2 Five students attempted five different games, and penalty points were given for any mistakes that they made. The table shows the penalty points incurred by the students.
| Game 1 | Game 2 | Game 3 | Game 4 | Game 5 |
| Ali | 5 | 7 | 3 | 8 | 8 |
| Beth | 8 | 6 | 4 | 8 | 7 |
| Cat | 6 | 1 | 2 | 10 | 3 |
| Di | 4 | 4 | 3 | 10 | 7 |
| Ell | 4 | 6 | 4 | 7 | 9 |
Using the Hungarian algorithm, each of the five students is to be allocated to a different game so that the total number of penalty points is minimised.
- By reducing the rows first and then the columns, show that the new table of values is
| 2 | 4 | 0 | 2 | 3 |
| 4 | 2 | 0 | 1 | 1 |
| 5 | 0 | 1 | \(k\) | 0 |
| 1 | 1 | 0 | 4 | 2 |
| 0 | 2 | 0 | 0 | 3 |
- Show that the zeros in the table in part (a) can be covered with three lines, and use augmentation to produce a table where five lines are required to cover the zeros.
- Hence find the possible ways of allocating the five students to the five games with the minimum total of penalty points.
- Find the minimum possible total of penalty points.
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