D1 January 2013 Q5
5.

[The weight of the network is 379]
Figure 5 represents the roads in a highland wildlife conservation park. The vertices represent warden stations. The number on each arc gives the length, in km, of the corresponding road.
During the winter months the park is closed. It is only necessary to ensure road access to the warden stations.
At the end of winter, Ben inspects all the roads before the park re-opens. He needs to travel along each road at least once. He will start and finish at A, and wishes to minimise the length of his route.
If Ben starts and finishes his inspection route at different warden stations, a shorter inspection route is possible.
| Scheme | Marks |
|---|---|
| AC (32) CF (14) DF (12) EF (17); BE (15) FI(18); IJ (10) GJ (9) DH (19) | M1 A1; A1 |
| (3) |
Notes
Accept the weight of each arc to represent the arcs (as each value is unique).
a1M1 First four arcs correctly chosen or first five nodes correctly chosen (A, C, F, D, E, …). Any rejections seen during selection scores M0. Order of nodes may be seen at the top of a matrix.
a1A1 First six arcs correctly chosen or all nodes correctly chosen (A, C, F, D, E, B, I, J, G, H). Order of nodes may be seen at the top of a matrix.
a2A1 CSO (must be considering arcs for this final mark).
Misread in (a): Starting at a node other than A scores M1 only – must have the first four arcs (or five nodes or numbers) correct.
| Starting at | Minimum Arcs required for M1 only | Nodes | Order |
|---|---|---|---|
| B | BE,EF,DF,CF | B, E, F, D, C | (10)15423(8967) |
| C | CF,DF,EF,BE | C, F, D, E, B | (10)51342(8967) |
| D | DF,CF,EF,BE | D, F, C, E, B | (10)53142(8967) |
| E | BE,EF,DF,CF | E, B, F, D, C | (10)25413(8967) |
| F | DF,CF,EF,BE | F, D, C, E, B | (10)53241(8967) |
| G | GJ,IJ,FI,DF | G, J, I, F, D | (10)(86)5(7)41(9)32 |
| H | DH,DF,CF,EF | H, D, F, C, E | (10)(6)4253(9)1(78) |
| I | IJ,GJ,FI,DF | I, J, G, F, D | (10)(86)5(7)43(9)12 |
| J | GJ,IJ,FI,DF | J, G, I, F, D | (10)(86)5(7)42(9)31 |
| Scheme | Marks |
|---|---|
| \(146 \times 80 =\) (£) 11 680 | M1 A1 |
| (2) |
Notes
b1M1 \(80 \times\) their MST weight. Accept a value in the interval \([114, 178] \times 80\) for this mark. If no working is seen then M0 unless answer is correct.
b1A1 CAO (11680 with no working scores both marks).
| Scheme | Marks |
|---|---|
| BF + GH = 32 + 40 = 72 BG + FH = 39 + 25 = 64* BH + FG = 57 + 37 = 94 | M1 A3,2,1,0 |
| Roads BE, EG and FH need repeating | A1ft A1 |
| (6) |
Notes
c1M1 Three distinct pairings of their four odd nodes.
c1A1 Any one row correct including pairing and total.
c2A1 Any two rows correct including pairing and total.
c3A1 All three rows correct including pairing and total.
c4A1ft Their smallest arcs repeated (e.g. accept BEG or BG via E but not just BG). BEG (or e.g. BG via E) could appear in their working.
c5A1 CAO BE, EG and FH. Accept BEG or BG via E (could appear in working) but not just BG.
| Scheme | Marks |
|---|---|
| 379 + 64 = 443 (km) | B1ft |
| (1) |
Notes
d1B1ft correct answer of 443 or 379 + their least out of a choice of at least two totals given in part (c).
| Scheme | Marks |
|---|---|
| Ben should choose to repeat FH (25) since this is the shortest. | M1 |
| He should choose B and G as his start and finish vertices | A1 |
| Route length is 379 + 25 = 404 (km) | A1 |
| (3) | |
| (15 marks) |
Notes
e1M1 FH (or 25) specifically identified as least.
e1A1 B and G identified as the start and finish nodes.
e2A1 404 CAO (condone lack of (or incorrect) units throughout).