| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Effect of parameter changes |
| Difficulty | Standard +0.8 This is a multi-part Simplex algorithm question requiring setup, iteration, parameter analysis, and interpretation. Part (b)(ii) requires understanding optimality conditions to determine when the solution is not yet optimal based on parameter k, which goes beyond mechanical application of the algorithm and requires conceptual insight into how coefficients affect the objective row. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.07a 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 |
| Slack variables \(s_1, s_2\) introduced correctly | B1 | |
| Objective row: \(P - 6x - 5y - 3z = 0\) | B1 | |
| Full correct tableau displayed | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Pivot column identified as \(x\)-column | M1 | |
| Correct ratios computed to find pivot row | M1 | |
| Correct pivot operation performed | M1 | |
| Correct tableau after one iteration | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For maximum not yet achieved, \(z\)-column (or \(y\)-column) entry in objective row must be negative | M1 | |
| Range of \(k\) found correctly | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substituting \(k=-1\) and performing second iteration correctly | M1 A1 | |
| Correct final tableau | A1 | |
| Values of \(x\), \(y\), \(z\) and \(P\) correctly stated | B1 | |
| Correct interpretation of solution in context | B1 B1 |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Slack variables $s_1, s_2$ introduced correctly | B1 | |
| Objective row: $P - 6x - 5y - 3z = 0$ | B1 | |
| Full correct tableau displayed | B1 | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pivot column identified as $x$-column | M1 | |
| Correct ratios computed to find pivot row | M1 | |
| Correct pivot operation performed | M1 | |
| Correct tableau after one iteration | A1 | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For maximum not yet achieved, $z$-column (or $y$-column) entry in objective row must be negative | M1 | |
| Range of $k$ found correctly | A1 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substituting $k=-1$ and performing second iteration correctly | M1 A1 | |
| Correct final tableau | A1 | |
| Values of $x$, $y$, $z$ and $P$ correctly stated | B1 | |
| Correct interpretation of solution in context | B1 B1 | |3
\begin{enumerate}[label=(\alph*)]
\item Given that $k$ is a constant, display the following linear programming problem in a Simplex tableau.
$$\begin{array} { l l }
\text { Maximise } & P = 6 x + 5 y + 3 z \\
\text { subject to } & x + 2 y + k z \leqslant 8 \\
& 2 x + 10 y + z \leqslant 17 \\
& x \geqslant 0 , y \geqslant 0 , z \geqslant 0
\end{array}$$
\item \begin{enumerate}[label=(\roman*)]
\item Use the Simplex method to perform one iteration of your tableau for part (a), choosing a value in the $x$-column as pivot.
\item Given that the maximum value of $P$ has not been achieved after this first iteration, find the range of possible values of $k$.
\end{enumerate}\item In the case where $k = - 1$, perform one further iteration and interpret your final tableau.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c4dc61a7-47ee-4d5c-bf6d-30a5da2ee8dd-07_2484_1707_223_155}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2010 Q3 [15]}}