Critical path analysis and scheduling (QCE General Mathematics Unit 4)

Syllabus dot point

“Construct a project network diagram (activity on arc, excluding dummy activities) from an activity table, use forward and backward scanning to determine the earliest starting time (EST) and latest starting time (LST) of each activity, locate the critical path, determine the minimum completion time, calculate float times for non-critical activities, and solve small-scale practical problems”

15 min readUpdated 23 Sep 2026

Quick answer

Critical path analysis turns an activity table into an activity-on-arc network (no dummies). A forward scan gives each activity's earliest starting time (take the maximum at merges), and its final value is the minimum completion time. A backward scan from that time gives latest starting times (take the minimum at branches). Float $=$ LST $-$ EST; activities with zero float form the critical path, the longest path through the network. Delaying a critical activity delays the project; delaying a non-critical one only matters if the delay exceeds its float.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

A project, such as building a shed, running an event or cooking a meal, is made up of activities that take time and must happen in a certain order. Some can run at the same time; others must wait for earlier ones to finish. Critical path analysis (CPA) finds the shortest possible time to complete the project and identifies which activities must not be delayed.

The QCAA General Mathematics 2025 syllabus (Unit 4, Topic 4, Networks and decision mathematics 1, "Project planning and scheduling using critical path analysis") asks you to

construct a project network diagram (activity on arc) that shows the durations and interdependencies of the activities, excluding dummy activities,
use forward and backward scanning to determine the earliest starting time (EST) and latest starting time (LST) of each activity,
use the ESTs and LSTs to locate the critical path(s),
use the critical path to determine the minimum completion time,
calculate float times for non-critical activities, and
solve small-scale practical problems.

The answer

The activity table

A project is described by an activity table (precedence table): each activity has a letter, a duration and its immediate predecessors, the activities that must be finished immediately before it can start.

Activity	Description	Duration (days)	Immediate predecessors
A	Clear site	2	none
B	Order materials	3	none
C	Lay slab	4	A
D	Build frame	5	B, C
E	Install roof	3	D
F	Fit door and windows	2	D
G	Paint	3	E
H	Landscape	6	A
I	Clean up	1	F, G, H

Drawing the network (activity on arc)

In an activity-on-arc network, each arc (directed edge) is an activity, labelled with its letter and duration, and each node is an event: the moment when the activities coming in have all finished and the activities going out can start.

Start with one start node. Activities with no predecessors (A and B) leave it.
An activity leaves the node where all of its immediate predecessors end. D needs B and C, so B and C must end at the same node, and D leaves that node.
Two activities with exactly the same predecessors leave the same node (E and F both leave the node where D ends).
Activities that nothing depends on end at a single finish node.
Check every row of the table against the diagram before scanning.

The syllabus excludes dummy activities, so exam networks are designed to be drawn without them. If you find you need one, recheck the table: you have probably merged two nodes that should be separate.

Forward scanning: earliest starting times

Work from the start node towards the finish. At each node write the earliest time all incoming activities can be finished:

\text{node's earliest time} = \max(\text{EST of incoming activity} + \text{its duration}).

Every activity leaving a node has that node's earliest time as its EST. The start node is 0. Take the maximum because an activity cannot begin until all of its predecessors are finished. At the node where D starts, B could finish at $0 + 3 = 3$ but C at $2 + 4 = 6$ , so D's EST is 6.

The earliest time at the finish node is the minimum completion time.

Backward scanning: latest starting times

Work from the finish node back to the start. The finish node's latest time equals the minimum completion time. At each earlier node write the latest time it can be reached without delaying the project:

\text{node's latest time} = \min(\text{latest time at the end of an outgoing activity} - \text{its duration}).

An activity's LST is the latest time at its end node minus its duration. Take the minimum because the node must be reached in time for every activity that leaves it. The start node's latest time should come back to 0; if it does not, there is an arithmetic slip.

EST, LST and float

The float of an activity is how long it can be delayed without delaying the whole project:

\text{float} = \text{LST} - \text{EST}.

Activity	B	C	D	E	F	G	H	I
EST	0	2	6	11	11	14	2	17
LST	3	2	6	11	15	14	11	17
Float	3	0	0	0	4	0	9	0

For example, F ends at the node with latest time 17, so its LST is $17 - 2 = 15$ and its float is $15 - 11 = 4$ days.

The critical path and minimum completion time

The critical activities are those with zero float (EST = LST). Joined from start to finish they form the critical path: A-C-D-E-G-I. Its length, $2 + 4 + 5 + 3 + 3 + 1 = 18$ days, is the minimum completion time. The critical path is the longest path through the network, because every path's activities must all be done and the longest one sets the finish time. A project can have more than one critical path when two paths tie for the longest length.

Delays and shortening activities

Delaying a critical activity delays the whole project by the same amount.
Delaying a non-critical activity by no more than its float has no effect. Delaying it by more adds the excess to the project time, and the critical path may change.
Float along a chain of non-critical activities can be shared: if one activity uses it up, the next one in that chain may have none left.
Shortening a non-critical activity never reduces the completion time. Shortening a critical activity reduces it only until another path becomes critical.

Key fact

Forward scan: EST $= \max(\text{EST} + \text{duration})$ over incoming activities; the finish value is the minimum completion time. Backward scan: latest time $= \min(\text{latest at end} - \text{duration})$ over outgoing activities; LST $=$ latest time at the activity's end node minus its duration. Float $=$ LST $-$ EST. Critical activities have zero float and form the longest path.

Worked examples: scanning, float and delays

1. Forward scan for the shed

Using the shed activity table, find every EST and the minimum completion time.

Start. A and B: EST 0.

C and H follow A: EST $0 + 2 = 2$ .

D follows B and C: $\max(0 + 3, 2 + 4) = \max(3, 6) = 6$ .

E and F follow D: $6 + 5 = 11$ .

G follows E: $11 + 3 = 14$ .

I follows F, G and H: $\max(11 + 2, 14 + 3, 2 + 6) = \max(13, 17, 8) = 17$ .

Finish. $17 + 1 = 18$ . The minimum completion time is 18 days.

2. Backward scan for the shed

Finish latest time 18, so I's LST is $18 - 1 = 17$ and the node where F, G and H end has latest time 17.

G: $17 - 3 = 14$ . H: $17 - 6 = 11$ . F: $17 - 2 = 15$ .
E: ends where G starts (latest 14): $14 - 3 = 11$ .
D: the node where E and F start must be reached by $\min(11, 15) = 11$ , so D's LST is $11 - 5 = 6$ .
B: $6 - 3 = 3$ . C: $6 - 4 = 2$ .
A: the node where C and H start must be reached by $\min(2, 11) = 2$ , so A's LST is $2 - 2 = 0$ . The start node comes back to 0, as it should.

Marker's note: show the max and min comparisons at the merge and branch nodes. Those are where marks are won and lost.

3. The effect of delays

In the shed project, the windows (F) arrive late and F takes 7 days instead of 2. Separately, landscaping (H) is delayed by 10 days. What happens in each case?

F delayed by 5 days. F's float is 4 days, so the project is delayed by $5 - 4 = 1$ day: path A-C-D-F-I becomes $2 + 4 + 5 + 7 + 1 = 19$ days, the new critical path.

H delayed by 10 days. H's float is 9 days, so the project is delayed by 1 day: A-H-I becomes $2 + 16 + 1 = 19$ days.

B delayed by 2 days (float 3) would have no effect.

4. Two critical paths

Activities: P (3, none), Q (4, none), R (5, P), S (4, Q), T (2, R and S). Find the critical path(s).

Forward scan: P, Q: EST 0. R: 3. S: 4. T: $\max(3 + 5, 4 + 4) = 8$ . Finish: 10.
Backward scan: T: $10 - 2 = 8$ . R: $8 - 5 = 3$ . S: $8 - 4 = 4$ . P: $3 - 3 = 0$ . Q: $4 - 4 = 0$ .
Floats: All zero, so there are two critical paths, P-R-T and Q-S-T, each 10 units long.
Consequence: Shortening R alone cannot finish the project sooner, because Q-S-T still takes 10. You would need to shorten both branches, or T, which is on both.

5. Shortening activities at least cost

In the shed project, C, E and F can each be shortened by 1 day for $400, $250 and $150. Find the cheapest way to finish in 17 days.

Only critical activities help: F is not critical (float 4), so shortening it changes nothing, even though it is cheapest.
Choose the cheaper critical activity: Shorten E by 1 day for $250: the critical path becomes 17 days, and no other path exceeds 17 (B-D-E-G-I is 14).
Answer: Shorten E; extra cost $250.

Common traps

Taking the minimum in the forward scan: An activity waits for all its predecessors, so take the largest incoming finishing time.
Taking the maximum in the backward scan: A node must be reached in time for every outgoing activity, so take the smallest value.
Reading the LST from the wrong node: An activity's LST is the latest time at its end node minus its duration, not the latest time at its start node.
Confusing the critical path with the longest single activity: The critical path is the longest chain from start to finish.
Counting float as available to every activity in a chain: Floats on consecutive non-critical activities are often shared.
Shortening a non-critical activity to speed up a project: It has no effect on the completion time.
Using dummy activities: The syllabus excludes them. Recheck which activities should end at the same node.

Exam technique

Draw the network large and neat, with each arc labelled "letter, duration" and each node showing two boxes (earliest | latest). Tick off each row of the activity table as you draw it. Do the forward scan in one colour and the backward scan in another. Then present a table of EST, LST and float for every activity: it earns marks even if your diagram is messy, and it makes the critical path and completion time easy to justify. In "effect of a delay" questions, compare the delay with the activity's float and state the new completion time.

Note

Imagine cooking a big dinner: some jobs (boiling rice, chopping vegetables) can happen at the same time, but some must wait (you cannot fry the curry before chopping it). The critical path is the chain of jobs with no spare time at all. If any of those runs late, dinner is late. The other jobs have "float": spare minutes you can use, as long as you do not use more than you have. Forward scanning works out the earliest each job can start; backward scanning works out the latest each job can start without making dinner late.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

The activity table for a project is shown. A (prereq none, 2 days); B (none, 4); C (A, 3); D (B, 6); E (D, 3); F (C and E, 4); G (D, 8); H (F and G, 4). a) Use the activity table to construct a network diagram, including earliest and latest starting times. [3 marks] b) Determine the critical path. [1 mark] c) Determine the shortest completion time for the project. [1 mark]

Show worked answer →

a) Network and scanning (3 marks). A and B leave the start node. C follows A; D follows B; E and G both leave the node where D ends; C and E end at the same node, from which F leaves; F and G end at the node from which H leaves (1 mark for the structure, 1 mark for letters and durations). Forward and backward scanning give (1 mark):

Activity	A	B	C	D	E	F	G	H
EST	0	0	2	4	10	13	10	18
LST	9	0	11	4	11	14	10	18

b) Critical path (1 mark). The activities with EST = LST are B, D, G and H, so the critical path is B-D-G-H.

c) Completion time (1 mark). $4 + 6 + 8 + 4 = 22$ days. Check: the other paths are shorter (B-D-E-F-H is 21 days, A-C-F-H is 13 days).

Practice5 marks

Jed is preparing and serving a meal. Tasks with duration (minutes) and prerequisite: A Assemble equipment 2 (none); B Boil rice 20 (A); C Prepare curry ingredients 6 (A); D Make naan bread 8 (A); E Simmer curry 40 (C); F Fry naan bread 4 (D); G Serve meal 2 (B, E, F). a) Construct a network diagram showing the sequence of tasks, labelling all tasks and durations, and use forward and backward scanning to show the earliest and latest starting times. [3 marks] b) Determine the critical activities and minimum completion time. [2 marks]

Show worked answer →

a) Network and scanning (3 marks). A leaves the start node; B, C and D all leave the node where A ends; E follows C; F follows D; B, E and F end at the node from which G leaves (1 mark for the structure, 1 mark for labels). Scanning gives (1 mark):

Task	A	B	C	D	E	F	G
EST	0	2	2	2	8	10	48
LST	0	28	2	36	8	44	48

b) Critical activities and minimum time (2 marks). A, C, E and G have zero float, so they are critical (1 mark). The minimum completion time is $2 + 6 + 40 + 2 = 50$ minutes (1 mark).

The rice (float 26 minutes) and the naan (float 34 minutes for each of D and F, shared along that branch) can slip without delaying the meal.

Original6 marks

A school is setting up for a fete. Activity (duration in hours, immediate predecessors): A book equipment (3, none); B recruit volunteers (5, none); C deliver equipment (4, A); D set up stalls (6, C); E set up sound system (2, C); F brief volunteers (2, B); G stock stalls (3, D and F); H sound check (1, E); I open gates (1, G and H). (a) Describe the activity-on-arc network. (b) Use forward and backward scanning to find the EST and LST of every activity. (c) State the critical path and minimum completion time. (d) Find the float of B, E and H.

Show worked answer →

(a) A and B leave the start node. C follows A. D and E both leave the node where C ends. F follows B. D and F end at the same node, from which G leaves. H follows E. G and H end at the node from which I leaves. No dummies are needed. (1 mark)

(b) (2 marks)

Activity	A	B	C	D	E	F	G	H	I
EST	0	0	3	7	7	5	13	9	16
LST	0	6	3	7	13	11	13	15	16

For example, G waits for D (finishes at $7 + 6 = 13$ ) and F (finishes at $5 + 2 = 7$ ), so its EST is $\max(13, 7) = 13$ .

(c) Critical path A-C-D-G-I; minimum completion time $3 + 4 + 6 + 3 + 1 = 17$ hours. (2 marks)

(d) Float $=$ LST $-$ EST: B has $6 - 0 = 6$ hours, E has $13 - 7 = 6$ hours and H has $15 - 9 = 6$ hours. (1 mark) The 6 hours on B is shared with F: if B uses all of it, F has none left.

Original3 marks

For the fete in the previous question, the volunteers take 8 hours longer to recruit than planned (B takes 13 hours). (a) What is the effect on the minimum completion time? (b) Would the same delay to activity E have any effect? Explain.

Show worked answer →

(a) B has 6 hours of float, so an 8-hour delay is 2 hours more than it can absorb. The path B-F-G-I becomes $13 + 2 + 3 + 1 = 19$ hours, so the fete setup now takes 19 hours, 2 hours longer, and B-F-G-I becomes the critical path. (2 marks)

(b) E also has 6 hours of float, so an 8-hour delay to E also delays the project by 2 hours: path A-C-E-H-I becomes $3 + 4 + 10 + 1 + 1 = 19$ hours. A delay of 6 hours or less to E would have no effect. (1 mark)

Original3 marks

A garden shed project has activities A (2 days, none), B (3, none), C (4, A), D (5, B and C), E (3, D), F (2, D), G (3, E), H (6, A) and I (1, F, G and H); its minimum completion time is 18 days. Activities C, E and F can each be shortened by 1 day, at extra costs of $400, $250 and $150 respectively. The owner wants the shed finished in 16 days. Which activities should be shortened, and what is the least extra cost?

Show worked answer →

The critical path is A-C-D-E-G-I (18 days). Only shortening critical activities can reduce the completion time, so F (float 4 days), although cheapest, would have no effect. (1 mark)

Shorten C and E by 1 day each. The critical path becomes $2 + 3 + 5 + 2 + 3 + 1 = 16$ days, and the other paths are no longer: B-D-E-G-I is 14 days, A-C-D-F-I is 13 days and A-H-I is 9 days. (1 mark)

The least extra cost is $400 + 250 = 650$ , so $650. (1 mark)

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Activity K has EST 7 days and LST 11 days. (a) Find its float. (b) Is K on the critical path?

Show worked solution →

(a) Float $= 11 - 7 = 4$ days. (1 mark)

(b) No. Critical activities have zero float (EST = LST). (1 mark)

foundation2 marks

Activity P (4 days) and activity Q (6 days) are the only immediate predecessors of activity R. P has EST 3 and Q has EST 2. Find the EST of R.

Show worked solution →

P finishes at $3 + 4 = 7$ and Q finishes at $2 + 6 = 8$ . (1 mark)

R must wait for both, so its EST is $\max(7, 8) = 8$ days. (1 mark)

foundation2 marks

Activity M (duration 5) is followed only by activities N and P, which have LSTs of 14 and 12. Find the LST of M.

Show worked solution →

M must finish by the earlier of the successors' latest starts: $\min(14, 12) = 12$ . (1 mark)

LST of M $= 12 - 5 = 7$ . (1 mark)

core3 marks

Activities: P (3, none), Q (4, none), R (5, P), S (4, Q), T (2, R and S). Find the minimum completion time and all critical paths.

Show worked solution →

Paths: P-R-T $= 3 + 5 + 2 = 10$ and Q-S-T $= 4 + 4 + 2 = 10$ . (1 mark)

Minimum completion time 10. (1 mark) Both paths are critical: every activity has EST = LST, so there are two critical paths, P-R-T and Q-S-T. (1 mark)

core2 marks

For the project in the previous question, would shortening R by 1 unit reduce the completion time? Explain.

Show worked solution →

No. P-R-T would become 9, but Q-S-T is still 10, so the project still takes 10. (1 mark) With two critical paths, both must be shortened to finish sooner, for example R and S, or T alone (which is on both). (1 mark)

exam4 marks

Activities (duration in days, immediate predecessors): A (2, none), B (3, none), C (4, A), D (5, B and C), E (3, D), F (2, D), G (3, E), H (6, A), I (1, F, G and H). (a) Find the EST and LST of every activity. (b) State the critical path and minimum completion time. (c) Find the floats of B, F and H.

Show worked solution →

(a) (2 marks)

Activity	A	B	C	D	E	F	G	H	I
EST	0	0	2	6	11	11	14	2	17
LST	0	3	2	6	11	15	14	11	17

(b) A-C-D-E-G-I, 18 days. (1 mark)

(c) B: 3 days; F: 4 days; H: 9 days. (1 mark)

exam2 marks

In the project from the previous question, activity F is delayed by 5 days and activity B by 2 days. Find the new minimum completion time.

Show worked solution →

B has 3 days of float, so a 2-day delay has no effect. (1 mark)

F has only 4 days of float, so a 5-day delay adds 1 day: path A-C-D-F-I becomes $2 + 4 + 5 + 7 + 1 = 19$ days. The project now takes 19 days. (1 mark)

Practise this

Sources & how we know this

General Mathematics General senior syllabus 2025 — QCAA