Vogel’s Method for Initiliztion of the Transportation Problem
Recall that the transportation problem has the form of the following problem.
Minimize 6x11 + 7x12 + 8x13 + 15x21 + 80x22 + 78x23
Subject to
x11 + x12 + x13 = 10
x21 + x22 + x23 = 15
x11 + x21 = 15
x12 + x22 = 5
x23 + x23 = 5
x ij ≥ 0.
We write this problem as
6
15
15
7
80
5
8
78
5
10
15 .
The Northwest Corner Rule of the notes fills in a basic feasible solution by starting at the upper left hand
corner. This is somewhat arbitrary and we could make some other choices about filling in the basic feasible solution
taking into account the cost coefficients. This is desirable since we would like to produce an initial basic feasible
solution close to the basic optimal solution, i.e., differing from the basic optimal by only a few simplex iterations.
One method of producing such an initial basic feasible solution is the following.
Vogel's Approximation Method (VAM)
The VAM assigns penalties to bad choices by assigning to each row (respectively, column) the penalty equal to
the difference of the two smallest cost coefficients in that row (respectively, column). In the preceding problem this is
demand
column penalty
6
15
15
15 - 6 = 9
7
80
5
80 - 7 = 73
8
78
5
78 - 8 = 70
supply
10
15
row penalty
7-6=1
78 - 15 = 63
.
Now one chooses the largest penalty, in this case 73, in a row or a column, in this case in a column, and then locates
the smallest cost in the chosen row or column, in this case 7. Then using the position of the smallest cost, one fills in
the largest possible supply or demand, in this case 5 in the 1,2 position. This satisfies the demand constraint in column
or row, in the case column 2, and part of the constraint in the unchosen row or column, here 5 out of 10 in the first
row. So problem is now updated to give
Vogel’s Method for Initialization of the Transportation Problem page 1
supply
6
75
8
10 - 5 = 5
15
80
78
15
demand 15 5 - 5 = 0
removed
5
X
One now computes the penalties for the remaining rows and columns as
6
15
demand
15
column penalty 15 - 6 = 9
removed
75
80
0
X
85
78
5-5=0
78 - 8 = 70
X
supply
row penalty
removed
5-5=0
15
8-6=2
78 - 15 = 63
X
.
Note that having a supply or demand of 0 effectively removes the row or column. Now only one choice is left,
viz. 15 in the 2,1 position. So we have the basic feasible index set consists of {(2, 1), (1,2), (1, 3)}plus one more
additional position (since we need m + n – 1 = 2 + 3 – 1 = 4 indices). We add an arbitrary index which is given as
(without the inessential 0's which clutter)
15
5 5 .
The Northwest Corner Rule gives
10
.
5 5 5
Comparing the objective values, we have that zVAM = 0⋅6 + 5⋅7 + 5⋅8 + 15⋅15 = 300 and zNWC = 10⋅6 + 5⋅15 +
5⋅80 + 5⋅78 = 925.
Vogel’s Method for Initialization of the Transportation Problem page 2