Critical path analysis

Count the arcs first. There are eight activities: wash face, get dressed, pack bag, charge phone, eat breakfast, brush teeth, listen to music and travel to school.

Listen to music leaves the node where pack bag and charge phone arrive. After pack bag, three arcs come before travel to school: eat breakfast, brush teeth and listen to music. So (D) is correct.

(A) is false because every activity in a project gets done. The critical path only decides the minimum time. (B) is false because charge phone runs alongside get dressed and pack bag. (C) is the careful one. Wash face, get dressed, pack bag and charge phone must all finish before listening to music can start. That is 4 of 8, exactly half, not less than half.

82.0% of students chose (D).

Forward scan. The central node takes the larger of the two arrivals, B on its own or A then C:

$$\begin{aligned}\text{EST} &= \max(5,\ 4+3) \\ &= 7\end{aligned}$$

Then E finishes at $7+8=15$ and F at $7+2+5=14$, so the fifth node is $\max(15,14)=15$. G finishes at $15+4=19$.

The critical path is ACEG, $4+3+8+4=19$ days. Listing the paths gives the same answer: ACEG 19, ACDFG 18, BEG 17, BDFG 16.

(A) 16 is the shortest path, and 25.1% of students chose it. “Minimum time to complete” is the longest path, because the project is not done until its slowest chain is done. (B) 17 takes 5 at the central node and ignores A then C. (D) 31 adds every duration, as if nothing could run at the same time. 61.4% chose (C).

The minimum time is the length of the critical path:

$$\begin{aligned}\text{time} &= 38+32+34+26 \\ &= 130\ \text{min} \\ &= 2\ \text{h}\ 10\ \text{min}\end{aligned}$$

Then $11{:}00\ \text{am} + 2\ \text{h}\ 10\ \text{min} = 1{:}10\ \text{pm}$.

(A) treats 130 minutes as 1 h 30 min. (C) reads 130 as the clock time 1:30. (D) writes the duration, 2 h 10 min, as if it were the finishing time. Q, T and U are red herrings, since they are not on the critical path. 79.9% of students chose (B).

Q cannot start until P has finished. P starts at 0 and takes 3, so the EST of Q is 3.

Q is critical, so its float is zero and its LST equals its EST, 3. You do not need a backward scan at all.

(B) has the right EST but takes Q’s duration as its LST, and 27.3% of students chose it. (C) uses the duration for both. (D) takes Q’s duration as its EST and the project’s finishing time, $3+6=9$, as its LST, which mixes up starting and finishing times. Only 49.1% chose (A).

Watch the news can start at 3 at the earliest. It has to be finished by 33, the latest time at the node where it arrives, so it can start as late as $33-10=23$.

$$\begin{aligned}\text{float} &= \text{LST} - \text{EST} \\ &= 23 - 3 \\ &= 20\ \text{min}\end{aligned}$$

Another way to see it: the news runs alongside turn on oven, prepare ingredients and cook food, which take $5+8+17=30$ minutes. The news takes 10, so it has 20 minutes to spare.

Notice that the 3|3 node shows 3 on the right, not 23. Two activities leave it, and the circle shows the tighter one, turn on oven. That is exactly why reading the right-hand number at the tail of an arc as that activity’s LST goes wrong.

(A) 10 is the duration of the news, and 33.9% of students chose it, almost as many as the key. (C) 23 is the LST itself. (D) 30 is $33-3$, which forgets to subtract the duration. Only 36.7% chose (B), the lowest result for any critical path item from 2021 to 2025.

The network is already scanned. The critical path is ABDFG, $6+5+3+3+5=22$ days. Only C and E are off it.

$$\begin{aligned}\text{C:}\ (12 - 4) - 6 &= 2 \\ \text{E:}\ 12 - 10 &= 2\end{aligned}$$

For C, the latest time at its end node is 12, so its LST is $12-4=8$ and its EST is 6. For E, the key tells you directly: EST 10, LST 12. Both have 2 days, so the greatest float is 2 days.

C and E are in a chain, so they share those 2 days. If C runs 2 days late, E has no float left. (B) 4 adds the two floats, which double counts. (C) 5 is the duration of E. (D) 12 is the LST of E. The 2020 subject report did not publish multiple choice results.

Both starting times are given, so this is one subtraction:

$$\begin{aligned}\text{float} &= \text{LST} - \text{EST} \\ &= 4 - 0 \\ &= 4\ \text{min}\end{aligned}$$

The critical path is XYZ, 9 minutes. W only has to be finished before Z starts at 5, and it takes 1 minute, so it can start anywhere from 0 to 4.

(A) is W’s EST, or the float of a critical activity. (B) is W’s duration, and 16.9% chose it. (D) is the LST plus the duration, a finishing time. 62.4% chose (C).

Scan the whole network first, then answer each part from it.

a) D starts when B finishes:

$$\begin{aligned}\text{EST}(D) &= 12 + 15 \\ &= 27\ \text{minutes}\end{aligned}$$

1 mark · correctly states EST for activity D

b) The lower node is $12+16=28$. Then take the largest at the right node and at the finish:

$$\begin{aligned}\text{right} &= \max(27+20,\ 28+14) \\ &= 47 \\ \text{finish} &= \max(47+8,\ 28+5) \\ &= 55\end{aligned}$$

$$\begin{aligned}\text{time} &= 12+15+20+8 \\ &= 55\ \text{minutes}\end{aligned}$$

1 mark · correctly determines minimum completion time

c) Work back from the finish. F must start by $55-8=47$, so E must start by

$$\begin{aligned}\text{LST}(E) &= 55 - 8 - 14 \\ &= 33\ \text{minutes}\end{aligned}$$

1 mark · correctly states LST for activity E

d) G starts at 28 at the earliest and must be finished by 55:

Float is the latest finish minus the earliest start minus the duration:

$$\begin{aligned}\text{float}(G) &= 55 - 28 - 5 \\ &= 22\ \text{minutes}\end{aligned}$$

1 mark · provides mathematical reasoning or working to support the answer

1 mark · calculates float time for activity G

The guide also accepts $\text{LST}(G) - \text{EST}(G) = 50 - 28$, and follow-through from earlier errors. Values written on the diagram count for a), b) and c) “only if labelled”.

Two slips to avoid: taking the EST of E as 27 (through B) instead of 28, and using 33, the right-hand number of the lower node, as if it were G’s latest start. That gives $55-33-5=17$. The node shows 33 because E is the tighter activity leaving it. G has its own LST of 50.

a) A and B have no prerequisite, so both leave the start node. C leaves where A arrives. D leaves where B arrives. E and G both need only D, so they leave the same node. F needs C and E, so C and E meet at one node. H needs F and G, so they meet at another. H arrives at the finish.

The node where C and E meet is the one to watch: $\max(2+3,\ 10+3) = 13$, not 5.

The start node shows 0 on the right because B is the tighter of the two activities leaving it. A’s own LST is $11-2=9$.

1 mark · correctly translates the information into a network diagram

1 mark · correctly labels each activity letter and duration

1 mark · provides evidence of forward and backward scanning

The guide notes “Accept alternative formats.” The subject report showed one: a sample that drew BDGH as a straight line along the bottom, with A and C above it.

b) The activities with zero float, in order:

$$B \to D \to G \to H$$

1 mark · determines critical path

Follow-through applies, and a path drawn on the network diagram is accepted.

c)

$$4 + 6 + 8 + 4 = 22\ \text{days}$$

1 mark · determines shortest time

Check it against the other paths: ACFH is 13 and BDEFH is 21. The longest, 22, matches the forward scan. Adding every duration (34) is the slip to avoid.

a) Only A has no prerequisite. B, C and D all need only A, so all three leave the node where A arrives. E follows C and F follows D. G needs B, E and F, so all three arrive at one node, and G runs from there to the finish.

The node where B, E and F meet is $\max(2+20,\ 8+40,\ 10+4) = 48$. Notice that B, the longest single arc out of the 2|2 node, is not on the critical path. C then E is longer.

The question asks for earliest and latest starting times for all tasks, and the node at the tail of B and D shows only the tightest LST, 2. Each task’s own LST is the latest time at its end node minus its duration:

1 mark · correctly constructs a network diagram showing the appropriate sequence for all tasks

1 mark · labels all tasks and durations on network diagram

1 mark · shows earliest and latest starting times for all tasks

The guide accepts alternative formats and allows follow-through.

b) Critical activities: A, C, E, G.

$$\begin{aligned}\text{time} &= 2 + 6 + 40 + 2 \\ &= 50\ \text{minutes}\end{aligned}$$

1 mark · determines critical activities

1 mark · determines minimum completion time, including units

The unit is part of the mark. The subject report’s sample was included “to show clear communication of the critical activities by writing a vertex sequence and the minimum completion time including minutes as the unit.”

This is a Paper 2 question with no parts, so the steps are yours to choose. It still asks for exactly what the five-mark Paper 1 questions ask for: draw, scan both ways, then find the float.

A then B run in a line. C and D both leave the node where B arrives. E follows C. F needs D and E, so they meet before F.

1 mark · correctly translates information into a network diagram showing all activities and durations

Forward scan: the node before F is $\max(8+15,\ 28+7) = 35$, and the project takes $35+30=65$ minutes.

1 mark · completes forward scanning to determine EST for each activity

Backward scan from 65: $65-30=35$, then $35-7=28$, $28-20=8$, $8-5=3$, $3-3=0$. D must start by $35-15=20$.

1 mark · completes backward scanning to determine LST for each activity

Every activity has EST equal to LST except D, which has EST 8 and LST 20. Activity D is a non-critical activity.

1 mark · identifies non-critical activity

$$\begin{aligned}\text{float}(D) &= 35 - 8 - 15 \\ &= 12\ \text{minutes}\end{aligned}$$

1 mark · determines float time for non-critical activity

The non-critical mark “may be implied by subsequent working”, and follow-through applies. The subject report’s sample was chosen to show “a start node and a finish node”, scanning values written “within the nodes”, and clear identification of the non-critical activity and its float.

The 2020 subject report labels this question complex unfamiliar. The critical path analysis is only the first half.

Step 1 · The network and both scans.

1 mark · correctly translates the information into a network

1 mark · determines LST for each activity

1 mark · determines EST for each activity

Step 2 · The minimum time. The critical path is ACEHJ:

$$3 + 2 + 5 + 1 + 3 = 14\ \text{days}$$

1 mark · determines minimum completion time

Step 3 · Turn it into a staffing question. Nothing in the question prompts this step. Finishing in 14 days means every critical task runs exactly on time and every other task finishes within its float. So the question becomes: can three people cover every task that has to be running at the same time?

• •One employee is tied to the critical chain A, C, E, H, J for all 14 days.
• •D takes 8 days and has 1 day of float, so it must start on day 6 or 7 and run to the end. A second employee is tied to D from then on.
• •That leaves one employee for F, G and I once B is finished, at the end of day 4 at the earliest. They need $4+3+2=9$ days of work, and F must be finished by the end of day 10 and I by the end of day 11.
• •From the end of day 4 to the end of day 11 is only 7 days. One person cannot fit 9 days of work into 7.

Checking every allowable start time confirms it: at some point four tasks have to run at once. The marking guide’s allocation for four workers shows how it does fit:

1 mark · determines whether three workers are sufficient

Step 4 · The verdict. The owner’s belief is not reasonable. With three employees, D could not be completed by day 14. At least four employees are needed to finish in the minimum 14 days.

1 mark · evaluates reasonableness of the claim

1 mark · shows logical organisation, communicating key steps

The guide’s note for that last mark lists “appropriate mathematical vocabulary, symbols and conventions, e.g. connecting statements, diagrams.” Follow-through is allowed for the LSTs, ESTs, minimum time and the sufficiency of three workers.