The stepping stone method is an algorithm for swapping the routes between sites that supply and receive goods.
We start with the matrix below, representing the costs of transporting from each quarry to each site, together with the supply capacity of each quarry and the receiving capacity of each site.
Site 1 
Site 2 
Site 3 
Site 4 
Supply of Gravel (Tons) 

Quarry A 
150 
200 
140 
160 
23 
Quarry B 
130 
110 
190 
220 
16 
Quarry C 
120 
170 
180 
100 
19 
Demand for Gravel (Tons) 
12 
13 
11 
22 
58 
The ‘North West Corner Method’ gives the starting solution shown below.
Site 1 
Site 2 
Site 3 
Site 4 
Supply of Gravel (Tons) 

Quarry A 
12 
11 
23 

Quarry B 
2 
14 
16 

Quarry C 
1 
18 
19 

Demand for Gravel (Tons) 
12 
13 
15 
18 
58 
To find an improved solution, we must find the shadow costs associated with each quarry and building site. The process is:

Set the cost associated with the source (quarry) to zero. The destination cost is then the total cost of transportation.

Move along the row to any other non empty squares and find the destination costs in the same way, by setting the source costs to zero.

When all possible destination costs for that row have been determined, go to the start of the next row.

Move along the row to any non empty squyares and use the destination costs found earlier to establish the source cost for the row. This done, find any other unknown destination costs.

Repeat 3 and 4 above until all source and destination costs have been found.
For example, destination and source costs are shown for the table at the top of the page in the table below, only the costs corresponding to the routes in the north west corner solution being considered.
C(1) 
C(2) 
C(3) 
C(4) 

C(A) 
150 
200 

C(B) 
110 
190 

C(C) 
180 
100 
We have
C(A)+C(1)=150 (1)
C(A)+C(2)=200 (2)
C(B)+C(2)=110 (3)
C(B)+C(3)=190 (4)
C(C)+C(3)=180 (5)
C(C)+C(4)=100 (6)
Setting the cost associated with quarry A, C(A) to zero gives C(1)=150.
Then from (2), C(2)=200
From (3), C(B)=110C(2)=110200=90
From (4), C(3)=190—C(B)=190 90=280
From (5), C(C)=180C(3)=180 280=100
From (6), C(4)=100C(C)=100 100=200.
We can put the costs for each site in the matrix.
C(1) 150 
C(2) 200 
C(3) 280 
C(4) 200 

C(A) 0 
150 
200 

C(B) 90 
110 
190 

C(C) 100 
180 
100 
Now find the improvement indices for routes not included in the initial north west corner solution (the empty squares above).
I(B1)=130C(B)C(1)=130—90150=70
I(C1)=120C(C)C(1)=120—100150=70
I(C2)=170C(C)C(2)=170—100200=70
I(A3)=140C(A)C(3)=1400280=140
I(A4)=160C(A)C(4)=1600200=40
I(B4)=220C(B)C(4)=220—80200=100
The most negative is I(A3)= 140 so we can reduce the cost by introducing route A3. Put %theta in the box A3 and make the rows and columns satisfy supply and demand constraints.
Site 1 
Site 2 
Site 3 
Site 4 
Supply of Gravel (Tons) 

Quarry A 
12 
11%theta 
%theta 
23 

Quarry B 
2+%theta 
14%theta 
16 

Quarry C 
1 
18 
19 

Demand for Gravel (Tons) 
12 
13 
15 
18 
58 
Maximise %theta that leave all the matrix entries positive. Inspection gives %theta = 11. We have the table below.
Site 1 
Site 2 
Site 3 
Site 4 
Supply of Gravel (Tons) 

Quarry A 
12 
0 
11 
23 

Quarry B 
13 
3 
16 

Quarry C 
1 
18 
19 

Demand for Gravel (Tons) 
12 
13 
15 
18 
58 
Compute the new improvement indices.
I(B1)=130C(B)C(1)=130—90150=70
I(C1)=120C(C)C(1)=120—100150=70
I(C2)=170C(C)C(2)=170—100200=70
I(A2)=200C(A)C(2)=0
I(A4)=160C(A)C(4)=1600200=40
I(B4)=220C(B)C(4)=220—80200=100
The most negative is I(A4)=40 so we can reduce the cost by introducing route A4. The %theta in the corresponding matrix cell.
Site 1 
Site 2 
Site 3 
Site 4 
Supply of Gravel (Tons) 

Quarry A 
12 
0 
11 – %theta 
%theta 
23 
Quarry B 
13 
3 
16 

Quarry C 
1+ %theta 
18 %theta 
19 

Demand for Gravel (Tons) 
12 
13 
15 
18 
58 
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