| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Perform one Simplex iteration |
| Difficulty | Moderate -0.5 This is a routine Simplex algorithm question requiring standard mechanical steps: identifying the pivot, performing row operations, and interpreting the result. While it involves multiple parts and careful arithmetic, it requires no problem-solving insight—just application of a learned algorithm. Slightly easier than average because the procedure is completely algorithmic once learned. |
| Spec | 7.07a Simplex tableau: initial setup in standard format7.07b 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 | 0 | 5 | 2 | 1 | -3 | 0 | 10 |
| \(x\) | 1 | 2 | 3 | 0 | 1 | 0 | 18 |
| \(t\) | 0 | 1 | -1 | 0 | 4 | 1 | 3 |
| \(P\) | 0 | 3 | -4 | 0 | 1 | 0 | 7 |
| Answer | Marks |
|---|---|
| \(x\) was increased first; it has the most positive value in the objective row (largest coefficient) | B1 |
| Answer | Marks |
|---|---|
| Pivot on \(y\) column; minimum ratio: \(10/5=2\), \(18/2=9\), \(3/1=3\); pivot row is \(r\) row | M1 |
| New \(r\) row: divide by 5 | M1 A1 |
| Row operations on remaining rows stated | M1 |
| Correct new tableau | A1 |
| Answer | Marks |
|---|---|
| Profit equation from tableau P row (read off value) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| T is not optimal / is optimal because there are still negative / no negative values in the P row | B1 B1 | Must reference P row |
## Question 4:
**(a)**
$x$ was increased first; it has the most positive value in the objective row (largest coefficient) | B1 |
**(b)**
Pivot on $y$ column; minimum ratio: $10/5=2$, $18/2=9$, $3/1=3$; pivot row is $r$ row | M1 |
New $r$ row: divide by 5 | M1 A1 |
Row operations on remaining rows stated | M1 |
Correct new tableau | A1 |
**(c)**
Profit equation from tableau P row (read off value) | B1 |
**(d)**
T is not optimal / is optimal because there are still negative / no negative values in the P row | B1 B1 | Must reference P row
---4. 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 is obtained after the first iteration.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Basic Variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
r & 0 & 5 & 2 & 1 & -3 & 0 & 10 \\
\hline
$x$ & 1 & 2 & 3 & 0 & 1 & 0 & 18 \\
\hline
$t$ & 0 & 1 & -1 & 0 & 4 & 1 & 3 \\
\hline
$P$ & 0 & 3 & -4 & 0 & 1 & 0 & 7 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item State which variable was increased first, giving a reason for your answer.
\item Perform one complete iteration of the simplex algorithm, to obtain a new tableau, T. Make your method clear by stating the row operations you use.
\item Write down the profit equation given by T .
\item State whether T is optimal. You must use your answer to (c) to justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2016 Q4 [9]}}