PAMANTASAN NG LUNGSOD NG MAYNILA GRADUATE SCHOOL OF ENGINEERING GEM 805
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OPTIMIZATION TECHNIQUES
Transportation Problem - Steppi Stepping ng Ston Stone e Metho Method d-
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 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 Transportation ransport ation Table DESTINATIONS
1 4
2 6
3 8
4 8
A
40 10 6
SOURCES
30 8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
SUPPLY
20
40 30
50
50
Z = 4x10+6x30+6x50+7x10+5x1 4x10+6x30+6x50+7x10+5x10+8x40 0+8x40 = 960
150
Optimum Solution: Stepping-Stone Method 1. Starting at an unused/empty cell, trace 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 consecutiv consecutive e used/occupied used/occupied cells lies lies in either the same row or column b. No three consecutiv consecutive e used/occupied used/occupied cells cells lie in the same row row or column
c. The first and last last cells of a sequence lies in the the same row or column d. No cell appears appears more more than once in a sequence sequence (i.e. (i.e. no duplication) e. Only horizon horizontal tal and vertica verticall moves allowed allowed and can can only change directions directions at used/occupie used/occupied d cells cells
Optimum Solution: Stepping-Stone Method Example:
At Cell A3, A3->B3->B4->C4->C1 A3->B3->B4->C4->C1->A1->A3 >A1->A3
DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method Example:
At Cell A4,
A4->C4->C1->A1->A4
DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method Example: At Cell B1,
B1->B4->C4->C1->B1 DESTINATIONS
1 4
2 6
3 8
4 8
A
40 10
SOURCES
6
30 8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
SUPPLY
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method Example:
At Cell B2,
B2->B4->C4->C1->A1->A2->B2 DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method Example:
At Cell C2,
C2->C1->A1->A2->C2
DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method Example:
At Cell C3,
C3->B3->B4->C4->C3 DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 50 5
7
6
10 8
C
50 10
DEMAND
20
40 30
50
50
150
Optimum Solution: Stepping-Stone Method 2. For every traced traced path or loop, loop, begin begin with with a plus plus (+) sign at the starting unused cell and alternately place a minus (-) and plus (+) sign at each used cell A3, A3->B3->B4->C4 A3->B3->B4->C4->C1->A1 ->C1->A1->A3 ->A3 Example: At Cell A3,
-
DESTINATIONS
1
4
2 6
3
4
+
8
SUPPLY
8
A
40 10 6
SOURCES
-
30 8
6
B
7
50 5 C
7
6
8
+
10 DEMAND
20
10
40 30
50
50
+
-
60
50
150
Optimum Solution: Stepping-Stone Method 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. Example: At Cell A3, A3->B3->B4->C4->C1->A1->A3 A3->B3->B4->C4->C1->A1->A3
-
DESTINATIONS
1
4 -4
2 6
3
4
+
8 8
SUPPLY
8
A
40 10 6
SOURCES
-
30 8
6 -6
B
7 +7
50 5 +5 C
7
10 8 -8
6
+
10 DEMAND
40
20
30
IA3 =
50
50
=2
+
-
60
50
150
Optimum Solution: Stepping-Stone Method Iteration Iteration #1 - Computing Computing for the Improveme Improvement nt Index: Index: At A3, A3->B3->B4->C4->C1->A1; A3->B3->B4->C4->C1->A1; I A3 = +8-6+7-8+5-4 = 2 At A4, A4->C4->C1->A1;
I A4 = +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; 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 are greater than or equal to zero, 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.
Optimum Solution: Stepping-Stone Method Example: At Cell C3, C3->B3->B4->C4
IC3 = +6-6+7-8 = -1
DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
-
7
+
B 50 5
7
6
+
10
-
8
C 10 DEMAND
20
60
50
40 30
50
50
150
Optimum Solution: Stepping-Stone Method To further improve 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. DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
B
50 40 50- 40
5
7
6
+
7
+
10 1 +040
-
8
C 10 DEMAND
20
40 30
50
60
50
4040 - 40 50
150
Optimum Solution: Stepping-Stone Method 5. Then, we have a new basic feasible solution… DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10
30
SOURCES
6
8
6
7
B
60 10 5
7
6
50 8
C
50 10
DEMAND
20
40 30
50
50
150
…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.
Optimum Solution: Stepping-Stone Method Iteration Iteration #2 - Computing Computing for the Improveme Improvement nt Index: Index: At A3, A3->C3->C1->A1;
I A3 = +8-6+5-4 = 3
At A4, A4->B4->B3->C3->C1->A1; A4->B4->B3->C3->C1->A1; I A4 = +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; 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; 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 b een reached.
Optimum Solution: Stepping-Stone Method …and computing the objective function Z: DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 10 6
SOURCES
30 8
6
7
B
60 10 5
7
6
50 8
C
50 10
DEMAND
20
40 30
50
50
Z = 4x10+6x30+6x10+7x50+5x10 4x10+6x30+6x10+7x50+5x10+6x40 +6x40 = 920
150
Optimum Solution: Stepping-Stone Method However, However, in checking the calculation in Iteration #2, there is an zero. This means that there is an improvement index equal to zero. ALTERNA ALTERNATE TE optimum solution: In Iteration #2 : At A3, A3->C3->C1->A1;
I A3 = +8-6+5-4 = 3
At A4, A4->B4->B3->C3->C1->A1; A4->B4->B3->C3->C1->A1; I A4 = +8-7+6-6+5-4 = 2 At B1, B1->B3->C3->C1; B1->B3->C3->C1;
IB1 = +6-6+6-5 = 1
At B2, B2->B3->C3->C1-> B2->B3->C3->C1->A1->A2; 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; C3->B3->B4;
IC3 = +8-6+6-7 = 1
Optimum Solution: Stepping-Stone Method 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. Hence, at C2->C1->A1->A2,
DESTINATIONS
1 4 A
+
2
-
6
101+ 0 10
3 8
4
SUPPLY
8 40
303-0 10
SOURCES
6
8
6
7
B
60 10 5
C
1010 - 10
DEMAND
20
+
7
6
50 8 50
10 30
40 50
50
150
Optimum Solution: Stepping-Stone Method Then the alternate optimum solution with objective function Z: DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 20 6
SOURCES
20 8
6
7
B
60 10 5
7
6
50 8
C
DEMAND
50
20
10
40
30
50
50
Z = 4x20+6x20+6x10+7x50+7x1 4x20+6x20+6x10+7x50+7x10+6x40 0+6x40 = 920
150
Optimum Solution: Stepping-Stone Method DEGENERACY 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) m+n-1) the solution is called DEGENERATE Example:
m + n -1 = 3 + 4 -1 = 6 DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 20
20
SOURCES
6
8
6
7
B
60 10 5
7
50 6
8
C
50 50
DEMAND
20
30
50
50
150
Optimum Solution: Stepping-Stone Method 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 stepp ing-stone paths to be traced. Then, all stepping-stone paths can be closed and improvement indices computed. Example:
DESTINATIONS
1 4
2 6
3 8
4
SUPPLY
8
A
40 20
20
SOURCES
6
8
6
7
B
60 10 5
7
0
50 6
8
C
50 50
DEMAND
20
30
50
50
150
Optimum Solution: Stepping-Stone Method
QUESTIONS?
DIOS MABALOS PO! Cam on ! Shukriya ! Thank you! Merci! Gracias! Obrigado!
