# Question 2:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Hungarian algorithm **minimises** | E1 | |
| $15 - x$ gives measure of criteria NOT met, which need minimising in order to maximise scores | E1 | Idea of high becoming low, etc. **Total: 2** |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Array giving $15 - x$: $\begin{pmatrix}2&4&6&5&2\\0&3&3&4&3\\3&5&7&1&1\\4&3&2&1&5\\3&1&1&2&1\end{pmatrix}$ | B1 | |
| Reduce rows (or columns then rows): $\begin{pmatrix}0&2&4&3&0\\0&3&3&4&3\\2&4&6&0&0\\3&2&1&0&4\\2&0&0&1&0\end{pmatrix}$ | M1 | Reduce rows (or columns then rows) |
| Reduced array correct | A1 | |
| Zeros can be covered with only **4 lines**, so adjustment needed | E1 | Augmented array shown |
| Reduction by subtracting 1 from each uncovered element | M1 | |
| And adding 1 to each element at intersection of two lines | A1 | |
| Matching on particular zeros | M1 | |
| $\text{Alex} \leftrightarrow 5$, $\text{Don} \leftrightarrow 3$, $\text{Bill} \leftrightarrow 1$, $\text{Ed} \leftrightarrow 2$, $\text{Cath} \leftrightarrow 4$ | A1 | If adjustment not done correctly and matching made, award B1 for 3 correct and B1 for rest correct. **Total: 8** |

## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Deleting row 2 and column 4 either in final matrix or reworking | M1 | |
| Final solution: $A \leftrightarrow 1$, $C \leftrightarrow 5$ | A1 | |
| $D \leftrightarrow 3$, $E \leftrightarrow 2$ | A1 | If no method award B2 if matching is all correct. **Total: 3** |

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