Put allocation in one of the empty cells as zero o solve degeneracy in the transportation problem we have to

Put a small element epsilon in any one of the empty cells

Allocate the smallest element epsilon in such a cell, which will not form a closed-loop with other loaded cells

Allocate the smallest element epsilon in such a cell, which will form a closed loop with other loaded cells

Complex method

Graphical method

Vector method

Hungarian method

Dividing each row by the elements of the row above it

Subtracting the elements of the row from the elements of the row above it

Subtracting the smallest element from all other elements of the row

Subtracting all the elements of the row from the highest element in the matrix

Equal to each other

Equal to $\mathrm{m}\times \mathrm{n}$ (where m and n are numbers of rows and columns)

$\mathrm{m}+\mathrm{n}$ (where m and n are numbers of rows and columns)

Equal to the number of rows or columns

Rectangular matrix

Square matrix

Identity matrix

None of the above

Zeros

All negative elements

All infinity

All ones

Both are rectangular matrices

Both are square matrices

Both can be solved by graphical method

Both have objective function and non-negativity constraints

The allocations in assignment problem are one to one

Because we find row opportunity cost matrix

Because we find column opportunity matrix

Because make allocations, one has to draw horizontal and Vertical lines

There will be a tie while making allocations

The problem will get alternate solutions

The problem degenerates and we have to use epsilon to solve degeneracy

We cannot solve the assignment problem by transportation algorithm

Least cost method

Vogel’s approximation method

Modified distribution method

All of the above