A2 June 2022 Q1
1. Four workers, A, B, C and D, are to be assigned to four tasks, 1, 2, 3 and 4. Each task must be assigned to just one worker and each worker must do only one task.
The cost of assigning each worker to each task is shown in the table below.
The total cost is to be minimised.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| A | 32 | 45 | 34 | 48 |
| B | 37 | 39 | 50 | 46 |
| C | 46 | 44 | 40 | 42 |
| D | 43 | 45 | 48 | 52 |
| Scheme | Marks | AO |
|---|---|---|
| Reduce rows \(\begin{bmatrix} 0 & 13 & 2 & 16 \\ 0 & 2 & 13 & 9 \\ 6 & 4 & 0 & 2 \\ 0 & 2 & 5 & 9 \end{bmatrix}\) and then columns \(\begin{bmatrix} 0 & 11 & 2 & 14 \\ 0 & 0 & 13 & 7 \\ 6 & 2 & 0 & 0 \\ 0 & 0 & 5 & 7 \end{bmatrix}\) | M1 A1 | 2.1 1.1b |
| Followed by \(\begin{bmatrix} 0 & 11 & 0 & 12 \\ 0 & 0 & 11 & 5 \\ 8 & 4 & 0 & 0 \\ 0 & 0 & 3 & 5 \end{bmatrix}\) | M1 A1ft | 2.1 1.1b |
| A – 3, B – 1, C – 4, D – 2 or A – 3, B – 2, C – 4, D – 1 | A1ft | 2.2a |
| (5) |
Notes
M1: simplifying the initial matrix by reducing rows and then columns. (Allow up to 2 independent slips).
A1: CAO
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 (lines may be implied). If lines are drawn they must be correct.
A1ft: CAO following on from row and column reduction final table. (f/t from previous table with no further slips).
A1ft: correct allocation ft their optimal table (both previous M marks must have been awarded in (a)) (Must be fully written. Do not accept just indicated on zeros on final matrix).
| Scheme | Marks | AO |
|---|---|---|
| 158 | B1 | 1.1b |
| (1) | ||
| (6 marks) |
Notes
B1: CAO – solution of original problem