The Stepping Stone and the Modified Distribution Method (MODI)
Stepping Stone:
Procedure for finding optimal transportation tableau.
Given the s = 3 supply and d = 3 demands tableau below, first create
feasible tableau by using the Northwest, VAM, Minimum cell, or Russell
method. In an s x d tableau, the number of basic (allocated) cells
must be s + d – 1 even if a 0 must be placed into a cell to satisfy
requirement,
Initial tableau
From
To
1
2
3
Demand
A
6
7
4
200
B
8
11
5
100
Feasible tableau using Northwest Corner Rule
From
To
A
B
1
6 (150)
8
2
7
(50)
11 (100)
3
4
5
Demand
200
100
C
10
11
12
300
Supply
150
175
275
600
C
10
11 (25)
12 (275)
300
Supply
150
175
275
600
Notice that there are 3 + 3 – 1 = 5 allocated cells of the 9 cells; and 4 non-basic cells.
Closed loops are formed starting in an empty cell; go to an allocated cell, then another and another
until returning to empty cell. Never go diagonal and you may pass over cells. Never visit a cell
twice.
Value = 150 * 6 + 50 * 7 + 100 * 11 + 25 * 11 + 275 * 12 = 5925
Example
From
To
1
2
3
Demand
To evaluate the unallocated cell 3A, the loop is 3A Æ 3C Æ 2C Æ 2A with values
4 – 12 + 11 – 7 = –4
meaning that for every unit shifted (give and take) into 3A and 2C from 3C and 2A
results in a savings of 4. When shifting, feasibility constraints of supply and
demand must be maintained. Look at the giving cells 3C with 275 and 2A with 50.
Shift the minimum 50 into the receiving cells 3A and 2C to create the tableau
below.
A
(150)
7
4
(50)
200
6
B
8
11 (100)
5
100
C
10
11 (75)
12 (225)
300
Supply
150
175
275
600
Value = 150 * 6 + 50 * 4 + 100 * 11 + 75 * 11 + 225 * 12 = 5725.
The delta value of the tableaus is 5925 – 5725 = 200 = 50 * 4
The stepping stone evaluation is done for all empty cells and the best cell improvement is chosen
for the change. Notice cell 3A entered into the basis and cell 2A left.
Loops always contain the same number of + and – cells alternating as one completes the loop.
In evaluating cell 3B, the loop is 3B Æ 3C Æ 2C Æ 2B.
Let’s evaluate cell 2A with loop 2A Æ 2C Æ 3C Æ 3A with 7 – 11 + 12 – 4 = 4. Positive quantities
mean no improvement.
When all unallocated (non-basic) cells evaluate to positive
quantities, the tableau is optimal.
If the problem is one of maximization, use the same procedure except that positive quantities
imply improvement.
MODI
Paths or loops in this method are determined mathematically. The tableau is modified with
u (row) and v (column) variables. Allocated cell costs cij = ui + vj.
Feasible tableau using Northwest procedure
va
From/To
A
u1
1
6 (150)
u2
2
7
(50)
u3
3
4
Demand
200
vb
B
8
11 (100)
5
100
vc
C
10
11 (25)
12 (275)
300
Supply
150
175
275
600
Cell formulas for allocated or basis cells are: u1 + va = c1a;
u2 + va = c2a; u2 +vb = c2b; u2 + vc = c2c
u3 + vc = c3c
The c-values are the respective cells cost. With 5 equations and 6 unknowns, we arbitrarily assign
one of the unknowns, e.g., u1 = 0 (but more efficient to assign u2 to 0 since more allocated cells in
the u2 row)***
Evaluate Cells
1A:
u1 + va = 6 or 0 + va = 6 => va = 6
2A:
u2 + va = 7 or u2 + 6 = 7 => u2 = 1;
2B:
u2 + vb = 11 or 1 + vb = 11 => vb = 10;
2C:
u2 + vc = 11 or 1 + vc = 11 => vc = 10;
3C:
u3 + vc = 12 or u3 + 10 = 12 => u3 = 2
Evaluate empty cells with the formula Xij = cij – ui - vj
1B: X1b = c1b – u1 – vb = 8 – 0 – 10 = –2 same value using stepping stone for cell 1B
1C: X1c = c1c – u1 – vc = 10 – 0 – 10 = 0 same value using stepping stone for cell 1C
3A: X3a = c3a – u3 – va = 4 – 2 – 6 = –4 same value using stepping stone for cell 3A
3B: X3b = c3b – u3 – vb = 5 – 2 – 10 = –7 same value using stepping stone for cell 3B
Re-visit stepping stone evaluations
1B:
1B Æ 1A Æ 2A Æ 2B => 8 – 6 + 7 – 11 = –2
1C:
1C Æ 1A Æ 2A Æ 2C =>10 – 6 + 7 – 11 = 0
3A:
3A Æ 3C Æ 2C Æ 2A => 4 – 12 + 11 – 7 = –4
3B:
3B Æ 3C Æ 2C Æ 2B => 5 – 12 + 11 – 11 = –7
Thus cells 3B and 1B can improve solution.
evaluations are positive.
The procedure is then repeated until all empty cell
*** Returning to initial tableau and using a different arbitrary assignment leads to the following:
Feasible tableau using Northwest procedure
vb=11
vc=11
va= 7
From/To
A
B
C
Supply
u1 = -1
1
6 (150)
8
10
150
u2 = 0
2
7
(50)
11 (100)
11 (25)
175
u3 = 1
3
4
5
12 (275)
275
Demand
200
100
300
600
Cell formulas for allocated or basis cells are: u1 + va = c1a;
u2 + va = c2a; u2 +vb = c2b; u2 + vc = c2c
u3 + vc = c3c
The c-values are the respective cells cost. With 5 equations and 6 unknowns, we arbitrarily assign
one of the unknowns, e.g. u2 = 0.
Then u2 = 0 => va = 7; vb = 11, vc = 11, u1 + va = 6 => u1 = –1; u3 + vc = 12 => u3 = 1.
Evaluating unallocated cells
1B => X1b = 8 –7 –(–1) = 2
1C => 10 – (–1) – 11 = 0
3A => 4 – 1 – 7 = –4
3B => 5 – 1 – 11 = –7
show the same as stepping stone and the same as assigning u1 to 0.
Alternate optimal solutions
Notice alternate solution is available with cell 1C = 0, but with the same optimal value.
Degeneracy
If number of unallocated cells is less than the number of rows R + the number of columns C – 1,
degeneracy has occurred (R + C – 1). The solution is to choose an empty cell and assign it a very
small value. The selection is a bit tricky to ensure the evaluation of the other empty cells.
For example, try evaluating any empty cell in the tableau below using the stepping stone approach.
Similarly try the MODI. Note that R + C – 1 = 3 + 3 – 1 = 6 ≠ 5.
From/To
1
2
3
Demand
A
6
7
4 (200
200
B
8
11 (100)
5
100
C
10 (225)
11
12 (75)
300
Supply
150
175
275
600
Prohibited Assignments
There may be times when a specified Supply is prohibited from satisfying a specified Demand. We
can revert to the Big M method for the cost
When Supply ≠ Demand
Dummy supply or demand can be set up for the excess and whatever is assigned to the dummy
demand does not get received, and to the dummy supply does not get shipped.
Example
From
To
1
2
3
Demand
Example
Supply > Demand tableau
A
B
7
6
12
8
8
10
0
100
Demand > Supply tableau
To
A
1
7
2
12
3
8
Dummy
0
Demand
200
From
C
11
9
12
300
B
6
8
10
0
175
Dummy
0
0
0
50
C
11
9
12
0
300
Supply
150
175
275
650
Supply
250
150
275
100
675
Maximizing
When maximizing rather than minimizing, similar to the assignment algorithm, subtract all cell
profits from the maximum overall profit and use the same minimizing procedures.
Transshipments
Each transshipment locality becomes both a supply and a demand center.