| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| 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 systematic application of the pivot procedure through multiple iterations. While it involves fractional arithmetic and careful bookkeeping across several tableaux, it follows the routine algorithmic process taught in D2 with no conceptual surprises or problem-solving insight required—making it slightly easier than average for A-level maths overall, though typical for D2 Simplex questions. |
| 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 |
| \(r\) | 5 | 3 | \(- \frac { 1 } { 2 }\) | 1 | 0 | 0 | 2500 |
| \(s\) | 3 | 2 | 1 | 0 | 1 | 0 | 1650 |
| \(t\) | \(\frac { 1 } { 2 }\) | - 1 | 2 | 0 | 0 | 1 | 800 |
| \(P\) | - 40 | - 50 | - 35 | 0 | 0 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | First tableau with correct pivot located, attempt to divide row. If choosing negative pivot no marks. | M1 A1 B1 |
| Second tableau with correct row operations. | M1 A1 ft | a3M1: Their correct pivot located, attempt to divide row. If choosing negative pivot M0M0. a3A1ft: Pivot row correct on follow through including change of b.v. |
| Third tableau with correct row operations. | M1 A1 | a2B1: All row operations CAO – allow if given in terms of old row 3. a4M1: (ft) The correct row operations used correctly at least once from their pivot, column x, s, t or value 'correct'. a4A1: CAO on numbers (ignore row operations and b.v.) |
| (b) | \(P = 47750 \enspace x = 0 \enspace y = 500 \enspace z = 650 \enspace r = 1325 \enspace s = t = 0\) | B1 ft B1 |
| 12 marks |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | First tableau with correct pivot located, attempt to divide row. If choosing negative pivot no marks. | M1 A1 B1 | a1M1: Correct pivot located, attempt to divide row. If choosing negative pivot M0M0. a1A1: CAO pivot row correct including change of b.v. a1B1: All row operations CAO – allow if given in terms of old row 2. a2M1: (ft) The correct row operations used correctly at least once from their pivot, column x, z, s or value 'correct'. a2A1: CAO on numbers (ignore row operations and b.v.) |
| | Second tableau with correct row operations. | M1 A1 ft | a3M1: Their correct pivot located, attempt to divide row. If choosing negative pivot M0M0. a3A1ft: Pivot row correct on follow through including change of b.v. |
| | Third tableau with correct row operations. | M1 A1 | a2B1: All row operations CAO – allow if given in terms of old row 3. a4M1: (ft) The correct row operations used correctly at least once from their pivot, column x, s, t or value 'correct'. a4A1: CAO on numbers (ignore row operations and b.v.) |
| (b) | $P = 47750 \enspace x = 0 \enspace y = 500 \enspace z = 650 \enspace r = 1325 \enspace s = t = 0$ | B1 ft B1 | b1B1ft: Their correct values stated for at least P, x, y, z from their 'optimal' iteration. No negatives. Two M marks in (a) must have been awarded. Allow implicit stating of $P$ e.g. $P+43x+27s+4t = 47750$ with $x, s, t = 0$. b2B1: CAO For all 7 variables correct and given explicitly. |
| | | **12 marks** | |
---3. 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
$r$ & 5 & 3 & $- \frac { 1 } { 2 }$ & 1 & 0 & 0 & 2500 \\
\hline
$s$ & 3 & 2 & 1 & 0 & 1 & 0 & 1650 \\
\hline
$t$ & $\frac { 1 } { 2 }$ & - 1 & 2 & 0 & 0 & 1 & 800 \\
\hline
$P$ & - 40 & - 50 & - 35 & 0 & 0 & 0 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\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 2014 Q3 [12]}}