| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| 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 iterations with specified pivot column, and recognizing optimality. While it involves multiple steps (7 marks total), it requires no problem-solving insight—just careful execution of a learned algorithm, 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 | Mark | Guidance |
| Introducing slack variables | M1 | |
| Initial tableau with \(P, x, y, z, r, s\) columns: Row 1: \(1, -3, -2, -4, 0, 0, 0\); Row 2: \(0, 1, 4, 2, 1, 0, 8\); Row 3: \(0, 2, 7, 3, 0, 1, 21\) | A2 | \(-1\) EE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Choosing correct pivot in \(z\)-column | M1 | and perhaps dividing by 2 |
| Row operations giving: Row 1: \(1, -1, 6, 0, 2, 0, 16\); Row 2: \(0, \frac{1}{2}, 2, 1, \frac{1}{2}, 0, 4\) | M1 | row operations |
| Row 3: \(0, \frac{1}{2}, 1, 0, -\frac{3}{2}, 1, 9\) | A1 | correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need to use \(x\)-column for pivot | M1 | |
| Choosing correct pivot | A1 | |
| Top row: \(1, 0, 10, 2, 3, 0, 24\) | M1, A1 | row operations, top row |
| Row 2: \(0, 1, 4, 2, 1, 0, 8\) | A1 | |
| Row 3: \(0, 0, -1, -1, -2, 1, 5\) | A1 | 5 marks total, third row |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Yes optimal | \(\text{B1}\checkmark\) | |
| No negative values in top row | E1 | 2 marks total |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Introducing slack variables | M1 | |
| Initial tableau with $P, x, y, z, r, s$ columns: Row 1: $1, -3, -2, -4, 0, 0, 0$; Row 2: $0, 1, 4, 2, 1, 0, 8$; Row 3: $0, 2, 7, 3, 0, 1, 21$ | A2 | $-1$ EE |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Choosing correct pivot in $z$-column | M1 | and perhaps dividing by 2 |
| Row operations giving: Row 1: $1, -1, 6, 0, 2, 0, 16$; Row 2: $0, \frac{1}{2}, 2, 1, \frac{1}{2}, 0, 4$ | M1 | row operations |
| Row 3: $0, \frac{1}{2}, 1, 0, -\frac{3}{2}, 1, 9$ | A1 | correct |
### Part (c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Need to use $x$-column for pivot | M1 | |
| Choosing correct pivot | A1 | |
| Top row: $1, 0, 10, 2, 3, 0, 24$ | M1, A1 | row operations, top row |
| Row 2: $0, 1, 4, 2, 1, 0, 8$ | A1 | |
| Row 3: $0, 0, -1, -1, -2, 1, 5$ | A1 | 5 marks total, third row |
### Part (c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Yes optimal | $\text{B1}\checkmark$ | |
| No negative values in top row | E1 | 2 marks total |
---5
\begin{enumerate}[label=(\alph*)]
\item Display the following linear programming problem in a Simplex tableau.
$$\begin{array} { l c }
\text { Maximise } & P = 3 x + 2 y + 4 z \\
\text { subject to } & x + 4 y + 2 z \leqslant 8 \\
& 2 x + 7 y + 3 z \leqslant 21 \\
& x \geqslant 0 , y \geqslant 0 , z \geqslant 0
\end{array}$$
\item Use the Simplex method to perform one iteration of your tableau for part (a), choosing a value in the $z$-column as pivot.
\item \begin{enumerate}[label=(\roman*)]
\item Perform one further iteration.
\item State whether or not this is the optimal solution, and give a reason for your answer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D2 2006 Q5 [13]}}