A2 June 2022 Q4
4.

Figure 1 shows a capacitated, directed network of pipes. The uncircled number on each arc represents the capacity of the corresponding pipe. The numbers in circles represent an initial flow.
You must state your route. (1)
| Scheme | Marks | AO |
|---|---|---|
| AE, BE, BF, BG, DB, DT, SB | B1 | 1.1b |
| (1) |
Notes
B1: CAO
| Scheme | Marks | AO |
|---|---|---|
| 95 | B1 | 1.1b |
| (1) |
Notes
B1: CAO
| Scheme | Marks | AO |
|---|---|---|
| The maximum feasible flow into F is 22 (from BF and DF) but the maximum feasible flow out of F is 24 so therefore FT cannot be full to capacity | B1 | 2.4 |
| (1) |
Notes
B1: Correct reasoning – argument must be numerical in nature (e.g. as a minimum comparison of 22 with 24)
| Scheme | Marks | AO |
|---|---|---|
| \(C_1\,(= 33 + 41 + 30) = 104\) \(C_2\,(= 53 + 30 + 14 + 0 + 17) = 114\) | B1 B1 | 1.1b 1.1b |
| (2) |
Notes
B1: CAO for \(C_1\)
B1: CAO for \(C_2\)
| Scheme | Marks | AO |
|---|---|---|
| SABDFT | B1 | 1.1b |
| (1) |
Notes
B1: CAO
| Scheme | Marks | AO |
|---|---|---|
| Use of max-flow min-cut theorem Identification of cut through AE, BE, BG, BF, DF and DT, Value of cut = 98, Value of flow = 98 Therefore it follows that flow is maximal | M1 A1 A1 | 2.1 3.1a 2.2a |
| (3) | ||
| (9 marks) |
Notes
M1: Construct argument based on max-flow min-cut theorem (e.g. attempt to find a cut through saturated arcs – must contain source on one side and sink on the other). Allow a cut drawn on the diagram (need not be the correct one)
A1: Use appropriate process of finding a minimum cut – AE, BE, BG, BF, DF and DT plus value correct and value of flow through the network stated correctly (98)
A1: Correct deduction that the flow is maximal – must use all four words ‘maximum’, ‘flow’, ‘minimum’ and ‘cut’ (allow abbreviations for maximum and minimum) dependent on previous A1.