transporation problem - stepping stone method
- 1. Transportation Problem - Stepping Stone Method - PAMANTASAN NG LUNGSOD NG MAYNILA GRADUATE SCHOOL OF ENGINEERING GEM 805 – OPTIMIZATION TECHNIQUES
- 2. Stepping Stone Method >>> 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. Optimum Solution: Stepping-Stone Method 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Z = 4x10+6x30+6x50+7x10+5x10+8x40 = 960 Transportation Table
- 4. 1. 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 Optimum Solution: Stepping-Stone Method
- 5. Example: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method A3->B3->B4->C4->C1->A1->A3At Cell A3,
- 6. Example: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method At Cell A4, A4->C4->C1->A1->A4
- 7. 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 Optimum Solution: Stepping-Stone Method B1->B4->C4->C1->B1 SOURCES DESTINATIONS Example: At Cell B1,
- 8. Example: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method At Cell B2, B2->B4->C4->C1->A1->A2->B2
- 9. Example: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method At Cell C2, C2->C1->A1->A2->C2
- 10. Example: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method At Cell C3, C3->B3->B4->C4->C3
- 11. 2. For every traced path or loop, begin with a plus (+) sign at the starting unused cell and alternately place a minus (-) and plus (+) sign at each used cell 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS - - - + + + Example: Optimum Solution: Stepping-Stone Method At Cell A3, A3->B3->B4->C4->C1->A1->A3
- 12. 3. Calculate an Improvement Index by first adding the unit-cost figures found in each cell containing a plus sign and subtracting the unit costs in each square containing a minus sign. 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS - - - + + + Example: At Cell A3, A3->B3->B4->C4->C1->A1->A3 Optimum Solution: Stepping-Stone Method IA3 = 2 -8 -6 +7 +5 -4 = 8
- 13. Optimum Solution: Stepping-Stone Method Iteration #1 - Computing for the Improvement Index: At A3, A3->B3->B4->C4->C1->A1; IA3 = +8-6+7-8+5-4 = 2 At A4, A4->C4->C1->A1; IA4 = +8-8+5-4 = 1 At B1, B1->B4->C4->C1; IB1 = +6-7-8-5 = 2 At B2, B2->B4->C4->C1->A1->A2; IB2 = +8-7+8-5+4-6 = 2 At C2, Loop C2->C1->A1->A2; IC2 = +7-5+4-6 = 0 At C3, C3->B3->B4->C4; IC3 = +6-6+7-8 = -1 4. If all indices calculated are greater than or equal to zero, then, an optimal solution had been reached. If not, select the path/loop that has the most negative value and use this to further improve the solution. Note: Should there be two or more “most” negative values, select arbitrarily.
- 14. Example: At Cell C3, C3->B3->B4->C4 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method + + - - IC3 = +6-6+7-8 = -1
- 15. To further improve the current solution, select the “smallest” number found in the path/loop C3->B3->B4->C4 containing minus(-) signs. This number is added to all cells on the closed path/loop with plus(+) signs and subtracted from all cells on the path assigned with minus(-) signs. Optimum Solution: Stepping-Stone Method 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 50 10 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS + + - - 40 50 - 40 10 + 40 40 - 40
- 16. 5. Then, we have a new basic feasible solution… 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 10 50 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method …and repeat steps 1 though 4 to calculate an Improvement Index for all unused squares in order to test whether an optimal solution has been reached.
- 17. Optimum Solution: Stepping-Stone Method Iteration #2 - Computing for the Improvement Index: At A3, A3->C3->C1->A1; IA3 = +8-6+5-4 = 3 At A4, A4->B4->B3->C3->C1->A1; IA4 = +8-7+6-6+5-4 = 2 At B1, B1->B3->C3->C1; IB1 = +6-6+6-5 = 1 At B2, B2->B3->C3->C1->A1->A2; IB2 = +8-6+6-5+4-6 = 1 At C2, C2->C1->A1->A2; IC2 = +7-5+4-6 = 0 At C4, C3->B3->B4; IC3 = +8-6+6-7 = 1 Since the results of all indices calculated are greater than or equal to zero, then, an optimal solution had been reached.
- 18. …and computing the objective function Z: 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 10 50 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method Z = 4x10+6x30+6x10+7x50+5x10+6x40 = 920
- 19. In Iteration #2 : At A3, A3->C3->C1->A1; IA3 = +8-6+5-4 = 3 At A4, A4->B4->B3->C3->C1->A1; IA4 = +8-7+6-6+5-4 = 2 At B1, B1->B3->C3->C1; IB1 = +6-6+6-5 = 1 At B2, B2->B3->C3->C1->A1->A2; IB2 = +8-6+6-5+4-6 = 1 At C2, C2->C1->A1->A2; IC2 = +7-5+4-6 = 0 At C4, C3->B3->B4; IC3 = +8-6+6-7 = 1 Optimum Solution: Stepping-Stone Method However, in checking the calculation in Iteration #2, there is an improvement index equal to zero. This means that there is an ALTERNATE optimum solution:
- 20. To calculate for the alternate optimum solution, again select the “smallest” number found in this path/loop containing minus(-) signs. This number is added to all cells on the closed path/loop with plus(+) signs and subtracted from all cells on the path assigned with minus(-) signs. Optimum Solution: Stepping-Stone Method 1 2 3 4 SUPPLY A 4 6 8 8 40 10 30 B 6 8 6 7 60 10 50 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS + + - - 10 30 - 1010 + 10 10 - 10 Hence, at C2->C1->A1->A2,
- 21. Then the alternate optimum solution with objective function Z: 1 2 3 4 SUPPLY A 4 6 8 8 40 20 20 B 6 8 6 7 60 10 50 C 5 7 6 8 50 10 40 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS Optimum Solution: Stepping-Stone Method Z = 4x20+6x20+6x10+7x50+7x10+6x40 = 920
- 22. 1 2 3 4 SUPPLY A 4 6 8 8 40 20 20 B 6 8 6 7 60 10 50 C 5 7 6 8 50 50 DEMAND 20 30 50 50 150 SOURCES DESTINATIONS When the number of empty/occupied cells in any solution (either initial or later) of the transportation table is not equal to the number of rows plus the number of columns minus 1 (i.e. m+n-1) the solution is called DEGENERATE Optimum Solution: Stepping-Stone Method Example: m + n -1 = 3 + 4 -1 = 6 DEGENERACY
- 23. DEGENERACY To handle degenerate problems, artificially create an occupied cell by placing a zero (representing a fake shipment) in one of the unused cells. Treating this cell as if it were occupied, it must be chosen in such a position as to allow all stepping-stone paths to be traced. Then, all stepping-stone paths can be closed and improvement indices computed. Optimum Solution: Stepping-Stone Method 1 2 3 4 SUPPLY A 4 6 8 8 40 20 20 B 6 8 6 7 60 10 50 C 5 7 6 8 50 50 DEMAND 20 30 50 50 150 SOURCES 0 Example: DESTINATIONS
- 25. DIOS MABALOS PO! Cam on ! Shukriya ! ありがとうございます！ Thank you! Merci! Gracias! Obrigado! 謝謝！