D1 June 2013 (R) Q2
2.
| A | B | C | D | E | F | |
|---|---|---|---|---|---|---|
| A | – | 85 | 110 | 160 | 225 | 195 |
| B | 85 | – | 100 | 135 | 180 | 150 |
| C | 110 | 100 | – | 215 | 200 | 165 |
| D | 160 | 135 | 215 | – | 235 | 215 |
| E | 225 | 180 | 200 | 235 | – | 140 |
| F | 195 | 150 | 165 | 215 | 140 | – |
The table shows the average journey time, in minutes, between six towns, A, B, C, D, E and F.
Kruskal’s algorithm may also be used to find a minimum spanning tree.
| Scheme | Marks |
|---|---|
| AB(85), BC(100), BD(135); BF(150), EF(140). | M1 A1; A1 |
| (3) |
Notes
a1M1 Prim’s – first three arcs correctly chosen or first four nodes correctly chosen, in order.{A, B, C, D,….}. Any rejections seen during selection is M0. Order of nodes may be seen across the top of the matrix {1, 2, 3, 4, –, –}
a1A1 First four arcs correctly chosen or all six nodes correctly chosen {A, B, C, D, F, E}. Order of nodes may be seen across the top of the matrix {1, 2, 3, 4, 6, 5}
a2A1 CSO (must be considering arcs for this final mark).
Misread: Starting at a node other than A scores M1 only – must have the first three arcs (or four nodes or numbers) correct.
| Starting at | Minimum arcs required for M1 | Nodes | order |
|---|---|---|---|
| A | AB BC BD | ABCD(FE) | 1234(65) |
| B | AB BC BD | BACD(FE) | 2134(65) |
| C | BC AB BD | CBAD(FE) | 3214(65) |
| D | BD AB BC | DBAC(FE) | 3241(65) |
| E | EF BF AB | EFBA(CD) | 43(56)12 |
| F | EF BF AB | FEBA(CD) | 43(56)21 |
| Scheme | Marks |
|---|---|
![]() | B1 |
| (1) |
Notes
b1B1 CAO (weights on arcs not required)
| Scheme | Marks |
|---|---|
| 610 (minutes) | B1 |
| (1) |
Notes
c1B1 CAO (condone lack of/incorrect units)
| Scheme | Marks |
|---|---|
E.g. (any three)
| B1 B1 B1 |
| (3) | |
| (8 marks) |
Notes
d1B1 One correct statement.
d2B1 A second correct statement.
d3B1 A third correct statement.
In part (d) all technical language must be correct (so do not condone point for vertex/node etc.)
