| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Write constraints from tableau |
| Difficulty | Standard +0.3 This is a standard simplex algorithm question requiring students to read equations from a tableau, perform routine pivot operations following a given rule (most negative in profit row), and state the final solution. While it involves multiple steps and careful arithmetic, it follows a completely algorithmic procedure taught directly in D2 with no problem-solving insight required, making it slightly easier than average. |
| Spec | 7.07a Simplex tableau: initial setup in standard format7.07b Simplex iterations: pivot choice and row operations |
| Basic variable | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | Value |
| \(r\) | 7 | 10 | 10 | 1 | 0 | 0 | 3600 |
| \(s\) | 6 | 9 | 12 | 0 | 1 | 0 | 3600 |
| \(t\) | 2 | 3 | 4 | 0 | 0 | 1 | 2400 |
| \(P\) | -35 | -55 | -60 | 0 | 0 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(7x + 10y + 10z + r = 3600\), \(6x + 9y + 12z + s = 3600\), \(2x + 3y + 4z + t = 2400\), \(P – 35x – 55y – 60z = 0\) | B2,1,0 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| b.v. | x | y |
| r | 2 | \(\frac{5}{2}\) |
| z | \(\frac{1}{2}\) | \(\frac{3}{4}\) |
| t | 0 | 0 |
| P | –5 | –10 |
| A1 | M1 | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| b.v. | x | y |
| y | \(\frac{4}{5}\) | 1 |
| z | \(-\frac{1}{10}\) | 0 |
| t | 0 | 0 |
| P | 3 | 0 |
| M1 | M1 | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(P = 20400\), \(x = 0\), \(y = 240\), \(z = 120\), \(r = 0\), \(s = 0\), \(t = 1200\) | M1 | A2ft, A1ft, 0 |
## Part (a)
**Answer:** $7x + 10y + 10z + r = 3600$, $6x + 9y + 12z + s = 3600$, $2x + 3y + 4z + t = 2400$, $P – 35x – 55y – 60z = 0$ | B2,1,0 | 4
## Part (b)(i)
**Answer:**
| b.v. | x | y | z | r | s | t | value | Row ops |
|---|---|---|---|---|---|---|---|---|
| r | 2 | $\frac{5}{2}$ | 0 | 1 | $-\frac{5}{6}$ | 0 | 600 | $R_1 – 10R_2$ |
| z | $\frac{1}{2}$ | $\frac{3}{4}$ | 1 | 0 | $\frac{1}{12}$ | 0 | 300 | $R_2 + 12$ |
| t | 0 | 0 | 0 | 0 | $-\frac{1}{3}$ | 1 | 1200 | $R_3 – 4R_2$ |
| P | –5 | –10 | 0 | 0 | 5 | 0 | 1800 | $R_4 + 60R_2$ |
| A1 | M1 | A1ft | B1 | 5 |
## Part (b)(ii)
**Answer:**
| b.v. | x | y | z | r | s | t | value | Row ops |
|---|---|---|---|---|---|---|---|---|
| y | $\frac{4}{5}$ | 1 | 0 | $\frac{2}{5}$ | $-\frac{1}{3}$ | 0 | 240 | $R_1 + \frac{5}{2}$ |
| z | $-\frac{1}{10}$ | 0 | 1 | $-\frac{3}{10}$ | $\frac{1}{3}$ | 0 | 120 | $R_2 – \frac{3}{4}R_1$ |
| t | 0 | 0 | 0 | 0 | $-\frac{1}{3}$ | 1 | 1200 | $R_3$ set |
| P | 3 | 0 | 0 | 4 | $\frac{5}{3}$ | 0 | 20400 | $R_4 + 10R_1$ |
| M1 | M1 | A1ft | M1 | A1 | 4 |
## Part (c)
**Answer:** $P = 20400$, $x = 0$, $y = 240$, $z = 120$, $r = 0$, $s = 0$, $t = 1200$ | M1 | A2ft, A1ft, 0 | 2
---The tableau below is the initial tableau for a maximising linear programming problem.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & 7 & 10 & 10 & 1 & 0 & 0 & 3600 \\
\hline
$s$ & 6 & 9 & 12 & 0 & 1 & 0 & 3600 \\
\hline
$t$ & 2 & 3 & 4 & 0 & 0 & 1 & 2400 \\
\hline
$P$ & -35 & -55 & -60 & 0 & 0 & 0 & 0 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Write down the four equations represented in the initial tableau above. [4]
\item Taking the most negative number in the profit row to indicate the pivot column at each stage, solve this linear programming problem. State the row operations that you use. [9]
\item State the values of the objective function and each variable. [3]
\end{enumerate}
(Total 16 marks)
\hfill \mbox{\textit{Edexcel D2 2006 Q8 [16]}}