D2 June 2011 Q1
1.

The network in Figure 1 shows the distances, in km, between six towns, A, B, C, D, E and F. Mabintou needs to visit each town. She will start and finish at A and wishes to minimise the total distance travelled.
(a) By inspection, complete the two copies of the table of least distances in your answer book. (3)
(b) Starting at A, use the nearest neighbour algorithm to find an upper bound for the length of Mabintou’s route. Write down the route which gives this upper bound. (3)
(c) Starting by deleting A, and all of its arcs, find a lower bound for the route length. (4)
| Scheme | Marks | |||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B3, 2, 1, 0 | |||||||||||||||||||||||||||||||||||||||||||||||||
| (3) |
Notes
The entries to be found are shown in bold.
1B1 One double entry correct
2B1 Two double entries correct
3B1 Three double entries correct
| Scheme | Marks |
|---|---|
| A C B E F D A 11 8 17 17 31 23 = 107 | M1 A1 A1 |
| (3) |
Notes
M1 NN route, each letter appearing once, condone lack of return vertex
1A1 CAO
2A1 CAO
| Scheme | Marks |
|---|---|
Delete A![]() | M1 |
| RMST weight = 61 | A1 |
| Lower bound = 61 + 11 + 19 = 91 km | M1 A1 |
| (4) | |
| (10 marks) |
Notes
1M1 Finding my RMST – accept 61 for both marks
1A1 Either 8 + 9 + 17 + 27 or 61 seen
2M1 Adding on two least arcs, accept 11 and 19 or AC and AB
2A1 91 CAO
