| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2001 |
| Session | January |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Set up initial Simplex tableau |
| Difficulty | Moderate -0.3 This is a standard D1 Simplex algorithm question with routine setup and execution. Part (a) is straightforward inequality interpretation, parts (b-c) follow textbook Simplex procedure with no unusual pivoting challenges, and parts (d-e) involve basic coordinate finding and standard interpretation. While it's a complete 20-mark question requiring multiple steps, each component is procedural and typical for D1, making it slightly easier than average A-level maths overall. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | B1 | Cotton: \(x + 2y \leq 70\) (available) |
| B1(2) | Wool: \(3x + 2y \leq 90\) (available). Non-negativity: \(x \geq 0, y \geq 0\) | |
| (b) | M1 | Income \(\frac{3}{2}P\) where \(P = 30x + 40y\). Objective to maximize P. Adding slack variables T and S: \(x + 2y + r = 70\) and \(3x + 2y + S = 90\). So initial tableau is: |
| Basic Var | \(x\) | \(y\) |
| T | 1 | 2 |
| A1(3) | S | |
| P | -30 | -40 |
| (c) | M1 A1 | θ values row 1: \(70/2 = 35\); row 2: \(90/2 = 45\). So mixed ② is pivot. Second tableau is: |
| Basic Var | \(x\) | \(y\) |
| M1 A1 | \(y\) | |
| S | ② | 0 |
| P | -10 | 0 |
| M1 A1(4) | θ values row 1: \(35/\frac{1}{2} = 70\); row 2: \(20/2 = 10\). So mixed ② is pivot. Third tableau is: | |
| Basic Var | \(x\) | \(y\) |
| B1 | \(y\) | |
| B1 | \(x\) | |
| B1(3) | P | |
| (d) | M1 A1(4) | So \(x = 10, y = 30, P = 1500\). Line \(x + 2y = 70\) passes through \((0, 35)\) and \((70, 0)\). Line \(3x + 2y = 90\) passes through \((0, 45)\) and \((30, 0)\). So A is \((0, 35)\), C is given by \(x + 2y = 70\) and \(3x + 2y = 90\), so \(x = 10\) and \(y = 30\). So A is \((0, 35)\), D is \((30, 0)\) |
| (e) | B1 | Initial tableau: relative to O \((x = 0, y = 0), P = 0\) |
| B1 | Second tableau: relative to A \((x = 0, y = 35), P = 1400\) | |
| B1(3) | Third tableau: relative to C \((x = 10, y = 30), P = 1500\) |
(a) | B1 | Cotton: $x + 2y \leq 70$ (available) |
| B1(2) | Wool: $3x + 2y \leq 90$ (available). Non-negativity: $x \geq 0, y \geq 0$ |
(b) | M1 | Income $\frac{3}{2}P$ where $P = 30x + 40y$. Objective to maximize P. Adding slack variables T and S: $x + 2y + r = 70$ and $3x + 2y + S = 90$. So initial tableau is: |
| | | Basic Var | $x$ | $y$ | $r$ | $S$ | Value |
| | | T | 1 | 2 | 1 | 0 | 70 |
| A1(3) | | S | 3 | 2 | 0 | 1 | 90 |
| | | P | -30 | -40 | 0 | 0 | 0 |
(c) | M1 A1 | θ values row 1: $70/2 = 35$; row 2: $90/2 = 45$. So mixed ② is pivot. Second tableau is: |
| | | Basic Var | $x$ | $y$ | $r$ | $S$ | Value |
| M1 A1 | | $y$ | $\frac{1}{2}$ | 1 | $\frac{1}{2}$ | 0 | 35 |
| | | S | ② | 0 | -1 | 1 | 20 |
| | | P | -10 | 0 | 20 | 0 | 1400 |
| M1 A1(4) | θ values row 1: $35/\frac{1}{2} = 70$; row 2: $20/2 = 10$. So mixed ② is pivot. Third tableau is: |
| | | Basic Var | $x$ | $y$ | $r$ | $S$ | Value |
| B1 | | $y$ | 0 | 1 | $\frac{3}{4}$ | $-\frac{1}{4}$ | 30 |
| B1 | | $x$ | 1 | 0 | $-\frac{1}{2}$ | $\frac{1}{2}$ | 10 |
| B1(3) | | P | 0 | 0 | 15 | 5 | 1500 |
(d) | M1 A1(4) | So $x = 10, y = 30, P = 1500$. Line $x + 2y = 70$ passes through $(0, 35)$ and $(70, 0)$. Line $3x + 2y = 90$ passes through $(0, 45)$ and $(30, 0)$. So A is $(0, 35)$, C is given by $x + 2y = 70$ and $3x + 2y = 90$, so $x = 10$ and $y = 30$. So A is $(0, 35)$, D is $(30, 0)$ |
(e) | B1 | Initial tableau: relative to O $(x = 0, y = 0), P = 0$ |
| B1 | Second tableau: relative to A $(x = 0, y = 35), P = 1400$ |
| B1(3) | Third tableau: relative to C $(x = 10, y = 30), P = 1500$ |
---A tailor makes two types of garment, A and B. He has available 70 m² of cotton fabric and 90 m² of woollen fabric. Garment A requires 1 m² of cotton fabric and 3 m² of woollen fabric. Garment B requires 2 m² of each fabric.
The tailor makes $x$ garments of type A and $y$ garments of type B.
\begin{enumerate}[label=(\alph*)]
\item Explain why this can be modelled by the inequalities
$$x + 2y \leq 70,$$
$$3x + 2y \leq 90,$$
$$x \geq 0, y \geq 0.$$
[2 marks]
\end{enumerate}
The tailor sells type A for £30 and type B for £40. All garments made are sold. The tailor wishes to maximise his total income.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Set up an initial Simplex tableau for this problem. [3 marks]
\item Solve the problem using the Simplex algorithm. [8 marks]
\end{enumerate}
Figure 4 shows a graphical representation of the feasible region for this problem.
\includegraphics{figure_4}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Obtain the coordinates of the points A, C and D. [4 marks]
\item Relate each stage of the Simplex algorithm to the corresponding point in Fig. 4. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2001 Q7 [20]}}