D2 June 2019 Q2
2. Table 1 shows the cost, in pounds, of transporting one unit of stock from each of three supply points, A, B and C, to each of four demand points, 1, 2, 3 and 4. It also shows the stock held at each supply point and the stock required at each demand point. A minimum cost solution is required.
| 1 | 2 | 3 | 4 | Supply | |
|---|---|---|---|---|---|
| A | 17 | 20 | 23 | 14 | 25 |
| B | 16 | 15 | 19 | 22 | 29 |
| C | 19 | 14 | 11 | 15 | 32 |
| Demand | 28 | 17 | 23 | 18 |
Table 1
Table 2 shows an initial solution given by the north-west corner method.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| A | 25 | |||
| B | 3 | 17 | 9 | |
| C | 14 | 18 |
Table 2
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
giving
| M1 A1 | ||||||||||||||||||||||||||||||||||||||||
| (2) |
Notes
a1M1: A valid route, only one empty square A4 used, thetas balance – some candidates are verifying that A4 is the entering cell (which is fine). For those that start at an incorrect entering cell then the M marks only are available in subsequent parts (unless recovered to the answers given in the scheme)
a1A1: Correct route, up to an improved solution (six numbers no zeros) – if there is a zero in cell B3 then A0 unless corrected in part (a)
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | ||||||||||||||||||||||||||||||||||||||||
giving
Entering cell C2, Exiting cell C4 | M1 A1 | ||||||||||||||||||||||||||||||||||||||||
| (4) |
Notes
b1M1: Finding 7 shadow costs and at least 6 improvement indices. Condone extra zero IIs.
b1A1: Shadow costs and improvement indices CAO (No extra zeros).
b2M1: A valid route, their most negative II chosen, only one empty square used, thetas balance
b2A1: CSO (for part (b)) – so all previous marks in this part must have been awarded – including exiting and entering cells stated correctly (entering is C2 and exiting is C4) – six numbers no zeros
| Scheme | Marks | ||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | ||||||||||||||||||||||||||||||
| Optimal as there are no negative improvement indices | A1 | ||||||||||||||||||||||||||||||
| (3) |
Notes
c1M1: Finding 7 shadow costs and all 6 IIs. Condone extra zero IIs.
c1A1: Shadow costs and 6 IIs CAO
c2A1: CSO (for part (c)) + optimal + reason
| Scheme | Marks |
|---|---|
| Cost = (£) 1206 | B1 |
| (1) | |
| 10 marks |
Notes
d1B1: CAO