| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Complete Simplex solution |
| Difficulty | Standard +0.3 This is a standard Simplex algorithm question requiring two iterations with clearly defined pivot selection rules. While it involves fractional arithmetic and multiple row operations, it follows a completely mechanical procedure taught directly in D1 with no problem-solving or insight required—making it slightly easier than average for A-level maths overall. |
| Spec | 7.07b Simplex iterations: pivot choice and row operations7.07c Interpret simplex: values of variables, slack, and objective |
| Basic variable | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | Value |
| \(r\) | 4 | \(\frac { 7 } { 3 }\) | \(\frac { 5 } { 2 }\) | 1 | 0 | 0 | 64 |
| \(s\) | 1 | 3 | 0 | 0 | 1 | 0 | 16 |
| \(t\) | 4 | 2 | 2 | 0 | 0 | 1 | 60 |
| \(P\) | -5 | \(\frac { - 7 } { 2 }\) | -4 | 0 | 0 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| b.v | x | y |
| r | 4 | 7 |
| \(\frac{3}{4}\) | \(\frac{2}{5}\) | |
| s | 1 | 3 |
| t | 4 | 2 |
| P | -5 | 7 |
| \(-\frac{2}{5}\) | ||
| b.v | x | y |
| r | 0 | \(\frac{1}{3}\) |
| s | 0 | \(\frac{5}{2}\) |
| x | 1 | \(\frac{1}{2}\) |
| P | 0 | \(-1\) |
| b.v | x | y |
| z | 0 | \(\frac{2}{3}\) |
| s | 0 | \(\frac{17}{6}\) |
| x | 1 | \(\frac{1}{6}\) |
| P | 0 | 0 |
| M1 A1 | (all three tableaux) | |
| M1 A1ft A1 | (9 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) There is still a negative number in the profit row. | B1 | (1 mark) |
**(a)**
| b.v | x | y | z | R | s | t | value |
|-----|---|---|---|---|---|---|-------|
| r | 4 | 7 | 5 | 1 | 0 | 0 | 64 |
| | | $\frac{3}{4}$ | $\frac{2}{5}$ | | | | |
| s | 1 | 3 | 0 | 0 | 1 | 0 | 16 |
| t | 4 | 2 | 2 | 0 | 0 | 1 | 60 |
| P | -5 | 7 | -4 | 0 | 0 | 0 | 0 |
| | | | $-\frac{2}{5}$ | | | | |
| b.v | x | y | z | R | s | t | value | Row ops |
|-----|---|---|---|---|---|---|-------|---------|
| r | 0 | $\frac{1}{3}$ | $\frac{1}{3}$ | 1 | 0 | -1 | 4 | $R_1 - 4R_3$ |
| s | 0 | $\frac{5}{2}$ | $-\frac{1}{2}$ | 0 | 1 | $-\frac{1}{4}$ | 1 | $R_2 - R_3$ |
| x | 1 | $\frac{1}{2}$ | $\frac{1}{2}$ | 0 | 0 | $\frac{1}{4}$ | 15 | $R_3 \div 4$ |
| P | 0 | $-1$ | $-\frac{3}{2}$ | 0 | 0 | $\frac{5}{4}$ | 75 | $R_4 + 5R_3$ |
| b.v | x | y | z | R | s | t | value | Row ops |
|-----|---|---|---|---|---|---|-------|---------|
| z | 0 | $\frac{2}{3}$ | 1 | 2 | 0 | -2 | 8 | $R_1 \div \frac{1}{3}$ |
| s | 0 | $\frac{17}{6}$ | 0 | 1 | 1 | $-\frac{5}{4}$ | 5 | $R_2 + \frac{1}{2}R_1$ |
| x | 1 | $\frac{1}{6}$ | 0 | -1 | 0 | $\frac{5}{4}$ | 11 | $R_3 - \frac{1}{2}R_1$ |
| P | 0 | 0 | 0 | 3 | 0 | $-\frac{7}{4}$ | 87 | $R_4 + \frac{3}{2}R_1$ |
| M1 A1 | (all three tableaux) |
| M1 A1ft A1 | (9 marks) |
**Guidance:** M1: CAO. A1ft: Follow through from their I value, condone lack of units here.
**(b)** There is still a negative number in the profit row. | B1 | (1 mark) |
**Total for Q6: 10 marks**
---6. The tableau below is the initial tableau for a maximising linear programming problem in $x , y$ and $z$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & 4 & $\frac { 7 } { 3 }$ & $\frac { 5 } { 2 }$ & 1 & 0 & 0 & 64 \\
\hline
$s$ & 1 & 3 & 0 & 0 & 1 & 0 & 16 \\
\hline
$t$ & 4 & 2 & 2 & 0 & 0 & 1 & 60 \\
\hline
$P$ & -5 & $\frac { - 7 } { 2 }$ & -4 & 0 & 0 & 0 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Taking the most negative number in the profit row to indicate the pivot column at each stage, perform two complete iterations of the simplex algorithm. State the row operations you use.\\
(9)
\item Explain how you know that your solution is not optimal.\\
(1)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2008 Q6 [10]}}