AS June 2024 Q2
2. A team of 5 players, A, B, C, D and E, competes in a quiz. Each player must answer one of 5 rounds, P, Q, R, S and T.
Each player must be assigned to exactly one round, and each round must be answered by exactly one player.
Player B cannot answer round Q, player D cannot answer round T, and player E cannot answer round R.
The number of points that each player is expected to earn in each round is shown in the table.
| P | Q | R | S | T | |
|---|---|---|---|---|---|
| A | 32 | 40 | 35 | 41 | 37 |
| B | 38 | – | 40 | 27 | 33 |
| C | 41 | 28 | 37 | 36 | 35 |
| D | 35 | 33 | 38 | 36 | – |
| E | 40 | 38 | – | 39 | 34 |
The team wants to maximise its total expected score.
The Hungarian algorithm is to be used to find the maximum total expected score that can be earned by the 5 players.
| Scheme | Marks | AO |
|---|---|---|
| Subtract each entry from a constant (e.g. 50) to convert from maximisation problem to minimisation | B1 | 1.1b |
| Add a large value (e.g. 100) to cells BQ, DT and ER so that they cannot be selected | B1 | 2.4 |
| (2) |
Notes
B1: Valid statement regarding converting a maximisation problem to a minimisation problem
B1: Explain the need to add a large value to avoid cells being selected – it must be clear which order the changes are applied
| Scheme | Marks | AO | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
e.g.
| B1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Reducing rows
| M1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
3 lines required (Row A, Columns P, R) so augment by 1
| M1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
4 lines required (e.g. Row A, E Column P, R) so augment by 1
4 lines required (e.g. Row A, C, E Column R) so augment by 1
| M1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Allocation A – T, B – R, C – P, D – S, E – Q | A1 | 2.2a | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Maximum Score: 192 | B1 | 2.2a | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| (6) | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| (8 marks) |
Notes
B1: Mark awarded when both steps complete (subtraction and addition of large values) accept one slip in values
M1: Simplifying the initial matrix by reducing rows and then columns (accept one additional slip)
M1: Develop an improved solution – need to see one double covered +e; one uncovered –e; and one single covered unchanged. 3 lines needed to 4 lines needed
M1: Develop an improved solution – need to see one double covered +e; one uncovered –e; and one single covered unchanged. 4 lines needed to 5 lines needed (so getting to the optimal table)
A1: CSO on final table (so must have scored all previous marks in this part ) + deduction of the correct allocation (note there are different fully correct tables)
B1: Maximum score stated
If subtracting from 41
| P | Q | R | S | T | |
|---|---|---|---|---|---|
| A | 9 | 1 | 6 | 0 | 4 |
| B | 3 | 100 | 1 | 14 | 8 |
| C | 0 | 13 | 4 | 5 | 6 |
| D | 6 | 8 | 3 | 5 | 100 |
| E | 1 | 3 | 100 | 2 | 7 |