Question 3 - A-Level Maths

AQA D2 2007 January — Question 3 13 marks

Exam Board	AQA
Module	D2 (Decision Mathematics 2)
Year	2007
Session	January
Marks	13
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 tableau setup, pivot selection justification, two full iterations with row operations, and interpretation of results. While mechanically procedural, it demands careful arithmetic across multiple steps, understanding of pivot selection rules (theta ratios), and recognition of optimality conditions—significantly more demanding than routine single-technique questions but standard for D2 module content.
Spec	7.07a Simplex tableau: initial setup in standard format 7.07b Simplex iterations: pivot choice and row operations 7.07c Interpret simplex: values of variables, slack, and objective

Display the following linear programming problem in a Simplex tableau. $$\begin{array} { l l } \text { Maximise } & P = 5 x + 8 y + 7 z \\ \text { subject to } & 3 x + 2 y + z \leqslant 12 \\ & 2 x + 4 y + 5 z \leqslant 16 \\ & x \geqslant 0 , y \geqslant 0 , z \geqslant 0 \end{array}$$
The Simplex method is to be used by initially choosing a value in the $y$-column as a pivot.
1. Explain why the initial pivot is 4 .
2. Perform two iterations of your tableau from part (a) using the Simplex method.
3. State the values of $P , x , y$ and $z$ after your second iteration.
4. State, giving a reason, whether the maximum value of $P$ has been achieved.

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Question 3:

Part (a)

Answer	Marks	Guidance
Answer	Marks	Guidance
Initial tableau: $\begin{array}{ccccccc} P & x & y & z & s & t & \text{Value}\\ 1 & -5 & -8 & -7 & 0 & 0 & 0\\ 0 & 3 & 2 & 1 & 1 & 0 & 12\\ 0 & 2 & \boxed{4} & 5 & 0 & 1 & 16 \end{array}$	M1, A2	SCA; $-1$ EE. Total: 3

Part (b)(i)

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{12}{2} = 6$; $\frac{16}{4} = 4$ and $4 < 6$	E1	Total: 1

Part (b)(ii)

Answer	Marks	Guidance
Answer	Marks	Guidance
First pivot step using 4: $\begin{array}{ccccccc}1&-1&0&3&0&2&32\\0&\boxed{2}&0&-1\frac{1}{2}&1&-\frac{1}{2}&4\\0&\frac{1}{2}&1&1\frac{1}{4}&0&\frac{1}{4}&4\end{array}$	M1, A1, A1	Using 4 as pivot and possibly dividing third row by 4; top row correct; second row correct (may have $0\ 2\ 4\ 5\ 0\ 1\ 16$)
Choice of pivot from $x$-column	M1	$\text{pivot} = \boxed{2}$ identified and used
$\begin{array}{ccccccc}1&0&0&2\frac{1}{4}&\frac{1}{2}&1\frac{3}{4}&34\\0&1&0&-\frac{3}{4}&\frac{1}{2}&-\frac{1}{4}&2\\0&0&1&1\frac{5}{8}&-\frac{1}{4}&\frac{3}{8}&3\end{array}$	m1, A1	Row operations; correct or scaled up (may have $0\ 0\ 4\ 6\frac{1}{2}\ -1\ 1\frac{1}{2}\ 12$). Total: 6

Part (b)(iii)

Answer	Marks	Guidance
Answer	Marks	Guidance
$\max P = 34$	$\text{B1}\sqrt{}$
$x = 2,\ y = 3,\ z = 0$	B1	All correct. Total: 2

Part (b)(iv)

Answer	Marks	Guidance
Answer	Marks	Guidance
Yes — no negative values in first row	$\text{E1}\sqrt{}$	No — if negatives in top row. Total: 1

# Question 3:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial tableau: $\begin{array}{ccccccc} P & x & y & z & s & t & \text{Value}\\ 1 & -5 & -8 & -7 & 0 & 0 & 0\\ 0 & 3 & 2 & 1 & 1 & 0 & 12\\ 0 & 2 & \boxed{4} & 5 & 0 & 1 & 16 \end{array}$ | M1, A2 | SCA; $-1$ EE. **Total: 3** |

## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{12}{2} = 6$; $\frac{16}{4} = 4$ and $4 < 6$ | E1 | **Total: 1** |

## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| First pivot step using 4: $\begin{array}{ccccccc}1&-1&0&3&0&2&32\\0&\boxed{2}&0&-1\frac{1}{2}&1&-\frac{1}{2}&4\\0&\frac{1}{2}&1&1\frac{1}{4}&0&\frac{1}{4}&4\end{array}$ | M1, A1, A1 | Using 4 as pivot and possibly dividing third row by 4; top row correct; second row correct (may have $0\ 2\ 4\ 5\ 0\ 1\ 16$) |
| Choice of pivot from $x$-column | M1 | $\text{pivot} = \boxed{2}$ identified and used |
| $\begin{array}{ccccccc}1&0&0&2\frac{1}{4}&\frac{1}{2}&1\frac{3}{4}&34\\0&1&0&-\frac{3}{4}&\frac{1}{2}&-\frac{1}{4}&2\\0&0&1&1\frac{5}{8}&-\frac{1}{4}&\frac{3}{8}&3\end{array}$ | m1, A1 | Row operations; correct or scaled up (may have $0\ 0\ 4\ 6\frac{1}{2}\ -1\ 1\frac{1}{2}\ 12$). **Total: 6** |

## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\max P = 34$ | $\text{B1}\sqrt{}$ | |
| $x = 2,\ y = 3,\ z = 0$ | B1 | All correct. **Total: 2** |

## Part (b)(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Yes — no negative values in first row | $\text{E1}\sqrt{}$ | No — if negatives in top row. **Total: 1** |

Show LaTeX source

3
\begin{enumerate}[label=(\alph*)]
\item Display the following linear programming problem in a Simplex tableau.

$$\begin{array} { l l } 
\text { Maximise } & P = 5 x + 8 y + 7 z \\
\text { subject to } & 3 x + 2 y + z \leqslant 12 \\
& 2 x + 4 y + 5 z \leqslant 16 \\
& x \geqslant 0 , y \geqslant 0 , z \geqslant 0
\end{array}$$
\item The Simplex method is to be used by initially choosing a value in the $y$-column as a pivot.
\begin{enumerate}[label=(\roman*)]
\item Explain why the initial pivot is 4 .
\item Perform two iterations of your tableau from part (a) using the Simplex method.
\item State the values of $P , x , y$ and $z$ after your second iteration.
\item State, giving a reason, whether the maximum value of $P$ has been achieved.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA D2 2007 Q3 [13]}}

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