D2 June 2016 Q5
5. The table below shows the cost of transporting one unit of stock from each of four supply points, 1, 2, 3 and 4, to each of three demand points, A, B and C. It also shows the stock held at each supply point and the stock required at each demand point. A minimal cost solution is required.
| A | B | C | Supply | |
|---|---|---|---|---|
| 1 | 18 | 23 | 20 | 15 |
| 2 | 22 | 17 | 25 | 36 |
| 3 | 24 | 21 | 19 | 28 |
| 4 | 21 | 22 | 17 | 20 |
| Demand | 40 | 20 | 25 |
After one iteration of the stepping-stone method the table becomes
| A | B | C | D | |
|---|---|---|---|---|
| 1 | 15 | |||
| 2 | 19 | 17 | ||
| 3 | 3 | 25 | ||
| 4 | 6 | 14 |
| Scheme | Marks |
|---|---|
| (total) supply > (total) demand | B1 |
| (1) |
Notes
a1B1: CAO (or to make demand = supply or because demand \(\neq\) supply (oe))
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B1 | ||||||||||||||||||||||||||||||||||||
| (1) |
Notes
b1B1: CAO
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B1 | ||||||||||||||||||||||||||||||||||||
| (1) |
Notes
c1B1: CAO
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
giving
Exiting cell is B3 | M1 A1 | ||||||||||||||||||||||||||||||||||||||||||||||||||
Entering cell is C4 | M1 A1 | ||||||||||||||||||||||||||||||||||||||||||||||||||
giving
Exiting Cell is D4 | M1 A1 | ||||||||||||||||||||||||||||||||||||||||||||||||||
| (6) |
Notes
d1M1: A valid route, only one empty square, D3 used, \(\theta\)’s balance
d1A1: Correct route, up to an improved solution (seven numbers no zeros)
d2M1: Finding 8 shadow costs and 9 improvement indices
d2A1: Shadow costs [Alt: A(0), B(−5), C(−2), D(−21), 1(18), 2(22), 3(21), 4(21)] and II CAO
d3M1: A valid route, their most negative II chosen, only one empty square used, \(\theta\)’s balance
d3A1: CSO (for part d) – so all previous marks in this part must have been awarded - including exiting and entering cells stated correctly
| Scheme | Marks | ||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | ||||||||||||||||||||||||||||||||||||
| Optimal since no negative improvement indices | A1 | ||||||||||||||||||||||||||||||||||||
| (3) | |||||||||||||||||||||||||||||||||||||
| 12 marks |
Notes
e1M1: Finding 8 shadow costs and all 9 improvement indices or at least 1 negative II found
e1A1: CAO for the shadow costs [Alt: A(0), B(−5), C(−4), D(−23), 1(18), 2 (22), 3(23), 4(21)] and the 9 positive IIs
e2A1: CSO (for part e) + reason + optimal