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Networks and decision mathematics (QCE General Mathematics Unit 4)

Syllabus dot point

“Choose and apply the right network technique for a practical problem: minimum spanning trees for connection (Topic 4), critical path analysis for project scheduling (Topic 4), maximum flow and minimum cut for flow networks (Topic 5) and the Hungarian algorithm for assignment (Topic 5), distinguishing them from shortest path problems (Topic 3)”

14 min read

Quick answer

Networks and decision mathematics 1 (Topic 4) covers minimum spanning trees, the least-weight way to connect every vertex (v−1v - 1 edges, by Prim's or Kruskal's algorithm), and critical path analysis, where forward and backward scans give earliest and latest starting times, float is LST −- EST, and the critical path is the longest path. Networks and decision mathematics 2 (Topic 5) covers flow networks, where the maximum flow equals the minimum cut capacity, and assignment problems solved by inspection or the Hungarian algorithm. Recognise the technique from what is being optimised, and do not confuse any of them with shortest paths.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Exam-style questions
  4. Practice questions

What this dot point is asking

The last two topics of Unit 4 in the QCAA General Mathematics 2025 syllabus use networks to make decisions:

  • Topic 4: Networks and decision mathematics 1. Trees and minimum connector problems (minimum spanning trees), and project planning and scheduling using critical path analysis.
  • Topic 5: Networks and decision mathematics 2. Flow networks (cuts, minimum cut and maximum flow), and assigning order with bipartite graphs and the Hungarian algorithm.

They build on Topic 3: Graphs and networks (terminology, adjacency matrices, planar graphs, Euler's formula, Eulerian and Hamiltonian routes, and shortest paths by trial and error).

Each technique has its own page. This page is the overview: it helps you recognise which technique a question needs, shows one compact example of each, and highlights the confusions that cost marks. In the external assessment these appear as short-response items and as extended problems where you must choose the method yourself.

The answer

Choosing the technique

The question asks for... Network features Technique Topic
the least total length or cost to connect every vertex undirected, weighted minimum spanning tree 4
the least time to finish a project of dependent activities, and which activities cannot be delayed directed, activities with durations and predecessors critical path analysis 4
the greatest amount that can pass from a source to a sink directed, edges with capacities maximum flow and minimum cut 5
the best one-to-one allocation of workers to tasks bipartite graph or square matrix assignment (Hungarian algorithm) 5
the least-weight route between two vertices weighted shortest path (trial and error) 3

Key words help: "connect all", "cable", "pipeline to every house" suggest a minimum spanning tree; "project", "activities", "earliest start", "delay" suggest critical path analysis; "capacity", "per hour", "source", "sink" suggest flow; "allocate", "assign", "each worker one task" suggest the Hungarian algorithm; "quickest route from A to B" is a shortest path.

Minimum spanning trees (Topic 4)

A tree is connected with no cycles; a tree on vv vertices has v−1v - 1 edges. A minimum spanning tree connects every vertex with the least total weight. With Prim's algorithm, start anywhere and repeatedly add the smallest edge from the tree to a new vertex. For edges P-Q 4, P-R 6, Q-R 3, Q-S 7, R-S 5, R-T 8 and S-T 2, starting at P: P-Q (4), Q-R (3), R-S (5), S-T (2), total 14. Kruskal's algorithm (smallest edges first, skipping any that form a cycle) gives the same total and is a good check.

Critical path analysis (Topic 4)

Draw the project as an activity-on-arc network (no dummy activities). A forward scan gives each activity's earliest starting time (EST), taking the maximum at merges; a backward scan from the finish time gives latest starting times (LST), taking the minimum at branches. Float == LST −- EST. Activities with zero float form the critical path, the longest path, whose length is the minimum completion time.

For A (4, none), B (3, A), C (6, A) and D (2, B and C): D's EST is max⁡(4+3,4+6)=10\max(4 + 3, 4 + 6) = 10, so the project takes 10+2=1210 + 2 = 12. The critical path is A-C-D, and B has float (10−3)−4=3(10 - 3) - 4 = 3.

Flow networks (Topic 5)

A flow network is a directed network from a source to a sink, with each edge's weight a capacity. A cut separates the source from the sink; its capacity is the sum of the capacities of edges crossing it from the source side to the sink side (edges crossing the other way are not counted). The maximum flow equals the capacity of the minimum cut. The syllabus limits problems to at most 8 possible cuts, so you can list them all.

A flow network and a minimum cutDirected edges with capacities: S to A 10, S to B 8, A to B 4, A to C 6, B to C 3, B to T 7 and C to T 9. A dashed cut line crosses only B to T and C to T, with capacity 7 plus 9, which is 16, a minimum cut, so the maximum flow is 16.10846379SABCTcut: 7 + 9 = 16sourcesink

With three vertices between S and T there are 23=82^3 = 8 cuts. Their capacities are 16, 16, 18, 18, 20, 21, 26 and 27, so the minimum cut is 16 (two different cuts achieve it) and the maximum flow is 16.

Assignment (Topic 5)

An allocation of workers to tasks can be shown as a bipartite graph or a cost matrix. Small problems can be solved by inspection (list the allocations). For 3 x 3 to 5 x 5 matrices, the Hungarian algorithm reduces rows, then columns, covers the zeros with the fewest lines, adjusts if the number of lines is less than the matrix size, then assigns one zero per row and column and totals the original entries. To maximise, first subtract every entry from the largest.

For costs X: 9, 6, 8; Y: 7, 4, 9; Z: 6, 5, 7, row then column reduction gives X (2, 0, 0), Y (2, 0, 3), Z (0, 0, 0), which needs 3 lines. Y takes job 2, X job 3 and Z job 1: total 8+4+6=188 + 4 + 6 = 18.

Common confusions

  • Minimum spanning tree versus shortest path. A minimum spanning tree minimises the total of all connections; a shortest path minimises one route. The route between two vertices along a minimum spanning tree is not necessarily the shortest route.
  • Critical path versus shortest path. The critical path is the longest path through a project network.
  • Minimum cut versus the smallest edge. A cut must separate source from sink; one small edge does not limit the flow unless every route passes through it.
  • Assignment versus minimum spanning tree. An assignment pairs members of two groups one-to-one; it does not connect vertices into a network.
Key fact

Topic 4 (Networks and decision mathematics 1): minimum spanning trees (v−1v - 1 edges, least total weight, Prim's or Kruskal's algorithm) and critical path analysis (forward scan for EST using maximums, backward scan for LST using minimums, float == LST −- EST, critical path == longest path). Topic 5 (Networks and decision mathematics 2): maximum flow == minimum cut (count only edges from the source side to the sink side) and the Hungarian algorithm for assignments (convert maximisation first; total the original matrix).

Worked examples: one of each

1. Connecting sites (minimum spanning tree)

Edges (km): P-Q 4, P-R 6, Q-R 3, Q-S 7, R-S 5, R-T 8, S-T 2. Find the minimum length of cable to connect all five sites.

Prim's from P
P-Q 4. From {P, Q}: Q-R 3. From {P, Q, R}: R-S 5 (not P-R 6, which would form a cycle, or Q-S 7). From {P, Q, R, S}: S-T 2.
Check
5−1=45 - 1 = 4 edges.
Answer
4+3+5+2=144 + 3 + 5 + 2 = 14 km.

2. Scheduling (critical path analysis)

Activities: A (3, none), B (5, none), C (2, A), D (4, A), E (3, B and C), F (2, D and E). Find the minimum completion time, the critical path(s) and the float of D.

Forward scan
A, B: 0. C, D: 3. E: max⁡(0+5,3+2)=5\max(0 + 5, 3 + 2) = 5. F: max⁡(3+4,5+3)=8\max(3 + 4, 5 + 3) = 8. Finish: 10.
Backward scan
F: 10−2=810 - 2 = 8. E: 8−3=58 - 3 = 5. D: 8−4=48 - 4 = 4. C: 5−2=35 - 2 = 3. B: 5−5=05 - 5 = 0. A: min⁡(3−3,4−3)=0\min(3 - 3, 4 - 3) = 0.
Answer
10 days. D has float 4−3=14 - 3 = 1; every other activity has zero float, so there are two critical paths: A-C-E-F and B-E-F.

3. Capacity (maximum flow)

For the flow network in the figure, find the maximum flow.

List the cuts
With A, B and C each on the source or sink side, there are 8 cuts. For example {S} | {A, B, C, T} has capacity 10+8=1810 + 8 = 18; {S, A, B} | {C, T} has capacity 6+3+7=166 + 3 + 7 = 16; {S, A, B, C} | {T} has capacity 7+9=167 + 9 = 16.
Minimum
The smallest capacity is 16.
Check with a flow
Send 7 along S-B-T, 6 along S-A-C-T, and 3 along S-A-B-C-T: every edge is within capacity and 16 reaches T.
Answer
The maximum flow is 16.

4. Allocation (Hungarian algorithm)

Costs X: 9, 6, 8; Y: 7, 4, 9; Z: 6, 5, 7. Find the minimum total cost.

Reduce
Rows minus 6, 4, 5: X (3, 0, 2), Y (3, 0, 5), Z (1, 0, 2). Columns minus 1, 0, 2: X (2, 0, 0), Y (2, 0, 3), Z (0, 0, 0).
Cover
Three lines are needed (for example column 2, row X and row Z), which equals the matrix size.
Assign
Y's only zero is job 2. X then takes job 3 and Z takes job 1.
Answer
8+4+6=188 + 4 + 6 = 18.
Common traps
Using the wrong technique
Read what is being optimised: total connection (tree), project time (critical path), throughput (flow), allocation (assignment), or one route (shortest path).
Calling the critical path the shortest path
It is the longest path.
Counting backward edges in a cut
Only edges from the source side to the sink side count.
Stopping at one cut
The maximum flow is the minimum over all cuts; list them all (at most 8 in this syllabus).
Creating a cycle in a spanning tree
A spanning tree on vv vertices has exactly v−1v - 1 edges.
Totalling the reduced matrix
Assignment totals come from the original matrix.
Taking the minimum in a forward scan
An activity waits for all its predecessors: take the maximum.
Exam technique

For an extended network problem, start with a sentence naming the technique and why ("every town must be connected at least cost, so find a minimum spanning tree"). Then show the algorithm step by step in a table, check a structural fact (the number of edges in a tree, the start node returning to 0 in a backward scan, a flow that achieves the minimum cut, the number of lines in the Hungarian test), and finish with the answer in context and units. Where a question asks you to "evaluate" a claim, compare your result with the claim explicitly and comment on any assumptions.

Note

Network maths is a toolkit of four ideas. To link every place with the least cable, use a minimum spanning tree. To plan a project so it finishes as soon as possible, find the critical path: the chain of jobs with no spare time. To find how much can flow through pipes or roads, find the narrowest "cut" through the network. To give each person one job at the least total cost, use the Hungarian method. The skill is spotting which tool a question needs.

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 leave the node where D ends; C and E end at the node that F leaves; F and G end at the node that H leaves. Forward and backward scanning give EST/LST: A 0/9, B 0/0, C 2/11, D 4/4, E 10/11, F 13/14, G 10/10, H 18/18.
b) Critical path (1 mark)
The activities with zero float are B, D, G and H: the critical path is B-D-G-H.
c) Completion time (1 mark)
4+6+8+4=224 + 6 + 8 + 4 = 22 days. The other chains are shorter (B-D-E-F-H is 21 days and A-C-F-H is 13 days).
Practice4 marks
The network diagram shows the flow of water from the tank (source) to the kitchen tap (sink), with dashed lines L1, L2, L3 and L4 drawn across it. a) Explain which dashed line is not a valid cut. [1 mark] b) Calculate the capacity of each dashed line that is a valid cut. [3 marks] (L1 crosses edges of capacity 20, 22, 15; L2 crosses 18, 19, 22, 15; L3 crosses 18, 8, 10.)
Show worked answer →

The original diagram is not reproduced here; the explainer uses the capacities listed in the question.

a) Invalid cut (1 mark). A cut must separate the source from the sink, so that every path from tank to tap crosses it. L4 is not valid because it leaves the tank and the tap on the same side.

b) Capacities (3 marks). Add the capacities of the edges the cut crosses from the source side to the sink side: L1 =20+22+15=57= 20 + 22 + 15 = 57; L2 =18+19+22+15=74= 18 + 19 + 22 + 15 = 74; L3 =18+8+10=36= 18 + 8 + 10 = 36.

The maximum flow cannot exceed the smallest of these, 36. It equals 36 only if no other possible cut (including ones not drawn) has a smaller capacity; to be sure, check every cut or find a flow of 36.

Original4 marks
State which network technique (minimum spanning tree, shortest path, critical path analysis, maximum flow, or assignment) is needed in each situation, with a reason. (a) A council wants the least total length of fibre-optic cable to connect six libraries. (b) Four builders must each be given one of four jobs to minimise the total cost. (c) A courier wants the quickest route from the depot to one customer. (d) A water authority wants to know how many megalitres per day its pipe network can deliver from the dam to the town.
Show worked answer →

(a) Minimum spanning tree: every library must be connected with the least total length. (1 mark)

(b) Assignment (Hungarian algorithm): a one-to-one allocation minimising total cost. (1 mark)

(c) Shortest path: the least-weight route between two particular vertices. (1 mark)

(d) Maximum flow: the greatest flow from a source to a sink through capacitated pipes, found from the minimum cut. (1 mark)

Original4 marks
A flow network has source S, sink T and directed edges with capacities: S to A 10, S to B 8, A to B 4, A to C 6, B to C 3, B to T 7, C to T 9. (a) Find the capacity of the cut that separates {S, A} from {B, C, T}. (b) Find the capacity of the cut that separates {S, B} from {A, C, T}, explaining why one crossing edge is not counted. (c) Determine the maximum flow.
Show worked answer →

(a) Edges from the source side to the sink side: S to B (8), A to B (4), A to C (6). Capacity 8+4+6=188 + 4 + 6 = 18. (1 mark)

(b) From the source side to the sink side: S to A (10), B to C (3), B to T (7): capacity 20. A to B also crosses the cut, but from the sink side to the source side, so it is not counted. (1 mark)

(c) Checking all 8 cuts, the smallest capacity is 16 (for example the cut crossing B to T and C to T, 7+9=167 + 9 = 16). (1 mark) A flow of 16 is possible (S to A 9, S to B 7, A to C 6, A to B 3, B to T 7, B to C 3, C to T 9), so the maximum flow is 16. (1 mark)

Original4 marks
A project has activities A (3 days, no predecessors), B (5, none), C (2, A), D (4, A), E (3, B and C) and F (2, D and E). (a) Find the minimum completion time. (b) Identify all critical paths. (c) Find the float of D.
Show worked answer →

(a) Forward scan: A, B: EST 0. C, D: 3. E: max⁡(0+5,3+2)=5\max(0 + 5, 3 + 2) = 5. F: max⁡(3+4,5+3)=8\max(3 + 4, 5 + 3) = 8. Finish 8+2=108 + 2 = 10 days. (2 marks)

(b) Two critical paths, each 10 days: A-C-E-F (3+2+3+23 + 2 + 3 + 2) and B-E-F (5+3+25 + 3 + 2). (1 mark)

(c) D's LST is 8−4=48 - 4 = 4 and its EST is 3, so its float is 1 day. (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
Match each term to its technique: (a) float (b) cut (c) Prim's algorithm (d) Hungarian algorithm.
Show worked solution →

(a) Critical path analysis. (b) Flow networks (maximum flow). (c) Minimum spanning trees. (d) Assignment problems. (2 marks, lose 1 for each error)

foundation2 marks
Edges P-Q 4, P-R 6, Q-R 3, Q-S 7, R-S 5, R-T 8, S-T 2. Find the minimum spanning tree and its weight.
Show worked solution →

Prim's from P: P-Q (4), Q-R (3), R-S (5), S-T (2). (1 mark)

Weight 4+3+5+2=144 + 3 + 5 + 2 = 14. (1 mark)

core3 marks
Workers X, Y and Z and jobs 1, 2 and 3 have costs X: 9, 6, 8; Y: 7, 4, 9; Z: 6, 5, 7. Use the Hungarian algorithm to find the minimum total cost.
Show worked solution →

Rows minus 6, 4, 5: X (3, 0, 2), Y (3, 0, 5), Z (1, 0, 2). Columns 1 and 3 minus 1 and 2: X (2, 0, 0), Y (2, 0, 3), Z (0, 0, 0). (1 mark)

Three lines are needed (row Z, column 2, and row X or column 3), so assign. Y: 2 (only zero); X: 3; Z: 1. (1 mark)

Total 8+4+6=188 + 4 + 6 = 18. (1 mark)

core2 marks
Activities: A (4, none), B (3, A), C (6, A), D (2, B and C). Find the minimum completion time and the float of B.
Show worked solution →

A-C-D =4+6+2=12= 4 + 6 + 2 = 12 is the longest path, so 12 time units. (1 mark)

B: EST 4; D starts at 10, so B's LST is 10−3=710 - 3 = 7; float 7−4=37 - 4 = 3. (1 mark)

exam3 marks
A student says: "The critical path is the shortest path through the project network, because it gives the minimum completion time." Evaluate this claim.
Show worked solution →

The claim is wrong. The minimum completion time is the length of the longest path, because every activity must be completed and the longest chain of dependent activities cannot be shortened by working in parallel. (2 marks)

The critical path is therefore the longest path; shortest paths belong to travel problems, not project scheduling. (1 mark)

exam3 marks
A spanning tree connecting 7 towns has total weight 48 km and a minimum spanning tree has weight 41 km. Roads cost $15 000 per km to build. (a) How many roads are in each tree? (b) How much is saved by building the minimum spanning tree instead?
Show worked solution →

(a) Each spanning tree of 7 vertices has 7−1=67 - 1 = 6 edges. (1 mark)

(b) (48−41)×15 000=105 000(48 - 41) \times 15\,000 = 105\,000, so $105 000 is saved. (2 marks)

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