1 A furniture manufacturer is planning a production run. He will be making wardrobes, drawer units and desks. All can be manufactured from the same wood.

He has available $200 \mathrm {~m} ^ { 2 }$ of wood for the production run. Allowing for wastage, a wardrobe requires $5 \mathrm {~m} ^ { 2 }$, a drawer unit requires $3 \mathrm {~m} ^ { 2 }$, and a desk requires $2 \mathrm {~m} ^ { 2 }$.

He has 200 hours available for the production run. A wardrobe requires 4.5 hours, a drawer unit requires 5.2 hours, and a desk requires 3.8 hours.

The completed furniture will have to be stored at the factory for a short while before being shipped. The factory has $50 \mathrm {~m} ^ { 3 }$ of storage space available. A wardrobe needs $1 \mathrm {~m} ^ { 3 }$, a drawer unit needs $0.75 \mathrm {~m} ^ { 3 }$, and a desk needs $0.5 \mathrm {~m} ^ { 3 }$.

The manufacturer needs to know what he should produce to maximise his income. He sells the wardrobes at $\pounds 80$ each, the drawer units at $\pounds 65$ each and the desks at $\pounds 50$ each.\\
(i) Formulate the manufacturer's problem as an LP.\\
(ii) Use the Simplex algorithm to solve the LP problem.\\
(iii) Interpret the results.\\
(iv) An extra $25 \mathrm {~m} ^ { 2 }$ of wood is found and is to be used. The new optimal solution is to make 44 wardrobes, no drawer units and no desks. However, this leaves some of each resource (wood, hours and space) left over. Explain how this can be possible.\\
(i) Given that $x$ and $y$ are propositions, draw a 4-line truth table for $x \Rightarrow y$, allowing $x$ and $y$ to take all combinations of truth values.

If $x$ is false and $x \Rightarrow y$ is true, what can be deduced about the truth value of $y$ ?

A story has it that, in a lecture on logic, the philosopher Bertrand Russell (1872-1970) mentioned that a false proposition implies any proposition.

A student challenged this, saying "In that case, given that $1 = 0$, prove that you are the Pope."\\
Russell immediately replied, "Add 1 to both sides of the equation: then we have $2 = 1$. The set containing just me and the Pope has 2 members. But $2 = 1$, so the set has only 1 member; therefore, I am the Pope."

Russell's string of statements is an example of a deductive sequence. Let $a$ represent " $1 = 0$ ", $b$ represent " $2 = 1$ ", $c$ represent "Russell and the Pope are 2" and $d$ represent "Russell and the Pope are 1". Then Russell's deductive sequence can be written as $( a \wedge ( a \Rightarrow b ) \wedge c ) \Rightarrow d$.\\
(ii) Assuming that $a$ is false, $b$ is false, $a \Rightarrow b$ is true, $c$ is true, and that $d$ can take either truth value, draw a 2-line truth table for $( a \wedge ( a \Rightarrow b ) \wedge c ) \Rightarrow d$.\\
(iii) What does the table tell you about $d$ with respect to the false proposition $a$ ?\\
(iv) Explain why Russell introduced propositions $b$ and $c$ into his argument.\\
(v) Russell could correctly have started a deductive sequence:\\
$a \wedge [ a \Rightarrow ( ( 0.5 = - 0.5 ) \Rightarrow ( 0.25 = 0.25 ) ) ]$.\\
Had he have done so could he correctly have continued it to end at $d$ ?\\
Justify your answer.\\
(vi) Draw a combinatorial circuit to represent $( a \wedge ( a \Rightarrow b ) \wedge c ) \Rightarrow d$.

3 Floyd's algorithm is applied to the incomplete network on 4 nodes drawn below. The weights on the arcs represent journey times.\\
\includegraphics[max width=\textwidth, alt={}, center]{4b5bc097-1052-4e44-8623-a84ceaab0289-4_400_558_347_751}

The final matrices are shown below.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{final time matrix}
\begin{tabular}{ | l | r | r | r | r | }
\cline { 2 - 5 }
\multicolumn{1}{c|}{} & \multicolumn{1}{c|}{$\mathbf { 1 }$} & $\mathbf { 2 }$ & $\mathbf { 3 }$ & \multicolumn{1}{c|}{$\mathbf { 4 }$} \\
\hline
$\mathbf { 1 }$ & 6 & 5 & 3 & 10 \\
\hline
\end{tabular}
\end{center}

\hfill \mbox{\textit{OCR MEI D2 2015 Q1 [16]}}