D2 June 2013 (R) Q5
5. A three-variable linear programming problem in \(x\), \(y\) and \(z\) is to be solved. The objective is to maximise the profit, \(P\).
The following tableau is obtained.
| Basic variable | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | Value |
|---|---|---|---|---|---|---|---|
| \(r\) | \(\frac{1}{2}\) | \(-\frac{1}{2}\) | 0 | 1 | 0 | \(-\frac{1}{2}\) | 10 |
| \(s\) | \(1\frac{1}{2}\) | \(2\frac{1}{2}\) | 0 | 0 | 1 | \(-\frac{1}{2}\) | 5 |
| \(z\) | \(\frac{1}{2}\) | \(\frac{1}{2}\) | 1 | 0 | 0 | \(\frac{1}{2}\) | 5 |
| \(P\) | –5 | –10 | 0 | 0 | 0 | 20 | 220 |
| Scheme | Marks | |||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 M1 A1 A1 | |||||||||||||||||||||||||||||||||||||||||||||
| (5) |
Notes
a1M1: correct pivot located, attempt to divide row. If choosing negative pivot M0M0.
a1A1: pivot row correct including change of b.v.
a2M1: (ft) One row (excluding the pivot row) correct or one column either the value, \(x\), \(s\) or \(t\) column correct.
a2A1ft: Correct row operations used at least once. One column either the value, \(x\), \(s\) or \(t\) column correct on the ft.
a3A1: CAO.
| Scheme | Marks |
|---|---|
| \(P + x + 4s + 18t = 240\) | B1 |
| (1) |
Notes
b1B1: CAO
| Scheme | Marks |
|---|---|
| \(P = 240 - x - 4s - 18t\) and at present \(x\), \(s\) and \(t\) are zero. If we increase any of these the profit will decrease. | B2, 1, 0 |
| (2) | |
| 8 marks |
Notes
c1B1: Using their profit equation to make a pertinent statement. Maybe muddled, if bod give this mark only. No ‘negatives’ in their profit equation.
c2B1: Good explanation – dependent on the correct equation being stated in (b).