| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2001 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Complete Simplex solution |
| Difficulty | Moderate -0.3 This is a standard textbook Simplex algorithm question with explicit instructions on which variables to pivot at each iteration. While it requires careful arithmetic and understanding of the method, it involves no problem-solving or decision-making since the pivot choices are given. The explanation parts are routine recall of Simplex concepts. Slightly easier than average due to the guided nature. |
| 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 |
| \(r\) | 1 | 0 | 1 | 1 | 0 | 0 | 4 |
| \(s\) | 1 | 4 | 2 | 0 | 1 | 0 | 6 |
| \(t\) | 1 | 1 | 2 | 0 | 0 | 1 | 12 |
| \(P\) | - 3 | - 6 | - 4 | 0 | 0 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Slack variables used to enable us to write inequalities as equalities, all \(\geq 0\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P - 3x - 6y - 4z = 0\) | B1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Initial tableau set up correctly | M1 | |
| \(R_2 \div 4\) performed correctly | A1 (R2) | |
| Pivot operation \(R_3 - R_2\) | M1 | |
| Row reduction correct | A1 | |
| \(R_4 + 6R_2\) | A1 | |
| A1ft (pivot) | ||
| Second pivot: \(R_2 - \frac{1}{4}R_1\) | M1 | A1 |
| \(R_3 - \frac{3}{4}R_1\) | A1 | |
| \(R_4 + \frac{3}{2}R_1\) | A1 | (10) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum profit is \(£15\) | M1 A1ft | |
| when \(x = 4\) kg, \(y = \frac{1}{2}\) kg, \(z = 0\) kg | A1ft | |
| The first and second constraints have no slack | B1ft | |
| There is a slack of \(7\frac{1}{2}\) in the third constraint | (4) |
## Question 7:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Slack variables used to enable us to write inequalities as equalities, all $\geq 0$ | M1 A1 | |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P - 3x - 6y - 4z = 0$ | B1 | **(3)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Initial tableau set up correctly | M1 | |
| $R_2 \div 4$ performed correctly | A1 (R2) | |
| Pivot operation $R_3 - R_2$ | M1 | |
| Row reduction correct | A1 | |
| $R_4 + 6R_2$ | A1 | |
| A1ft (pivot) | | |
| Second pivot: $R_2 - \frac{1}{4}R_1$ | M1 | A1 |
| $R_3 - \frac{3}{4}R_1$ | A1 | |
| $R_4 + \frac{3}{2}R_1$ | A1 | **(10)** |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum profit is $£15$ | M1 A1ft | |
| when $x = 4$ kg, $y = \frac{1}{2}$ kg, $z = 0$ kg | A1ft | |
| The first and second constraints have no slack | B1ft | |
| There is a slack of $7\frac{1}{2}$ in the third constraint | | **(4)** |
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7. This question is to be answered on the sheet provided in the answer booklet.
A chemical company makes 3 products $X , Y$ and $Z$. It wishes to maximise its profit $\pounds P$. The manager considers the limitations on the raw materials available and models the situation with the following Linear Programming problem.
Maximise
$$\begin{gathered}
P = 3 x + 6 y + 4 z \\
x \quad + \quad z \leq 4 \\
x + 4 y + 2 z \leq 6 \\
x + y + 2 z \leq 12 \\
x \geq 0 , \quad y \geq 0 , \quad z \geq 0
\end{gathered}$$
subject to\\
where $x , y$ and $z$ are the weights, in kg , of products $X , Y$ and $Z$ respectively.\\
A possible initial tableau is
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & 1 & 0 & 1 & 1 & 0 & 0 & 4 \\
\hline
$s$ & 1 & 4 & 2 & 0 & 1 & 0 & 6 \\
\hline
$t$ & 1 & 1 & 2 & 0 & 0 & 1 & 12 \\
\hline
$P$ & - 3 & - 6 & - 4 & 0 & 0 & 0 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain
\begin{enumerate}[label=(\roman*)]
\item the purpose of the variables $r , s$ and $t$,
\item the final row of the tableau.
\end{enumerate}\item Solve this Linear Programming problem by using the Simplex alogorithm. Increase $y$ for your first iteration and than increase $x$ for your second iteration.
\item Interpret your solution.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2001 Q7 [17]}}