quantitative analysis -transport method 1
DESCRIPTION
MBA student class presentationTRANSCRIPT
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Transportation Methods
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Developing Initial Solutions
• Northwest Method• Intuitive Method• Vogel Approximation Method
Evaluating Solutions
• Stepping Stone Method• MODI Method
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5 9 3
100
2 6
100
200
200 30 90 80
4
1
B CA
2
Warehouse F
act
ory
Supply
Demand
Example
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1
B CA
2
X1 X2 X3
X4 X5 X6
OBJ. FUNCTION : Minimize C = 5X1 + 9X2 + 3X3 + 4X4 + 2X5 + 6X6
Constraints :
X1 +X2 + X3 = 100
X4 +X5 + X6 = 100
X1 +X4 = 80
X2 + X5 = 90
X3 +X6 = 30
Conditions :
X1 , X2 , X3 , X4 , X5 , X6 = > 0
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Northwest corner rule
• Allocate the available supply to the cells starting from the northwest corner (Upper left) and moving down vertically or horizontally, satisfying the demand.
• Observed that the number of cells used must be equal to the number of rows + the number of columns – one.
• Compute for the cost of transportation summing up the products of the cost of transport and amount to be transported to the destination.
• Proceed to check for areas of improvement using any of the methods stated in the next number.
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5 9 3
80
2 6
100
200
200 30 90 80
4
1
B CA
2
Northwest Method
100
N
E
S
W
20
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5 9 3
80
2 6
100
200
200 30 90 80
4
1
B CA
2
100 20 20
Northwest Method
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5 9 3
80
2 6
100
200
200 30 90 80
4
1
B CA
2
100 20
70
20
30
Northwest Method
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5 9 3
80
2 6
100
200
200 30 90 80
4
1
B CA
2
100 20
70 30
TC =80(5)+20(9) + 70(2) + 30(6) = 900Completed Cells = R + C - 1 = 2 + 3 - 1 = 4
20
30
Northwest Method
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B CA
2
Factory
80 20
70 30
-
-
+
+
1B 2C
2A
3
1A -5 +4
+3____ ____
+ -
2A
-2
+7 -7
+ 6
Stepping Stone Method
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5 9 3
100
2 6
100
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200 30 90 80
4
1
B CA
2
1B
2C -6 +3
+2____ ____
+ -
1C
-9
+5 -15
-10
-
-
+
+
80 20
70 30 2B
1C
Stepping Stone Method
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5 9 3
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B CA
2
80 20
90 10
-
TC =80(5)+20(3) + 90(2) + 10(6) = 700
Stepping Stone Method
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5 9 3
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1
B CA
2
80 20
90 10
+
+
-
-
2B
1C -3 +9
+6____ ____
+ -
1B
-2
+15 -5
2C
1B
+10
Stepping Stone Method
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5 9 3
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1
B CA
2
80 20
90 10
-
-
+
+
2C
1A -5 +4
+2____ ____
+ -
2A
-6
+5 -11
-4
1C
2A
Stepping Stone Method
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5 9 3
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1
B CA
2
80 20
90 10
-
-
+
+
2C
1A -5 +4
+2____ ____
+ -
2A
-6
+5 -11
-4
1C
2A
Stepping Stone Method
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5 9 3
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1
B CA
2
70 30
90 10
Stepping Stone Method
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1
B CA
2
70 30
90 10
- +
+ -
2B
1A -5 +9
+4____ ____
+ -
1B
-2
+13 -7
+ 6
2A
1B
Stepping Stone Method
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1
B CA
2
70 30
90 10
- +
+ -
2B
1C -3 +5
+6____ ____
+ -
2C
-4
+11 -7
+ 4
2A
1A
Stepping Stone Method
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B CA
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70 30
90 10
TC =70(5)+30(3) + 10(4) + 90(2) = 660 SOLUTION IS OPTIMAL
Stepping Stone Method
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Least Cost Method (Intuitive Method)
• Start from the cell with the least cost. Then work your way to the other cells always considering the least transportation cost.
• Compute the cost of transport and improve using any of the two methods in no. 4.
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1
B CA
2
Intuitive Method
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5 9 3
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1
B CA
2
Intuitive Method
90
10
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5 9 3
100
2 6
100
30 90 80
4
1
B CA
2
Intuitive Method
30
5 9 3
100
2 6
100
200
200 30 90 80
4
1
B CA
2 90
10
70
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5 9 3
100
2 6
100
30 90 80
4
1
B CA
2
Intuitive Method
30
5 9 3
100
2 6
100
200
200 30 90 80
4
1
B CA
2 90 10
70
10
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5 9 3
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2 6
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1
B CA
2
Intuitive Method
30 70
90 10
70
10
Completed Cells = R + C - 1 = 2 + 3 - 1 = 4 TC =70(5)+30(3) + 10(4) + 90(2) = 660
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2 6
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1
B CA
2
MODI Method
30 70
90 10
Index
0
5 3
-1
3
Cell Cost - ( Row Index + Column Index )
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5 9 3
100
2 6
100
200
200 30 90 80
4
1
B CA
2
MODI Method
30 70
90 10
Index
0
5 3
-1
3
1B: 9 - ( 0 + 3 ) = + 6
2C: 6 - (-1 + 3 ) = + 4
SOLUTION IS OPTIMAL
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VAM - Vogel Approximation Method
Based on the concept of minimizing opportunitycost.
The opportunity cost of a given row or column is the difference between the lowest cost and the second lowest cost alternative.
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Procedure :
1. For each row an column, select the lowest and second lowest alternatives from those not already allocated and calculate the opportunity cost.
2. Scan the opportunity cost figures and identify the row or columns with the largest opportunity cost.
3. Allocate as many units as possible to this row or column in the square with the least cost.
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100
100 45 25 30
VAM - Vogel Approximation Method
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
Row/Column 2nd Lowest Cost - Lowest Cost = Opportunity Cost
Row W 8 5 3Row X 6 4 2Row Y 7 6 1Column A 7 6 1Column B 8 6 2Column C 5 4 1
Largest
25
VAM - Vogel Approximation Method
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
Row/Column 2nd Lowest Cost - Lowest Cost = Opportunity Cost
Row X 6 4 2Row Y 7 6 1Column A 7 6 1Column B 8 6 2Column C 9 4 5 Largest
25
20
VAM - Vogel Approximation Method
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
Row/Column 2nd Lowest Cost - Lowest Cost = Opportunity Cost
Row X 8 6 2Row Y 7 6 1Column A 7 6 1Column B 8 6 2
25
20
Largest
Largest
25
VAM - Vogel Approximation Method
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15
VAM - Vogel Approximation Method
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Thank You!
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TC =25 (5)+15(6) + 20(4) + 15(7) + 25(6) = 550
B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15
7 6 5
0
-1
0
WA: 9 - ( 0 + 7 ) = + 2
WB: 8 - ( 0 + 6 ) = + 2
XB: 8 - ( -1 + 6 ) = + 3
YC: 9 - ( 0 + 5 ) = + 4
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15
+
+
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15 +
+
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15 +
+
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B CA
W
X
Y
9
6
7
8 5
8 4
6 9
25
35
40
100 100
45
25
30
25
20
25 15
15
+
+
+