This document summarizes the Stepping Stone Method used in Operations Research to determine an optimal initial basic feasible solution for transportation problems. The method treats the transportation table as a pond that can only be crossed using occupied cells as stepping stones. It involves forming closed loops through unused and occupied cells, calculating improvement indices, and reallocating units to find an optimal solution. The example problem shows applying the method to find the optimal allocation of products from warehouses to destinations.

1. PRESENTATION
ON
“STEPPING STONE METHOD”
OPERATIONS RESEARCH
SUBMITTED BY:
HIMANSHI GUPTA (140120119057)/ ME/A1
GUIDED BY:
PROF. PARANJAY PATEL
(2171901)
GANDHINAGAR INSTITUTE OF TECHNOLOGY

2. INTRODUCTION
• This is a one of the methods used to determine optimality of an initial
basic feasible solution (i.e. Northwest Corner Rule, Least Cost or
Vogel’s Approximation).
• The method is derived from the analogy of crossing a pond using
stepping stones.
• This means that the entire transportation table is assumed to be a pond
and the occupied cells are the stones needed to make certain
movements within the pond.

3. STEPS INVOLVED
• Starting at an unused/empty cell, trace a closed path or loop back to the
original cell via cells that are currently being used and/or occupied.
• Note: A closed path or loop is a sequence of cells in the transportation
table such that the first cell is unused/empty and all the other cells are
used/occupied with the following conditions:
a) Each pair of consecutive used/occupied cells lies in either the same
row or column.
b) No three consecutive used/occupied cells lie in the same row or
column.
c) The first and last cells of a sequence lies in the same row or column.
d) No cell appears more than once in a sequence (i.e. no duplication).
e) Only horizontal and vertical moves allowed and can only change
directions at used/occupied cells

4. PROBLEM
Q. This is the transportation table for a company from its different
warehouses. Find the most optimum solution.

5. SOLUTION STEP 1
• By VAM we find the initial basic feasible solution.
• Here number of allotments is 6 which is equal to n+m-1

6. SOLUTION STEP 2
• Now for every unoccupied cell form a loop
Δ=+8-6+7-8+5-4=2

7. SOLUTION STEP 3
Δ=+8-8+5-4=1

8. SOLUTION STEP 4
Δ=+6-7+8-5=2

9. SOLUTION STEP 5
Δ=+8-7+8-5+4-6=2

10. SOLUTION STEP 6
Δ=+4-6+7-5=0

11. SOLUTION STEP 7
Δ=+7-8+6-6=-1

12. SOLUTION STEP 8
• Each negative cost indicates the amount by which
transportation cost can be reduced by allocating one unit of
product were to be shipped from that source
• Choose the source which has most negative amount
• We can find maximum possible units which can be transported
by forming a loop
• Now for cell C-3 the value is negative so we will have to form
a loop and do reallocation of units

13. SOLUTION STEP 9
• The maximum quantity that can be shipped on the new
money-saving route can be found by referring to the closed
path of plus signs and minus signs drawn for the route and
selecting the smallest number found in those squares
containing minus signs
• To find the new solution add the smallest number to all the
squares with positive sign and subtract it from cells with
negative sign

14. SOLUTION STEP 10

15. • We have a new basic feasible solution:

16. FINAL SOLUTION

17. • Repeat all the above steps to calculate an Improvement Index for all
unused squares in order to test whether an optimal solution has been
reached.
• Since the results of all indices calculated are greater than or equal to
zero, then, an optimal solution had been reached.
• In second iteration, there is an improvement index equal to zero. This
means that there is an ALTERNATE optimum solution.