Decision Mathematics 2 June 2015, Question 3 12 marks
Also involves Minimum spanning trees
| A | B | C | D | E | F | G | |
| A | – | x | 41 | 43 | 38 | 21 | 30 |
| B | x | – | 27 | 38 | 19 | 29 | 51 |
| C | 41 | 27 | – | 24 | 37 | 35 | 40 |
| D | 43 | 38 | 24 | – | 44 | 52 | 25 |
| E | 38 | 19 | 37 | 44 | – | 20 | 28 |
| F | 21 | 29 | 35 | 52 | 20 | – | 49 |
| G | 30 | 51 | 40 | 25 | 28 | 49 | – |
The network represented by the table shows the least distances, in km, between seven theatres, A, B, C, D, E, F and G. Jasmine needs to visit each theatre at least once starting and finishing at A. She wishes to minimise the total distance she travels. The least distance between A and B, is x km, where 21 < x < 27
Using Prim’s algorithm, starting at A, obtain a minimum spanning tree for the network. You should list the arcs in the order in which you consider them.
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Use your answer to (a) to determine an initial upper bound for the length of Jasmine’s route.
Use the nearest neighbour algorithm, starting at A, to find a second upper bound for the length of the route.
The nearest neighbour algorithm starting at F gives a route of F – E – B – A – G – D – C – F.
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State which of these two nearest neighbour routes gives the better upper bound. Give a reason for your answer.
Starting by deleting A, and all of its arcs, a lower bound of 159 km for the length of the route is found.
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Find x, making your method clear.
Write down the smallest interval that you can be confident contains the optimal length of Jasmine’s route. Give your answer as an inequality.
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