| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Effect of parameter changes |
| Difficulty | Standard +0.8 This is a Further Maths Decision Mathematics question requiring understanding of the Simplex algorithm with a parameter k, performing iterations, and interpreting optimality conditions for different parameter values. While mechanical in execution, it requires solid understanding of pivot selection, tableau updates, and optimality criteria—more demanding than standard A-level but routine for D2 students. |
| 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 |
| \(\boldsymbol { P }\) | \(\boldsymbol { x }\) | \(\boldsymbol { y }\) | \(\boldsymbol { Z }\) | \(s\) | \(\boldsymbol { t }\) | value |
| 1 | \(- k\) | -6 | -5 | 0 | 0 | 0 |
| 0 | ||||||
| 0 |
| \(\boldsymbol { P }\) | \(\boldsymbol { x }\) | \(\boldsymbol { y }\) | \(\boldsymbol { Z }\) | \(\boldsymbol { s }\) | \(\boldsymbol { t }\) | value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Constraint 1 row: \(0, 2, 1, 4, 1, 0, 11\) | B1 | |
| Constraint 2 row: \(0, 1, 3, 6, 0, 1, 18\) | B1 | Both rows correct |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Pivot on \(y\)-column: smallest ratio \(= \min(11/1, 18/3) = 6\), so row 2 is pivot row | M1 | Correct pivot chosen |
| New row 2 = old row 2 \(\div 3\) | M1 | |
| New row 1 = old row 1 \(- 1\times\) new row 2 | M1 | |
| New objective row updated correctly | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(k=1\): all entries in objective row \(\geq 0\) (no negatives) | M1 | Correct reasoning |
| Maximum value of \(P = 36\) | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(k=3\): \(x\)-column still negative in objective row, so further pivot needed | M1 | Identify correct pivot column |
| Perform correct pivot operation | M1 A1 | |
| New tableau correct | A1 | |
| Interpretation: state values of \(x, y, z\) and \(P\) | B1 B1 | |
| Total: | 6 |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Constraint 1 row: $0, 2, 1, 4, 1, 0, 11$ | B1 | |
| Constraint 2 row: $0, 1, 3, 6, 0, 1, 18$ | B1 | Both rows correct |
| Total: | **2** | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pivot on $y$-column: smallest ratio $= \min(11/1, 18/3) = 6$, so row 2 is pivot row | M1 | Correct pivot chosen |
| New row 2 = old row 2 $\div 3$ | M1 | |
| New row 1 = old row 1 $- 1\times$ new row 2 | M1 | |
| New objective row updated correctly | A1 | |
| Total: | **4** | |
## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $k=1$: all entries in objective row $\geq 0$ (no negatives) | M1 | Correct reasoning |
| Maximum value of $P = 36$ | A1 | |
| Total: | **2** | |
## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $k=3$: $x$-column still negative in objective row, so further pivot needed | M1 | Identify correct pivot column |
| Perform correct pivot operation | M1 A1 | |
| New tableau correct | A1 | |
| Interpretation: state values of $x, y, z$ and $P$ | B1 B1 | |
| Total: | **6** | |3
\begin{enumerate}[label=(\alph*)]
\item Given that $k$ is a constant, complete the Simplex tableau below for the following linear programming problem.
Maximise
$$P = k x + 6 y + 5 z$$
subject to
$$\begin{gathered}
2 x + y + 4 z \leqslant 11 \\
x + 3 y + 6 z \leqslant 18 \\
x \geqslant 0 , y \geqslant 0 , z \geqslant 0
\end{gathered}$$
\item Use the Simplex method to perform one iteration of your tableau for part (a), choosing a value in the $\boldsymbol { y }$-column as pivot.
\item \begin{enumerate}[label=(\roman*)]
\item In the case when $k = 1$, explain why the maximum value of $P$ has now been reached and write down this maximum value of $P$.
\item In the case when $k = 3$, perform one further iteration and interpret your new tableau.
\section*{Answer space for question 3}
(a)
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
$\boldsymbol { P }$ & $\boldsymbol { x }$ & $\boldsymbol { y }$ & $\boldsymbol { Z }$ & $s$ & $\boldsymbol { t }$ & value \\
\hline
1 & $- k$ & -6 & -5 & 0 & 0 & 0 \\
\hline
0 & & & & & & \\
\hline
0 & & & & & & \\
\hline
\end{tabular}
\end{center}
(b)
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
$\boldsymbol { P }$ & $\boldsymbol { x }$ & $\boldsymbol { y }$ & $\boldsymbol { Z }$ & $\boldsymbol { s }$ & $\boldsymbol { t }$ & value \\
\hline
& & & & & & \\
\hline
& & & & & & \\
\hline
& & & & & & \\
\hline
\end{tabular}
\end{center}
\section*{Answer space for question 3}
(c)(i) $\_\_\_\_$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D2 2012 Q3 [14]}}