transporation problem - stepping stone method

Transportation Problem - Stepping Stone Method -

PAMANTASAN NG LUNGSOD NG MAYNILAGRADUATE SCHOOL OF ENGINEERING

GEM 805 – OPTIMIZATION TECHNIQUES

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.

Optimum Solution:Stepping-Stone Method

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS

Z = 4x10+6x30+6x50+7x10+5x10+8x40 = 960

Transportation Table

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


Example:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


A3->B3->B4->C4->C1->A1->A3

At Cell A3,

Example:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


At Cell A4, A4->C4->C1->A1->A4

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150


B1->B4->C4->C1->B1

SOURCES

DESTINATIONS

Example: At Cell B1,

Example:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


At Cell B2,

B2->B4->C4->C1->A1->A2->B2

Example:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


At Cell C2, C2->C1->A1->A2->C2

Example:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


At Cell C3,

C3->B3->B4->C4->C3

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

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 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

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

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 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


IA3 = 2

-8

-6 +7

+5

-4

=

8


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 = -14. 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.

Example: At Cell C3, C3->B3->B4->C4

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


+

+

-

-

IC3 = +6-6+7-8 = -1

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.


1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 50 10

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS

+

+

-

-

40

50 - 40 10 + 40

40 - 40

5. Then, we have a new basic feasible solution…

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 10 50

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


…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.


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 = 1Since the results of all indices calculated are greater than or equal to zero, then, an optimal solution had been reached.

…and computing the objective function Z:

1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 10 50

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


Z = 4x10+6x30+6x10+7x50+5x10+6x40 = 920

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


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:

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.


1 2 3 4 SUPPLY

A4 6 8 8

40 10 30

B6 8 6 7

60 10 50

C5 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,

Then the alternate optimum solution with objective function Z:

1 2 3 4 SUPPLY

A4 6 8 8

40 20 20

B6 8 6 7

60 10 50

C5 7 6 8

50 10 40

DEMAND 20 30 50 50 150

SOURCES

DESTINATIONS


Z = 4x20+6x20+6x10+7x50+7x10+6x40 = 920

1 2 3 4 SUPPLY

A4 6 8 8

40 20 20

B6 8 6 7

60 10 50

C5 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


Example: m + n -1 = 3 + 4 -1 = 6

DEGENERACY

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.


1 2 3 4 SUPPLY

A4 6 8 8

40 20 20

B6 8 6 7

60 10 50

C5 7 6 8

50 50

DEMAND 20 30 50 50 150

SOURCES

0

Example: DESTINATIONS

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transporation problem - stepping stone method