A2 October 2020 Q7
7. A maximisation linear programming problem in \(x\), \(y\) and \(z\) is to be solved using the two-stage simplex method.
The partially completed initial tableau is shown below.
| Basic variable | \(x\) | \(y\) | \(z\) | \(s_1\) | \(s_2\) | \(s_3\) | \(a_1\) | \(a_2\) | Value |
|---|---|---|---|---|---|---|---|---|---|
| \(s_1\) | 1 | 2 | 3 | 1 | 0 | 0 | 0 | 0 | 45 |
| \(a_1\) | 3 | 2 | 0 | 0 | \(-1\) | 0 | 1 | 0 | 9 |
| \(a_2\) | \(-1\) | 0 | 4 | 0 | 0 | \(-1\) | 0 | 1 | 4 |
| \(P\) | \(-2\) | \(-1\) | \(-3\) | 0 | 0 | 0 | 0 | 0 | 0 |
| \(A\) |
[Table 1 in the answer book is a copy of the partially completed initial tableau above.]
The following tableau is obtained after two iterations of the first stage of the two-stage simplex method.
| Basic variable | \(x\) | \(y\) | \(z\) | \(s_1\) | \(s_2\) | \(s_3\) | \(a_1\) | \(a_2\) | Value |
|---|---|---|---|---|---|---|---|---|---|
| \(s_1\) | 0 | \(\frac{5}{6}\) | 0 | 1 | \(\frac{7}{12}\) | \(\frac{3}{4}\) | \(-\frac{7}{12}\) | \(-\frac{3}{4}\) | \(\frac{147}{4}\) |
| \(x\) | 1 | \(\frac{2}{3}\) | 0 | 0 | \(-\frac{1}{3}\) | 0 | \(\frac{1}{3}\) | 0 | 3 |
| \(z\) | 0 | \(\frac{1}{6}\) | 1 | 0 | \(-\frac{1}{12}\) | \(-\frac{1}{4}\) | \(\frac{1}{12}\) | \(\frac{1}{4}\) | \(\frac{7}{4}\) |
| \(P\) | 0 | \(\frac{5}{6}\) | 0 | 0 | \(-\frac{11}{12}\) | \(-\frac{3}{4}\) | \(\frac{11}{12}\) | \(\frac{3}{4}\) | \(\frac{45}{4}\) |
| \(A\) | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
| Scheme | Marks | AO |
|---|---|---|
| Maximise \((P =)\ 2x + y + 3z\) | B1 | 3.4 |
| \(x + 2y + 3z \leqslant 45\) | B1 | 3.4 |
| \(3x + 2y \geqslant 9\) | B1 | 3.4 |
| \(-x + 4z \geqslant 4\) | B1 | 1.1b |
| (4) |
Notes
(a) B1: CAO – including maximise (or max)
B1: CAO (oe)
B1: CAO (oe)
B1: CAO (oe)
| Scheme | Marks | AO | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(A = -(a_1 + a_2) \Rightarrow -(9 - 3x - 2y + s_2 + 4 + x - 4z + s_3)\) | M1 | 2.1 | ||||||||||
\(A - 2x - 2y - 4z + s_2 + s_3 = -13\) therefore bottom row of the table is
| A1 | 2.2a | ||||||||||
| (2) |
Notes
(b) M1: Setting up the new objective and substituting for \(a_1\) and \(a_2\)
A1: Correct values substituted into Table 1
| Scheme | Marks | AO |
|---|---|---|
| (i) In the given tableau the value of the objective \(A\) is equal to zero indicating that a basic feasible solution has been found | B1 | 2.4 |
| (ii) \(x = 3,\ y = 0,\ z = \frac{7}{4},\ s_1 = \frac{147}{4},\ s_2 = s_3 = 0\) | B1 B1 | 3.4 1.1b |
| (3) |
Notes
(c) B1: CAO – mention that \(A = 0\)
B1: At least three values stated correctly
B1: All six values correct (ignore values stated for \(a_1, a_2\) and \(P\))
| Scheme | Marks | AO | |||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 M1 A1 A1 | 2.1 1.1b 1.1b 1.1b 1.1b | |||||||||||||||||||||||||||||||||||||||||||||
| (5) |
Notes
(d) M1: Correct pivot located, attempt to divide row
A1: Pivot row correct including change of b.v.
M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(y\), \(s_1, s_3\) or value) correct following through their choice of pivot from column \(s_2\) or \(s_3\)
A1: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(y\), \(s_1, s_3\) or value) correct
A1: CAO all values and row operations correctly stated – allow if row operations given in terms of old row 1 – ignore b.v. column for this mark
| Scheme | Marks | AO |
|---|---|---|
| (i) Yes, an optimal solution has been found as there are no negative values in the objective (\(P\)) row | B1 | 2.4 |
| (ii) \(P = 69\) | B1ft | 3.4 |
| (iii) \(s_2 = 63,\ x = 24,\ z = 7\) | B1ft | 3.4 |
| (3) | ||
| (17 marks) |
Notes
(e)(i) B1: Correct reasoning of why solution is optimal or using \(P = 69 - \frac{15}{7}y - \frac{11}{7}s_1 - \frac{3}{7}s_3\) and mentioning increasing \(y, s_1, s_3\) would decrease \(P\) (oe)
(e)(ii) B1ft: their value of \(P\) – dependent on both M marks in (d)
(e)(iii) B1ft: their values of the basic variables only – dependent on both M marks in (d)