| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Identify pivot and justify |
| Difficulty | Moderate -0.3 This is a standard Simplex algorithm question requiring mechanical application of taught procedures: identifying the pivot using the minimum ratio test, performing row operations, and reading off the solution. While it involves multiple steps and careful arithmetic, it requires no problem-solving insight or novel thinking—just following the algorithm exactly as taught in D2. |
| 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 objective7.07d Simplex terminology: basic feasible solution, basic/non-basic variable |
| \(\boldsymbol { P }\) | \(x\) | \(y\) | \(\boldsymbol { Z }\) | \(\boldsymbol { s }\) | \(\boldsymbol { t }\) | \(\boldsymbol { u }\) | \(v\) | value |
| 1 | 0 | -12 | 0 | 5 | -3 | 0 | 0 | 37 |
| 0 | 1 | -8 | 0 | 1 | 2 | 0 | 0 | 16 |
| 0 | 0 | 4 | 0 | 0 | 3 | 0 | 1 | 20 |
| 0 | 0 | 2 | 0 | -3 | 2 | 1 | 0 | 14 |
| 0 | 0 | 1 | 1 | 2 | 5 | 0 | 0 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Choose value 4 from \(y\)-column | B1 | |
| Reason: all other values in \(y\)-column are negative or zero, so 4 is the only positive value giving a valid pivot (or: most negative coefficient in objective row is \(-12\) so \(y\) enters; pivot must be positive) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New pivot row = old row \(\div 4\) | M1 | |
| Eliminate \(y\) from all other rows | M1 | |
| Correct new tableau produced | A2 | A1 for at least two correct rows |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| All values in \(P\)-row (objective row) are \(\geq 0\), so no further improvement possible | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Read off basic variables; state maximum \(P\) value | B1 | |
| State corresponding values of \(x\), \(y\), \(z\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State values of slack variables \(s\), \(t\), \(u\), \(v\) | B1 | |
| Number of constraints with zero slack stated (i.e. binding constraints) | B1 |
# Question 4:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Choose value 4 from $y$-column | B1 | |
| Reason: all other values in $y$-column are negative or zero, so 4 is the only positive value giving a valid pivot (or: most negative coefficient in objective row is $-12$ so $y$ enters; pivot must be positive) | B1 | |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New pivot row = old row $\div 4$ | M1 | |
| Eliminate $y$ from all other rows | M1 | |
| Correct new tableau produced | A2 | A1 for at least two correct rows |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| All values in $P$-row (objective row) are $\geq 0$, so no further improvement possible | B1 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Read off basic variables; state maximum $P$ value | B1 | |
| State corresponding values of $x$, $y$, $z$ | B1 | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| State values of slack variables $s$, $t$, $u$, $v$ | B1 | |
| Number of constraints with zero slack stated (i.e. binding constraints) | B1 | |
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4 A linear programming problem consists of maximising an objective function $P$ involving three variables $x , y$ and $z$. Slack variables $s , t , u$ and $v$ are introduced and the Simplex method is used to solve the problem. Several iterations of the method lead to the following tableau.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
$\boldsymbol { P }$ & $x$ & $y$ & $\boldsymbol { Z }$ & $\boldsymbol { s }$ & $\boldsymbol { t }$ & $\boldsymbol { u }$ & $v$ & value \\
\hline
1 & 0 & -12 & 0 & 5 & -3 & 0 & 0 & 37 \\
\hline
0 & 1 & -8 & 0 & 1 & 2 & 0 & 0 & 16 \\
\hline
0 & 0 & 4 & 0 & 0 & 3 & 0 & 1 & 20 \\
\hline
0 & 0 & 2 & 0 & -3 & 2 & 1 & 0 & 14 \\
\hline
0 & 0 & 1 & 1 & 2 & 5 & 0 & 0 & 8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item The pivot for the next iteration is chosen from the $\boldsymbol { y }$-column. State which value should be chosen and explain the reason for your choice.
\item Perform the next iteration of the Simplex method.
\end{enumerate}\item Explain why your new tableau solves the original problem.
\item State the maximum value of $P$ and the values of $x , y$ and $z$ that produce this maximum value.
\item State the values of the slack variables at the optimum point. Hence determine how many of the original inequalities still have some slack when the optimum is reached.
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2008 Q4 [11]}}