D2 June 2016 Q2
2.

Figure 1 shows a capacitated, directed network of pipes. The number on each arc represents the capacity of the corresponding pipe. The numbers in circles represent an initial flow.
(a) List the saturated arcs. (2)
(b) State the value of the initial flow. (1)
(c) State the capacities of the cuts \(C_1\) and \(C_2\) (2)
(d) By inspection, find a flow-augmenting route to increase the flow by three units. You must state your route. (1)
(e) Prove that the new flow is maximal. (2)
| Scheme | Marks |
|---|---|
| Saturated arcs: SB, SC, AE, DT, FT | M1 A1 |
| (2) |
Notes
a1M1: All correct – accept one omission and/or one extra arc
a1A1: CAO
| Scheme | Marks |
|---|---|
| 59 | B1 |
| (1) |
Notes
b1B1: CAO (59)
| Scheme | Marks |
|---|---|
| \(C_1 = 72\), \(C_2 = 86\) | B1 B1 |
| (2) |
Notes
c1B1: CAO (72)
c2B1: CAO (86)
| Scheme | Marks |
|---|---|
| SABCFET | B1 |
| (1) |
Notes
d1B1: CAO (accept, SA, AB, BC, CF, FE, ET)
| Scheme | Marks |
|---|---|
| The cut through DT, AE and CF (or DT, AE, BC and SC) has a value 62 | M1 |
| Value of the flow is 62, so by max flow – min cut theorem, flow is maximal | A1 |
| (2) | |
| 8 marks |
Notes
e1M1: The arcs of the correct cut stated or a correct cut drawn on their diagram in the answer book – please check carefully for this
e1A1: Must have stated the (maximum) flow as 62 and a conclusion based on the max flow min cut theorem - they must use all four words, ‘max’, ‘flow’, ‘min’, ‘cut’ in their conclusion