Question

DI 9.2-2. * Consider the transportation problem having the fol- lowing parameter table: Destination 1 2 3 4 5 Supply Source 1 - 2 3 4 2 7 8 0 4 4 6 7 0 4 6 3 5 0 2 5 M 2 0 5 7 4 5 0 5 Demand

Answer 1

TOTAL number of demand constraints : 5 and supply is 4
Problem Table is

	D1	D2	D3	D4	D5		Supply
S1	2	4	6	5	7		4
S2	7	6	3	999999	4		6
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	4	4	2	5	5

Table-1

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6	3	999999	4		6
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	0	4	2	5	5

The rim values for S2=6 and D1=0 are compared.

The smaller of the two i.e. min(6,0) = 0 is assigned to S2 D1

This meets the complete demand of D1 and leaves 6 - 0 = 6 units with S2

Table-2

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6	3	999999	4		6
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	0	4	2	5	5

The rim values for S2=6 and D2=4 are compared.

The smaller of the two i.e. min(6,4) = 4 is assigned to S2 D2

This meets the complete demand of D2 and leaves 6 - 4 = 2 units with S2

Table-3

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3	999999	4		2
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	0	0	2	5	5

The rim values for S2=2 and D3=2 are compared.

The smaller of the two i.e. min(2,2) = 2 is assigned to S2 D3

This exhausts the capacity of S2 and leaves 2 - 2 = 0 units with D3

Table-4

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3(2)	999999	4		0
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	0	0	0	5	5

The rim values for S3=6 and D3=0 are compared.

The smaller of the two i.e. min(6,0) = 0 is assigned to S3 D3

This meets the complete demand of D3 and leaves 6 - 0 = 6 units with S3

Table-5

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3(2)	999999	4		0
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	0	0	0	5	5

The rim values for S3=6 and D4=5 are compared.

The smaller of the two i.e. min(6,5) = 5 is assigned to S3 D4

This meets the complete demand of D4 and leaves 6 - 5 = 1 units with S3

Table-6

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3(2)	999999	4		0
S3	8	7	5	2(5)	5		1
S4	0	0	0	0	0		4
Demand	0	0	0	0	5

The rim values for S3=1 and D5=5 are compared.

The smaller of the two i.e. min(1,5) = 1 is assigned to S3 D5

This exhausts the capacity of S3 and leaves 5 - 1 = 4 units with D5

Table-7

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3(2)	999999	4		0
S3	8	7	5	2(5)	5(1)		0
S4	0	0	0	0	0		4
Demand	0	0	0	0	4

The rim values for S4=4 and D5=4 are compared.

The smaller of the two i.e. min(4,4) = 4 is assigned to S4 D5

Table-8

	D1	D2	D3	D4	D5		Supply
S1	2(4)	4	6	5	7		0
S2	7	6(4)	3(2)	999999	4		0
S3	8	7	5	2(5)	5(1)		0
S4	0	0	0	0	0(4)		0
Demand	0	0	0	0	0

Initial feasible solution is

	D1	D2	D3	D4	D5		Supply
S1	2 (4)	4	6	5	7		4
S2	7	6 (4)	3 (2)	999999	4		6
S3	8	7	5	2 (5)	5 (1)		6
S4	0	0	0	0	0 (4)		4
Demand	4	4	2	5	5

The minimum total transportation cost =2×4+6×4+3×2+2×5+5×1+0×4=53

Here, the number of allocated cells = 6, which is two less than to m + n - 1 = 4 + 5 - 1 = 8
∴ This solution is degenerate

vogel approximation rules.

TOTAL number of supply constraints : 4
TOTAL number of demand constraints : 5
Problem Table is

	D1	D2	D3	D4	D5		Supply
S1	2	4	6	5	7		4
S2	7	6	3	999999	4		6
S3	8	7	5	2	5		6
S4	0	0	0	0	0		4
Demand	4	4	2	5	5

Table-1

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2	4	6	5	7		4	2=4-2
S2	7	6	3	999999	4		6	1=4-3
S3	8	7	5	2	5		6	3=5-2
S4	0	0	0	0	0		4	0=0-0
Demand	4	4	2	5	5
Column Penalty	2=2-0	4=4-0	3=3-0	2=2-0	4=4-0

The maximum penalty, 4, occurs in column D2.

The minimum cij in this column is c42 = 0.

The maximum allocation in this cell is min(4,4) = 4.
It satisfy supply of S4 and demand of D2.

Table-2

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2	4	6	5	7		4	3=5-2
S2	7	6	3	999999	4		6	1=4-3
S3	8	7	5	2	5		6	3=5-2
S4	0	0(4)	0	0	0		0	--
Demand	4	0	2	5	5
Column Penalty	5=7-2	--	2=5-3	3=5-2	1=5-4

The maximum penalty, 5, occurs in column D1.
The minimum cij in this column is c11 = 2.
The maximum allocation in this cell is min(4,4) = 4.
It satisfy supply of S1 and demand of D1.

Table-3

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2(4)	4	6	5	7		0	--
S2	7	6	3	999999	4		6	1=4-3
S3	8	7	5	2	5		6	3=5-2
S4	0	0(4)	0	0	0		0	--
Demand	0	0	2	5	5
Column Penalty	--	--	2=5-3	999997=999999-2	1=5-4

The maximum penalty, 999997, occurs in column D4.

The minimum cij in this column is c34 = 2.

The maximum allocation in this cell is min(6,5) = 5.
It satisfy demand of D4 and adjust the supply of S3 from 6 to 1 (6 - 5 = 1).

Table-4

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2(4)	4	6	5	7		0	--
S2	7	6	3	999999	4		6	1=4-3
S3	8	7	5	2(5)	5		1	0=5-5
S4	0	0(4)	0	0	0		0	--
Demand	0	0	2	0	5
Column Penalty	--	--	2=5-3	--	1=5-4

The maximum penalty, 2, occurs in column D3.

The minimum cij in this column is c23 = 3.

The maximum allocation in this cell is min(6,2) = 2.
It satisfy demand of D3 and adjust the supply of S2 from 6 to 4 (6 - 2 = 4).

Table-5

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2(4)	4	6	5	7		0	--
S2	7	6	3(2)	999999	4		4	4
S3	8	7	5	2(5)	5		1	5
S4	0	0(4)	0	0	0		0	--
Demand	0	0	0	0	5
Column Penalty	--	--	--	--	1=5-4

The maximum penalty, 5, occurs in row S3.

The minimum cij in this row is c35 = 5.

The maximum allocation in this cell is min(1,5) = 1.
It satisfy supply of S3 and adjust the demand of D5 from 5 to 4 (5 - 1 = 4).

Table-6

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2(4)	4	6	5	7		0	--
S2	7	6	3(2)	999999	4		4	4
S3	8	7	5	2(5)	5(1)		0	--
S4	0	0(4)	0	0	0		0	--
Demand	0	0	0	0	4
Column Penalty	--	--	--	--	4

The maximum penalty, 4, occurs in row S2.
The minimum cij in this row is c25 = 4.
The maximum allocation in this cell is min(4,4) = 4.
It satisfy supply of S2 and demand of D5.

Initial feasible solution is

	D1	D2	D3	D4	D5		Supply	Row Penalty
S1	2(4)	4	6	5	7		4	2 | 3 | -- | -- | -- | -- |
S2	7	6	3(2)	999999	4(4)		6	1 | 1 | 1 | 1 | 4 | 4 |
S3	8	7	5	2(5)	5(1)		6	3 | 3 | 3 | 0 | 5 | -- |
S4	0	0(4)	0	0	0		4	0 | -- | -- | -- | -- | -- |
Demand	4	4	2	5	5
Column Penalty	2 5 -- -- -- --	4 -- -- -- -- --	3 2 2 2 -- --	2 3 999997 -- -- --	4 1 1 1 1 4

The minimum total transportation cost =2×4+3×2+4×4+2×5+5×1+0×4=45
Here, the number of allocated cells = 6, which is two less than to m + n - 1 = 4 + 5 - 1 = 8
∴ This solution is degenerate