The problem solved by the Stepping Stone algorithm is as follows:
Let’s take an example to show how the algorithm works. Let’s take four origins and five applicants with the costs and quantity following the table:
c_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

7

12

1

5

6

12

O_{2}

15

3

12

6

14

11

O_{3}

8

16

10

12

7

14

O_{4}

18

8

17

11

16

8

demande

10

11

15

5

4

It is possible to perform the algorithm using two tables (one for costs, one for flows). It is also possible to display the two values in the same box since only the flow will vary.
Step 1: To obtain a basic solution
By Northwest corner
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

10

2

–

–

–

12

O_{2}

–

9

2

–

–

11

O_{3}

–

–  13 
1

–

14

O_{4}

–

–

–

4

4

8

demande

10

11

15

5

4

The total cost of the basic solution is: 10*7+2*12+9*3+2*12+13*10+12+4*11+4*16 = 395.
By BalasHammer
This method uses the transport cost difference between the two best choices for supply and demand. The basic solution is often very close to the optimal solution.
The first iteration of the method gives: D_{O1} = 4, D_{O2} = 3, D_{O3} = 1, D_{O4} = 3, D_{D1} = 1, D_{D2} = 5, D_{D3} = 9, D_{D4} = 1, D_{D5} = 2. The column D_{3} have the biggest cost difference, its lowest cost is 1 so we saturate this element O_{1} D_{3} with the flow min(12, 15). The row O_{1} is saturated.
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

X

X

12

X

X

12

O_{2}

–

–

–

–

–

11

O_{3}

–

–  – 
–

–

14

O_{4}

–

–

–

–

–

8

demande

10

11

15

5

4

For the new cost difference calculation, we will no longer consider the values of raw O_{1}. We get after 5 iterations to the following configuration:
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

–

–

12

–

–

12

O_{2}

–

11

–

–

–

11

O_{3}

10  –  – 
–

4

14

O_{4}

–

–

3

5

–

8

demande

10

11

15

5

4

Step 2: Calculating potentials MODI
Once you have a basic solution, the idea is to modify the solution in order to improve the objective function. That is to say that the flow must be modified. For that, we will choose a flow lowering at most the total cost of transport. The first step in determining this flow is to calculate the potentials. The potentials are calculated ONLY on the boxes having a nonzero flow!
We can then calculate other potentials. The potentials are calculated gradually. In our case, we calculated the potential of raw 1 from c_{12}, so it is possible to calculate the potential of column 1 or column 2.
We can solve for column 1, p_{D1} = c_{11} + P_{O1} = 7. For raw 2, p_{D2} = c_{12} + P_{O1} = 12. Dnd so on: p_{O2} = p_{D2} – c_{22} = 12 3 = 9; p_{D3} = 21; p_{03} = 11; P_{D4} = 23: P_{O4} = 12; P_{D5} = 28.
Step 3: Calculation of unit cost variation
v_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

p_{O}

O_{1}

–

–

20

18

19

0

O_{2}

17

– 
–

8

5

9

O_{3}

12  15  – 
–

10

11

O_{4}

23

8

8

–

–

12

p_{D}

7

12

21

23

28

Step 4: Calculation of the maximum amount of flow
For example, the element 0_{1}D_{3} takes the following circuit f_{13} > f_{12} > f_{22} > f_{23} > f_{13}with a minimum flow of 2.
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

p_{O}

O_{1}

–

–

2 
1

1

0

O_{2}

–  – 
–

1

1

9

O_{3}

–  –  – 
–

1

11

O_{4}

–

–

–

–

–

12

p_{D}

7

12

21

23

28

By multiplying f_{ij}*v_{ij}, we know the variation of the total cost by the modification by the flow f_{ij}. We choose the element with the largest f_{ij}*v_{ij}, here it is O_{1}D_{3}
Step 5: To update the table
The computation of update of the waves is done by the rule of the « + – » without counting the return to the box of origin. Thus in the circuit: f_{13} += 2, f_{12} =2, f_{22} += 2 and f_{23} = 2. The table is as follows:
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

10

22 = –

0+2 = 2 
–

–

12

O_{2}

–  9+2 = 11 
22 = –

–

–

11

O_{3}

–  –  13 
1

–

14

O_{4}

–

–

–

4  4 
8

demande

10

11

15

5

4

The total cost of the basic solution is: 10 * 7 + 2 * 12 + 9 * 3 + 2 * 12 + 13 * 10 + 12 + 4 * 11 + 4 * 16 = 395. The cost of this solution is: 10 * 7 + 11 * 3 + 2 * 1 + 13 * 10 + 12 + 4 * 11 + 4 * 16 = 355. This is the basic solution minus f_{13}*v_{13} = 2*20 = 40.
In our example, we have in the end the following table:
f_{ij}

D_{1}

D_{2}

D_{3}

D_{4}

D_{5}

offer

O_{1}

–

–

12 
–

–

12

O_{2}

–  11 
–

–

–

11

O_{3}

10  –  – 
–

4

14

O_{4}

–

–

3

5  – 
8

demande

10

11

15

5

4

With a total cost of: 10*8+11*3+12*1+3*17+5*11+4*7 = 259.
Aside
We talk about flow because the problem is solved as a problem of min cost flow (maximum flow at minimum cost) in a complete bipartite graph – a set of sources connected to a set of sinks (hence the rule of « + – » since we go once in the direction of the flow then in the wrong direction in the chosen circuit).