The Northwest
corner solution assigns flow to cells starting in the upper
left corner. As each cell is considered, flow is assigned to
use up all of the supply for a row or all the demand for a
column. A complete description is in the textbook. The graphic
shows the feasible flow that is determined by the Northwest
Corner rule. The flows appear in the green cells.
The dual variables u and v shown
as the right most column and bottom row, respectively.
The reduced costs are in the yellow cells.
The dual variables are associated with the rows and columns
of the tableau. They are assigned so that the reduced costs
of all basic cells are zero.
The formula for the reduced cost is
d(i,j) = c(i, j)  u(i)  v(j).
In this formula, i is the index of the
row and j is the index of the column. c(i, j) is
the cost for the cell. u(i) and v(j) are the
dual variables associated with row i and column j.
This notation as well as the complete algorithm are discussed
in the textbook.
One of the dual variables is arbitrary and we assign that
one the value of zero. For the case shown, we have assigned
the 0 value to v(3). Given the first assignment it is
always possible to assign the others based on the requirement
that the reduced costs for basic cells are 0. Once v(3) is
assigned, then the values of u(3) and u(4) are
determined by the expressions:
u(3) = v(3) + c(3,3) = 8
u(4) = v(3) + c(4,3) = 9.
The basic cells all have d(i,j)= 0, but
the other cells take on positive or negative values. The condition
for optimality for the transportation problem is that all cells
must have nonnegative reduced costs. This is clearly not true
for the tableau shown in the figure, because the solution is
not optimum.
The next page shows how to iteratively change
the basis until the optimum is determined.
