Edexcel International A Level Maths · D1 · Decision Mathematics 1
Critical path analysis
28 free, high-quality past-paper-style questions on Critical path analysis for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 224 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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A charity fun run is being organised. The table shows the activities involved, their durations in days and their immediate predecessors.
Activity
Immediately preceded by
Duration (days)
A
−
3
B
−
5
C
A
4
D
A, B
2
E
C
2
F
D, E
1
(a)
Draw an activity network for the project, using activity on arc and exactly one dummy.
(4)
(b)
Explain why a dummy is needed in your network.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A and B leave a single start event
B1
C leaves the end of A, and D leaves the end of B
B1
a dummy from the end of A to the end of B, with the arrow pointing to the end of B
B1
E follows C; D and E end at the same event, from which F leads to a single finish event; all arcs directed and labelled
B1
Part (b)
D depends on both A and B but C depends on A only, so C and D cannot leave the same event; the dummy lets D wait for A without making C wait for B
B1
Model sketch
Figure: one possible activity network (dummy dashed, durations in days in brackets)
Common error: Pointing the dummy from the end of B to the end of A. That would make C wait for B, which the table does not say. Everything leaving the arrowhead's event waits for everything entering the tail's event.
Setting up a stall at a craft market is modelled by an activity network with events numbered 0 to 5. The table lists each arc of the network and its duration in hours; the dummy has duration 0.
Activity
From event
To event
Duration (hours)
A
0
1
4
B
0
2
6
C
1
2
3
D
1
3
7
E
2
4
5
dummy
3
4
0
F
3
5
3
G
4
5
3
Figure: activity network drawn from the table (events numbered, durations in hours in brackets, dummy dashed); not to scale
(a)
Find the early event time of each event.
(2)
(b)
Find the late event time of each event.
(2)
(c)
State the critical activities and the minimum time needed to complete the project.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass taking the largest value at events 2 and 4, e.g. event 2: max(6, 4 + 3) = 7, event 4: max(7 + 5, 11 + 0) = 12
M1
early times 0: 0, 1: 4, 2: 7, 3: 11, 4: 12, 5: 15
A1
Part (b)
backward pass taking the smallest value at events 3, 1 and 0, e.g. event 3: min(15 − 3, 12 − 0) = 12
M1
late times 0: 0, 1: 4, 2: 7, 3: 12, 4: 12, 5: 15
A1
Part (c)
critical activities A, C, E and G ft their times
A1
minimum completion time 15 hours
B1
Exam tip: Include the dummy in both passes: it is an arc of duration 0. Here it makes the early time of event 4 compare 12 with 11.
A garden shed is assembled from a kit. The activities, their durations in hours and their immediate predecessors are given in the table. Each activity requires one worker, and once started an activity is completed by the same worker without a break.
Activity
Immediately preceded by
Duration (hours)
A
−
2
B
−
9
C
A
7
D
A
9
E
B, D
2
F
C
5
G
E, F
2
(a)
Show that the minimum time in which the shed can be assembled is 16 hours.
(2)
(b)
Calculate a lower bound for the number of workers needed to assemble the shed in 16 hours.
(2)
(c)
Only two workers are available. Explain why the shed cannot be assembled in 16 hours.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
longest route through the precedences: A, C, F, G = 2 + 7 + 5 + 2 = 16
M1
comparing with the other routes, e.g. A, D, E, G = 15 and B, E, G = 13, so the minimum time is 16 hours
A1
Part (b)
total of durations = 36 hours; 36 ÷ 16 = 2.25
M1
lower bound = 3 workers (rounded up)
A1
Part (c)
two workers can complete at most 2 × 16 = 32 hours of work in 16 hours, but 36 hours are needed; so at least 36 ÷ 2 = 18 hours
B1
Common error: Rounding 2.25 down to 2. A lower bound for workers is always rounded UP: two workers cannot fit 36 hours of work into 16 hours.
The refit of a café is modelled by the activity network in Figure 1. Dummies are shown as dashed arcs.
(a)
Write down the immediate predecessors of each of the activities D, F and G.
(3)
(b)
Explain why the dummy from event 3 to event 4 is needed.
(1)
(c)
State all the activities that could be in progress at the same time as activity C.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
D: A and B
B1
F: D and E
B1
G: C and F
B1
Part (b)
D and E have the same predecessors (A, B) and the same successor (F); without the dummy they would both run between the same two events, so they could not each be identified by a unique pair of events
B1
Part (c)
B, D, E and F, since none of them has to come before or after C
B1
Exam tip: Read predecessors by tracing BACK from the activity's start event along every arc, dummies included. D starts at event 2, and the dummy from event 1 brings A in as well as B.
Producing the first episodes of a podcast series is modelled by the activity network in Figure 1. The number in brackets on each arc is the duration of the activity in hours.
(a)
Find the early event time of each event.
(2)
(b)
Find the late event time of each event.
(2)
(c)
State the critical activities and the minimum time needed to complete the project.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass taking the largest value where arcs meet, e.g. event 2: max(7, 4 + 2) = 7, event 3: max(4 + 9, 7 + 5) = 13
Common error: Taking the largest value in the backward pass. At event 1 the late time is the SMALLER of 8 − 2 and 13 − 9; using 6 would let A and D slip and delay the project.
Figure 1 shows the activity network for landscaping a garden, with durations in days. The early and late event times have been completed.
(a)
Write down the critical path.
(1)
(b)
Find the total float of each of the activities D, E and H.
(3)
(c)
Activity E takes 3 days longer than planned. Find the new minimum time for the landscaping.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A – C – F – I (21 days)
B1
Part (b)
D: 11 − 3 − 4 = 4 days
B1
E: 14 − 9 − 4 = 1 day
B1
H: 21 − 5 − 13 = 3 days
B1
Part (c)
E has total float 1, so the project is delayed by 3 − 1 = 2 days: 23 days
B1
Exam tip: Total float = late time at the end event − duration − early time at the start event. For E that is 14 − 9 − 4, not 13 − 9 − 4: the early time at the end of an arc plays no part.
Figure 1 is the cascade (Gantt) chart for producing a school yearbook, drawn to scale with times in days. Each non-critical activity is drawn starting at its earliest start time, and its total float is shown as a dashed box.
(a)
Write down the critical activities and the minimum time needed to produce the yearbook.
(2)
(b)
Write down the total float of activity D and of activity G.
(2)
(c)
State which activities must be in progress at time 5.5 days.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A, C, F and H
B1
15 days
B1
Part (b)
D: 5 days
B1
G: 1 day
B1
Part (c)
C and E
B1
Exam tip: An activity MUST be in progress at time t when it cannot have finished yet even if started as early as possible, and cannot have started later: latest start ≤ t < earliest finish. Slide each dashed bar to the right end of its float to see the latest start.
The launch of a mobile app is planned as the activities in the table, with durations in days.
Activity
Immediately preceded by
Duration (days)
A
−
2
B
−
7
C
−
6
D
A, B
9
E
B, C
5
F
D
8
G
E
3
H
F, G
5
(a)
Draw an activity network for the project, using activity on arc and exactly two dummies.
(4)
(b)
Find the minimum time needed to complete the project, showing your early event times.
(2)
(c)
State the critical path.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A, B and C leave a single start event
B1
a dummy from the end of B to the end of A, with D leaving the end of A
B1
a dummy from the end of B to the end of C, with E leaving the end of C
B1
F follows D and G follows E; F and G end at one event, followed by H to a single finish event
B1
Part (b)
early times: end of B 7, start of D max(2, 7) = 7, start of E max(6, 7) = 7, end of D 16, end of E 12, start of H max(16 + 8, 12 + 3) = 24
M1
minimum completion time 29 days
A1
Part (c)
B – D – F – H (with the dummy from the end of B) ft
A1
Model sketch
Figure: one possible activity network (dummies dashed, durations in days in brackets)
Exam tip: B appears in both D's and E's predecessor lists, while A and C each appear in only one. That is the signal for two dummies, both starting at the end of B.
The restoration of a vintage car is modelled by an activity network with events numbered 0 to 6. The arcs are listed in the table.
Activity
From event
To event
A
0
1
B
0
2
dummy
1
2
C
1
3
D
2
3
H
2
6
F
3
4
E
3
5
dummy
4
5
G
5
6
Figure: activity network drawn from the table (dummies dashed); not to scale
(a)
Write down the immediate predecessors of each of the activities C, D, E, F, G and H.
(4)
(b)
Explain why the dummy from event 1 to event 2 is needed.
(1)
(c)
Explain why the dummy from event 4 to event 5 is needed.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
C: A, and D: A, B
B1
E: C, D and F: C, D
B1
G: E, F
B1
H: A, B
B1
Part (b)
D and H depend on both A and B, but C depends on A only; the dummy makes the activities leaving event 2 wait for A as well as B
B1
Part (c)
E and F have the same predecessors and the same successor (G); without the dummy both would run from event 3 to event 5, so they could not be identified uniquely by their events
B1
Exam tip: To read a predecessor list off a network, trace back from the activity's start event through every arc, including dummies, until you reach real activities.
A school drama club models the production of its play by an activity network with events numbered 0 to 6. The arcs and their durations in days are shown in the table.
Activity
From event
To event
Duration (days)
A
0
1
4
B
0
2
3
C
0
3
6
dummy
1
2
0
D
1
4
3
E
2
4
9
F
3
5
9
H
4
5
8
G
4
6
9
I
5
6
5
Figure: activity network drawn from the table (durations in days in brackets, dummy dashed); not to scale
(a)
Carry out a forward pass and a backward pass to find the early and late event times.
(4)
(b)
State the critical path and the length of the project.
(1)
(c)
Find the total float of each of the activities B, D and G.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 2, 4 and 5, e.g. event 4: max(4 + 3, 4 + 9) = 13
M1
early times 0: 0, 1: 4, 2: 4, 3: 6, 4: 13, 5: 21, 6: 26
A1
backward pass with minimum at events 4, 1 and 0, e.g. event 4: min(26 − 9, 21 − 8) = 13
M1
late times 0: 0, 1: 4, 2: 4, 3: 12, 4: 13, 5: 21, 6: 26
A1
Part (b)
A – (dummy) – E – H – I, length 26 days ft
A1
Part (c)
B: 4 − 3 − 0 = 1 day
B1
D: 13 − 3 − 4 = 6 days
B1
G: 26 − 9 − 13 = 4 days
B1
Common error: Ignoring the dummy from event 1 to event 2 in the forward pass. E cannot start until A is finished, so event 2 has early time 4, not 3.
A school is organising a science fair. The table shows the activities, their durations in days, their immediate predecessors and, from the early and late event times of an activity network, the earliest start time and latest finish time of each activity. The project takes 20 days.
Activity
Immediately preceded by
Duration (days)
Earliest start
Latest finish
A
−
2
0
2
B
−
3
0
4
C
A
2
2
4
D
A
2
2
6
E
B, C
5
4
9
F
D
5
4
11
G
E
2
9
11
H
F, G
9
11
20
(a)
Find the total float of each of the activities B, D and F.
(2)
(b)
Draw a cascade (Gantt) chart for the project.
(3)
(c)
State which activities must be happening at time 6.5 days.
(1)
(d)
At time 9.5 days, state which activity must be happening and which activity could be happening.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
B: 4 − 3 − 0 = 1, D: 6 − 2 − 2 = 2
M1
F: 11 − 5 − 4 = 2; so B 1 day, D 2 days, F 2 days
A1
Part (b)
critical activities A, C, E, G, H on one row from 0 to 20, end to end
B1
B, D and F on separate rows, each bar starting at its earliest start (0, 2, 4) with the correct length
B1
floats shown, e.g. dotted, ending at the latest finish times 4, 6 and 11
B1
Part (c)
E and F
B1
Part (d)
G must be happening and F could be happening
B1
Model sketch
Figure: cascade chart: critical activities on the top row, floats shown dashed
Exam tip: An activity must be happening at time t when latest start ≤ t < earliest finish. For F at 6.5: latest start 11 − 5 = 6 and earliest finish 4 + 5 = 9, so it must be in progress.
A caterer prepares food for a wedding. The activities, their durations in hours, their immediate predecessors and their latest finish times are shown. The minimum completion time with unlimited workers is 15 hours. Each activity needs one worker, and an activity once started is completed by the same worker without a break.
Activity
Immediately preceded by
Duration (hours)
Latest finish
A
−
8
11
B
−
2
2
C
−
7
12
D
A
1
12
E
B
5
7
F
C, D
2
14
G
E
7
14
H
F, G
1
15
(a)
Two workers are available. Find a lower bound for the time needed to complete the project.
(2)
(b)
Schedule the activities for the two workers. Whenever a worker is free, allocate the available activity with the smallest latest finish time. State the completion time.
(4)
(c)
Explain why this schedule finishes in the minimum possible time for two workers.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
total of durations = 33 hours; 33 ÷ 2 = 16.5
M1
lower bound 17 hours
A1
Part (b)
time 0: B (latest finish 2) and A (11) start; C (12) waits
M1
worker 1: B 0–2, E 2–7, C 7–14
A1
worker 2: A 0–8, D 8–9, G 9–16; then F 14–16 on worker 1 and H 16–17
A1
completion time 17 hours
A1
Part (c)
it finishes at 17 hours, which equals the lower bound found in part (a), so no schedule with two workers can be shorter
B1
Model sketch
Figure: one schedule for the two workers, finishing at 17 hours
Exam tip: Check each start time against the predecessors on BOTH workers' rows: F cannot start until C (worker 1) and D (worker 2) are both finished.
The rebuild of a company website is modelled by an activity network with events numbered 0 to 6, described arc by arc below. A joins event 0 to event 1 and lasts 5 days; B joins event 0 to event 2 and lasts 5 days; C joins event 1 to event 3 and lasts 2 days; E joins event 1 to event 4 and lasts 7 days; D joins event 2 to event 3 and lasts 4 days; F joins event 3 to event 5 and lasts 4 days; A dummy joins event 4 to event 5; G joins event 4 to event 6 and lasts 4 days; H joins event 5 to event 6 and lasts 7 days.
Figure: activity network drawn from the description (durations in days in brackets, dummy dashed); not to scale
(a)
Find the early and late event times.
(3)
(b)
State the critical path and the minimum completion time.
(1)
(c)
Activity C is found to need 4 more days than planned (all other durations unchanged). Find the new minimum completion time and the new critical path.
(2)
(d)
Instead, activity G is delayed by 3 days. Determine the effect on the project.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass correct with maximum at events 3, 5 and 6
M1
early times 0: 0, 1: 5, 2: 5, 3: 9, 4: 12, 5: 13, 6: 20
A1
late times 0: 0, 1: 6, 2: 5, 3: 9, 4: 13, 5: 13, 6: 20
A1
Part (b)
B – D – F – H, 20 days ft
A1
Part (c)
C has total float 9 − 2 − 5 = 2, so the project is delayed by 4 − 2 = 2 days: 22 days
B1
new critical path A – C – F – H
B1
Part (d)
total float of G = 20 − 4 − 12 = 4
M1
3 ≤ 4, so there is no effect: the project still takes 20 days
A1
Exam tip: Compare the delay with the activity's total float. Only the excess, delay − float, is added to the project, and when that happens the delayed activity becomes critical.
The activities needed to organise a school sports day and their immediate predecessors are shown in the table.
Activity
Immediately preceded by
A
−
B
−
C
A
D
A, B
E
B
F
C, D
G
E
Figure 1 shows a student's attempt to draw an activity network for the project, using activity on arc.
(a)
Explain two errors in the student's network.
(2)
(b)
Draw a correct activity network for the project, using activity on arc and exactly two dummies.
(4)
(c)
Explain why one dummy is not enough for this project.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the dummy from event 1 to event 2 makes E wait for A, but E depends on B only
B1
the network has two finish events (5 and 6); it must end at a single finish event
B1
Part (b)
A and B leave a single start event, with C leaving the end of A and E leaving the end of B
B1
D starts at its own event, reached by a dummy from the end of A and a dummy from the end of B
B1
C and D end at the same event, from which F leaves
B1
F and G end at a single finish event; all arcs directed and labelled
B1
Accept any layout and any event numbering that shows exactly these precedences; a network with more than two dummies, or with D starting at the end of A or of B, cannot score the second mark
—
Part (c)
D needs both A and B, but C needs only A and E needs only B, so D cannot start at the end of A or at the end of B; its start event needs a dummy from each
B1
Common error: Fixing one dependency with a dummy and creating another. Before drawing a dummy, check every activity that leaves its arrowhead event: each of them will now wait for everything that enters the tail event.
Figure 1 shows the activity network for refurbishing a library, with durations in days. At each event the left box holds the early event time and the right box the late event time. Four of the times have been replaced by the letters w, x, y and z.
(a)
Find the values of w and x.
(2)
(b)
Find the values of y and z.
(2)
(c)
State the critical path.
(1)
(d)
Find the total float of activity E and of activity H.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
w = max(3 + 10, 8 + 6) = 14
B1
x = max(14 + 5, 15 + 3) = 19
B1
Part (b)
y = min(8 − 4, 14 − 10) = 4
B1
z = 16 − 9 = 7
B1
Part (c)
B – F – I (19 days) ft
A1
Part (d)
E: 14 − 10 − 3 = 1 day ft
B1
H: 16 − 9 − 5 = 2 days ft
B1
Exam tip: An early time looks back along the arcs ENTERING an event (largest total); a late time looks forward along the arcs LEAVING it (smallest difference). For y that means arcs D and E, which both leave event 1.
Two fitters are booked to install a new kitchen. For each task the table gives its length in hours, the tasks that must be finished first, and the latest time by which it must end if the job is to take 14 hours, the shortest time possible. A task is done by one fitter from start to end without interruption.
Activity
Immediately preceded by
Duration (hours)
Latest finish
A
−
3
3
B
−
4
9
C
A
2
9
D
A
5
8
E
B, C
3
12
F
D
4
12
G
E, F
2
14
The site manager proposes the plan shown in Figure 1, in which the fitters are labelled Worker 1 and Worker 2.
(a)
Explain why the schedule in Figure 1 is not possible.
(2)
(b)
Explain why the kitchen cannot be fitted by two workers in less than 14 hours.
(2)
(c)
Schedule the activities for the two workers. Whenever a worker is free, allocate the available activity with the smallest latest finish time. State the completion time.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
E is shown starting at time 4
B1
but E needs C, which in the schedule does not finish until time 9 (and cannot finish before time 5)
B1
Part (b)
total of durations = 23, so two workers need at least 23 ÷ 2 = 11.5, i.e. 12 hours
M1
but the critical path takes 14 hours whatever the number of workers, so at least 14 hours
A1
Part (c)
time 0: A (3) and B (9) start; at time 3, D (8) is chosen before C (9)
M1
worker 1: A 0–3, D 3–8, F 8–12, G 12–14
A1
worker 2: B 0–4, C 4–6, E 6–9
A1
completion time 14 hours
A1
Common error: Starting an activity as soon as a worker is free without checking every predecessor. E needs B AND C; when worker 2 finishes B at time 4, C has not even started.
Figure 1 is the cascade (Gantt) chart for setting up a pop-up shop, drawn to scale with times in hours. Each non-critical activity starts at its earliest start time and its total float is shown as a dashed box. The immediate predecessors are: D follows A, E follows B, F follows C, and G follows D and E. Each activity needs one worker, and an activity once started is completed by the same worker without a break.
(a)
Write down the total float of activity C and of activity E.
(2)
(b)
Explain why three workers would be needed if every activity started at its earliest start time.
(1)
(c)
Find a lower bound for the number of workers needed to complete the project in 13 hours.
(2)
(d)
Show how two workers can complete the project in 13 hours.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
C: 8 hours
B1
E: 4 hours
B1
Part (b)
A, B and C would all be in progress between time 0 and time 2
B1
Part (c)
total of durations = 24; 24 ÷ 13 = 1.84…
M1
lower bound 2 workers
A1
Part (d)
critical activities at fixed times A 0–6, D 6–10, G 10–13, with B, C, E and F fitted into the workers' free time
M1
a complete valid schedule, e.g. worker 1: A 0–6, D 6–10, G 10–13; worker 2: B 0–2, E 2–6, C 6–9, F 9–11
A1
Accept any schedule that passes every check: no worker has two activities at once; each activity runs for its full duration; E starts at or after the end of B and ends by 10; F starts at or after the end of C; everything ends by 13 (e.g. worker 1: A, E 6–10, G; worker 2: B, C 2–5, D 6–10, F 10–12)
—
Exam tip: Float is what lets you level the workload. Delay a non-critical activity inside its dashed box and the project end time does not move.
Figure 1 shows the activity network for upgrading the heating in a village hall. The durations are in days; activity C takes x days.
(a)
Taking x = 3, find the early and late event times.
(3)
(b)
State the critical path and the minimum completion time when x = 3.
(1)
(c)
Find the greatest value of x for which the minimum completion time is still 21 days.
(2)
(d)
Given instead that x = 12, find the minimum completion time and the critical path.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 2, 3 and 4, e.g. event 3: max(5 + 3, 8 + 6) = 14
M1
early times 0: 0, 1: 5, 2: 8, 3: 14, 4: 18, 5: 21
A1
late times 0: 0, 1: 6, 2: 8, 3: 14, 4: 18, 5: 21
A1
Part (b)
B – E – G – H, 21 days ft
A1
Part (c)
total float of C = 14 − 3 − 5 = 6 when x = 3
M1
so x can rise by 6: greatest value x = 9
A1
Part (d)
route A – C – G – H = 5 + 12 + 4 + 3 = 24 days, longer than 21
B1
critical path A – C – G – H
B1
Exam tip: With an unknown duration, compare the routes through it with the longest route that avoids it. Here every route through C is A – C – G – H = x + 12, and the longest route avoiding C is 21, so the project takes longer as soon as x > 9 (at x = 9 both routes are critical).
Figure 1 shows the activity network for installing solar panels on a school roof. Durations are in days and dummies are shown as dashed arcs.
(a)
Explain the purpose of each of the two dummies.
(2)
(b)
Find the early and late event times.
(3)
(c)
State the critical path and the minimum completion time.
(1)
(d)
Find the total float of activity D.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
dummy from 1 to 2: E depends on A as well as B, while C and D depend on A only, so it shows that E must wait for A
B1
dummy from 3 to 4: C and D have the same predecessor (A) and the same successor (F), so it lets each be identified by its own pair of events
B1
Part (b)
forward pass with maximum at events 2, 4 and 5, e.g. event 4: max(4 + 7, 9 + 0) = 11
M1
early times 0: 0, 1: 4, 2: 6, 3: 9, 4: 11, 5: 15, 6: 17
A1
late times 0: 0, 1: 5, 2: 6, 3: 12, 4: 12, 5: 15, 6: 17
A1
Part (c)
B – E – G, 17 days ft
A1
Part (d)
12 − 5 − 4 = 3 days ft
B1
Common error: Leaving the dummies out of the passes. A dummy is an arc of duration 0: at event 2 the forward pass compares 6 with 4 + 0, and at event 1 the backward pass compares 12 − 5, 12 − 7 and 6 − 0.
An extension is being built on to a health clinic. The activities, their durations in weeks and their immediate predecessors are shown in the table.
Activity
Immediately preceded by
Duration (weeks)
A
−
9
B
−
7
C
−
6
D
A
4
E
B
4
F
C
2
G
D, E
7
H
E, F
9
I
G
3
J
H, I
7
(a)
Draw an activity network for the project, using activity on arc and exactly two dummies.
(4)
(b)
Find the early and late event times for your network.
(4)
(c)
State the critical path and the minimum project duration.
(1)
(d)
Find the total float of each of the activities B, E and H.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
A, B and C leave a single start event, followed by D, E and F respectively
B1
a dummy from the end of E to the start of G, which also follows D
B1
a dummy from the end of E to the start of H, which also follows F
B1
I follows G; H and I end at one event, followed by J to a single finish event
B1
Part (b)
forward pass with maximum at the starts of G, H and J, e.g. start of G: max(9 + 4, 7 + 4) = 13
M1
early times correct, finishing at 30
A1
backward pass with minimum at the end of E: min(13 − 0, 14 − 0) = 13
M1
late times correct, e.g. start of H 14, end of C 12, end of B 9
A1
Part (c)
A – D – G – I – J, 30 weeks
A1
Part (d)
B: 9 − 7 − 0 = 2
B1
E: 13 − 4 − 7 = 2
B1
H: 23 − 9 − 11 = 3 weeks
B1
Model sketch
Figure: one possible activity network (dummies dashed, durations in weeks in brackets)
Exam tip: E appears as a predecessor of both G and H, but D and F appear only once each. When one activity feeds two different starts that each have another predecessor, expect a dummy from its end event to each start.
A stage is built for a music festival. The project is modelled by an activity network with events numbered 0 to 7. For each event, the arcs leaving it are listed with their durations. Each activity needs one worker. Leaving event 0: B to event 1 (2 hours), A to event 2 (6 hours), C to event 3 (2 hours). Leaving event 1: a dummy to event 2, E to event 5 (9 hours). Leaving event 2: D to event 4 (6 hours). Leaving event 3: F to event 5 (8 hours). Leaving event 4: G to event 6 (8 hours). Leaving event 5: H to event 6 (8 hours), I to event 7 (9 hours). Leaving event 6: J to event 7 (4 hours).
Figure: activity network drawn from the description (durations in hours in brackets, dummy dashed); not to scale
(a)
Find the early and late event times.
(3)
(b)
State the critical path.
(1)
(c)
Draw a cascade (Gantt) chart for the project.
(4)
(d)
Use your cascade chart to state which activities must be happening at time 13.5 hours, and which activity could be happening.
(2)
(e)
Find a lower bound for the number of workers needed to build the stage in the minimum time.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 2, 5, 6 and 7
M1
early times 0: 0, 1: 2, 2: 6, 3: 2, 4: 12, 5: 11, 6: 20, 7: 24
A1
late times 0: 0, 1: 3, 2: 6, 3: 4, 4: 12, 5: 12, 6: 20, 7: 24
A1
Part (b)
A – D – G – J (24 hours) ft
A1
Part (c)
critical activities A, D, G, J on one row: 0–6, 6–12, 12–20, 20–24
B1
B 0–2 (float 1), C 0–2 (float 2), E 2–11 (float 1), F 2–10 (float 2)
B1
H 11–19 (float 1), I 11–20 (float 4)
B1
each non-critical bar starting at its earliest start, with its float shown to its latest finish
B1
Part (d)
G and H must be happening
B1
I could be happening
B1
Part (e)
total of durations = 62; 62 ÷ 24 = 2.58…
M1
lower bound 3 workers
A1
Model sketch
Figure: cascade chart: critical activities on the top row, floats shown dashed
Exam tip: On a cascade chart, put all the critical activities on the top row and each non-critical activity on its own row; the dotted float box always ends at the late time of the activity's end event.
A small ferry is refitted in dry dock. The refit is modelled by an activity network with events numbered 0 to 6; each arc is listed as (start event, end event) with its activity and duration. Each activity needs one worker, and an activity once started is completed by the same worker without a break.
(0, 1): activity A, 2 days
(0, 2): activity C, 3 days
(0, 3): activity B, 9 days
(1, 4): activity D, 3 days
(2, 3): dummy
(2, 5): activity F, 6 days
(3, 5): activity E, 2 days
(4, 5): dummy
(4, 6): activity G, 6 days
(5, 6): activity H, 5 days
Figure: activity network drawn from the list (durations in days in brackets, dummies dashed); not to scale
(a)
Find the early and late event times.
(3)
(b)
State the critical path and the minimum duration of the refit.
(1)
(c)
Find a lower bound for the number of workers needed to complete the refit in the minimum time.
(2)
(d)
Schedule the activities for three workers so that the refit is completed in the minimum time. Whenever a worker is free, allocate the available activity with the smallest latest finish time.
(4)
(e)
Only two workers are available. Show that the refit cannot be completed in less than 18 days.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 3 and 5
M1
early times 0: 0, 1: 2, 2: 3, 3: 9, 4: 5, 5: 11, 6: 16
A1
late times 0: 0, 1: 7, 2: 5, 3: 9, 4: 10, 5: 11, 6: 16
A1
Part (b)
B – E – H, 16 days ft
A1
Part (c)
total of durations = 36; 36 ÷ 16 = 2.25
M1
lower bound 3 workers
A1
Part (d)
latest finish times used: A 7, B 9, C 5, D 10, E 11, F 11, G 16, H 16
M1
worker 1: C 0–3, F 3–9, E 9–11, H 11–16
A1
worker 2: A 0–2, D 2–5, G 5–11; worker 3: B 0–9
A1
finishes at 16 days, every activity after all its predecessors
A1
accept E 9–11 and H 11–16 on worker 3 instead of worker 1
—
Part (e)
two workers can do at most 2 days of work per day
M1
36 ÷ 2 = 18, so at least 18 days are needed
A1
Model sketch
Figure: one schedule for the three workers, finishing at 16 days
Common error: Missing a predecessor that arrives along a dummy. H needs D as well as E and F, because of the dummy from event 4 to event 5. Trace back through every dummy before starting an activity.
A small business conference is being organised. The activities, their durations in days and their immediate predecessors are shown in the table. Each activity needs one worker, and an activity once started is completed by the same worker without a break.
Activity
Immediately preceded by
Duration (days)
A
−
3
B
−
5
C
A
7
D
A, B
5
E
C
5
F
C
4
G
E, F
2
H
D, G
7
(a)
Draw an activity network for the project, using activity on arc and exactly two dummies.
(4)
(b)
Find the early and late event times.
(3)
(c)
State the critical path and the minimum completion time.
(1)
(d)
Find a lower bound for the number of workers needed to complete the project in the minimum time.
(1)
(e)
Schedule the activities for two workers, allocating the available activity with the smallest latest finish time whenever a worker is free, and show that they complete the project in 24 days.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
A and B leave a single start event; C leaves the end of A
B1
a dummy from the end of A to the end of B, with D leaving the end of B
B1
E and F both leave the end of C; one of them ends at an extra event joined by a dummy to the end of the other, where G starts
B1
D and G end at one event, followed by H to a single finish event
B1
Part (b)
forward pass with maximum where the dummies meet, e.g. start of G: max(10 + 5, 14 + 0) = 15, start of H: max(5 + 5, 15 + 2) = 17
M1
early times correct, finishing at 24
A1
late times correct, e.g. end of B 12, end of F 15
A1
Part (c)
A – C – E – G – H, 24 days
A1
Part (d)
38 ÷ 24 = 1.58…, so 2 workers
B1
Part (e)
worker 1: A 0–3, C 3–10, E 10–15, G 15–17, H 17–24
M1
worker 2: B 0–5, D 5–10, F 10–14; finish at 24 days
A1
Model sketch
Figure: one possible activity network (dummies dashed, durations in days in brackets)
Exam tip: E and F have the same predecessor (C) and the same successor (G). That is exactly the situation for a uniqueness dummy.
Figure 1 shows the activity network for replacing the roof of a sports hall, with durations in weeks. The boxes at each event are for the early and late event times.
(a)
Find the early and late event times.
(4)
(b)
State the critical path and the minimum project duration.
(1)
(c)
Find the total float of each of the activities A, E and H.
(3)
(d)
An extra activity K is needed. K must follow D and E, and I cannot start until K is finished; it is drawn as an arc from event 4 to event 5. Find the greatest duration of K for which the project still takes 19 weeks.
(2)
(e)
K takes 5 weeks. Find the new minimum project duration and the new critical path.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 4, 5 and 6, e.g. event 4: max(4 + 6, 6 + 5) = 11
M1
early times 0: 0, 1: 4, 2: 6, 3: 3, 4: 11, 5: 14, 6: 19
A1
backward pass with minimum at events 2 and 0, e.g. event 2: min(13 − 5, 14 − 8) = 6
M1
late times 0: 0, 1: 7, 2: 6, 3: 10, 4: 13, 5: 14, 6: 19
A1
Part (b)
B – F – I, 19 weeks ft
A1
Part (c)
A: 7 − 4 − 0 = 3
B1
E: 13 − 5 − 6 = 2
B1
H: 19 − 6 − 11 = 2 weeks
B1
Part (d)
K runs from event 4 (early time 11) to event 5 (late time 14)
Exam tip: Before adding an arc, check it adds no hidden precedences: K from event 4 to event 5 makes K wait for everything entering event 4 (D and E), and makes I wait for K, which is exactly what is required.
Figure 1 shows the activity network for renovating a playground, with durations in days. Each activity needs one worker, and an activity once started is completed by the same worker without a break.
(a)
Find the early and late event times.
(3)
(b)
State the critical path and the minimum completion time, and find the total float of each of the activities B, C, E and G.
(3)
(c)
Find a lower bound for the number of workers needed to complete the project in 19 days.
(2)
(d)
Show that the project cannot be completed in 19 days by the number of workers found in part (c).
(2)
(e)
Three workers are available. Using the rule ‘whenever a worker is free, allocate the available activity with the smallest latest finish time’, state the activities done by each worker.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 3, 5 and 6, e.g. event 3: max(4, 3 + 0, 3 + 0) = 4
M1
early times 0: 0, 1: 3, 2: 3, 3: 4, 4: 14, 5: 16, 6: 19
A1
late times 0: 0, 1: 4, 2: 4, 3: 4, 4: 14, 5: 16, 6: 19
A1
Part (b)
A – D – F – H, 19 days ft
A1
B: 4 − 3 − 0 = 1 and C: 4 − 3 − 0 = 1
B1
E: 16 − 6 − 3 = 7 and G: 19 − 4 − 14 = 1
B1
Part (c)
total of durations = 35; 35 ÷ 19 = 1.84…
M1
lower bound 2 workers
A1
Part (d)
A, B and C cannot start before time 0 and must all be finished by time 4 (latest finish 4 for each), so all their work lies in the first 4 days
M1
that is 4 + 3 + 3 = 10 worker-days, but 2 workers can do only 2 × 4 = 8 in 4 days, so 19 days is impossible with 2 workers
A1
Alternative: B and C each have latest start 1 and earliest finish 3, and A runs 0–4, so A, B and C must all be in progress between times 1 and 3 [M1 A1]
—
Part (e)
A, B and C (latest finish 4 each) all start at time 0; E (the only available activity) starts at 3 on a worker freed by B or C; D 4–14
M1
F 14–16 and G 14–18 on two different free workers, H 16–19; finish at 19 days, e.g. worker 1: A 0–4, D 4–14, F 14–16, H 16–19; worker 2: B 0–3, E 3–9, G 14–18; worker 3: C 0–3
A1
Accept any assignment of these activities, at these times, to the three workers
—
Common error: Trusting the lower bound. Total work ÷ critical time only says the job cannot be done with FEWER workers; whether that many is enough depends on how the work is forced to overlap.
A council is rebuilding a footbridge over a river. The work is modelled by the activity-on-arc network in Figure 1, where the number beside each activity is its duration in days. Every activity is carried out by a single worker, who stays on it until it is complete.
(a)
Find the early and late event times.
(3)
(b)
State the critical path and the minimum completion time.
(1)
(c)
Find a lower bound for the number of workers needed to complete the project in 20 days.
(2)
(d)
Only two workers are available. Schedule the activities for the two workers: whenever a worker is free, allocate the available activity with the smallest latest finish time. State the completion time.
(4)
(e)
Explain why no schedule with two workers can finish in less than 23 days.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
forward pass with maximum at events 3, 4, 5 and 6, e.g. event 3: max(6 + 3, 3 + 7) = 10
M1
early times 0: 0, 1: 6, 2: 3, 3: 10, 4: 10, 5: 12, 6: 20
A1
late times 0: 0, 1: 7, 2: 3, 3: 10, 4: 13, 5: 12, 6: 20
A1
Part (b)
B – E – H – I, 20 days ft
A1
Part (c)
total of durations = 45; 45 ÷ 20 = 2.25
M1
lower bound 3 workers
A1
Part (d)
latest finish times used: A 7, B 3, C 10, D 13, E 10, F 12, G 20, H 12, I 20
M1
worker 1: B 0–3, E 3–10, H 10–12, D 12–16, G 16–23
A1
worker 2: A 0–6, C 6–9, F 9–14, I 14–22
A1
completion time 23 days
A1
Accept the two workers labelled the other way round
—
Part (e)
two workers can do at most 2 days of work per day, so they need at least 45 ÷ 2 = 22.5 days
M1
every duration is a whole number of days, so the two workers' workloads are whole numbers adding to 45; one worker has at least 23 days of work, so at least 23 days, which this schedule achieves
A1
Exam tip: At each decision time list only the activities whose predecessors have ALL finished. At time 6, C (latest finish 10), F (12) and D (13) are all available; worker 2 takes C, and D is not started until time 12.
Exam tip: D is the only activity that appears in two predecessor lists alongside different partners (C for G, E for H). That is why both dummies start at the end of D.
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