D1-4 Critical path analysis — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section D1-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

D1-4.1 · Modelling of a project by an activity network, from a precedence table.

Explanation

An activity network is drawn activity on arc:
each activity is an arc labelled with its name and duration, and each numbered event is a node marking the moment when every activity into it has finished. A precedence table lists, for each activity, only its immediate predecessors, so an activity that depends on $A$ through $B$ is recorded as depending on $B$ alone.
A dummy is a zero-duration arc, always drawn dashed, and it is needed for two reasons. The first is logical: when one activity needs $A$ and $B$ while another needs only $A$ , the two cannot leave the same event, so $A$ ends at its own event and a dummy carries its dependency forward.
The second is uniqueness: two activities must never share both their start event and their end event, since each activity has to be identified by its pair of events, so a dummy separates them.
Every network has exactly one start event, with no arcs entering, and one finish event, with no arcs leaving.

Worked example

A project has activities $A$ and $B$ with no predecessors, $C$ with immediate predecessor $A$ , and $D$ with immediate predecessors $A$ and $B$ . Explain why a dummy is needed and describe the network.

1.Both $A$ and $B$ leave the start event, event $1$ .
2. $A$ must end at its own event, event $2$ , because $C$ depends on $A$ alone.
3. $B$ ends at event $3$ , and $D$ leaves event $3$ .
4.A dummy from event $2$ to event $3$ carries the dependency of $D$ on $A$ without making $C$ depend on $B$ .

Answer: One dummy is needed, from the end of $A$ to the start of $D$ , because $D$ depends on both $A$ and $B$ while $C$ depends on $A$ only.

Common mistakes

Don't fall into the trap of listing a predecessor that is implied by another one, so that the table is not reduced to immediate predecessors.
Don't fall into the trap of drawing two activities between the same pair of events without separating them with a dummy.
Don't fall into the trap of giving a dummy a non-zero duration or drawing it with a solid line.

Exam tip

Work down the precedence table one activity at a time and ask which activities must be complete before it starts; a dummy is needed exactly when two activities have overlapping but unequal predecessor sets.

Tier 1 · Easy

1.
Explain what is meant by a dummy in an activity network, and state the duration of a dummy.
(3)
(Total for Question 1 is 3 marks)
2.
A project has activities $A$ , $B$ , $C$ , $D$ with $A$ and $B$ having no predecessors, $C$ having immediate predecessor $A$ , and $D$ having immediate predecessor $B$ . State how many dummies are needed and describe the activity network.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A project has the precedence table: $A$ and $B$ have no predecessors; $C$ has immediate predecessor $A$ ; $D$ has immediate predecessors $A$ and $B$ ; $E$ has immediate predecessor $C$ ; $F$ has immediate predecessor $C$ ; $G$ has immediate predecessors $D$ and $E$ ; $H$ has immediate predecessors $F$ and $G$ . Describe an activity network for this project, giving the event numbers of every arc.
(7)
(Total for Question 1 is 7 marks)
2.
Explain the two distinct reasons a dummy may be required in an activity network, and give a precedence table of three activities for which no dummy is required at all.
(5)
(Total for Question 2 is 5 marks)
3.
A precedence table is given as: $A$ has no predecessors; $B$ has immediate predecessor $A$ ; $C$ has immediate predecessors $A$ and $B$ ; $D$ has immediate predecessors $B$ and $C$ . Explain why this table is not correctly reduced to immediate predecessors, and write down the corrected table.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A project has activities $A$ to $F$ with the precedence table: $A$ and $B$ have no predecessors; $C$ has immediate predecessor $A$ ; $D$ has immediate predecessor $A$ ; $E$ has immediate predecessors $B$ and $C$ ; $F$ has immediate predecessors $D$ and $E$ . Describe an activity network, state the number of dummies you have used, and justify each one.
(7)
(Total for Question 1 is 7 marks)
2.
A project has activities $P$ , $Q$ , $R$ , $S$ where $P$ and $Q$ have no predecessors, and both $R$ and $S$ have immediate predecessors $P$ and $Q$ . Explain why exactly two dummies are required, and describe the resulting network.
(6)
(Total for Question 2 is 6 marks)
3.
Explain why an activity network must have exactly one start event and exactly one finish event, and describe how to modify a network in which three activities have no predecessors and two activities have no successors.
(6)
(Total for Question 3 is 6 marks)
4.
A project has activities $A$ to $G$ with the precedence table: $A$ , $B$ and $C$ have no predecessors; $D$ has immediate predecessors $A$ and $B$ ; $E$ has immediate predecessors $B$ and $C$ ; $F$ has immediate predecessor $D$ ; $G$ has immediate predecessors $D$ and $E$ . Determine the minimum number of dummies needed and describe the network.
(8)
(Total for Question 4 is 8 marks)
5.
A student draws an activity network in which an arc runs from event $5$ back to event $3$ , creating a directed cycle. Explain why this cannot represent a valid project, and describe the check that a precedence table must pass before a network can be drawn.
(6)
(Total for Question 5 is 6 marks)

D1-4.2 · Completion of the precedence table for a given activity network.

Explanation

Reading a precedence table off a network reverses the drawing process. For each activity, look at the event it leaves and list every activity that enters that event; those are its immediate predecessors.
A dummy is not an activity, so when a dummy enters the event, follow the dummy backwards and take the activities entering the dummy's own start event instead.
An activity leaving the start event has no predecessors.
Finally the table must be reduced: if an activity is listed and is also a predecessor of another listed activity, remove it, because a precedence table records only immediate predecessors.
A useful check is that every activity appears somewhere in the table, that the number of activities with no predecessors matches the number of arcs leaving the start event, and that no activity is listed as its own predecessor directly or through a chain.

Worked example

In a network, $A$ runs from event $1$ to event $2$ , $B$ runs from event $1$ to event $3$ , a dummy runs from event $2$ to event $3$ , $C$ leaves event $2$ and $D$ leaves event $3$ . Write down the precedence table.

1. $A$ and $B$ leave the start event $1$ , so neither has a predecessor.
2. $C$ leaves event $2$ ; only $A$ enters event $2$ , so $C$ depends on $A$ .
3. $D$ leaves event $3$ ; $B$ and the dummy enter event $3$ .
4.The dummy starts at event $2$ , which $A$ enters, so $D$ depends on $A$ and $B$ .

Answer: $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ and $B$ .

Common mistakes

Don't fall into the trap of recording the dummy itself as a predecessor instead of tracing back through it.
Don't fall into the trap of leaving in a predecessor that is implied by another entry, so the table is not reduced.
Don't fall into the trap of reading the activities leaving the event rather than the activities entering it.

Exam tip

Take the activities in the order they appear on the network and write the event each one leaves beside it; the table then follows from a single scan of the arcs entering those events.

Tier 1 · Easy

1.
In an activity network, activity $E$ leaves event $4$ , and the only activities entering event $4$ are $B$ and $C$ . Write down the immediate predecessors of $E$ , and state what you would do differently if a dummy also entered event $4$ .
(3)
(Total for Question 1 is 3 marks)
2.
An activity network has $A$ from event $1$ to event $2$ , $B$ from event $2$ to event $3$ , $C$ from event $3$ to event $4$ and $D$ from event $4$ to event $5$ . Write down the precedence table.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
An activity network on events $1$ to $6$ has: $A$ from $1$ to $2$ ; $B$ from $1$ to $3$ ; a dummy from $2$ to $3$ ; $C$ from $2$ to $4$ ; $D$ from $3$ to $4$ ; $E$ from $3$ to $5$ ; $F$ from $4$ to $5$ ; $G$ from $5$ to $6$ . Write down the complete precedence table.
(7)
(Total for Question 1 is 7 marks)
2.
A student reads a precedence table from a network and writes $G$ : $A$ , $C$ , $E$ , having found that $C$ enters the event $G$ leaves, that $A$ is a predecessor of $C$ , and that $E$ also enters that event. Explain the error and write down the corrected entry.
(4)
(Total for Question 2 is 4 marks)
3.
An activity network on events $1$ to $5$ has: $P$ from $1$ to $2$ ; $Q$ from $1$ to $3$ ; a dummy from $3$ to $2$ ; $R$ from $2$ to $4$ ; $S$ from $2$ to $5$ ; $T$ from $4$ to $5$ . Write down the precedence table and state which of the two reasons for a dummy applies here.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
An activity network on events $1$ to $6$ has: $A$ from $1$ to $2$ ; $B$ from $1$ to $3$ ; $C$ from $1$ to $4$ ; a dummy from $3$ to $2$ ; a dummy from $3$ to $4$ ; $D$ from $2$ to $5$ ; $E$ from $4$ to $6$ ; $F$ from $5$ to $6$ . Write down the precedence table, explaining how you treated each dummy, and state which event is the finish event.
(8)
(Total for Question 1 is 8 marks)
2.
Explain why the precedence table read from a network is unique, but the network drawn from a precedence table is not. Illustrate the second point with a table of four activities.
(6)
(Total for Question 2 is 6 marks)
3.
An activity network on events $1$ to $6$ has: $A$ from $1$ to $2$ ; $B$ from $1$ to $3$ ; $C$ from $2$ to $4$ ; $D$ from $2$ to $3$ ; $E$ from $3$ to $5$ ; $F$ from $4$ to $5$ ; $G$ from $5$ to $6$ ; $H$ from $4$ to $6$ . Write down the precedence table and identify every activity with no successor.
(7)
(Total for Question 3 is 7 marks)
4.
A precedence table read from a network gives $A$ : none; $B$ : $A$ ; $C$ : $A$ ; $D$ : $B$ and $C$ ; $E$ : $D$ ; $F$ : $D$ ; $G$ : $E$ and $F$ . Determine the number of arcs, including any dummies, in the smallest activity network that realises this table, and justify the count.
(7)
(Total for Question 4 is 7 marks)
5.
An examiner writes that in a precedence network, precedence tables will only show immediate predecessors. Explain what would go wrong if a table listed all predecessors instead, using a project of five activities in a single chain as an example.
(6)
(Total for Question 5 is 6 marks)

D1-4.3 · Algorithm for finding the critical path. Earliest and latest event times. Earliest and latest start and finish times for activities. Identification of critical activities and critical path(s).

Explanation

The forward pass sets the earliest event time of the start event to $0$ and then gives each event the largest value of earliest time plus duration over the activities entering it.
The backward pass sets the latest event time of the finish event equal to its earliest time and then gives each event the smallest value of latest time minus duration over the activities leaving it.
For an activity from event $i$ to event $j$ with duration $d$ , the earliest start is $e_i$ , the earliest finish is $e_i+d$ , the latest finish is $l_j$ and the latest start is $l_j-d$ .
An activity is critical when its earliest start equals its latest start, which happens exactly when $e_i=l_i$ , $e_j=l_j$ and $l_j-e_i=d$ ; a critical path is a chain of critical activities running from the start event to the finish event, and there may be more than one.
Since every activity needs one worker, the least number of workers that could finish the project in the critical time is $\left\lceil\dfrac{\text{sum of all durations}}{\text{project duration}}\right\rceil$ .

Worked example

A project has $A$ of duration $4$ with no predecessors, $B$ of duration $6$ with no predecessors, $C$ of duration $3$ after $A$ , and $D$ of duration $5$ after $B$ and $C$ . Find the project duration and the critical path.

1.Forward pass: earliest times are $0$ at the start, $4$ after $A$ , and $\max(6, 4+3)=7$ before $D$ .
2.The project duration is $7+5=12$ .
3.Backward pass: the latest time before $D$ is $12-5=7$ , and the latest time after $A$ is $7-3=4$ .
4. $A$ has $4=4$ and $C$ has $7-4=3$ , so both are critical; $B$ has latest start $7-6=1$ but earliest start $0$ , so it has float $1$ .

Answer: The project takes $12$ and the critical path is $A$ , $C$ , $D$ .

Common mistakes

Don't fall into the trap of taking the minimum on the forward pass or the maximum on the backward pass.
Don't fall into the trap of calling an activity critical because both of its events are critical, without checking that $l_j-e_i$ equals the duration.
Don't fall into the trap of assuming there is only one critical path when two chains have the same length.

Exam tip

Do the whole forward pass before starting the backward pass, and check that the latest time at the start event comes out as $0$ ; if it does not, there is an arithmetic error.

Tier 1 · Easy

1.
An activity runs from event $3$ to event $6$ and has duration $7$ . The earliest event time at event $3$ is $9$ and the latest event time at event $6$ is $20$ . Write down the earliest start, earliest finish, latest finish and latest start of the activity.
(4)
(Total for Question 1 is 4 marks)
2.
A project has activities $A$ of duration $5$ and $B$ of duration $4$ with no predecessors, $C$ of duration $6$ after $A$ , and $D$ of duration $3$ after $A$ and $B$ . Find the project duration.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A project has activities with durations $A=5$ , $B=4$ , $C=6$ , $D=3$ , $E=7$ , $F=2$ , $G=4$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ and $B$ ; $E$ : $C$ ; $F$ : $C$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ . Find the project duration and the critical path.
(7)
(Total for Question 1 is 7 marks)
2.
For the project with durations $A=5$ , $B=4$ , $C=6$ , $D=3$ , $E=7$ , $F=2$ , $G=4$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ and $B$ ; $E$ : $C$ ; $F$ : $C$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ , drawn on events $1$ to $7$ with $A$ from $1$ to $2$ , $B$ from $1$ to $3$ , a dummy from $2$ to $3$ , $C$ from $2$ to $4$ , $D$ from $3$ to $5$ , $E$ from $4$ to $5$ , $F$ from $4$ to $6$ , $G$ from $5$ to $6$ and $H$ from $6$ to $7$ , write down the earliest and latest time at every event.
(7)
(Total for Question 2 is 7 marks)
3.
A project has durations $A=4$ , $B=6$ , $C=3$ , $D=5$ , $E=2$ , $F=4$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ and $C$ ; $F$ : $D$ and $E$ . Find the project duration and all critical paths.
(7)
(Total for Question 3 is 7 marks)

Tier 3 · Hard

1.
A project has durations $A=5$ , $B=7$ , $C=4$ , $D=6$ , $E=3$ , $F=8$ , $G=2$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ ; $F$ : $C$ and $D$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ . Find the project duration, the critical path, and a lower bound for the number of workers needed to finish in that time.
(9)
(Total for Question 1 is 9 marks)
2.
Explain why an activity whose start and end events are both critical need not itself be critical, and give a numerical example.
(6)
(Total for Question 2 is 6 marks)
3.
A project has durations $A=6$ , $B=9$ , $C=7$ , $D=4$ , $E=5$ , $F=8$ , $G=3$ , $H=6$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ and $D$ ; $F$ : $C$ ; $G$ : $E$ ; $H$ : $F$ and $G$ . Find the project duration, the critical path, and the total float of every non-critical activity.
(9)
(Total for Question 3 is 9 marks)
4.
A project has duration $30$ days and its critical path consists of activities $P$ , $Q$ , $R$ , $S$ with durations $8$ , $6$ , $9$ and $7$ . The duration of $Q$ is reduced by $4$ days. Explain why the project duration need not fall by $4$ days, and state the smallest and largest possible new project durations.
(7)
(Total for Question 4 is 7 marks)
5.
For the project with durations $A=5$ , $B=7$ , $C=4$ , $D=6$ , $E=3$ , $F=8$ , $G=2$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ ; $F$ : $C$ and $D$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ , determine the largest amount by which the duration of $C$ could be increased without changing the project duration, and the effect on the critical path of increasing it by exactly that amount.
(8)
(Total for Question 5 is 8 marks)

D1-4.4 · Calculation of the total float of an activity. Construction of Gantt (cascade) charts.

Explanation

The total float of an activity from event $i$ to event $j$ with duration $d$ is $F(i, j)=l_j-e_i-d$ , where $e_i$ is the earliest time for event $i$ and $l_j$ is the latest time for event $j$ . Equivalently it is the latest start minus the earliest start, or the latest finish minus the earliest finish.
Critical activities are exactly those with zero total float. A Gantt or cascade chart draws time along the horizontal axis, one row per activity.
Critical activities are drawn as solid blocks in a single unbroken line across the top, since they cannot move. Each non-critical activity is drawn as a block starting at its earliest start, followed by a lightly shaded rectangle of length equal to its total float, showing the window in which the block may slide.
The chart makes it easy to read off which activities may be under way at a given time: draw a vertical line at that time and read every row it crosses.
Total floats may not be used independently, because two activities in the same chain can share the same float.

Worked example

An activity of duration $6$ runs from event $2$ to event $5$ . The earliest time at event $2$ is $9$ and the latest time at event $5$ is $20$ . Find its total float and describe its bar on a Gantt chart.

1.Total float $=l_5-e_2-d=20-9-6=5$ .
2.The earliest start is $9$ and the earliest finish is $15$ .
3.The latest start is $20-6=14$ and the latest finish is $20$ .
4.The bar is drawn from $9$ to $15$ , followed by a float rectangle from $15$ to $20$ .

Answer: The total float is $5$ , and the activity is drawn as a block from $9$ to $15$ with a float window extending to $20$ .

Common mistakes

Don't fall into the trap of using the latest time at the start event instead of at the end event in the float formula.
Don't fall into the trap of sliding two activities on the same chain by their full floats at once, which double counts shared float.
Don't fall into the trap of drawing the critical activities with gaps between them on the cascade chart.

Exam tip

Write $F=l_j-e_i-d$ at the top of your working and substitute the three numbers explicitly; nearly all float errors come from picking the wrong event time.

Tier 1 · Easy

1.
An activity of duration $4$ runs from event $2$ to event $6$ . The earliest time at event $2$ is $7$ and the latest time at event $6$ is $15$ . Calculate the total float of the activity.
(2)
(Total for Question 1 is 2 marks)
2.
State the total float of a critical activity, and explain how critical activities are drawn on a Gantt chart.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
For a project of duration $24$ with activities $A=5$ , $B=7$ , $C=4$ , $D=6$ , $E=3$ , $F=8$ , $G=2$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ ; $F$ : $C$ and $D$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ , calculate the total float of every activity.
(7)
(Total for Question 1 is 7 marks)
2.
For a project of duration $24$ in which the critical activities are $A$ from $0$ to $5$ , $D$ from $5$ to $11$ , $F$ from $11$ to $19$ and $H$ from $19$ to $24$ , and the non-critical activities have earliest starts and floats $B$ : $0$ and $7$ ; $C$ : $5$ and $2$ ; $E$ : $7$ and $7$ ; $G$ : $11$ and $6$ , with durations $B=7$ , $C=4$ , $E=3$ , $G=2$ , describe the Gantt chart, giving the start and end of every block and every float rectangle.
(7)
(Total for Question 2 is 7 marks)
3.
Using the Gantt chart of a project in which $A$ runs $0$ to $5$ critically, $B$ has block $0$ to $7$ with float to $14$ , $C$ has block $5$ to $9$ with float to $11$ , $D$ runs $5$ to $11$ critically and $E$ has block $7$ to $10$ with float to $17$ , list every activity that could be in progress at time $8$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For a project with durations $A=6$ , $B=9$ , $C=7$ , $D=4$ , $E=5$ , $F=8$ , $G=3$ , $H=6$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ and $D$ ; $F$ : $C$ ; $G$ : $E$ ; $H$ : $F$ and $G$ , calculate the total float of every activity and describe the Gantt chart.
(9)
(Total for Question 1 is 9 marks)
2.
In the project above, activities $D$ , $E$ and $G$ each have total float $3$ . Explain why the three floats may not be used independently, and state the largest total delay that can be shared among them without extending the project.
(7)
(Total for Question 2 is 7 marks)
3.
An activity $X$ has total float $5$ and duration $9$ , and its earliest start is $12$ . Write down its earliest finish, latest start and latest finish. The project manager delays $X$ by $2$ and then discovers that a later activity $Y$ , which immediately follows $X$ and has total float $5$ , must also be delayed. Explain what has happened to the float of $Y$ .
(7)
(Total for Question 3 is 7 marks)
4.
A project has duration $27$ . An activity $Z$ runs from event $4$ to event $7$ with duration $d$ , where the earliest time at event $4$ is $11$ and the latest time at event $7$ is $22$ . Find the values of $d$ for which $Z$ is critical, for which $Z$ has a float of exactly $4$ , and for which the given data would be impossible.
(7)
(Total for Question 4 is 7 marks)
5.
Explain why the critical activities on a Gantt chart form a single unbroken line across the whole width of the chart, and describe what it would mean if a gap appeared between two consecutive critical blocks.
(6)
(Total for Question 5 is 6 marks)

D1-4.5 · Construct resource histograms (including resource levelling) based on the number of workers required to complete each activity.

Explanation

A resource histogram plots time along the horizontal axis and the number of workers in use along the vertical axis. Here the number of workers required by each activity is given and need not be one.
To draw the histogram, first fix a schedule, usually every activity at its earliest start time, then for each time interval add up the workers of every activity in progress. The height of the histogram at a given time is that total,
and the area under the histogram equals the total number of worker-days in the project, whatever schedule is chosen.
Resource levelling means using the floats of non-critical activities to shift them within their windows so that the peak height falls and the profile is flatter, without extending the project.
Critical activities cannot move, so the critical profile is a floor beneath which no levelling can go. A useful check is that $\left\lceil\dfrac{\text{total worker-days}}{\text{project duration}}\right\rceil$ is a lower bound for the peak, but the true minimum peak is often larger because of precedence constraints.

Worked example

In a project of duration $6$ , activity $P$ needs $2$ workers on days $1$ to $3$ and activity $Q$ needs $3$ workers on days $2$ to $5$ . Find the histogram heights and the peak.

1.Day $1$ : only $P$ is running, so the height is $2$ .
2.Days $2$ and $3$ : both are running, so the height is $2+3=5$ .
3.Days $4$ and $5$ : only $Q$ is running, so the height is $3$ .
4.Day $6$ : nothing is running, so the height is $0$ .

Answer: The heights are $2, 5, 5, 3, 3, 0$ and the peak is $5$ workers, on days $2$ and $3$ .

Common mistakes

Don't fall into the trap of counting activities instead of workers when several activities need more than one worker each.
Don't fall into the trap of moving a critical activity while levelling.
Don't fall into the trap of assuming the peak can always be brought down to the worker-day lower bound.

Exam tip

Tabulate the interval each activity occupies before drawing anything; the histogram is then a column-by-column addition and mistakes are easy to spot.

Tier 1 · Easy

1.
In a project, activity $P$ needs $2$ workers and runs on days $1$ to $4$ , and activity $Q$ needs $3$ workers and runs on days $3$ to $6$ . Write down the height of the resource histogram on each of days $1$ to $6$ .
(3)
(Total for Question 1 is 3 marks)
2.
A project lasts $12$ days and requires $29$ worker-days in total. Calculate a lower bound for the number of workers needed, and explain why the actual peak of the histogram may be larger.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A project has activities $A$ of duration $3$ needing $2$ workers, $B$ of duration $4$ needing $1$ worker, $C$ of duration $2$ needing $3$ workers, $D$ of duration $5$ needing $2$ workers and $E$ of duration $3$ needing $1$ worker, with precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $B$ ; $E$ : $C$ and $D$ . Find the project duration and the critical activities.
(5)
(Total for Question 1 is 5 marks)
2.
For that project, with $A$ needing $2$ workers, $B$ needing $1$ , $C$ needing $3$ , $D$ needing $2$ and $E$ needing $1$ , draw the resource histogram with every activity at its earliest start time by writing down the height on each of days $1$ to $12$ , and state the peak.
(6)
(Total for Question 2 is 6 marks)
3.
For the same project, write down the histogram heights when every activity is scheduled at its latest start time, and verify that the total area is unchanged.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
For the project with $A$ of duration $3$ needing $2$ workers, $B$ of duration $4$ needing $1$ , $C$ of duration $2$ needing $3$ , $D$ of duration $5$ needing $2$ and $E$ of duration $3$ needing $1$ , and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $B$ ; $E$ : $C$ and $D$ , determine the smallest possible peak of the resource histogram if the project must still finish in $12$ days.
(9)
(Total for Question 1 is 9 marks)
2.
For that project the total requirement is $29$ worker-days over $12$ days, giving a lower bound of $3$ workers, yet the minimum peak is $5$ . Explain in detail why the bound is not attained.
(7)
(Total for Question 2 is 7 marks)
3.
Explain what resource levelling is, why critical activities cannot take part in it, and what the manager gains from a flatter histogram.
(6)
(Total for Question 3 is 6 marks)
4.
A project of duration $10$ days has a histogram with heights $4, 4, 6, 6, 6, 3, 3, 2, 2, 2$ when every activity starts as early as possible. The only non-critical activity is $R$ , which needs $3$ workers for $2$ days, currently runs on days $3$ and $4$ , and has total float $5$ . Determine the smallest peak achievable by moving $R$ , and state the days on which it should run.
(8)
(Total for Question 4 is 8 marks)
5.
Explain why the area under a resource histogram is the same for every schedule of the same project, and describe how that fact is used as a check when a histogram has been drawn.
(6)
(Total for Question 5 is 6 marks)

D1-4.6 · Scheduling the activities using the least number of workers required to complete the project.

Explanation

A scheduling diagram assigns each activity to a named worker and a time slot, so that no worker does two activities at once and every activity starts only after all of its predecessors have finished.
Start from the lower bound $\left\lceil\dfrac{\text{sum of the durations}}{\text{critical time}}\right\rceil$ , since each activity needs one worker and the project cannot be shorter than the critical path.
That bound is not always achievable, so the method is to try to construct a schedule with that many workers and, if the attempt provably fails, to argue that one more is needed and construct a schedule with that many.
A good construction puts the whole critical path on one worker, since those activities run end to end with no gaps, and then fits the non-critical activities into the remaining workers inside their float windows.
Always check each activity against its earliest start and latest finish, and state the schedule as a table of worker, activity and time interval.

Worked example

A project of critical time $10$ has activities totalling $18$ days of work, with critical path $P$ from $0$ to $6$ and $Q$ from $6$ to $10$ , and non-critical activities $R$ of duration $5$ with window $0$ to $8$ and $S$ of duration $3$ with window $2$ to $10$ . Find the least number of workers and give a schedule.

1.The lower bound is $\lceil18\div10\rceil=2$ workers.
2.Worker $1$ takes the critical path: $P$ from $0$ to $6$ and $Q$ from $6$ to $10$ .
3.Worker $2$ takes $R$ from $0$ to $5$ , which lies inside its window $0$ to $8$ .
4.Worker $2$ then takes $S$ from $5$ to $8$ , which lies inside its window $2$ to $10$ .

Answer: Two workers suffice, with worker $1$ on $P$ then $Q$ and worker $2$ on $R$ then $S$ .

Common mistakes

Don't fall into the trap of starting an activity before all of its predecessors have finished.
Don't fall into the trap of treating the lower bound as the answer without exhibiting a schedule that attains it.
Don't fall into the trap of splitting one activity between two workers, which is not allowed.

Exam tip

Give the critical path to a single worker first; that worker is then fully occupied for the whole project and the remaining activities are much easier to place.

Tier 1 · Easy

1.
A project has a critical time of $20$ days and its activities have durations totalling $54$ days. Each activity requires one worker. Calculate a lower bound for the number of workers needed to complete the project in $20$ days.
(2)
(Total for Question 1 is 2 marks)
2.
Explain why it is sensible to give the whole critical path to a single worker when constructing a schedule.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A project has durations $A=5$ , $B=4$ , $C=6$ , $D=3$ , $E=7$ , $F=2$ , $G=4$ , $H=5$ with precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ and $B$ ; $E$ : $C$ ; $F$ : $C$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ . The critical time is $27$ and the critical path is $A$ , $C$ , $E$ , $G$ , $H$ . Find a lower bound for the number of workers and construct a schedule attaining it.
(8)
(Total for Question 1 is 8 marks)
2.
Explain why a lower bound of $n$ workers does not guarantee that $n$ workers are enough, and describe the two things a complete answer to a scheduling question must contain.
(5)
(Total for Question 2 is 5 marks)
3.
A project has critical time $16$ and total activity duration $46$ days, with the critical path consisting of three activities. Calculate the lower bound for the number of workers, and calculate the total number of idle worker-days if that number of workers is used and the project finishes in $16$ days.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A project has durations $A=5$ , $B=7$ , $C=4$ , $D=6$ , $E=3$ , $F=8$ , $G=2$ , $H=5$ and precedence $A$ : none; $B$ : none; $C$ : $A$ ; $D$ : $A$ ; $E$ : $B$ ; $F$ : $C$ and $D$ ; $G$ : $D$ and $E$ ; $H$ : $F$ and $G$ . The critical time is $24$ with critical path $A$ , $D$ , $F$ , $H$ . Find the least number of workers needed and give a full schedule.
(9)
(Total for Question 1 is 9 marks)
2.
In the schedule for that project, activity $C$ is placed from time $7$ to time $11$ although its earliest start is $5$ . Explain why the delay is necessary given that worker $2$ also carries out $B$ , and state what would happen if $C$ were instead started at time $5$ .
(7)
(Total for Question 2 is 7 marks)
3.
A project has durations $P=5$ , $R=4$ , $S=4$ , $Q=7$ , $T=3$ and precedence $P$ : none; $R$ : none; $S$ : none; $Q$ : $P$ , $R$ and $S$ ; $T$ : $Q$ . Each activity needs one worker. Find the critical time, the lower bound for the number of workers, and the least number of workers actually required, justifying your answer.
(9)
(Total for Question 3 is 9 marks)
4.
Explain why the number of workers needed to complete a project in the critical time can exceed the lower bound obtained by dividing total duration by critical time, and describe how to argue rigorously that a given number of workers is not enough.
(7)
(Total for Question 4 is 7 marks)
5.
For the project with durations $A=5$ , $B=4$ , $C=6$ , $D=3$ , $E=7$ , $F=2$ , $G=4$ , $H=5$ , critical time $27$ and critical path $A$ , $C$ , $E$ , $G$ , $H$ , the client now insists the project finish in $25$ days. Explain why no number of workers can achieve this, and state the smallest reduction in the duration of a single critical activity that would make it possible.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

D1-4.1 · Modelling of a project by an activity network, from a precedence table.

Tier 1 · Easy

Mark scheme for D1-4.1 Tier 1 · Easy
Question	Scheme	Marks
1	A dummy is an arc that represents no work It is used to show a dependency that could not otherwise be drawn, or to give two activities distinct pairs of events Its duration is $0$	3
(3 marks)		3
Notes
In an activity-on-arc network every arc is an activity, so a dependency that cannot be shown by the shape of the network alone is carried by an arc that consumes no time. That arc is the dummy, drawn dashed with duration $0$ . It also separates two activities that would otherwise run between the same two events, since each activity must be identified by a unique pair of event numbers.
2	No dummies are needed $A$ runs from event $1$ to event $2$ and $C$ runs from event $2$ to the finish event $B$ runs from event $1$ to event $3$ and $D$ runs from event $3$ to the finish event	3
(3 marks)		3
Notes
The project splits into two independent chains, $A$ then $C$ and $B$ then $D$ . Each chain can be drawn as two arcs in series from the common start event to the common finish event, with its own intermediate event. No activity depends on a mixture of predecessors, so no dependency has to be carried by a zero-duration arc and no two activities share both of their events.

Tier 2 · Standard

Mark scheme for D1-4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$A$ runs from event $1$ to event $2$ $B$ runs from event $1$ to event $3$ A dummy runs from event $2$ to event $3$ $C$ runs from event $2$ to event $4$ $D$ runs from event $3$ to event $5$ $E$ runs from event $4$ to event $5$ $F$ runs from event $4$ to event $6$ $G$ runs from event $5$ to event $6$ , and $H$ runs from event $6$ to event $7$	7
(7 marks)		7
Notes
Since $C$ needs $A$ only but $D$ needs $A$ and $B$ , activity $A$ must finish at its own event $2$ , with a dummy from event $2$ to event $3$ so that $D$ , leaving event $3$ , waits for both $A$ and $B$ . Activities $E$ and $F$ both follow $C$ alone, so they both leave event $4$ , the end of $C$ . Then $G$ needs $D$ and $E$ , so both end at event $5$ ; and $H$ needs $F$ and $G$ , so both end at event $6$ . Finally $H$ runs to the single finish event $7$ .
2	The first reason is logical: an activity depending on a proper subset of another activity's predecessors cannot leave the same event, so a dummy carries the extra dependency The second reason is uniqueness: two activities must not share both their start event and their finish event, so a dummy separates them An example needing no dummy is $A$ with no predecessors, $B$ with immediate predecessor $A$ , and $C$ with immediate predecessor $B$ This is a single chain, so each activity has its own pair of events	5
(5 marks)		5
Notes
The logical use arises when, say, $X$ depends on $P$ and $Q$ while $Y$ depends on $P$ only: drawing $P$ and $Q$ into a shared event would wrongly force $Y$ to wait for $Q$ , so $P$ ends separately and a dummy passes its completion on to $X$ . The uniqueness use arises when two activities have identical predecessor and successor sets, which would place them between the same two events; since an activity is named by that pair, one of them is redrawn through a new event joined by a dummy. A simple chain $A$ , then $B$ , then $C$ triggers neither situation.
3	$C$ lists $A$ , but $A$ is already a predecessor of $B$ , so waiting for $B$ automatically means waiting for $A$ $D$ lists $B$ , but $B$ is already a predecessor of $C$ , so waiting for $C$ automatically means waiting for $B$ Corrected table: $A$ has no predecessors, $B$ has immediate predecessor $A$ , $C$ has immediate predecessor $B$ , $D$ has immediate predecessor $C$ The project is a single chain $A$ , $B$ , $C$ , $D$	5
(5 marks)		5
Notes
A precedence table records only immediate predecessors, so any activity that is reachable through another listed predecessor must be removed. Here $A$ precedes $B$ , so listing $A$ against $C$ adds nothing once $B$ is listed; the same argument removes $B$ from the entry for $D$ once $C$ is listed. Stripping the implied entries leaves a chain, which needs no dummies and takes the total of the four durations.

Tier 3 · Hard

Mark scheme for D1-4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$A$ runs from event $1$ to event $2$ and $B$ runs from event $1$ to event $3$ $C$ runs from event $2$ to event $3$ $D$ runs from event $2$ to event $4$ $E$ runs from event $3$ to event $4$ $F$ runs from event $4$ to event $5$ No dummies are needed $C$ and $D$ both follow $A$ alone, so both may leave event $2$ ; $E$ needs $B$ and $C$ , which both end at event $3$ ; and $F$ needs $D$ and $E$ , which both end at event $4$	7
(7 marks)		7
Notes
Work down the table. Activities $C$ and $D$ share the single predecessor $A$ , so they leave the event at which $A$ finishes. Activity $E$ needs exactly $B$ and $C$ , and no other activity depends on $B$ or $C$ separately, so $B$ and $C$ may both be drawn into one event. Likewise $F$ needs exactly $D$ and $E$ , and nothing depends on just one of them, so they may share their finish event. Since no activity depends on a proper subset of another activity's predecessors, and no two activities share both events, this network needs no dummy at all.
2	$P$ and $Q$ share the start event and the merge event, and $R$ and $S$ share the merge event and the finish event, so both pairs risk being drawn between the same two events $P$ runs from event $1$ to event $2$ and $Q$ runs from event $1$ to event $3$ A dummy from event $3$ to event $2$ carries the completion of $Q$ to the start of $R$ and $S$ , so $P$ and $Q$ no longer share both events $R$ runs from event $2$ to event $4$ and $S$ runs from event $2$ to event $5$ , where event $5$ is the single finish event A second dummy from event $4$ to event $5$ brings $R$ into that finish event without giving $R$ and $S$ the same pair of events Two dummies are required, and both are needed for uniqueness rather than for logic	6
(6 marks)		6
Notes
Both $R$ and $S$ wait for the same pair $P$ and $Q$ , so logically $P$ and $Q$ could end at a shared merge event and $R$ and $S$ could both leave it. The obstacle is that an activity is identified by its start and finish events, so $P$ and $Q$ cannot both run from event $1$ to that merge event. Giving $Q$ its own finish event and joining it to the merge point by a zero-duration dummy solves this without changing any dependency. The same clash appears again at the other end: neither $R$ nor $S$ has a successor, so both must enter the one finish event, and they cannot both run from the merge event to it. Sending $R$ into its own event and joining that to the finish event by a second dummy separates them. Neither dummy can be dropped, so the minimum is two.
3	The start event represents the single moment the project begins, and every activity with no predecessor leaves it The finish event represents the single moment the project ends, and every activity with no successor enters it Two start events would leave the relative timing of the two groups undefined The three activities with no predecessors are all drawn leaving the one start event The two activities with no successors are both drawn entering the one finish event A dummy is added only if that would give two activities the same pair of events	6
(6 marks)		6
Notes
The forward pass sets the earliest event time of the start node to zero, and the backward pass sets the latest event time of the finish node to the project duration; both passes require a unique node to anchor them, so the network is drawn with one source and one sink. Activities with no predecessors are ready at time zero and therefore all leave the start event, and activities with no successors all end at the finish event. The only complication is uniqueness: if two of the parallel activities would then run between the same two events, one of them is redirected through a fresh event and joined by a dummy.
4	$B$ is a predecessor of both $D$ and $E$ , while $A$ feeds only $D$ and $C$ feeds only $E$ $A$ runs from event $1$ to event $2$ and $D$ leaves event $2$ $C$ runs from event $1$ to event $3$ and $E$ leaves event $3$ $B$ runs from event $1$ to event $4$ A dummy runs from event $4$ to event $2$ , and a second dummy runs from event $4$ to event $3$ $D$ runs from event $2$ to event $5$ and $E$ runs from event $3$ to event $6$ A dummy runs from event $5$ to event $6$ , since $G$ needs $D$ and $E$ while $F$ needs $D$ alone Three dummies are needed, and $F$ runs from event $5$ to the finish while $G$ runs from event $6$ to the finish	8
(8 marks)		8
Notes
Activity $B$ is shared between two different merge points, so it cannot end at either of them directly: it finishes at its own event $4$ and two dummies carry its completion to the start of $D$ and to the start of $E$ . On the output side, $D$ feeds both $F$ , which needs $D$ alone, and $G$ , which needs $D$ and $E$ ; so $D$ ends at its own event $5$ , from which $F$ leaves, and a third dummy carries the completion of $D$ to event $6$ where $E$ also ends and $G$ begins. Each of the three dummies removes a dependency that the shape of the network cannot express, so none can be dropped.
5	Following the cycle, each activity on it must finish before the next one starts Going once round the cycle shows that an activity must finish before itself starts No activity could ever begin, so the project has no schedule The forward pass would never terminate, since each event time would keep increasing The precedence relation must be acyclic Equivalently, the activities must be capable of being listed in an order in which every activity appears after all of its predecessors	6
(6 marks)		6
Notes
An arc from event $5$ to event $3$ means the activities leaving event $3$ wait for something that itself waits, through the chain from event $3$ to event $5$ , on those very activities. Chasing the earliest event times round the loop increases them without limit, so the forward pass has no solution and the project duration is undefined. The test a precedence table must pass is that the dependencies form a directed acyclic graph, which is exactly the condition that the activities can be arranged in a linear order with every activity following all of its predecessors.

D1-4.2 · Completion of the precedence table for a given activity network.

Tier 1 · Easy

Mark scheme for D1-4.2 Tier 1 · Easy
Question	Scheme	Marks
1	The immediate predecessors of $E$ are $B$ and $C$ A dummy is not an activity, so it would not be listed Instead you would follow the dummy back to its start event and add the activities entering that event	3
(3 marks)		3
Notes
The activities entering the event an activity leaves are precisely the ones that must finish first, so $E$ waits for $B$ and $C$ . A dummy carries a dependency but represents no work, so it never appears in a precedence table; the dependency it carries is found by tracing to the event at the dummy's tail and reading the activities that enter there.
2	$A$ : no predecessors $B$ : $A$ $C$ : $B$ $D$ : $C$	3
(3 marks)		3
Notes
The four arcs form a single chain from the start event $1$ to the finish event $5$ . Activity $A$ leaves the start event so has no predecessor; each later activity leaves an event entered by exactly one activity, namely the one before it in the chain. The table therefore records a strict sequence with no branching and no dummies.

Tier 2 · Standard

Mark scheme for D1-4.2 Tier 2 · Standard
Question	Scheme	Marks
1	$A$ : no predecessors $B$ : no predecessors $C$ : $A$ $D$ : $A$ and $B$ $E$ : $A$ and $B$ $F$ : $C$ and $D$ $G$ : $E$ and $F$	7
(7 marks)		7
Notes
Activities $A$ and $B$ leave the start event $1$ , so neither has a predecessor. Activity $C$ leaves event $2$ , which only $A$ enters. Activities $D$ and $E$ both leave event $3$ , entered by $B$ and by the dummy from event $2$ ; tracing that dummy back gives $A$ , so each of $D$ and $E$ depends on $A$ and $B$ . Activity $F$ leaves event $4$ , entered by $C$ and $D$ . Activity $G$ leaves event $5$ , entered by $E$ and $F$ , and runs to the finish event $6$ , from which nothing leaves.
2	A precedence table lists only immediate predecessors $A$ is a predecessor of $C$ , so waiting for $C$ already implies waiting for $A$ $A$ must therefore be removed from the entry Corrected entry: $G$ : $C$ and $E$	4
(4 marks)		4
Notes
Only activities that enter the event $G$ leaves are immediate predecessors, and here those are $C$ and $E$ . The activity $A$ reaches $G$ only through $C$ , so it is a predecessor but not an immediate one; including it makes the table redundant and, if the network were redrawn from it, would suggest an extra arc that does not exist. Removing $A$ gives the reduced entry $C$ and $E$ .
3	$P$ : no predecessors $Q$ : no predecessors $R$ : $P$ and $Q$ $S$ : $P$ and $Q$ $T$ : $R$ Both $R$ and $S$ need exactly $P$ and $Q$ , so the dummy is needed for uniqueness, to stop $P$ and $Q$ sharing the same pair of events	6
(6 marks)		6
Notes
Activities $P$ and $Q$ leave the start event, so neither has a predecessor. Activities $R$ and $S$ leave event $2$ , entered by $P$ and by the dummy from event $3$ ; tracing the dummy back to event $3$ finds $Q$ , so both $R$ and $S$ depend on $P$ and $Q$ . Activity $T$ leaves event $4$ , entered only by $R$ . Since $R$ and $S$ have identical predecessor sets, no dependency is lost by merging, and the dummy exists purely so that $P$ and $Q$ do not both run from event $1$ to event $2$ .

Tier 3 · Hard

Mark scheme for D1-4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$A$ , $B$ and $C$ all leave the start event $1$ , so none has a predecessor $D$ leaves event $2$ , entered by $A$ and by the dummy from event $3$ Tracing that dummy back to event $3$ finds $B$ , so $D$ : $A$ and $B$ $E$ leaves event $4$ , entered by $C$ and by the dummy from event $3$ , so $E$ : $B$ and $C$ $F$ leaves event $5$ , entered only by $D$ , so $F$ : $D$ Nothing leaves event $6$ , so event $6$ is the finish event Full table: $A$ : none; $B$ : none; $C$ : none; $D$ : $A$ , $B$ ; $E$ : $B$ , $C$ ; $F$ : $D$	8
(8 marks)		8
Notes
Each dummy is traced backwards to the activities entering its tail event. Activity $B$ is needed by both $D$ and $E$ , while $A$ is needed only by $D$ and $C$ only by $E$ , so $B$ finishes at its own event $3$ and the two dummies deliver its completion to the merge points at events $2$ and $4$ . That is why $B$ appears against both $D$ and $E$ . Activity $F$ leaves event $5$ , which only $D$ enters, and both $E$ and $F$ run into event $6$ , from which nothing leaves, making it the single finish event. No entry needs reducing, since none of $A$ , $B$ , $C$ or $D$ is a predecessor of another activity in the same list.
2	Reading the table is mechanical: for each activity the set of activities entering its start event is determined by the network, then reduced in only one way Drawing the network involves free choices of event numbering Extra dummies may be inserted without changing any dependency For the table $A$ : none; $B$ : none; $C$ : $A$ and $B$ ; $D$ : $A$ and $B$ , one drawing merges $A$ and $B$ at one event using a dummy for uniqueness Another drawing uses a different dummy direction or a different event numbering and is equally valid Both networks give back the same precedence table	6
(6 marks)		6
Notes
The predecessor set of an activity is fixed by which arcs enter the event it leaves, and reduction to immediate predecessors removes exactly the activities reachable through another entry, so the table is determined. Going the other way, nothing in the table fixes how events are numbered, which of two parallel activities is redirected through a dummy, or whether a redundant dummy is present, so many networks satisfy the same table. The test of correctness is therefore always to read the table back off the drawn network and compare it with the original.
3	$A$ : none; $B$ : none $C$ : $A$ $D$ : $A$ $E$ : $B$ and $D$ $F$ : $C$ $G$ : $E$ and $F$ $H$ : $C$ The activities with no successor are $G$ and $H$ , both of which enter the finish event $6$	7
(7 marks)		7
Notes
Activities $A$ and $B$ leave the start event. Activities $C$ and $D$ leave event $2$ , entered only by $A$ . Activity $E$ leaves event $3$ , entered by $B$ and by $D$ , and neither is a predecessor of the other, so both stay. Activities $F$ and $H$ leave event $4$ , entered only by $C$ . Activity $G$ leaves event $5$ , entered by $E$ and $F$ . Nothing leaves event $6$ , so the activities entering it, $G$ and $H$ , have no successors; note that $D$ here is a genuine activity, not a dummy, so it is listed against $E$ in the normal way.
4	There are $7$ activities, so at least $7$ arcs are needed $B$ and $C$ share the single predecessor $A$ and the single successor $D$ Drawing both directly between the end of $A$ and the start of $D$ would give them the same pair of events One dummy is needed to separate $B$ and $C$ $E$ and $F$ likewise share predecessor $D$ and successor $G$ , so a second dummy is needed No dependency requires a dummy, so no further dummies are needed The smallest network has $7+2=9$ arcs	7
(7 marks)		7
Notes
Every activity is an arc, giving seven. No activity depends on a proper subset of another activity's predecessors, so no dummy is required for logical reasons. The uniqueness rule bites twice: the pair $B$ , $C$ has identical predecessor and successor sets, as does the pair $E$ , $F$ , and in each case one of the two must be routed through an extra event joined by a zero-duration arc. That is two dummies, and since each pair needs at least one and no other pair is affected, nine arcs is the minimum.
5	In a chain $A$ , $B$ , $C$ , $D$ , $E$ the immediate-predecessor table has one entry per activity after the first A table of all predecessors would list $E$ : $A$ , $B$ , $C$ , $D$ Redrawing from that table would suggest four separate arcs entering the event $E$ leaves The network would carry dependencies that are already implied, so it would contain redundant dummies Different readers would draw different networks from the same table The reduced table is the unique minimal description, which is why the convention is fixed	6
(6 marks)		6
Notes
For a chain the only genuine constraints are that each activity follows the one before it, four constraints in all. Listing every predecessor records ten constraints, six of which are consequences of the others. A drawing that honoured all ten would need extra arcs or dummies to deliver each redundant dependency to the merge point, inflating the network without changing any earliest or latest time. Fixing the convention at immediate predecessors makes the table the smallest complete description of the project and makes reading it back off a network a well-defined operation.

D1-4.3 · Algorithm for finding the critical path. Earliest and latest event times. Earliest and latest start and finish times for activities. Identification of critical activities and critical path(s).

Tier 1 · Easy

Mark scheme for D1-4.3 Tier 1 · Easy
Question	Scheme	Marks
1	Earliest start $=9$ Earliest finish $=9+7=16$ Latest finish $=20$ Latest start $=20-7=13$	4
(4 marks)		4
Notes
The earliest an activity can begin is the earliest its start event can occur, so the earliest start is $9$ and the earliest finish is $9+7=16$ . The latest it may finish is the latest event time at its end event, $20$ , and working backwards through the duration gives a latest start of $20-7=13$ .
2	$A$ finishes at $5$ and $B$ finishes at $4$ $C$ runs from $5$ to $11$ $D$ cannot start until both $A$ and $B$ are done, so it runs from $5$ to $8$ Project duration $=\max(11, 8)=11$	4
(4 marks)		4
Notes
The forward pass gives earliest finishes of $5$ for $A$ and $4$ for $B$ . Activity $C$ waits only for $A$ , so it starts at $5$ and finishes at $11$ . Activity $D$ waits for both, so it starts at $\max(5, 4)=5$ and finishes at $8$ . The project is complete when both $C$ and $D$ are done, at time $11$ .

Tier 2 · Standard

Mark scheme for D1-4.3 Tier 2 · Standard
Question	Scheme	Marks
1	Earliest finishes: $A=5$ , $B=4$ , $C=11$ , $D=8$ , $E=18$ , $F=13$ $G$ starts at $\max(8, 18)=18$ and finishes at $22$ $H$ starts at $\max(13, 22)=22$ and finishes at $27$ Project duration $=27$ Critical path: $A$ , $C$ , $E$ , $G$ , $H$ Check: $5+6+7+4+5=27$	7
(7 marks)		7
Notes
The forward pass gives $A=5$ and $B=4$ ; then $C$ runs $5$ to $11$ and $D$ runs $5$ to $8$ ; then $E$ runs $11$ to $18$ and $F$ runs $11$ to $13$ ; then $G$ waits for $D$ and $E$ so runs $18$ to $22$ ; and $H$ waits for $F$ and $G$ so runs $22$ to $27$ . The backward pass gives latest finishes $H=27$ , $G=22$ , $F=22$ , $E=18$ , $D=18$ , $C=11$ , $B=15$ , $A=5$ , so the activities with zero float are $A$ , $C$ , $E$ , $G$ and $H$ , and their durations sum to the project duration, confirming the critical path.
2	Event $1$ : earliest $0$ , latest $0$ Event $2$ : earliest $5$ , latest $5$ Event $3$ : earliest $5$ , latest $15$ Event $4$ : earliest $11$ , latest $11$ Event $5$ : earliest $18$ , latest $18$ Event $6$ : earliest $22$ , latest $22$ Event $7$ : earliest $27$ , latest $27$	7
(7 marks)		7
Notes
Forward: event $2$ takes $0+5=5$ ; event $3$ takes $\max(0+4, 5+0)=5$ using the dummy; event $4$ takes $5+6=11$ ; event $5$ takes $\max(5+3, 11+7)=18$ ; event $6$ takes $\max(11+2, 18+4)=22$ ; event $7$ takes $22+5=27$ . Backward: event $6$ is $27-5=22$ ; event $5$ is $22-4=18$ ; event $4$ is $\min(18-7, 22-2)=11$ ; event $3$ is $18-3=15$ ; event $2$ is $\min(11-6, 15-0)=5$ ; event $1$ is $\min(5-5, 15-4)=0$ , which confirms the arithmetic.
3	Earliest finishes: $A=4$ , $B=6$ , $C=7$ , $D=9$ $E$ starts at $\max(6, 7)=7$ and finishes at $9$ $F$ starts at $\max(9, 9)=9$ and finishes at $13$ Project duration $=13$ $A$ , $C$ , $E$ , $F$ has length $4+3+2+4=13$ $A$ , $D$ , $F$ has length $4+5+4=13$ There are two critical paths, and $B$ is the only activity with float, of $1$	7
(7 marks)		7
Notes
The forward pass gives $A=4$ , $B=6$ , then $C$ from $4$ to $7$ and $D$ from $4$ to $9$ ; $E$ waits for $B$ and $C$ so runs $7$ to $9$ ; $F$ waits for $D$ and $E$ , both of which finish at $9$ , so it runs $9$ to $13$ . The backward pass gives latest starts $F=9$ , $E=7$ , $D=4$ , $C=4$ , $B=1$ , $A=0$ , so every activity except $B$ has zero float. Two distinct chains of critical activities run from start to finish, and each has total duration $13$ .

Tier 3 · Hard

Mark scheme for D1-4.3 Tier 3 · Hard
Question	Scheme	Marks
1	Earliest finishes: $A=5$ , $B=7$ , $C=9$ , $D=11$ , $E=10$ $F$ starts at $\max(9, 11)=11$ and finishes at $19$ $G$ starts at $\max(11, 10)=11$ and finishes at $13$ $H$ starts at $\max(19, 13)=19$ and finishes at $24$ Project duration $=24$ Critical path: $A$ , $D$ , $F$ , $H$ , of length $5+6+8+5=24$ Sum of all durations $=5+7+4+6+3+8+2+5=40$ Lower bound $=\left\lceil\dfrac{40}{24}\right\rceil=2$ workers	9
(9 marks)		9
Notes
The forward pass gives $A=5$ , $B=7$ , $C$ from $5$ to $9$ , $D$ from $5$ to $11$ , $E$ from $7$ to $10$ , then $F$ from $11$ to $19$ , $G$ from $11$ to $13$ and $H$ from $19$ to $24$ . The backward pass gives latest starts $H=19$ , $G=17$ , $F=11$ , $E=14$ , $D=5$ , $C=7$ , $B=7$ , $A=0$ , so the zero-float activities are $A$ , $D$ , $F$ and $H$ , forming the single critical path of length $24$ . Every activity needs one worker, so the total labour is $40$ worker-days over $24$ days, and at least $\lceil40/24\rceil=2$ workers are needed.
2	An event is critical when its earliest and latest times are equal An activity from event $i$ to event $j$ is critical only when $l_j-e_i$ equals its duration If $l_j-e_i$ exceeds the duration, the activity has float even though both events are critical For example, take $e_i=l_i=6$ and $e_j=l_j=15$ with an activity of duration $7$ The float is $15-6-7=2$ , so the activity is not critical Another chain of activities between the same two events must be taking the full $9$	6
(6 marks)		6
Notes
Critical events say only that those two moments cannot move; they say nothing about how the time between them is filled. If a longer chain runs in parallel between the same pair of events, a shorter activity spanning them has spare time and is not critical. The correct test is the one on the activity itself: the window $l_j-e_i$ must be exactly the duration, so the float $l_j-e_i-d$ is zero. In the example the window is $9$ and the duration is $7$ , leaving float $2$ .
3	Earliest finishes: $A=6$ , $B=9$ , $C=13$ , $D=10$ $E$ starts at $\max(9, 10)=10$ and finishes at $15$ $F$ runs $13$ to $21$ and $G$ runs $15$ to $18$ $H$ starts at $\max(21, 18)=21$ and finishes at $27$ Project duration $=27$ , critical path $A$ , $C$ , $F$ , $H$ of length $6+7+8+6=27$ Float of $B=4$ Float of $D=3$ Float of $E=3$ Float of $G=3$	9
(9 marks)		9
Notes
The forward pass gives $A=6$ , $B=9$ , $C$ from $6$ to $13$ , $D$ from $6$ to $10$ , $E$ from $10$ to $15$ , $F$ from $13$ to $21$ , $G$ from $15$ to $18$ and $H$ from $21$ to $27$ . The backward pass gives latest finishes $H=27$ , $G=21$ , $F=21$ , $E=18$ , $D=13$ , $C=13$ , $B=13$ , $A=6$ . Subtracting duration from latest finish gives latest starts $G=18$ , $E=13$ , $D=9$ , $B=4$ , so the floats are $18-15=3$ for $G$ , $13-10=3$ for $E$ , $9-6=3$ for $D$ and $4-0=4$ for $B$ , while $A$ , $C$ , $F$ and $H$ have zero float.
4	The critical path was the longest chain, of length $30$ Reducing $Q$ shortens every chain through $Q$ by $4$ , so the critical chain becomes $26$ A chain that does not pass through $Q$ keeps its old length and may now be the longest Such a chain had length $30-f$ , where $f\geqslant1$ is its float The new duration is $\max(26,\;30-f_{\min})$ , taken over the chains that avoid $Q$ If the smallest such float is $4$ or more, the new duration is $26$ If the smallest such float is $1$ , the new duration is $29$ So the new duration lies between $26$ and $29$ inclusive	7
(7 marks)		7
Notes
Project duration is the length of the longest chain from start to finish, so shortening one activity only helps until a rival chain takes over. Every chain containing $Q$ falls by $4$ , taking the old critical chain from $30$ to $26$ . A chain that avoids $Q$ is untouched and had length $30-f$ for its float $f$ , which is at least $1$ because a float of $0$ would have made it critical as well. The new duration is therefore the larger of $26$ and the longest untouched chain, which lies between $26$ and $29$ ; it equals $26$ exactly when every chain avoiding $Q$ had float $4$ or more.
5	Project duration $=24$ with critical path $A$ , $D$ , $F$ , $H$ $C$ has earliest start $5$ and earliest finish $9$ The latest finish of $C$ is the latest start of $F$ , which is $11$ Float of $C=11-9=2$ $C$ may be increased by at most $2$ , to a duration of $6$ With $C=6$ the chain $A$ , $C$ , $F$ , $H$ also has length $5+6+8+5=24$ There are then two critical paths, $A$ , $D$ , $F$ , $H$ and $A$ , $C$ , $F$ , $H$ $C$ becomes critical, so any further increase lengthens the project	8
(8 marks)		8
Notes
The forward and backward passes give a project duration of $24$ with $A$ , $D$ , $F$ , $H$ critical, and $C$ running from $5$ to $9$ against a latest finish of $11$ , since $F$ has latest start $11$ . That leaves a float of $2$ , so $C$ may grow to duration $6$ before the chain through it matches the critical length. At exactly $6$ the two chains $A$ , $C$ , $F$ , $H$ and $A$ , $D$ , $F$ , $H$ both total $24$ , so both are critical and $C$ now has zero float; increasing $C$ beyond $6$ makes its chain the unique longest and pushes the project past $24$ .

D1-4.4 · Calculation of the total float of an activity. Construction of Gantt (cascade) charts.

Tier 1 · Easy

Mark scheme for D1-4.4 Tier 1 · Easy
Question	Scheme	Marks
1	$F=l_6-e_2-d$ $F=15-7-4=4$	2
(2 marks)		2
Notes
The total float is the latest time at the end event minus the earliest time at the start event minus the duration. Substituting gives $15-7-4=4$ , so the activity may be delayed by up to $4$ time units without extending the project.
2	A critical activity has total float $0$ The critical activities are drawn as solid blocks along the top of the chart They form one unbroken line from time $0$ to the end of the project, with no float rectangles	3
(3 marks)		3
Notes
An activity is critical when its earliest and latest start times coincide, so its total float is zero and it cannot be delayed at all. On a cascade chart there is therefore nothing to shade beside it, and because the critical activities form a chain from the start event to the finish event they join end to end and fill the whole width of the chart.

Tier 2 · Standard

Mark scheme for D1-4.4 Tier 2 · Standard
Question	Scheme	Marks
1	$A$ : $0$ $B$ : $7$ $C$ : $2$ $D$ : $0$ $E$ : $7$ $F$ : $0$ $G$ : $6$ $H$ : $0$	7
(7 marks)		7
Notes
The forward pass gives earliest starts $A=0$ , $B=0$ , $C=5$ , $D=5$ , $E=7$ , $F=11$ , $G=11$ , $H=19$ , and the backward pass gives latest starts $A=0$ , $B=7$ , $C=7$ , $D=5$ , $E=14$ , $F=11$ , $G=17$ , $H=19$ . Subtracting earliest start from latest start gives the floats $0$ , $7$ , $2$ , $0$ , $7$ , $0$ , $6$ , $0$ respectively, so $A$ , $D$ , $F$ and $H$ are critical.
2	Critical row: $A$ from $0$ to $5$ , $D$ from $5$ to $11$ , $F$ from $11$ to $19$ , $H$ from $19$ to $24$ , drawn solid and unbroken $B$ : block $0$ to $7$ , float rectangle $7$ to $14$ $C$ : block $5$ to $9$ , float rectangle $9$ to $11$ $E$ : block $7$ to $10$ , float rectangle $10$ to $17$ $G$ : block $11$ to $13$ , float rectangle $13$ to $19$	7
(7 marks)		7
Notes
Each non-critical activity is drawn from its earliest start for its own duration, and the float rectangle then extends for the length of its total float, ending at its latest finish. So $B$ runs $0$ to $7$ with float to $14$ ; $C$ runs $5$ to $9$ with float to $11$ ; $E$ runs $7$ to $10$ with float to $17$ ; and $G$ runs $11$ to $13$ with float to $19$ . The critical activities occupy a single row across the full width, since their blocks meet end to end and total the project duration of $24$ .
3	$B$ could be in progress, since its window runs from $0$ to $14$ and $8$ lies inside it $C$ could be in progress, since its window runs from $5$ to $11$ $D$ must be in progress, since it is critical and runs from $5$ to $11$ $E$ could be in progress, since its window runs from $7$ to $17$ $A$ could not, since it must finish by $5$	4
(4 marks)		4
Notes
Draw a vertical line at time $8$ and read every row it crosses. The line misses $A$ , whose window closes at $5$ . It crosses the window of $B$ , which spans $0$ to $14$ , the window of $C$ , which spans $5$ to $11$ , and the window of $E$ , which spans $7$ to $17$ ; in each case the activity may or may not be running at time $8$ depending on how its float is used. It crosses the solid block of $D$ , which has no float, so $D$ is certainly running.

Tier 3 · Hard

Mark scheme for D1-4.4 Tier 3 · Hard
Question	Scheme	Marks
1	Project duration $=27$ and the critical activities are $A$ , $C$ , $F$ , $H$ $A$ : block $0$ to $6$ ; $C$ : block $6$ to $13$ ; $F$ : block $13$ to $21$ ; $H$ : block $21$ to $27$ , all solid $B$ : float $4$ , block $0$ to $9$ , float rectangle $9$ to $13$ $D$ : float $3$ , block $6$ to $10$ , float rectangle $10$ to $13$ $E$ : float $3$ , block $10$ to $15$ , float rectangle $15$ to $18$ $G$ : float $3$ , block $15$ to $18$ , float rectangle $18$ to $21$	9
(9 marks)		9
Notes
The forward pass gives earliest starts $A=0$ , $B=0$ , $C=6$ , $D=6$ , $E=10$ , $F=13$ , $G=15$ , $H=21$ with a project duration of $27$ . The backward pass gives latest finishes $H=27$ , $F=21$ , $G=21$ , $E=18$ , $C=13$ , $D=13$ , $B=13$ , $A=6$ , so the floats are $B: 13-9-0=4$ , $D: 13-6-4=3$ , $E: 18-10-5=3$ , $G: 21-15-3=3$ and zero for $A$ , $C$ , $F$ , $H$ . Each non-critical block is drawn at its earliest start with a float rectangle of the stated length, and the four critical blocks join end to end across the full $27$ .
2	$D$ precedes $E$ and $E$ precedes $G$ , so the three lie on one chain The chain $A$ , $D$ , $E$ , $G$ , $H$ has length $6+4+5+3+6=24$ The project duration is $27$ , so the chain has $3$ days of spare time in total Delaying $D$ by $3$ uses all of it, leaving $E$ and $G$ with no float The total float of an activity is calculated as if every other activity were at its earliest time The largest total delay that can be shared among $D$ , $E$ and $G$ is $3$	7
(7 marks)		7
Notes
Total float measures how far one activity could slip if nothing else moved, so quoting it for each of several activities on a chain counts the same slack more than once. Here the chain from the start event through $D$ , $E$ and $G$ to the finish is $3$ shorter than the critical path, so exactly $3$ days of slack exist along the whole chain. Any delays applied to $D$ , $E$ and $G$ add up along the chain, so their total must not exceed $3$ , however it is distributed.
3	Earliest finish $=12+9=21$ Latest start $=12+5=17$ Latest finish $=17+9=26$ $X$ and $Y$ lie on the same chain, so they share the same $5$ units of slack Delaying $X$ by $2$ consumes $2$ of the shared slack The remaining float available to $Y$ is $5-2=3$ Delaying $Y$ by more than $3$ would now extend the project	7
(7 marks)		7
Notes
The float shifts the whole window: $X$ may start anywhere from $12$ to $17$ and must finish by $26$ . When $X$ is delayed by $2$ , every activity that follows it on the same chain has its earliest start pushed back by $2$ while its latest times are unchanged, so each one loses $2$ units of float. Since $Y$ immediately follows $X$ and had float $5$ computed on the assumption that $X$ started as early as possible, $Y$ now has only $3$ left.
4	$F=22-11-d=11-d$ $Z$ is critical when $11-d=0$ , that is $d=11$ $Z$ has float $4$ when $11-d=4$ , that is $d=7$ A float cannot be negative, so $d>11$ is impossible A duration must be positive, so the data requires $0<d\leqslant11$	7
(7 marks)		7
Notes
The float formula gives $F=l_7-e_4-d=22-11-d=11-d$ . Setting $F=0$ makes $Z$ critical at $d=11$ , and setting $F=4$ gives $d=7$ . Since the latest finish must be at least the earliest finish, the float cannot be negative, so any duration above $11$ would mean the activity could not fit between its events and the stated event times would be inconsistent with the project duration of $27$ .
5	Critical activities have zero float, so each is drawn at a fixed position The critical path runs from the start event to the finish event Consecutive critical activities meet because the earliest finish of one equals the earliest start of the next Their durations sum to the project duration, so together they span the full width A gap would mean a stretch of time on the critical path with no critical activity running That is impossible: the time either belongs to a critical activity or the chain is not critical, so a gap indicates an arithmetic error	6
(6 marks)		6
Notes
On the critical path every activity starts the instant its predecessor finishes, because any delay would push the project end back. The chain begins at time $0$ and ends at the project duration, and the sum of the critical durations equals that duration, so the blocks tile the whole axis without overlap or gap. If a chart shows a gap, either a critical activity has been mislabelled or the forward and backward passes disagree, and the usual check is that the latest time at the start event has come out as $0$ .

D1-4.5 · Construct resource histograms (including resource levelling) based on the number of workers required to complete each activity.

Tier 1 · Easy

Mark scheme for D1-4.5 Tier 1 · Easy
Question	Scheme	Marks
1	Days $1$ and $2$ : height $2$ Days $3$ and $4$ : height $5$ Days $5$ and $6$ : height $3$	3
(3 marks)		3
Notes
On each day add the workers of every activity in progress. Only $P$ runs on days $1$ and $2$ , giving $2$ ; both run on days $3$ and $4$ , giving $2+3=5$ ; and only $Q$ runs on days $5$ and $6$ , giving $3$ .
2	Lower bound $=\left\lceil\dfrac{29}{12}\right\rceil=\lceil2.41\dot{6}\rceil=3$ workers The bound assumes the work can be spread perfectly evenly Precedence constraints may force several activities to overlap, raising the peak above $3$	3
(3 marks)		3
Notes
The area under the histogram is the total worker-days, $29$ , spread over $12$ days, so the average height is $29/12$ and the peak is at least the next whole number, $3$ . That calculation ignores the order in which the activities must be done: if two heavy activities are both forced into the same interval by their predecessors, the histogram must rise higher than the average at that moment.

Tier 2 · Standard

Mark scheme for D1-4.5 Tier 2 · Standard
Question	Scheme	Marks
1	Earliest finishes: $A=3$ , $B=4$ , $C=5$ , $D=9$ $E$ starts at $\max(5, 9)=9$ and finishes at $12$ Project duration $=12$ days Critical activities: $B$ , $D$ , $E$ Check: $4+5+3=12$	5
(5 marks)		5
Notes
The forward pass gives $A$ from $0$ to $3$ and $B$ from $0$ to $4$ , then $C$ from $3$ to $5$ and $D$ from $4$ to $9$ , then $E$ from $9$ to $12$ . The backward pass gives latest starts $E=9$ , $D=4$ , $C=7$ , $B=0$ , $A=4$ , so $B$ , $D$ and $E$ have zero float while $A$ and $C$ each have float $4$ . The chain $B$ , $D$ , $E$ has total duration $12$ , which confirms it as the critical path.
2	$A$ occupies days $1$ to $3$ , $B$ days $1$ to $4$ , $C$ days $4$ to $5$ , $D$ days $5$ to $9$ , $E$ days $10$ to $12$ Days $1$ to $3$ : height $2+1=3$ Day $4$ : height $1+3=4$ Day $5$ : height $3+2=5$ Days $6$ to $9$ : height $2$ Days $10$ to $12$ : height $1$ The peak is $5$ workers on day $5$	6
(6 marks)		6
Notes
At earliest starts, $A$ runs on days $1$ to $3$ , $B$ on days $1$ to $4$ , $C$ on days $4$ to $5$ , $D$ on days $5$ to $9$ and $E$ on days $10$ to $12$ . Adding the worker requirements day by day gives $3, 3, 3, 4, 5, 2, 2, 2, 2, 1, 1, 1$ . The maximum of that list is $5$ , occurring on day $5$ where $C$ with $3$ workers overlaps $D$ with $2$ .
3	Latest starts: $A$ on days $5$ to $7$ , $B$ on days $1$ to $4$ , $C$ on days $8$ to $9$ , $D$ on days $5$ to $9$ , $E$ on days $10$ to $12$ Days $1$ to $4$ : height $1$ Days $5$ to $7$ : height $2+2=4$ Days $8$ and $9$ : height $3+2=5$ Days $10$ to $12$ : height $1$ Total area $=4(1)+3(4)+2(5)+3(1)=4+12+10+3=29$ worker-days This equals $3(2)+4(1)+2(3)+5(2)+3(1)=6+4+6+10+3=29$ , so the area is unchanged	6
(6 marks)		6
Notes
Activity $A$ has float $4$ , so its latest start is day $5$ , and $C$ has float $4$ , so its latest start is day $8$ ; the critical activities $B$ , $D$ and $E$ do not move. Adding worker requirements gives heights $1, 1, 1, 1, 4, 4, 4, 5, 5, 1, 1, 1$ . The area under any histogram is the sum over activities of duration times workers, which does not depend on when the activities are scheduled, and both calculations give $29$ worker-days.

Tier 3 · Hard

Mark scheme for D1-4.5 Tier 3 · Hard
Question	Scheme	Marks
1	The critical activities $B$ , $D$ , $E$ are fixed: $B$ on days $1$ to $4$ at $1$ worker, $D$ on days $5$ to $9$ at $2$ , $E$ on days $10$ to $12$ at $1$ $C$ needs $3$ workers for $2$ days and must lie in the window from day $4$ to day $9$ Day $4$ has a critical base of $1$ and days $5$ to $9$ have a critical base of $2$ $C$ occupies two consecutive days, so at least one of them has base $2$ That day reaches $2+3=5$ So the peak is at least $5$ , and scheduling $A$ on days $1$ to $3$ and $C$ on days $4$ to $5$ attains it The smallest possible peak is $5$ workers	9
(9 marks)		9
Notes
The three critical activities cannot move, so they contribute a fixed base of $1$ on days $1$ to $4$ , $2$ on days $5$ to $9$ and $1$ on days $10$ to $12$ . Activity $C$ has float $4$ and must start no earlier than day $4$ , since $A$ takes three days, and finish by day $9$ ; whichever two consecutive days it takes, at least one lies in the range $5$ to $9$ where the base is already $2$ , so the histogram reaches $5$ there. Activity $A$ needs $2$ workers for three days and can be placed on days $1$ to $3$ , where the base is only $1$ , giving a height of $3$ . The profile $3, 3, 3, 4, 5, 2, 2, 2, 2, 1, 1, 1$ therefore attains the bound, and no schedule does better.
2	The bound $\left\lceil29/12\right\rceil=3$ assumes the work can be spread evenly across all $12$ days The critical activities fix $1$ worker on days $1$ to $4$ , $2$ on days $5$ to $9$ and $1$ on days $10$ to $12$ Activity $C$ alone needs $3$ workers simultaneously, so the peak is at least $3$ on the days it runs $C$ cannot start before day $4$ , because it waits for $A$ $C$ cannot finish after day $9$ , because $E$ waits for it and $E$ is critical In that window at least one day already carries $2$ critical workers, so the height reaches $5$ The bound ignores both precedence and the fact that an activity's workers must all be present at once	7
(7 marks)		7
Notes
A worker-day count treats labour as a fluid that can be poured into any day, which two features of the project forbid. First, the workers of a single activity are indivisible in time: $C$ needs three of them together for two consecutive days, not six worker-days spread thinly. Second, precedence pins $C$ into the window from day $4$ to day $9$ , which overlaps the stretch where the critical activity $D$ is already using two workers. Combining the two constraints forces a day of height $5$ , so the averaging bound of $3$ is unattainable here.
3	Resource levelling means shifting non-critical activities within their float windows to reduce the peak of the histogram The project duration is not allowed to change Critical activities have zero float, so moving one would extend the project The profile of the critical activities is therefore a fixed floor for the histogram A flatter histogram means a steadier workforce, so fewer workers need to be hired and released The peak determines the size of the team that must be available, so lowering it lowers cost	6
(6 marks)		6
Notes
Levelling exploits the only freedom the schedule has, namely the float of the non-critical activities, to move demand away from the busiest moments into quieter ones. Any attempt to move a critical activity immediately pushes the finish date back, which the exercise forbids, so the demand generated by the critical chain is untouchable. The benefit is practical: a contractor must engage enough workers to meet the peak for the whole period they are needed, so a schedule whose peak is five rather than eight employs a smaller team throughout and avoids paying for idle capacity in the troughs.
4	Removing $R$ leaves the base heights $4, 4, 3, 3, 6, 3, 3, 2, 2, 2$ $R$ may start on any day from $3$ to $8$ , since its float is $5$ Placing $R$ on days $3$ and $4$ gives peaks $6, 6$ there and $6$ on day $5$ Placing $R$ on days $6$ and $7$ gives $3+3=6$ on each Placing $R$ on days $7$ and $8$ gives $6$ on day $7$ and $5$ on day $8$ Placing $R$ on days $8$ and $9$ gives $2+3=5$ on each, and the overall peak becomes $6$ from day $5$ Day $5$ has base $6$ and is fixed, so the peak can never fall below $6$ The smallest peak is $6$ , attained for example with $R$ on days $8$ and $9$	8
(8 marks)		8
Notes
Subtracting the three workers of $R$ from days $3$ and $4$ gives the fixed base profile $4, 4, 3, 3, 6, 3, 3, 2, 2, 2$ , whose own maximum is $6$ on day $5$ . Since day $5$ is generated entirely by critical activities, no placement of $R$ can bring the peak below $6$ . Placing $R$ on days $8$ and $9$ raises those days only to $5$ , so the histogram becomes $4, 4, 3, 3, 6, 3, 3, 5, 5, 2$ with peak $6$ ; the bound is attained and the profile is markedly flatter than the original.
5	Each activity contributes its number of workers for the whole of its duration That contribution is duration multiplied by workers, which does not depend on when the activity starts The area under the histogram is the sum of those contributions over all activities So the area equals the total worker-days of the project whatever the schedule To check a drawn histogram, add the column heights and compare with the sum of duration times workers over the activity list A mismatch shows an activity has been drawn with the wrong length, the wrong height or in the wrong place	6
(6 marks)		6
Notes
Rescheduling moves an activity's block horizontally but changes neither its width nor its height, so its area is invariant; summing over activities shows the total area is fixed. That gives a cheap and complete arithmetic check on a histogram: total the heights column by column and compare with the independently computed worker-day total. Levelling therefore never reduces the amount of work, it only redistributes it, which is exactly why lowering the peak must fill in a trough somewhere else.

D1-4.6 · Scheduling the activities using the least number of workers required to complete the project.

Tier 1 · Easy

Mark scheme for D1-4.6 Tier 1 · Easy
Question	Scheme	Marks
1	Lower bound $=\left\lceil\dfrac{54}{20}\right\rceil$ $54\div20=2.7$ , so the lower bound is $3$ workers	2
(2 marks)		2
Notes
Each worker can supply at most $20$ worker-days in the $20$ days available, so $n$ workers supply at most $20n$ . Requiring $20n\geqslant54$ gives $n\geqslant2.7$ , and since $n$ is a whole number the lower bound is $3$ .
2	Critical activities have zero float, so their times are fixed Consecutive critical activities meet end to end with no gaps One worker can therefore carry out the whole chain without ever being idle or double booked	3
(3 marks)		3
Notes
Because each critical activity begins the moment its predecessor ends, the critical path forms an unbroken block of work from time $0$ to the project duration. Assigning it to one worker uses that worker fully and removes every fixed commitment from the rest of the schedule, leaving only the activities that have float to be fitted around the remaining workers.

Tier 2 · Standard

Mark scheme for D1-4.6 Tier 2 · Standard
Question	Scheme	Marks
1	Sum of durations $=5+4+6+3+7+2+4+5=36$ Lower bound $=\left\lceil\dfrac{36}{27}\right\rceil=2$ workers Worker $1$ : $A$ from $0$ to $5$ , $C$ from $5$ to $11$ , $E$ from $11$ to $18$ , $G$ from $18$ to $22$ , $H$ from $22$ to $27$ Worker $2$ : $B$ from $0$ to $4$ , $D$ from $5$ to $8$ , $F$ from $11$ to $13$ $D$ starts at $5$ , after both $A$ and $B$ finish, and its latest finish is $18$ $F$ starts at $11$ , after $C$ finishes, and its latest finish is $22$ Two workers are sufficient	8
(8 marks)		8
Notes
The durations total $36$ over a critical time of $27$ , so at least $\lceil36/27\rceil=2$ workers are needed. Giving the critical chain $A$ , $C$ , $E$ , $G$ , $H$ to worker $1$ occupies that worker for the whole $27$ days. Worker $2$ then has $B$ , $D$ and $F$ to place, totalling only $9$ days. Activity $B$ has no predecessors so runs from $0$ to $4$ ; $D$ needs $A$ and $B$ , both complete at $5$ , so runs from $5$ to $8$ , well inside its window ending at $18$ ; and $F$ needs $C$ , complete at $11$ , so runs from $11$ to $13$ , inside its window ending at $22$ . No worker does two activities at once, so the schedule is valid and the bound is attained.
2	The bound only compares total work with total time available It ignores precedence, which can force activities apart or together It also ignores the fact that an activity cannot be split between workers A complete answer must give a valid schedule with the claimed number of workers It must also justify that no smaller number is possible, usually by quoting the lower bound	5
(5 marks)		5
Notes
Dividing total work by the critical time asks only whether enough worker-days exist; a schedule must also respect the order of the activities and keep each activity on one worker for a single unbroken interval. Those extra constraints can make the bound unattainable, for instance when several long activities all become available at the same moment. The examiner therefore expects both halves of the argument: a table showing who does what and when, which proves sufficiency, and the arithmetic bound, which proves necessity.
3	Lower bound $=\left\lceil\dfrac{46}{16}\right\rceil=\lceil2.875\rceil=3$ workers Three workers over $16$ days supply $3\times16=48$ worker-days Work required $=46$ worker-days Idle time $=48-46=2$ worker-days	5
(5 marks)		5
Notes
The bound comes from requiring $16n\geqslant46$ , giving $n\geqslant2.875$ and hence $n=3$ . With three workers the total capacity over the sixteen days is $48$ worker-days, of which $46$ are used by the activities, so exactly $2$ worker-days are idle however the schedule is arranged. That small slack is a warning that the schedule will be tight and the bound may not in fact be attainable.

Tier 3 · Hard

Mark scheme for D1-4.6 Tier 3 · Hard
Question	Scheme	Marks
1	Sum of durations $=5+7+4+6+3+8+2+5=40$ Lower bound $=\left\lceil\dfrac{40}{24}\right\rceil=2$ workers Worker $1$ : $A$ from $0$ to $5$ , $D$ from $5$ to $11$ , $F$ from $11$ to $19$ , $H$ from $19$ to $24$ Worker $2$ : $B$ from $0$ to $7$ , $C$ from $7$ to $11$ , $E$ from $11$ to $14$ , $G$ from $14$ to $16$ $C$ needs $A$ , finished at $5$ , and must end by $11$ so that $F$ can start; $7$ to $11$ satisfies both $E$ needs $B$ , finished at $7$ , and must end by $17$ ; $11$ to $14$ satisfies both $G$ needs $D$ , finished at $11$ , and $E$ , finished at $14$ , and must end by $19$ ; $14$ to $16$ satisfies both Two workers are sufficient, so the least number of workers is $2$	9
(9 marks)		9
Notes
The activities total $40$ days over a critical time of $24$ , giving a bound of $2$ . Worker $1$ takes the critical chain and is busy for the whole project. Worker $2$ must fit $B$ , $C$ , $E$ and $G$ , totalling $16$ days, into $24$ . Taking them in the order $B$ , $C$ , $E$ , $G$ works because each becomes available in time: $B$ from the start; $C$ once $A$ finishes at $5$ , and it must be complete by $11$ for $F$ , which the interval $7$ to $11$ achieves exactly; $E$ once $B$ finishes at $7$ ; and $G$ once both $D$ and $E$ are done at $14$ , finishing at $16$ , comfortably before $H$ starts at $19$ . Every constraint holds, so $2$ workers suffice and the bound is attained.
2	Worker $2$ is occupied by $B$ from $0$ to $7$ A worker cannot carry out two activities at once, so $C$ cannot start before $7$ $C$ has earliest start $5$ and latest finish $11$ , giving a float of $2$ Starting at $7$ uses the whole float and finishes exactly at $11$ Starting $C$ at time $5$ would need a third worker, since $B$ is still running With two workers the schedule would then need $B$ moved, but $B$ has float $7$ and could start later, so a valid alternative exists Starting $C$ later than $7$ would push its finish past $11$ and delay $F$ , extending the project	7
(7 marks)		7
Notes
Two constraints act on $C$ at once. The precedence constraint says it cannot start before $A$ finishes at $5$ and must finish by $11$ so that $F$ can begin on time. The resource constraint says worker $2$ is busy with $B$ until $7$ . Since $C$ takes four days, the only interval satisfying both is exactly $7$ to $11$ , which consumes the whole of its float of $2$ . Insisting on the earliest start of $5$ would require a second worker to be free at that moment, which with only two workers means rearranging $B$ ; that is possible because $B$ has float $7$ , but it is not the schedule given.
3	$P$ , $R$ and $S$ all start at $0$ , so $Q$ starts at $\max(5,4,4)=5$ and finishes at $12$ $T$ runs from $12$ to $15$ , so the critical time is $15$ with critical path $P$ , $Q$ , $T$ Total duration $=5+4+4+7+3=23$ , so the lower bound is $\left\lceil\dfrac{23}{15}\right\rceil=2$ workers $R$ and $S$ have latest finish $5$ , since $Q$ has latest start $5$ , so both lie wholly in the interval $0$ to $5$ $P$ also lies wholly in that interval Work that must be done in the interval $0$ to $5$ is $5+4+4=13$ worker-units Two workers supply only $2\times5=10$ worker-units there, so $2$ workers cannot finish in $15$ With $3$ workers: worker $1$ does $P$ from $0$ to $5$ , then $Q$ from $5$ to $12$ , then $T$ from $12$ to $15$ ; worker $2$ does $R$ from $0$ to $4$ ; worker $3$ does $S$ from $0$ to $4$ The least number of workers is $3$	9
(9 marks)		9
Notes
The forward pass gives $Q$ an earliest start of $5$ and the project a duration of $15$ , with $P$ , $Q$ and $T$ critical. The averaging bound over the whole project is only $2$ , so the busy window must be examined separately. Since $Q$ cannot slip, $R$ and $S$ must both finish by time $5$ , and $P$ occupies the same window, so $13$ worker-units of work are trapped in an interval of length $5$ . Two workers supply $10$ worker-units there, which is not enough, so two workers cannot meet the critical time. Three workers supply $15$ worker-units in that window, and the schedule shown runs $P$ , $R$ and $S$ simultaneously from time $0$ , after which one worker carries the rest of the critical chain alone. Hence three workers are both necessary and sufficient.
4	The bound assumes work can be spread evenly across the whole project Precedence can force several activities into the same short window An activity cannot be split between workers or interrupted So the demand in a sub-interval can exceed what the workers can supply there To prove $n$ workers are not enough, choose an interval of length $t$ Show that the activities that must lie wholly inside it total more than $nt$ Then no schedule with $n$ workers can complete the project in the critical time	7
(7 marks)		7
Notes
The global bound compares total work with total capacity, but a schedule must also satisfy capacity in every sub-interval. The rigorous argument mirrors the global one locally: find a window into which some activities are forced by their earliest starts and latest finishes, add their durations together with any critical work that must run there, and compare with the $nt$ worker-units that $n$ workers supply in a window of length $t$ . If the demand exceeds the supply, the impossibility is proved and one then exhibits a schedule with $n+1$ workers to establish the exact answer.
5	The critical path has length $5+6+7+4+5=27$ Its activities must be done one after another, since each is a predecessor of the next No worker can shorten a chain of dependent activities by working in parallel So the project cannot finish before $27$ days however many workers are hired To reach $25$ days the critical path must be shortened by $2$ Reducing any one of $A$ , $C$ , $E$ , $G$ or $H$ by $2$ shortens that chain to $25$ This works only if no other chain then exceeds $25$ , which must be checked	7
(7 marks)		7
Notes
Adding workers can only allow independent activities to run at the same time; it cannot overlap two activities when one is a predecessor of the other. The critical path is precisely such a chain, so its total length of $27$ is a floor on the project duration for any workforce. Meeting a deadline of $25$ therefore requires the chain itself to be shortened by at least $2$ , which a $2$ -day reduction in any single critical activity achieves. The reduction must then be re-tested, because the next longest chain may become critical and hold the duration above $25$ .