| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Complete Simplex solution |
| Difficulty | Standard +0.8 This is a complete Simplex algorithm problem requiring multiple iterations, pivot identification with ratio tests, tableau operations, and interpretation. While mechanically procedural, it demands careful arithmetic across several steps, understanding of optimality conditions, and interpretation of slack variables. More demanding than standard single-topic questions but follows a well-defined algorithm 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 variable7.07e Graphical interpretation: iterations as edges of convex polygon7.07f Algebraic interpretation: explain simplex calculations |
| \(\boldsymbol { P }\) | \(\boldsymbol { x }\) | \(\boldsymbol { y }\) | \(\boldsymbol { z }\) | \(s\) | \(t\) | \(\boldsymbol { u }\) | value |
| 1 | -3 | -2 | -1 | 0 | 0 | 0 | 0 |
| 0 | -1 | 1 | 1 | 1 | 0 | 0 | 4 |
| 0 | 2 | 1 | 4 | 0 | 1 | 0 | 10 |
| 0 | 4 | 2 | 3 | 0 | 0 | 1 | 21 |
| Answer | Marks | Guidance |
|---|---|---|
| Pivot is \(4\) (row 3, \(x\)-column) | B1 | Correct pivot identified |
| Most negative value in objective row is \(-3\) so \(x\)-column chosen; \(\frac{4}{1}=4\), \(\frac{10}{2}=5\), \(\frac{21}{4}\) — smallest ratio is \(\frac{21}{4}\) so bottom row | M1 A1 | Smallest positive ratio \(21/4\) chosen |
| Answer | Marks | Guidance |
|---|---|---|
| New row 4 (pivot row) \(= \frac{1}{4} \times\) old row 4: \(0, 1, \frac{1}{2}, \frac{3}{4}, 0, 0, \frac{1}{4}, \frac{21}{4}\) | M1 | Dividing pivot row by pivot |
| Row operations to eliminate \(x\) from other rows | M1 | |
| Correct updated tableau | A2 | Award A1 for two correct rows |
| Negative values still in objective row, so optimum not reached | A1 | Correct reason stated |
| Answer | Marks | Guidance |
|---|---|---|
| Second pivot identified and correct iteration performed | M1 A1 | Correct pivot |
| Correct final tableau | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum \(P = \) [correct value from tableau] | B1 | Reading off optimal value |
| Values of \(x, y, z\) stated correctly | B1 | Reading off variables |
| Slack variable with non-zero value identified | B1 | Correct slack variable |
| Corresponding inequality stated as still having slack | B1 | Correct inequality stated |
## Question 4:
**(a)(i)**
Pivot is $4$ (row 3, $x$-column) | B1 | Correct pivot identified
Most negative value in objective row is $-3$ so $x$-column chosen; $\frac{4}{1}=4$, $\frac{10}{2}=5$, $\frac{21}{4}$ — smallest ratio is $\frac{21}{4}$ so bottom row | M1 A1 | Smallest positive ratio $21/4$ chosen
**(a)(ii)**
New row 4 (pivot row) $= \frac{1}{4} \times$ old row 4: $0, 1, \frac{1}{2}, \frac{3}{4}, 0, 0, \frac{1}{4}, \frac{21}{4}$ | M1 | Dividing pivot row by pivot
Row operations to eliminate $x$ from other rows | M1 |
Correct updated tableau | A2 | Award A1 for two correct rows
Negative values still in objective row, so optimum not reached | A1 | Correct reason stated
**(b)(i)**
Second pivot identified and correct iteration performed | M1 A1 | Correct pivot
Correct final tableau | A2 |
**(b)(ii)**
Maximum $P = $ [correct value from tableau] | B1 | Reading off optimal value
Values of $x, y, z$ stated correctly | B1 | Reading off variables
Slack variable with non-zero value identified | B1 | Correct slack variable
Corresponding inequality stated as still having slack | B1 | Correct inequality stated
4 The Simplex method is to be used to maximise $P = 3 x + 2 y + z$ subject to the constraints
$$\begin{aligned}
- x + y + z & \leqslant 4 \\
2 x + y + 4 z & \leqslant 10 \\
4 x + 2 y + 3 z & \leqslant 21
\end{aligned}$$
The initial Simplex tableau is given below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
$\boldsymbol { P }$ & $\boldsymbol { x }$ & $\boldsymbol { y }$ & $\boldsymbol { z }$ & $s$ & $t$ & $\boldsymbol { u }$ & value \\
\hline
1 & -3 & -2 & -1 & 0 & 0 & 0 & 0 \\
\hline
0 & -1 & 1 & 1 & 1 & 0 & 0 & 4 \\
\hline
0 & 2 & 1 & 4 & 0 & 1 & 0 & 10 \\
\hline
0 & 4 & 2 & 3 & 0 & 0 & 1 & 21 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item The first pivot is to be chosen from the $x$-column. Identify the pivot and explain why this particular value is chosen.
\item Perform one iteration of the Simplex method and explain how you know that the optimal value has not been reached.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Perform one further iteration.
\item Interpret the final tableau and write down the initial inequality that still has slack.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{172c5c92-4254-4593-b741-1caa83a1e833-11_2486_1714_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D2 2011 Q4 [15]}}