Assume that current basis is optimal.

$($a$)$ He knows that $c_B{B}^{-1}b$ is the objective function value. He says that if we change a non$-$basic

$c($for example, we change $c_4$ and $x_4$ is non$-$basic$),$ since we only have $c_B$ in that multiplication,

the optimal value cannot change. Explain why this may not be true by showing the cases

where he would be right or wrong.

$($b$)$ Now he argues that changing $c_B$ will change the objective function value since it is calculated

by $c_B{B}^{-1}b,$ however, the optimal solution will not change since it is calculated by only $B^{-1}b.$

Again explain the cases where he would be correct and would not be correct and hope that

he gets it$.$

$($c$)$ Suppose that $x_1$ is a non$-$basic variable. Someone changes both the coefficient column it

corresponds $(a_1)$ and its objective coefficient $(c_1).$ Thankfully, after these changes, he knows

that he should be checking the row$-0$ values. However, just when he started calculating

$c_B{B}^{-1}A-c,$ you stop him and remind him that there are $150344$ columns $($and only $3$ rows,

such an interesting problem$)$ and he cannot do it in a reasonable time. He immediately gives

up$.$ You say that there is a much more efficient way of doing this and now you explain:

$($d$)$ Hea is now very excited because you have taught him a very valuable trick! Now he will try

to do the same by changing $a_2$ and $c_2,$ however, you see that $x_2$ is a basic variable. Give him

the bad news...