Network Flow Topic Test
Network flow · MST-12-S2-06
Maths Standard Year 12 · All 3 lessons · MC checkpoint plus separate short-answer practice
L1, Terminology and Directed Diagrams
L2, Flow Capacity and Saturated Edges
L3, Max-Flow Min-Cut and Meeting Demand
25 MC
8 SA
~55 min
0/25
MC Checkpoint
Answer questions to see your score.
Recommended next step after MC checkpoint
Complete the 25 multiple choice questions to unlock a sharper next move. The short-answer section below is separate practice.
Part A, Multiple Choice (1 mark each, 25 marks total)
A has flow leaving the network
B has flow arriving and none leaving
C lies on the minimum cut
D has the largest capacity of any vertex
A, has flow leaving the network. The source is where flow begins.
A sends flow out into the network
B has equal inflow and outflow
C receives flow from the network
D can be any intermediate vertex
C, receives flow from the network. The sink is the destination.
A a dashed line
B an arrow
C a double line
D a line with no label
B, an arrow. Directed edges use arrows.
A the actual flow currently on that edge
B the physical length of that edge
C the number of paths using that edge
D the maximum flow allowed on that edge
D, the maximum flow allowed on that edge. Capacity is an upper limit.
A visit every vertex in the network
B use only the largest-capacity edges
C follow the direction of arrows
D avoid every saturated edge
C, follow the direction of arrows. Valid directed paths follow arrows.
A saturated
B unused
C over capacity
D certain to lie on the minimum cut
A, saturated. Flow equals capacity.
A $8$
B $12$
C $20$
D $4$
D, $4$. $12 - 8 = 4$.
B, $5$. The bottleneck is the smallest capacity.
A reduce remaining capacities on used edges
B increase the remaining capacities on used edges
C delete every vertex on that path
D reset all other paths to zero flow
A, reduce remaining capacities on used edges. Used capacity is no longer available.
A $6$
B $8$
C $14$
D $48$
C, $14$. $6 + 8 = 14$.
A the number of vertices in the network
B the capacity of the largest single edge
C the number of paths from source to sink
D any cut capacity separating source and sink
D, any cut capacity separating source and sink. Every cut is an upper bound.
A maximum flow equals the largest cut capacity
B maximum flow equals minimum cut capacity
C minimum flow equals the minimum cut capacity
D maximum flow equals the sum of all cut capacities
B, maximum flow equals minimum cut capacity. The theorem links max flow to min cut.
A $6$
B $7$
C $17$
D $11$
C, $17$. Add crossing capacities.
A $18$
B $9$
C $19$
D $36$
A, $18$. A cut gives an upper bound.
A The maximum flow must be more than 18
B The maximum flow is 18
C The cut cannot be a minimum cut
D Nothing, the two numbers are unrelated
B, The maximum flow is 18. The flow reaches the upper bound.
A the network has at least 20 edges
B every edge has capacity at least 20
C the minimum cut is at most 20
D maximum flow is at least 20
D, maximum flow is at least 20. Capacity must meet or exceed demand.
A $S \to A$ labelled 10
B $A \to S$ labelled 10
C an undirected edge $S$–$A$ labelled 10
D $S \to A$ labelled with its flow instead of 10
A, $S \to A$ labelled 10. The arrow follows the row direction.
A The source
B Every intermediate vertex
C The sink
D The vertex with the largest capacity
C, The sink. The sink is the final destination.
A inflow exceeds outflow
B inflow equals outflow
C outflow exceeds inflow
D inflow and outflow both equal the capacity
B, inflow equals outflow. Flow conservation balances intermediate vertices.
A No, the inflow is too large
B No, the outflow is too large
C Cannot tell without the edge capacities
D Yes
D, Yes. Outgoing total is $11$.
A possible, because 25 is close to 23
B possible if an extra path is drawn without changing any capacity
C impossible, because the maximum flow is only 23
D possible only if the flow is split evenly between routes
C, impossible. Max flow cannot exceed 23.
A from the source side to the sink side
B from the sink side to the source side
C crossing the cut in either direction
D that are already saturated
A, from the source side to the sink side. Directed cut capacity counts forward crossing edges.
A $15$
B $11$
C $35$
D $9$
D, $9$. The smallest capacity controls the path.
A is already the maximum
B is not maximum yet
C must be reduced to 9
D has exceeded the minimum cut
B, is not maximum yet. An augmenting path means more flow can be added.
A putting a list of numbers into order
B finding the shortest single route between two towns
C moving limited resources through connected routes
D scheduling tasks that must follow one another
C, moving limited resources through connected routes. They model capacity-limited movement.
Part B, Short Answer (separate practice)
(a) $S \to A$, $S \to B$, $A \to T$, $B \to T$.
(b) Source $S$, sink $T$.
(a) Bottleneck is 9.
(b) Remaining capacities are 3 and 0.
(a) Both totals are 12.
(b) Yes, flow is conserved.
(a) Cut capacity is 21.
(b) Maximum flow is at most 21.
(a) Maximum flow is 16.
(b) Flow equals the cut upper bound.
(a) No.
(b) At least 2 units.
(a) Total flow is 19.
(b) The paths must not compete for the same saturated edge.
(a) A path is a directed route from source to sink.
(b) A cut separates source from sink and limits flow.
Network Flow Complete
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