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\def\headlineaux{\sc Math 540 T: Algorithm Design and Analysis
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\begin{titlepage}\begin{centerline}{\Huge \framebox{Burton Rosenberg}}
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\section*{Test 1 \hfill{\parbox{3in}
    {\small\sc\begin{flushright} Februrary 24, 5:00--6:15 \end{flushright}}}}

There are four problems each counting equally.
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{\Large Name: \hrulefill}
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\begin{center}{
{\Large
\begin{tabular}{c|c}
Problem & Credit \\ \hline\hline
 1 & \\ \hline 2 & \\ \hline 3 & \\ 
 \hline 4 & \\ \hline\hline
  Total & \end{tabular} 
}}
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\em On my honor, I have neither given nor received
aid on this examination-assignment.\\
~\\
Signature: \hrulefill
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\newpage
\begin{enumerate}
\item {\sc [Simplex Method]}

Solve the Following LP showing step-by-step the simplex method:
\[\begin{array}{lcccccccc}
   \max & & x_1 & + & 2 x_2 & + & x_3 \\
\\
   \mbox{s.t.} & & x_1 & + & x_2 &+& x_3 &\le&2\\
 & & x_1 &+& x_2 & & &\le& 1 \\
\\
  & & x_1, & x_2, & x_3 &\ge&0\\
\end{array}\]
\newpage

\item {\sc [Duality]}

\begin{itemize}
\item[(a)] Give the Dual of the previous LP problem.
\item[(b)] Find the optimal dual solution, using whatever method
  you wish.
\item[(c)] Demonstrate the Complementary Slackness conditions
  for your optimal dual/primal solution pair.
  That is, what should be true and what is true for each of the
  5 variable-inequality pairings.
\end{itemize}
\newpage

\item {\sc [LU Decomposition]}

\begin{itemize}
\item[(a)]
Use Gaussian Elimination with partial pivoting to decompose,
\[ A = \left [ \begin{array}{ccc} 1&1&1\\
  2&0&2\\ 0&3&3\\ \end{array}\right],\]
into the product,
\[ L_3P_3L_2P_2L_1P_1A=U,\]
where $L_i$ are column $i$ eta-matrices, $P_i$ are permutation
matrices, and $U$ is upper triangular with 1's down the diagonal.

\item[(b)] Use back substitution and your decomposition to
find $x_1,x_2,x_3$ real numbers which satisfy,
\[ \begin{array}{ccccccc}
   x_1 &+& x_2 &+& x_3 &=& 1 \\
  x_1 & & &+& x_3 &=& 1/2 \\
   & & x_2 &+& x_3 &=& 1/3 \\
\end{array}\]
\end{itemize}
\newpage

\item {\sc [Theory]}

Prove that the product $AB$ of two square matrices is nonsingular 
if and only if both $A$ and $B$ are nonsingular.
\newpage

\end{enumerate}
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