| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2011 |
| 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 mechanical application of the pivot procedure with clear instructions (use most negative in profit row). While it involves fractions and multiple iterations, it's a routine textbook exercise testing procedural fluency rather than problem-solving or insight. Slightly easier than average due to explicit guidance on pivot selection. |
| 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 |
| \(r\) | \(- \frac { 1 } { 2 }\) | 0 | 2 | 1 | \(- \frac { 1 } { 2 }\) | 0 | 10 |
| \(y\) | \(\frac { 1 } { 2 }\) | 1 | \(\frac { 3 } { 4 }\) | 0 | \(\frac { 1 } { 4 }\) | 0 | 5 |
| \(t\) | \(\frac { 1 } { 2 }\) | 0 | 1 | 0 | \(- \frac { 1 } { 4 }\) | 1 | 4 |
| \(P\) | - 7 | 0 | 1 | 0 | 4 | 0 | 320 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P - 7x + z + 4s = 320\) | B1 | |
| Or equivalently: \(P = 7x - z - 4s + 320\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Pivot column: \(x\) (most negative \(= -7\)) | M1 | |
| Ratios: \(r\): \(10 \div \frac{1}{2} = 20\); \(y\): \(5 \div \frac{1}{2} = 10\); \(t\): \(4 \div \frac{1}{2} = 8\) — pivot row is \(t\) | M1 | |
| New \(t\)-row \(= \) old \(t\)-row \(\times 2\) | A1 | Row operations stated |
| Updated rows correct | A1 | |
| New tableau correct | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 8\), \(y = 1\), \(z = 0\) | A1 | |
| \(r = 0\), \(s = 0\), \(t = 0\) (or non-zero as appropriate) | A1 | |
| \(P = 376\) | A1 | cao |
# Question 3:
## Part (a) – Profit equation
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P - 7x + z + 4s = 320$ | B1 | |
| Or equivalently: $P = 7x - z - 4s + 320$ | B1 | |
## Part (b) – Simplex iterations
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pivot column: $x$ (most negative $= -7$) | M1 | |
| Ratios: $r$: $10 \div \frac{1}{2} = 20$; $y$: $5 \div \frac{1}{2} = 10$; $t$: $4 \div \frac{1}{2} = 8$ — pivot row is $t$ | M1 | |
| New $t$-row $= $ old $t$-row $\times 2$ | A1 | Row operations stated |
| Updated rows correct | A1 | |
| New tableau correct | A1 | |
## Part (c) – Final values
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 8$, $y = 1$, $z = 0$ | A1 | |
| $r = 0$, $s = 0$, $t = 0$ (or non-zero as appropriate) | A1 | |
| $P = 376$ | A1 | cao |
---3. 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.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & $- \frac { 1 } { 2 }$ & 0 & 2 & 1 & $- \frac { 1 } { 2 }$ & 0 & 10 \\
\hline
$y$ & $\frac { 1 } { 2 }$ & 1 & $\frac { 3 } { 4 }$ & 0 & $\frac { 1 } { 4 }$ & 0 & 5 \\
\hline
$t$ & $\frac { 1 } { 2 }$ & 0 & 1 & 0 & $- \frac { 1 } { 4 }$ & 1 & 4 \\
\hline
$P$ & - 7 & 0 & 1 & 0 & 4 & 0 & 320 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Write down the profit equation represented in the tableau.
\item Taking the most negative number in the profit row to indicate the pivot column at each stage, solve this linear programming problem. Make your method clear by stating the row operations you use.
\item State the value of the objective function and of each variable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2011 Q3 [10]}}