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\begin{document}

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QUESTION

\begin{description}

\item[(a)]
Describe briefly some features of a computer implementation of the
network simplex method that would improve the efficiency of a
standard version of the algorithm.

\item[(b)]
Show that the following linear programming problem can be
formulated as a minimum cost network flow problem.

\begin{tabular}{ll}
Minimize&$z = 8x_1 + 7x_2 + 2x_3 + 6x_4 + 2x_5 + 6x_6$\\&$+ 5x_7 +
8x_8 + 7x_9 +9x_{10} + 8x_{11}$\\subject to&$x_1,\ldots,x_{11}
\geq 0$\\&$x_1 + x_2 + x_3 = 20$\\& $x_3 + x_4 = 16$\\&$x_4 + x_5
= 25$\\&$x_6 + x_7 + x_8 = 10$\\&$x_8 + x_9+ x_{10} =
30$\\&$x_{10} + x_{11} =32$\\&$x_1 + x_{6} \leq 23$
\end{tabular}

Starting with a solution in which $x_2$, $x_4$, $x_5$, $x_7$,
$x_9$ and $x_{11}$ take positive values, and the last constraint
is satisfied as a strict inequality, use the network simplex
method to solve the problem.

\end{description}

ANSWER

\begin{description}

\item[(a)]
The main points are:

\begin{itemize}

\item
compute dual variables by updating their values from the previous
iteration, rather than performing a complete recalculation

\item
compute reduced costs for a subset of arcs, and if there are
negative reduced costs, then choose the entering arc from this
subset.

\end{itemize}

\item[(b)]
The constraints can be written as

\begin{eqnarray*}
x_1+x_2+x_3&=&20\\ -x_3-x_4&=&-16\\ x_4+x_5&=&25\\
x_6+x_7+x_8&=&10\\ -x_8-x_9+x_{10}&=&-30\\ x_{10}+x_{11}&=&32\\
-x_1-x_6-s&=&-23\\ -x_2-x_5-x_7+x_9-x_{11}+s&=&-18
\end{eqnarray*}

where $s$ is a slack variable, and the last redundant constraint
is deduced from the others.

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Non basic&$y_i+c_{ij}-y_j$\\ \hline (1,2)&$-9$&(Entering arc
(1,2))\\ (1,7)&1\\ (4,5)&1\\ (4,7)&$-4$\\ (6,5)&$-6$
\end{tabular}

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$\theta=16$ leaving arc (3,2)

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Entering arc (6,5)

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Leaving arc (8,5)

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Non-basic&$y_i+c_{ij}-y)j$\\ (1,7)&1\\ (3,2)&9\\ (4,5)&2\\
(4,7)&1\\ (8,5)&6
\end{tabular}

Thus we have an optimal solution.

$x_3=16\ x_2=4\ x_5=25\ x_{11}=2\ x_{10}=30\ x_7=10\
x_1=x_4=x_6=x_8=x_9=0\ z=446$

\end{description}




\end{document}
