👉Chandigarh Call Girls 👉9878799926👉Just Call👉Chandigarh Call Girl In Chandiga...
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.
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
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.
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
