Assume that current basis is optimal.

$($

a

$)$

$$He knows that

$$

$$

$$

$-$

$1$

$$

$$is the objective function value. He says that if we change a non

$-$

basic

$$

$($

for example, we change

$$

$4$

$$and

$$

$4$

$$is non

$-$

basic

$)$

$,$

$$since we only have

$$

$$

$$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

$$

$$

$$will change the objective function value since it is calculated

by

$$

$$

$$

$-$

$1$

$$

$,$

$$however$,$ the optimal solution will not change since it is calculated by only

$$

$-$

$1$

$$

$.$

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

he gets it

$.$

$($

c

$)$

$$Suppose that

$$

$1$

$$is a non

$-$

basic variable. Someone changes both the coefficient column it

corresponds

$($

$$

$1$

$)$

$$and its objective coefficient

$($

$$

$1$

$)$

$.$

$$Thankfully$,$ after these changes, he knows

that he should be checking the row

$-$

$0$

$$values$.$ However, just when he started calculating

$$

$$

$$

$-$

$1$

$$

$-$

$$

$,$

$$you stop him and remind him that there are

$1$

$5$

$0$

$3$

$4$

$4$

$$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

$$

$2$

$$and

$$

$2$

$,$

$$however$,$ you see that

$$

$2$

$$is a basic variable. Give him

the bad news...