the activity lives in a box and an arrow says only that one box precedes another: it carries a dependency and nothing else, so every precedence table can be drawn with plain arrows alone. B1 for the arrow carrying only the dependency.

Question 14

A project has J (5 days, no predecessors), K (3, none), L (8, after J), N (6, after K), P (4, after L), Q (7, after N) and R (2, after P and Q). Find the project duration and the critical path. The duration of L is then cut from 8 days to 5. Find the new duration and the new critical path.

Accepted Answer

Forward pass: J finishes at 5 and K at 3. L runs 5 to 13 and N runs 3 to 9. P runs 13 to 17 and Q runs 9 to 16. R waits for the later of P and Q, so it runs 17 to 19. M1 for the forward pass, A1 for 19 days.
J, L, P, R take 5 + 8 + 4 + 2 = 19, which is the project duration, so the critical path is J, L, P, R. A1. The other route, K, N, Q, R, takes 3 + 6 + 7 + 2 = 18 and so carries one day of float.
With L cut to 5 days: L now runs 5 to 10 and P runs 10 to 14, while Q still finishes at 16. R therefore starts at 16 and the project takes 18 days. M1 for a fresh forward pass, A1 for 18.
The critical path is now K, N, Q, R, with 3 + 6 + 7 + 2 = 18. A1.
Three days were cut from a critical activity and the project came down by one. Once L had given up the single day of float that K, N, Q, R was carrying, the second route became the longest and set the duration, and the remaining two days of the cut changed nothing.

Question 15

A project has P (no predecessors), Q (none), R (after P and Q), S (after P and Q), W (after P) and T (after R, S and W). Draw the activity-on-arc network using as few dummies as possible, and justify each dummy you use.

Accepted Answer

P and Q leave the start event. W needs P alone, so W leaves the event at the end of P. B1 for P, Q and W placed.
R and S both need P and Q, so both must leave an event that P and Q have reached. Let Q arrive at an event that R and S leave from, and run a dummy from the end of P into it, carrying P's dependency to R and to S at once. M1 for that dummy, A1 for the reason: without it, either R and S would be free of P, or W would be tied to Q as well.
R and S now share a starting event, and both are needed before T, so both would arrive at the event T leaves from. Two activities may not run between the same pair of events, because an activity is named by its two events, so a second dummy takes S from a fresh event of its own into the event T leaves from. M1 for that dummy, A1 for the reason.
Test a drawing by reading the precedence table back off it: an activity's predecessors are exactly those that reach its starting event by some route, dummies counted in.

Question 16

The activity network shows a project as an activity-on-arc network, each arc labelled with its activity letter and its duration in days. Carry out the forward pass and the backward pass to find the early and late event times, and hence state the duration of the project and its critical activities. Find also the total float of activity D.

Accepted Answer

Forward pass, taking the largest arriving time at each event, since an event cannot happen until every activity into it has finished. Event 1 is at 0. Event 2 is at 0 + 4 = 4 and event 3 at 0 + 6 = 6. Event 4 has only C arriving, 4 + 7 = 11. Event 5 is a merge: D arrives at 4 + 3 = 7 and E at 6 + 5 = 11, so the early time is the larger, 11. M1 for taking the larger at the merge, A1 for event 4 at 11 and event 5 at 11.
Event 6 is also a merge: F arrives at 11 + 5 = 16 and G at 11 + 8 = 19, so its early time is 19. A1 for 19.
Backward pass, taking the smallest leaving time at each event. Event 6 is at 19. Event 5 must let G run, so it is at 19 − 8 = 11; event 4 must let F run, so it is at 19 − 5 = 14. M1 for taking the smaller at a burst.
Event 2 is a burst, with C needing 14 − 7 = 7 and D needing 11 − 3 = 8, so its late time is the smaller, 7. Event 3 gives 11 − 5 = 6, and event 1 gives the smaller of 7 − 4 = 3 and 6 − 6 = 0, that is 0, which is the check that the passes agree. A1 for events 2 at 7 and 3 at 6.
The project takes 19 days. An activity is critical when its two events have equal early and late times and the gap between them is exactly its duration. Events 1, 3, 5 and 6 all have equal times, and 6 − 0 = 6 is B's duration, 11 − 6 = 5 is E's and 19 − 11 = 8 is G's, so the critical activities are B, E and G. A1 for the critical activities and the duration.
D runs from event 2 to event 5, so its total float is 11 − 4 − 3 = 4 days. B1 for the total float of D.