AQA D2 2011 January — Question 4 15 marks

Exam BoardAQA
ModuleD2 (Decision Mathematics 2)
Year2011
SessionJanuary
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicThe Simplex Algorithm
TypeComplete Simplex solution
DifficultyStandard +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.
Spec7.07a 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

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.
\(\boldsymbol { P }\)\(\boldsymbol { x }\)\(\boldsymbol { y }\)\(\boldsymbol { z }\)\(s\)\(t\)\(\boldsymbol { u }\)value
1-3-2-10000
0-1111004
021401010
042300121
    1. The first pivot is to be chosen from the \(x\)-column. Identify the pivot and explain why this particular value is chosen.
    2. Perform one iteration of the Simplex method and explain how you know that the optimal value has not been reached.
    1. Perform one further iteration.
    2. Interpret the final tableau and write down the initial inequality that still has slack.
      \includegraphics[max width=\textwidth, alt={}]{172c5c92-4254-4593-b741-1caa83a1e833-11_2486_1714_221_153}

Question 4:
(a)(i)
AnswerMarks 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 rowM1 A1 Smallest positive ratio \(21/4\) chosen
(a)(ii)
AnswerMarks 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 rowsM1
Correct updated tableauA2 Award A1 for two correct rows
Negative values still in objective row, so optimum not reachedA1 Correct reason stated
(b)(i)
AnswerMarks Guidance
Second pivot identified and correct iteration performedM1 A1 Correct pivot
Correct final tableauA2
(b)(ii)
AnswerMarks Guidance
Maximum \(P = \) [correct value from tableau]B1 Reading off optimal value
Values of \(x, y, z\) stated correctlyB1 Reading off variables
Slack variable with non-zero value identifiedB1 Correct slack variable
Corresponding inequality stated as still having slackB1 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]}}