D2 June 2010 Q2
2. A team of four workers, Harry, Jess, Louis and Saul, are to be assigned to four tasks, 1, 2, 3 and 4. Each worker must be assigned to one task and each task must be done by just one worker.
Jess cannot be assigned to task 4.
The amount, in pounds, that each person would earn while assigned to each task is shown in the table below.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| Harry | 18 | 24 | 22 | 17 |
| Jess | 20 | 25 | 19 | - |
| Louis | 25 | 24 | 27 | 22 |
| Saul | 19 | 26 | 23 | 14 |
| Scheme | Marks |
|---|---|
| Since maximising, subtract all elements from some \(n \geqslant 27\) \(\begin{bmatrix}12 & 6 & 8 & 13 \\ 10 & 5 & 11 & 60 \\ 5 & 6 & 3 & 8 \\ 11 & 4 & 7 & 16\end{bmatrix}\) | 1M1 2M1 |
| Reduce rows \(\begin{bmatrix}6 & 0 & 2 & 7 \\ 5 & 0 & 6 & 55 \\ 2 & 3 & 0 & 5 \\ 7 & 0 & 3 & 12\end{bmatrix}\) then columns \(\begin{bmatrix}4 & 0 & 2 & 2 \\ 3 & 0 & 6 & 50 \\ 0 & 3 & 0 & 0 \\ 5 & 0 & 3 & 7\end{bmatrix}\) | 3M1 A1 |
| \(\begin{bmatrix}2 & 0 & 0 & 0 \\ 1 & 0 & 4 & 48 \\ 0 & 5 & 0 & 0 \\ 3 & 0 & 1 & 5\end{bmatrix}\) | 4M1 A1ft |
| \(\begin{bmatrix}2 & 1 & 0 & 0 \\ 0 & 0 & 3 & 47 \\ 0 & 6 & 0 & 0 \\ 2 & 0 & 0 & 4\end{bmatrix}\) | 5M1 A1 |
| (8) |
Notes
1M1: Subtracting from some \(n \geqslant 27\), condone up to two errors
2M1: Dealing with (Jess, 4) entry.
3M1: Reducing rows then columns
1A1: cao (pick up (J,4) value here)
4M1: Double covered +e; one uncovered – e; and one single covered unchanged. 2 lines needed to 3 lines needed.
2A1ft: ft correct - no errors
5M1: Double covered +e; one uncovered – e; and one single covered unchanged. 3 line to 4 line solution.
3A1: correct - no errors
| Scheme | Marks | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Three optimal allocations:
| M1 | ||||||||||||||||
| Total amount earned by team: £90 | A1 | ||||||||||||||||
| (2) | |||||||||||||||||
| (10 marks) |
Notes
1M1: A complete, correct solution.
1A1: cao
Q2 Special case (Minimises)
| Scheme | Marks |
|---|---|
| \(\begin{bmatrix}18 & 24 & 22 & 17 \\ 20 & 25 & 19 & 60 \\ 25 & 24 & 27 & 22 \\ 19 & 26 & 23 & 14\end{bmatrix}\xrightarrow{\text{row reduction}}\begin{bmatrix}1 & 7 & 5 & 0 \\ 1 & 6 & 0 & 41 \\ 3 & 2 & 5 & 0 \\ 5 & 12 & 9 & 0\end{bmatrix}\) | M0 M1 |
| \(\xrightarrow{\text{column reductions}}\begin{bmatrix}0^* & 5 & 5 & 0 \\ 0 & 4 & 0^* & 41 \\ 2 & 0^* & 5 & 0 \\ 4 & 10 & 9 & 0^*\end{bmatrix}\) | M1 A1 |
| M0 M0 | |
| Solution: Harry – 1 Jess – 3 Louis – 2 Saul – 4 | M1 |
| Total £75 | A1 |
| Maximum 5 marks |