A manufacturer of electronic instruments produces two types of timer: a standard and a precision model with net profits of \(\$ 10\) and \(\$ 15\), respectively. His work force cannot produce more than 50 instruments per day. Moreover, the four main components used in production are in short supply so that the following stock constraints hold: $$ \begin{array}{|c|c|c|c|} \hline & & \text { Number used } & \text { per timer } \\ \hline \text { Component } & \text { Stock } & \text { Standard } & \text { Precision } \\ \hline \mathrm{a} & 220 & 4 & 2 \\ \hline \mathrm{b} & 160 & 2 & 4 \\ \hline \mathrm{c} & 370 & 2 & 10 \\ \hline \mathrm{d} & 300 & 5 & 6 \\ \hline \end{array} $$ Graphically determine the point of optimum profit. If profits on the standard timer were to change, by how much could they change without altering the original solution?

Solve the following linear programming problem: Maximize \(\quad 6 \mathrm{~L}_{1}+11 \mathrm{~L}_{2}\) subject to: \(\quad 2 \mathrm{~L}_{1}+\mathrm{L}_{2} \leq 104\) \(\mathrm{L}_{1}+2 \mathrm{~L}_{2} \leq 76\) and \(L_{1} \geq 0, L_{2} \geq 0\)

Use the branch and bound method to solve the integer programming problem Maximize \(\quad P=2 x_{1}+3 x_{2}+x_{3}+2 x_{4}\) subject to $$ \begin{aligned} &5 \mathrm{x}_{1}+2 \mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4} \leq 15 \\ &2 \mathrm{x}_{1}+6 \mathrm{x}_{2}+10 \mathrm{x}_{3}+8 \mathrm{x}_{4} \leq 60 \\\ &\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4} \leq 8 \\ &2 \mathrm{x}_{1}+2 \mathrm{x}_{2}+3 \mathrm{x}_{3}+3 \mathrm{x}_{4} \leq 16 \\\ &\mathrm{x}_{1} \leq 3, \mathrm{x}_{2} \leq 7, \mathrm{x}_{3} \leq 5, \mathrm{x}_{4} \leq 5 \end{aligned} $$

Show through the simplex method and then graphically that the following linear program has no solution. Maximize \(\quad 2 \mathrm{x}+\mathrm{y}\) subject to $$ \begin{aligned} &-x+y \leq 1 \\ &x-2 y \leq 2 \end{aligned} $$ \(\mathrm{x}, \mathrm{y} \geq 0 .\)

The optimum solution to the problem Maximize $\quad \mathrm{P}=12 \mathrm{x}_{1}+9 \mathrm{x}_{2}$ subject to $$ \begin{aligned} &3 \mathrm{x}_{1}+2 \mathrm{x}_{2} \leq 7 \\ &3 \mathrm{x}_{1}+\mathrm{x}_{2} \leq 4 \\ &\mathrm{x}_{1} \geq 0, \mathrm{x}_{2} \geq 0 \end{aligned} $$ is \(P=9(7 / 2)=31(1 / 2)\). The solution to the dual is $\mathrm{y}_{1}=4(1 / 2), \mathrm{y}_{2}=0$ Now assume the first constraint of \((2)\) is changed from 7 to 8 , i.e., $$ 3 \mathrm{x}_{1}+2 \mathrm{x}_{2} \leq 8 $$ Find the increase in \(\mathrm{P}\). What is the dual for this new problem?

Consider the general \(2 \times 2\) matrix game: $$ \begin{array}{|l|l|l|l|} \hline & \mathrm{B}: & \mathrm{B}_{1} & \mathrm{~B}_{3} \\ \hline \mathrm{A}: & & & \\ \hline \mathrm{A}_{1} & & \mathrm{a}_{11} & \mathrm{a}_{12} \\ \hline \mathrm{A}_{2} & & \mathrm{a}_{21} & \mathrm{a}_{22} \\ \hline \end{array} $$ Show that if this game has a saddle point, it must have a dominance.