2 Five rooms, \(A , B , C , D , E\), in a building need to be connected to a computer network using expensive cabling. Rob wants to find the cheapest way to connect the rooms by finding a minimum spanning tree for the cable lengths. The length of cable, in metres, needed to connect each pair of rooms is given in the table below.
| \multirow{2}{*}{} | Room |
| | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) |
| \multirow{5}{*}{Room} | \(A\) | - | 12 | 30 | 15 | 22 |
| B | 12 | - | 24 | 16 | 30 |
| C | 30 | 24 | - | 20 | 25 |
| D | 15 | 16 | 20 | - | 10 |
| E | 22 | 30 | 25 | 10 | - |
- Apply Prim's algorithm in matrix (table) form, starting at vertex \(A\) and showing all your working. Write down the order in which arcs were added to the tree. Draw the resulting tree and state the length of cable needed.
A sixth room, \(F\), is added to the computer network. The distances from \(F\) to each of the other rooms are \(A F = 32 , B F = 29 , C F = 31 , D F = 35 , E F = 30\).
- Use your answer to part (i) to write down a lower bound for the length of the minimum tour that visits every vertex of the extended network, finishing where it starts. Apply the nearest neighbour method, starting from vertex \(A\), to find an upper bound for the length of this tour.