| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Simplex tableau interpretation |
| Difficulty | Standard +0.3 This is a standard simplex method question requiring tableau interpretation and iteration. Part (a) tests basic understanding of reading constraints and objective function from a tableau (routine recall). Part (b) involves mechanical application of the simplex algorithm with clear pivot selection rules given. While it requires multiple iterations and careful arithmetic, it follows a completely algorithmic procedure with no problem-solving insight needed. Slightly easier than average due to the prescriptive nature and clear instructions. |
| 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 |
| \(r\) | 15 | - 2 | 3 | 1 | 0 | 0 | 180 |
| \(s\) | 10 | 1 | 1 | 0 | 1 | 0 | 80 |
| \(t\) | 1 | 6 | - 2 | 0 | 0 | 1 | 100 |
| \(P\) | - 1 | - 2 | - 5 | 0 | 0 | 0 | 0 |
| Answer | Marks |
|---|---|
| \(P = x + 2y + 5z\) | B1 |
| Answer | Marks |
|---|---|
| \(15x – 2y + 3z \leq 180\) | |
| \(10x + y + z \leq 80\) | M1 A1 (3) |
| \(x + 6y – 2z \leq 100\) |
| Answer | Marks |
|---|---|
| M1 A1 | |
| M1 A1 | |
| M1 A1 1ft (8) |
| Answer | Marks |
|---|---|
| \(P = 364; x = 0; y = 12; z = 68; r = s = 0; t = 164\) | M1 A1 (2) |
**Part (a)(i):**
$P = x + 2y + 5z$ | B1
**Part (a)(ii):**
$15x – 2y + 3z \leq 180$ |
$10x + y + z \leq 80$ | M1 A1 (3)
$x + 6y – 2z \leq 100$ |
**Part (b):**
| M1 A1
| M1 A1
| M1 A1 1ft (8)
**Part (c):**
$P = 364; x = 0; y = 12; z = 68; r = s = 0; t = 164$ | M1 A1 (2)
**Notes for Question 5:**
- **ai1B1:** CAO - allow in any equivalent form e.g. $P – x – 2y – 5z = 0$ but not say $P = x + 2y + 5z = 0$
- **aii1M1:** Two inequalities (or equations with slack variables) correct
- **aii1A1:** CAO
- **b1M1:** Correct pivot located (3 in the z column), attempt to divide row. If choosing negative pivot then M0M0
- **b1A1:** CAO pivot row correct including change of b.v. (so r must be changed to z)
- **b2M1:** (ft) All values in one of the non-pivot rows correct or one of the non zero/one columns (x, y, r or value) correct following through their choice of pivot
- **b2A1:** CAO on all values for the first iteration – ignore row ops and b.v. column for this mark
- **b3M1:** Their correct pivot located following their first iteration, attempt to divide row. If choosing negative pivot M0M0 - however, allow recovery for the third and fourth M marks only if positive pivot chosen for the second iteration after a negative pivot chosen for the first iteration
- **b3A1ft:** Their pivot row correct including change of b.v. following their first iteration
- **b4M1:** (ft) All values in one of the non-pivot rows correct or one of the non zero/one columns (x, r, s or value) correct following through their choice of pivot
- **b4A1:** CAO for all values and row operations for both iterations - including all eight row operations stated correctly (ignore b.v. column for this mark)
- **c1M1:** Their correct values stated for at least P, x, y, z from their 'optimal' iteration so there must be no negatives in the profit row. Two M marks in (b) must have been awarded – the numerical value of P must be explicitly stated and not as part of an equation
- **c1A1:** CAO for all seven values explicitly stated
---5. The tableau below is the initial tableau for a three-variable linear programming problem in $x , y$ and $z$. The objective is to maximise the profit, $P$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Basic variable & $x$ & $y$ & $z$ & $r$ & $s$ & $t$ & Value \\
\hline
$r$ & 15 & - 2 & 3 & 1 & 0 & 0 & 180 \\
\hline
$s$ & 10 & 1 & 1 & 0 & 1 & 0 & 80 \\
\hline
$t$ & 1 & 6 & - 2 & 0 & 0 & 1 & 100 \\
\hline
$P$ & - 1 & - 2 & - 5 & 0 & 0 & 0 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Using the information in the tableau, write down
\begin{enumerate}[label=(\roman*)]
\item the objective function,
\item the three constraints as inequalities.
\end{enumerate}\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 final values of the objective function and each variable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2017 Q5 [13]}}