| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2017 |
| Session | Specimen |
| Marks | 13 |
| Topic | Linear Programming |
| Type | Simplex tableau interpretation |
| Difficulty | Standard +0.3 This is a straightforward linear programming question requiring students to translate a word constraint into a tableau row. Part (i) involves basic algebraic manipulation (red ≤ white + 50 becomes 25x + 40y + 20z ≤ 25x + 30y + 30z + 50, simplifying to 10y - 10z ≤ 50 or y - z ≤ 5) and understanding slack variables. Part (ii) asks to identify the pivot cell from comparing two tableaux, which is routine simplex method application. While this is Further Maths content, these are standard textbook exercises requiring only method application, not problem-solving insight, making it slightly easier than average. |
| Spec | 7.07a Simplex tableau: initial setup in standard format7.07b Simplex iterations: pivot choice and row operations |
| Cost (£) | Red | White | Yellow | Pink | |
| Pack A | 50 | 25 | 25 | 25 | 25 |
| Pack B | 48 | 40 | 30 | 30 | 0 |
| Pack C | 53 | 20 | 30 | 40 | 10 |
| Initial tableau | \(P\) | \(x\) | \(y\) | \(z\) | \(s\) | \(t\) | \(u\) | RHS |
| Row 1 | 1 | \(-1\) | \(-1\) | \(-1\) | 0 | 0 | 0 | 0 |
| Row 2 | 0 | 0 | 1 | \(-1\) | 1 | 0 | 0 | 5 |
| Row 3 | 0 | \(-5\) | 6 | 2 | 0 | 1 | 0 | 0 |
| Row 4 | 0 | 50 | 48 | 53 | 0 | 0 | 1 | 240 |
| After first iteration | \(P\) | \(x\) | \(y\) | \(z\) | \(s\) | \(t\) | \(u\) | RHS |
| Row 5 | 1 | 0 | \(-0.04\) | 0.06 | 0 | 0 | 0.02 | 4.8 |
| Row 6 | 0 | 0 | 1 | \(-1\) | 1 | 0 | 0 | 5 |
| Row 7 | 0 | 0 | 10.8 | 7.3 | 0 | 1 | 0.1 | 24 |
| Row 8 | 0 | 1 | 0.96 | 1.06 | 0 | 0 | 0.02 | 4.8 |
A garden centre sells tulip bulbs in mixed packs. The cost of each pack and the number of tulips of each colour are given in the table.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
& Cost (£) & Red & White & Yellow & Pink \\
\hline
Pack A & 50 & 25 & 25 & 25 & 25 \\
\hline
Pack B & 48 & 40 & 30 & 30 & 0 \\
\hline
Pack C & 53 & 20 & 30 & 40 & 10 \\
\hline
\end{tabular}
\end{center}
Dirk is designing a floral display in which he will need the number of red tulips to be at most 50 more than the number of white tulips, and the number of white tulips to be less than or equal to twice the number of pink tulips. He has a budget of £240 and wants to find out which packs to buy to maximise the total number of bulbs.
Dirk uses the variables $x$, $y$ and $z$ to represent, respectively, how many of pack A, pack B and pack C he buys. He sets up his problem as an initial simplex tableau, which is shown below.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Initial tableau & $P$ & $x$ & $y$ & $z$ & $s$ & $t$ & $u$ & RHS \\
\hline
Row 1 & 1 & $-1$ & $-1$ & $-1$ & 0 & 0 & 0 & 0 \\
\hline
Row 2 & 0 & 0 & 1 & $-1$ & 1 & 0 & 0 & 5 \\
\hline
Row 3 & 0 & $-5$ & 6 & 2 & 0 & 1 & 0 & 0 \\
\hline
Row 4 & 0 & 50 & 48 & 53 & 0 & 0 & 1 & 240 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Show how the constraint on the number of red tulips leads to one of the rows of the tableau. [3]
\end{enumerate}
The tableau that results after the first iteration is shown below.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
After first iteration & $P$ & $x$ & $y$ & $z$ & $s$ & $t$ & $u$ & RHS \\
\hline
Row 5 & 1 & 0 & $-0.04$ & 0.06 & 0 & 0 & 0.02 & 4.8 \\
\hline
Row 6 & 0 & 0 & 1 & $-1$ & 1 & 0 & 0 & 5 \\
\hline
Row 7 & 0 & 0 & 10.8 & 7.3 & 0 & 1 & 0.1 & 24 \\
\hline
Row 8 & 0 & 1 & 0.96 & 1.06 & 0 & 0 & 0.02 & 4.8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Which cell was used as the pivot? [1]
\item Explain why row 2 and row 6 are the same. [1]
\item \begin{enumerate}[label=(\alph*)]
\item Read off the values of $x$, $y$ and $z$ after the first iteration. [1]
\item Interpret this solution in terms of the original problem. [2]
\end{enumerate}
\item Identify the variable that has become non-basic. Use the pivot row of the initial tableau to eliminate $x$ algebraically from the equation represented by Row 1 of the initial tableau. [3]
\end{enumerate}
The feasible region can be represented graphically in three dimensions, with the variables $x$, $y$ and $z$ corresponding to the $x$-axis, $y$-axis and $z$-axis respectively. The boundaries of the feasible region are planes. Pairs of these planes intersect in lines and at the vertices of the feasible region these lines intersect.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{5}
\item The planes defined by each of the new basic variables being set equal to 0 intersect at a point. Show how the equations from part (v) are used to find the values $P$ and $x$ at this point. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2017 Q5 [13]}}