| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Complete Simplex solution |
| Difficulty | Standard +0.3 This is a standard textbook Simplex algorithm question requiring mechanical application of the method: setting up the initial tableau, performing two pivot operations as directed, and interpreting the final tableau. While it involves multiple steps, each step follows a well-defined algorithmic procedure taught in D2 with no problem-solving insight required, making it slightly easier than average. |
| Spec | 7.07a Simplex tableau: initial setup in standard format7.07b Simplex iterations: pivot choice and row operations7.07c Interpret simplex: values of variables, slack, and objective |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Introduce slack variables \(s_1, s_2\): \(x + 3y + 2z + s_1 = 11\), \(3x + 4y + 2z + s_2 = 21\) | B1 | Correct constraint rows |
| \(P - 3x - 6y - 2z = 0\) | B1 | Correct objective row with P |
| Answer | Marks | Guidance |
|---|---|---|
| BV | \(x\) | \(y\) |
| \(s_1\) | 1 | 3 |
| \(s_2\) | 3 | 4 |
| \(P\) | \(-3\) | \(-6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Pivot column is \(y\)-column; ratios \(11/3\) and \(21/4\); minimum is \(11/3\), so pivot row is \(s_1\) row | M1 | Correct ratios and identification of pivot element (3) |
| New \(s_1\) row \(\div 3\): \(\frac{1}{3}, 1, \frac{2}{3}, \frac{1}{3}, 0, \frac{11}{3}\) | A1 | Correct new pivot row |
| New \(s_2\) row: \(s_2 - 4 \times\) new row 1: \(\frac{5}{3}, 0, \frac{-2}{3}, \frac{-4}{3}, 1, \frac{19}{3}\) | A1 | Correct tableau after iteration |
| New \(P\) row: \(P + 6\times\) new row 1: \(-1, 0, 2, 2, 0, 22\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Pivot column is \(x\)-column (most negative = \(-1\)); ratios \(\frac{11/3}{1/3}=11\) and \(\frac{19/3}{5/3}=\frac{19}{5}\); pivot element is \(\frac{5}{3}\) | M1 | Correct identification of pivot |
| New \(s_2\) row \(\times \frac{3}{5}\): \(1, 0, \frac{-2}{5}, \frac{-4}{5}, \frac{3}{5}, \frac{19}{5}\) | A1 | Correct new pivot row |
| Updated \(y\) row and \(P\) row correct | A1 | Full correct tableau |
| Answer | Marks | Guidance |
|---|---|---|
| BV | \(x\) | \(y\) |
| \(y\) | 0 | 1 |
| \(x\) | 1 | 0 |
| \(P\) | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum value of \(P = \frac{129}{5} = 25.8\) | B1 | Correct value stated |
| \(x = \frac{19}{5}\), \(y = \frac{16}{5}\), \(z = 0\) | B1 | Correct values of basic variables |
| Slack variables: \(s_1 = 0\), \(s_2 = 0\) (both constraints are tight/binding) | B1 | Correct interpretation of slack variables |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Introduce slack variables $s_1, s_2$: $x + 3y + 2z + s_1 = 11$, $3x + 4y + 2z + s_2 = 21$ | B1 | Correct constraint rows |
| $P - 3x - 6y - 2z = 0$ | B1 | Correct objective row with P |
**Initial Tableau:**
| BV | $x$ | $y$ | $z$ | $s_1$ | $s_2$ | RHS |
|----|-----|-----|-----|-------|-------|-----|
| $s_1$ | 1 | 3 | 2 | 1 | 0 | 11 |
| $s_2$ | 3 | 4 | 2 | 0 | 1 | 21 |
| $P$ | $-3$ | $-6$ | $-2$ | 0 | 0 | 0 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pivot column is $y$-column; ratios $11/3$ and $21/4$; minimum is $11/3$, so pivot row is $s_1$ row | M1 | Correct ratios and identification of pivot element (3) |
| New $s_1$ row $\div 3$: $\frac{1}{3}, 1, \frac{2}{3}, \frac{1}{3}, 0, \frac{11}{3}$ | A1 | Correct new pivot row |
| New $s_2$ row: $s_2 - 4 \times$ new row 1: $\frac{5}{3}, 0, \frac{-2}{3}, \frac{-4}{3}, 1, \frac{19}{3}$ | A1 | Correct tableau after iteration |
| New $P$ row: $P + 6\times$ new row 1: $-1, 0, 2, 2, 0, 22$ | | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pivot column is $x$-column (most negative = $-1$); ratios $\frac{11/3}{1/3}=11$ and $\frac{19/3}{5/3}=\frac{19}{5}$; pivot element is $\frac{5}{3}$ | M1 | Correct identification of pivot |
| New $s_2$ row $\times \frac{3}{5}$: $1, 0, \frac{-2}{5}, \frac{-4}{5}, \frac{3}{5}, \frac{19}{5}$ | A1 | Correct new pivot row |
| Updated $y$ row and $P$ row correct | A1 | Full correct tableau |
**Resulting tableau:**
| BV | $x$ | $y$ | $z$ | $s_1$ | $s_2$ | RHS |
|----|-----|-----|-----|-------|-------|-----|
| $y$ | 0 | 1 | $\frac{4}{5}$ | $\frac{3}{5}$ | $\frac{-1}{5}$ | $\frac{16}{5}$ |
| $x$ | 1 | 0 | $\frac{-2}{5}$ | $\frac{-4}{5}$ | $\frac{3}{5}$ | $\frac{19}{5}$ |
| $P$ | 0 | 0 | $\frac{8}{5}$ | $\frac{6}{5}$ | $\frac{1}{5}$ | $\frac{129}{5}$ |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum value of $P = \frac{129}{5} = 25.8$ | B1 | Correct value stated |
| $x = \frac{19}{5}$, $y = \frac{16}{5}$, $z = 0$ | B1 | Correct values of basic variables |
| Slack variables: $s_1 = 0$, $s_2 = 0$ (both constraints are tight/binding) | B1 | Correct interpretation of slack variables |
---4
\begin{enumerate}[label=(\alph*)]
\item Display the following linear programming problem in a Simplex tableau.
$$\begin{array} { l c }
\text { Maximise } & P = 3 x + 6 y + 2 z \\
\text { subject to } & x + 3 y + 2 z \leqslant 11 \\
& 3 x + 4 y + 2 z \leqslant 21 \\
\text { and } & x \geqslant 0 , y \geqslant 0 , z \geqslant 0 .
\end{array}$$
\item The first pivot to be chosen is from the $y$-column.
Perform one iteration of the Simplex method.
\item Perform one further iteration.
\item Interpret the tableau obtained in part (c) and state the values of your slack variables.
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2014 Q4 [11]}}