transportation problem using row minima method calculator

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Solution

Solution provided by AtoZmath.com

1. A Company has 3 production facilities S1, S2 and S3 with production capacity of 7, 9 and 18 units (in 100's) per week of a product, respectively. These units are tobe shipped to 4 warehouses D1, D2, D3 and D4 with requirement of 5,6,7 and 14 units (in 100's) per week, respectively. The transportation costs (in rupees) per unit between factories to warehouses are given in the table below.

	D₁	D₂	D₃	D₄	Capacity
S₁	19	30	50	10	7
S₂	70	30	40	60	9
S₃	40	8	70	20	18
Demand	5	8	7	14	34

Find initial basic feasible solution for given problem by using
(a) North-West corner method
(b) Least cost method
(c) Vogel's approximation method
(d) obtain an optimal solution by MODI method
if the object is to minimize the total transportation cost.

2. Find an initial basic feasible solution for given transportation problem by using
(a) North-West corner method
(b) Least cost method
(c) Vogel's approximation method

	D₁	D₂	D₃	D₄	Supply
S₁	11	13	17	14	250
S₂	16	18	14	10	300
S₃	21	24	13	10	400
Demand	200	225	275	250

Example

1. Find Solution using Row minima method
D1 D2 D3 D4 Supply
S1 19 30 50 10 7
S2 70 30 40 60 9
S3 40 8 70 20 18
Demand 5 8 7 14

Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 4
Problem Table is

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10	7
`S_2`	70	30	40	60	9
`S_3`	40	8	70	20	18

Demand	5	8	7	14

In `1^(st)` row, The smallest transportation cost is 10 in cell `S_1 D_4`.

The allocation to this cell is min(7,14) = 7.
This exhausts the capacity of `S_1` and leaves 14 - 7 = 7 units with `D_4`

Table-1

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30	40	60	9
`S_3`	40	8	70	20	18

Demand	5	8	7	7

In `2^(nd)` row, The smallest transportation cost is 30 in cell `S_2 D_2`.

The allocation to this cell is min(9,8) = 8.
This satisfies the entire demand of `D_2` and leaves 9 - 8 = 1 units with `S_2`

Table-2

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30(8)	40	60	1
`S_3`	40	8	70	20	18

Demand	5	0	7	7

In `2^(nd)` row, The smallest transportation cost is 40 in cell `S_2 D_3`.

The allocation to this cell is min(1,7) = 1.
This exhausts the capacity of `S_2` and leaves 7 - 1 = 6 units with `D_3`

Table-3

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30(8)	40(1)	60	0
`S_3`	40	8	70	20	18

Demand	5	0	6	7

In `3^(rd)` row, The smallest transportation cost is 20 in cell `S_3 D_4`.

The allocation to this cell is min(18,7) = 7.
This satisfies the entire demand of `D_4` and leaves 18 - 7 = 11 units with `S_3`

Table-4

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30(8)	40(1)	60	0
`S_3`	40	8	70	20(7)	11

Demand	5	0	6	0

In `3^(rd)` row, The smallest transportation cost is 40 in cell `S_3 D_1`.

The allocation to this cell is min(11,5) = 5.
This satisfies the entire demand of `D_1` and leaves 11 - 5 = 6 units with `S_3`

Table-5

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30(8)	40(1)	60	0
`S_3`	40(5)	8	70	20(7)	6

Demand	0	0	6	0

In `3^(rd)` row, The smallest transportation cost is 70 in cell `S_3 D_3`.

The allocation to this cell is min(6,6) = 6.
Table-6

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10(7)	0
`S_2`	70	30(8)	40(1)	60	0
`S_3`	40(5)	8	70(6)	20(7)	0

Demand	0	0	0	0

Initial feasible solution is

	`D_1`	`D_2`	`D_3`	`D_4`	Supply
`S_1`	19	30	50	10 (7)	7
`S_2`	70	30 (8)	40 (1)	60	9
`S_3`	40 (5)	8	70 (6)	20 (7)	18

Demand	5	8	7	14

The minimum total transportation cost `= 10 xx 7 + 30 xx 8 + 40 xx 1 + 40 xx 5 + 70 xx 6 + 20 xx 7 = 1110`

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6
`:.` This solution is non-degenerate