If BV is no longer optimal, then the new optimal solution can be found by using the formulae torecreate the entire tableau for BV and then continuing the simplex algorithm with the BV tableau asthe starting tableau.There can be two reasons why a change in an LP$$s parameters causes BV to be no longer optimal.First, a variable $($or variables$)$in row $0$may have a negative coefficient. In this case, a better $($largerz$-$value$)$bfs can be obtained by pivoting in a non$-$basic variable with a negative coefficient in row $0.$If this occurs, it is said that BV is now a sub$-$optimal basis.Second, a constraint $($or constraints$)$may now have a negative right$-$hand side. In this case, at leastone member of BV will now be negative and BV will no longer yield a bfs$.$If this occurs, it is said thatBV is now an infeasible basis.Just like with columns of non$-$basic variables, ranges and changes can be applied and calculated forcolumns of basic variables. Please provide an example where we calculate a range for a column of a basic variable, a problem with $3$constraints$.$ Please povide context to your example and interpret your calculations so that I can easasily understand and use this to prepare for my test. thank you