| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Interpret optimal tableau |
| Difficulty | Moderate -0.3 This is a straightforward simplex tableau interpretation question requiring only recognition of optimality conditions (checking for non-negative profit row entries) and reading values directly from the tableau. Part (c) requires one simple calculation using the profit row coefficient. No actual simplex iterations or problem-solving insight needed—purely procedural tableau reading, making it slightly easier than average. |
| Spec | 7.06f Integer programming: branch-and-bound method7.07c Interpret simplex: values of variables, slack, and objective |
| Basic variable | \(x\) | \(y\) | \(Z\) | \(r\) | \(s\) | \(t\) | Value |
| \(s\) | 3 | 0 | 2 | 0 | 1 | \(-\frac{2}{3}\) | \(\frac{2}{3}\) |
| \(r\) | 4 | 0 | \(\frac{7}{2}\) | 1 | 0 | 8 | \(\frac{9}{2}\) |
| \(y\) | 5 | 1 | 7 | 0 | 0 | 3 | 7 |
| P | 3 | 0 | 2 | 0 | 0 | 8 | 63 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Yes, there are no negative values in the profit row | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(p = 63\), \(x = 0\), \(y = 7\), \(z = 0\), \(r = \frac{9}{2}\), \(s = \frac{2}{3}\), \(t = 0\) | M1, A1, A1, A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\frac{63}{7} = 9\) | M1, A1 | 2 marks |
### (a)
**Answer:** Yes, there are **no negative values** in the profit row | B1 | 1 mark
### (b)
**Answer:** $p = 63$, $x = 0$, $y = 7$, $z = 0$, $r = \frac{9}{2}$, $s = \frac{2}{3}$, $t = 0$ | M1, A1, A1, A1 | 3 marks
### (c)
**Answer:** $\frac{63}{7} = 9$ | M1, A1 | 2 marks
---A three-variable linear programming problem in $x$, $y$ and $z$ is to be solved. The objective is to maximise the profit P. The following tableau was obtained.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Basic variable & $x$ & $y$ & $Z$ & $r$ & $s$ & $t$ & Value \\
\hline
$s$ & 3 & 0 & 2 & 0 & 1 & $-\frac{2}{3}$ & $\frac{2}{3}$ \\
\hline
$r$ & 4 & 0 & $\frac{7}{2}$ & 1 & 0 & 8 & $\frac{9}{2}$ \\
\hline
$y$ & 5 & 1 & 7 & 0 & 0 & 3 & 7 \\
\hline
P & 3 & 0 & 2 & 0 & 0 & 8 & 63 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item State, giving your reason, whether this tableau represents the optimal solution. [1]
\item State the values of every variable. [3]
\item Calculate the profit made on each unit of $y$. [2]
\end{enumerate}
(Total 6 marks)
\hfill \mbox{\textit{Edexcel D2 2004 Q8 [6]}}