To convert the transportation problem into a maximisation model we

have to $($a$)$ write the inverse of the matrix, $($b$)$ Multiply the rim

requirements by $-1,($c$)$ To multiply the matrix by $-1,($d$)$ We cannot

convert the transportation problem into a maximisation problem, as it is

basically a minimisation problem. $()$

In a transportation problem where the demand or requirement is

equal to the available resource is known as $($a$)$ Balanced transportation

problem, $($b$)$ Regular transportation problem, $($c$)$ Resource allocation

transportation problem, $($d$)$ Simple transportation model. $()$

The total number of allocation in a basic feasible solution of

transportation problem of $mn$ size is equal to $($a$)mn,(b)(\frac{m}{n})-1,$

$($c$)m+n+1($d$)m+n-1.()$

When the total allocations in a transportation model of $mn$ size is

not equals to $m+n-1$ the situation is known as $($a$)$ Unbalanced

situation, $($b$)$ Tie situation, $($c$)$ Degeneracy, $($d$)$ None of the above. $()$

The opportunity cost of a row in a transportation problem is obtained

by: $($a$)$ Deducting the smallest element in the row from all other

elements of the row, $($b$)$ Adding the smallest element in the row to all

other elements of the row, $($c$)$ Deducting the smallest element in the row

from the next highest element of the row,