Least Cost Method
Vogel’s approximation method
+ example
Problem solution.
+ example
Problem solution.
Question
No 1:
Determine
the initial basic feasible solution to the following transportation problem
using the following two methods.
a)
Least Cost Method
b) Vogel’s approximation method
Order
|
1
|
2
|
3
|
4
|
Supply
|
A
|
10
|
2
|
20
|
11
|
15
|
B
|
12
|
7
|
9
|
20
|
25
|
C
|
4
|
14
|
16
|
18
|
10
|
Demand
|
5
|
15
|
15
|
15
|
50
|
a) Least Cost Method
Order
|
1
|
2
|
3
|
4
|
Supply
|
A
|
10
|
2(15)
|
20
|
11
|
15
|
B
|
12
|
7
|
9
|
20
|
25
|
C
|
4
|
14
|
16
|
18
|
10
|
Demand
|
5
|
15
|
15
|
15
|
Order
|
1
|
3
|
4
|
Supply
|
|
B
|
12
|
9
|
20
|
25
|
|
C
|
4 (5)
|
16
|
18
|
||
Demand
|
5
|
15
|
15
|
Order
|
3
|
4
|
Supply
|
||
B
|
9(15)
|
20
|
|||
C
|
16
|
18
|
5
|
||
Demand
|
15
|
15
|
Order
|
4
|
Supply
|
|||
B
|
20
|
10
|
|||
C
|
18(5)
|
5
|
|||
Demand
|
Order
|
4
|
Supply
|
|||
B
|
20(10)
|
10
|
|||
Demand
|
10
|
Cost of the allocation by the least cost method is:
2*15+4*5+9*15+18*5+20*10=30+20+135+90+200=475
b) Vogel’s
approximation method
Order
|
1
|
2
|
3
|
4
|
Supply
|
A
|
10
|
2
|
20
|
11
|
15
|
B
|
12
|
7
|
9
|
20
|
25
|
C
|
4
|
14
|
16
|
18
|
10
|
Demand
|
5
|
15
|
15
|
15
|
50
|
Order
|
1
|
2
|
3
|
4
|
Supply
|
Row difference
|
A
|
10
|
2
|
20
|
11
|
15
|
(8)
|
B
|
12
|
7
|
9
|
20
|
25
|
(2)
|
C
|
4(5)
|
14
|
16
|
18
|
(10)
|
|
Demand
|
5
|
15
|
15
|
15
|
||
Column difference
|
(6)
|
(5)
|
(7)
|
(7)
|
Order
|
2
|
3
|
4
|
Supply
|
Row difference
|
|
A
|
2(15)
|
20
|
11
|
15
|
(9)
|
|
B
|
7
|
9
|
20
|
25
|
(2)
|
|
C
|
14
|
16
|
18
|
5
|
(2)
|
|
Demand
|
15
|
15
|
15
|
|||
Column difference
|
(5)
|
(7)
|
(7)
|
Order
|
3
|
4
|
Supply
|
Row difference
|
||
B
|
9(15)
|
20
|
(11)
|
|||
C
|
16
|
18
|
5
|
(2)
|
||
Demand
|
15
|
15
|
||||
Column difference
|
(7)
|
(2)
|
Order
|
4
|
Supply
|
||||
B
|
20
|
10
|
||||
C
|
18(5)
|
5
|
||||
Demand
|
||||||
Order
|
4
|
Supply
|
|||
B
|
20
|
10
|
|||
Demand
|
10
|
Order
|
4
|
Supply
|
|||
B
|
20 (10)
|
10
|
|||
Demand
|
10
|
Cost of the allocation by the least cost method is:
5*4+2*15+15*9+18*5+20*10=20+30+135+90+200=475
Question
No 2:
Initial basic feasible solution of a transportation problem
using the Vogel’s approximation method is given in the following table
Warehouse
|
Customer
|
Supply
|
|||
1
|
2
|
3
|
4
|
||
A
|
1(20)
|
2
|
1(10)
|
4
|
30
|
B
|
3
|
3(20)
|
2(20)
|
1(10)
|
50
|
C
|
4
|
2(20)
|
5
|
9
|
20
|
Demand
|
20
|
40
|
30
|
10
|
|
UV-multiplier
process is used to find the optimality condition. Find
Warehouse
|
Customer
|
Supply
|
||||
1
|
2
|
3
|
4
|
|||
A
|
1(20)
|
2
|
1(10)
|
4
|
30
|
 |
B
|
3
|
3(20)
|
2(20)
|
1(10)
|
50
|
 |
C
|
4
|
2(20)
|
5
|
9
|
20
|
 |
Demand
|
20
|
40
|
30
|
10
|
||
 |
 |
 |
 |
|||
Warehouse
|
Customer
|
||||
1
|
2
|
3
|
4
|
||
A
|
1
|
-
|
1
|
-
|
|
B
|
-
|
3
|
2
|
1
|
|
C
|
4
|
-
|
5
|
9
|
|
|
|
|
|
||
Warehouse
|
Customer
|
||||
1
|
2
|
3
|
4
|
||
A
|
1
|
1
|
=-1 |
||
B
|
3
|
2
|
1
|
=0 |
|
C
|
4
|
5
|
9
|
=-1 |
|
| =2 |
=3 |
=2 |
=1 |
||
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