| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Interpret optimal tableau |
| Difficulty | Moderate -0.3 This is a straightforward interpretation question from the simplex algorithm requiring students to recognize optimality (all coefficients in P-row non-negative), read off the solution from the tableau, and explain why non-basic variables shouldn't increase. It tests understanding rather than computation, making it slightly easier than average but still requires knowledge of D2 simplex method concepts. |
| Spec | 7.07c Interpret simplex: values of variables, slack, and objective7.07d Simplex terminology: basic feasible solution, basic/non-basic variable |
| Basic variable | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | Value |
| \(r\) | 0 | 0 | \(1\frac{1}{3}\) | 1 | 0 | \(-\frac{1}{3}\) | \(\frac{5}{3}\) |
| \(y\) | 0 | 1 | \(3\frac{1}{3}\) | 0 | 1 | \(-\frac{1}{3}\) | \(\frac{1}{3}\) |
| \(x\) | 1 | 0 | \(-3\) | 0 | \(-1\) | \(\frac{1}{3}\) | 1 |
| \(P\) | 0 | 0 | 1 | 0 | 1 | 1 | 11 |
While solving a maximizing linear programming problem, the following tableau was obtained.
\begin{tabular}{c|ccccccc}
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & 0 & 0 & $1\frac{1}{3}$ & 1 & 0 & $-\frac{1}{3}$ & $\frac{5}{3}$ \\
$y$ & 0 & 1 & $3\frac{1}{3}$ & 0 & 1 & $-\frac{1}{3}$ & $\frac{1}{3}$ \\
$x$ & 1 & 0 & $-3$ & 0 & $-1$ & $\frac{1}{3}$ & 1 \\
$P$ & 0 & 0 & 1 & 0 & 1 & 1 & 11
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Explain why this is an optimal tableau. [1]
\item Write down the optimal solution of this problem, stating the value of every variable. [3]
\item Write down the profit equation from the tableau. Use it to explain why changing the value of any of the non-basic variables will decrease the value of $P$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q10 [6]}}