Here Total Demand = 75 is greater than Total Supply = 65. So We
add a dummy supply constraint with 0 unit cost and with allocation
10.
Now, The modified table is
A | B | C | Supply | |
1 | 10 | 25 | 13 | 20 |
2 | 20 | 25 | 8 | 10 |
3 | 12 | 30 | 16 | 35 |
Sdummy | 0 | 0 | 0 | 10 |
Demand | 20 | 30 | 25 |
The rim values for 1=20 and A=20 are compared.
The smaller of the two i.e. min(20,20) = 20 is
assigned to 1 A
This exhausts the capacity of 1 and leaves 20 - 20 = 0 units with
A
Table-1
A | B | C | Supply | |
1 |
10(20) |
25 | 13 | 0 0=20-20 |
2 |
20 |
25 | 8 | 10 |
3 |
12 |
30 | 16 | 35 |
Sdummy |
0 |
0 | 0 | 10 |
Demand | 0 0=20-20 | 30 | 25 |
The rim values for 2=10 and A=0 are compared.
The smaller of the two i.e. min(10,0) = 0 is
assigned to 2 A
This meets the complete demand of A and leaves 10 - 0 = 10
units with 2
Table-2
A | B | C | Supply | |
1 |
10(20) |
25 | 13 | 0 |
2 |
20 |
25 | 8 | 10 10=10-0 |
3 |
12 |
30 | 16 | 35 |
Sdummy |
0 |
0 | 0 | 10 |
Demand | 0 0=0-0 | 30 | 25 |
The rim values for 2=10 and B=30 are compared.
The smaller of the two i.e. min(10,30) = 10 is
assigned to 2 B
This exhausts the capacity of 2 and leaves 30 - 10 = 20 units with
B
Table-3
A | B | C | Supply | |
1 |
10(20) |
25 | 13 | 0 |
2 |
20 |
25(10) | 8 | 0 0=10-10 |
3 |
12 |
30 | 16 | 35 |
Sdummy |
0 |
0 | 0 | 10 |
Demand | 0 | 20 20=30-10 | 25 |
The rim values for 3=35 and B=20 are compared.
The smaller of the two i.e. min(35,20) = 20 is
assigned to 3 B
This meets the complete demand of B and leaves 35 - 20 =
15 units with 3
Table-4
A | B | C | Supply | |
1 |
10(20) |
25 |
13 | 0 |
2 |
20 |
25(10) |
8 | 0 |
3 |
12 |
30(20) |
16 | 15 15=35-20 |
Sdummy |
0 |
0 |
0 | 10 |
Demand | 0 | 0 0=20-20 | 25 |
The rim values for 3=15 and C=25 are compared.
The smaller of the two i.e. min(15,25) = 15 is
assigned to 3 C
This exhausts the capacity of 3 and leaves 25 - 15 = 10 units with
C
Table-5
A | B | C | Supply | |
1 |
10(20) |
25 |
13 | 0 |
2 |
20 |
25(10) |
8 | 0 |
3 |
12 |
30(20) |
16(15) | 0 0=15-15 |
Sdummy |
0 |
0 |
0 | 10 |
Demand | 0 | 0 | 10 10=25-15 |
The rim values for
Sdummy=10 and
C=10 are compared.
The smaller of the two i.e. min(10,10) = 10 is
assigned to
Sdummy
C
Table-6
A | B | C | Supply | |
1 |
10(20) |
25 |
13 |
0 |
2 |
20 |
25(10) |
8 |
0 |
3 |
12 |
30(20) |
16(15) |
0 |
Sdummy |
0 |
0 |
0(10) |
0 0=10-10 |
Demand | 0 | 0 | 0 0=10-10 |
The initial feasible solution is
A | B | C | Supply | |
1 | 10 (20) | 25 | 13 | 20 |
2 | 20 | 25 (10) | 8 | 10 |
3 | 12 | 30 (20) | 16 (15) | 35 |
Sdummy | 0 | 0 | 0 (10) | 10 |
Demand | 20 | 30 | 25 |
The minimum total transportation cost
=10×20+25×10+30×20+16×15+0×10=1290
Here, the number of allocated cells = 5, which is one less than to
m + n - 1 = 4 + 3 - 1 = 6
This solution is degenerate
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