transportation problem using least cost method Unbalanced supply and demand example

3. Unbalanced supply and demand example

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Home > Operation Research calculators > Least cost method example

2. least cost method example ( Enter your problem )

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Unbalanced supply and demand example
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$D_(dummy)$

Unbalanced supply and demand
If the total supply is not equal to the total demand then the problem is called unbalanced transportation problem.

It's solution :
1. If the total supply is more than the total demand, then we add a new column, with transportation cost 0
2. If the total demand is more than the total supply, then we add a new row, with transportation cost 0

Example
Find Solution using Least Cost method
D1 D2 D3 Supply
S1 4 8 8 76
S2 16 24 16 82
S3 8 16 24 77
Demand 72 102 41

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

	$D_1$	$D_2$	$D_3$	Supply
$S_1$	4	8	8	76
$S_2$	16	24	16	82
$S_3$	8	16	24	77

Demand	72	102	41

Here Total Demand = 215 is less than Total Supply = 235. So We add a dummy demand constraint with 0 unit cost and with allocation 20.
Now, The modified table is

The smallest transportation cost is 0 in cell

$S_1 D_(dummy)$

The allocation to this cell is min(76,20) = 20.
This satisfies the entire demand of

$D_(dummy)$ and leaves 76 - 20 = 56 units with

$S_1$

Table-1

The smallest transportation cost is 4 in cell

$S_1 D_1$

The allocation to this cell is min(56,72) = 56.
This exhausts the capacity of

$S_1$ and leaves 72 - 56 = 16 units with

$D_1$

Table-2

The smallest transportation cost is 8 in cell

$S_3 D_1$

The allocation to this cell is min(77,16) = 16.
This satisfies the entire demand of

$D_1$ and leaves 77 - 16 = 61 units with

$S_3$

Table-3

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4(56)	8	8	0(20)	0
$S_2$	16	24	16	0	82
$S_3$	8(16)	16	24	0	61

Demand	0	102	41	0

The smallest transportation cost is 16 in cell

$S_3 D_2$

The allocation to this cell is min(61,102) = 61.
This exhausts the capacity of

$S_3$ and leaves 102 - 61 = 41 units with

$D_2$

Table-4

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4(56)	8	8	0(20)	0
$S_2$	16	24	16	0	82
$S_3$	8(16)	16(61)	24	0	0

Demand	0	41	41	0

The smallest transportation cost is 16 in cell

$S_2 D_3$

The allocation to this cell is min(82,41) = 41.
This satisfies the entire demand of

$D_3$ and leaves 82 - 41 = 41 units with

$S_2$

Table-5

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4(56)	8	8	0(20)	0
$S_2$	16	24	16(41)	0	41
$S_3$	8(16)	16(61)	24	0	0

Demand	0	41	0	0

The smallest transportation cost is 24 in cell

$S_2 D_2$

The allocation to this cell is min(41,41) = 41.
Table-6

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4(56)	8	8	0(20)	0
$S_2$	16	24(41)	16(41)	0	0
$S_3$	8(16)	16(61)	24	0	0

Demand	0	0	0	0

Initial feasible solution is

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4 (56)	8	8	0 (20)	76
$S_2$	16	24 (41)	16 (41)	0	82
$S_3$	8 (16)	16 (61)	24	0	77

Demand	72	102	41	20

The minimum total transportation cost

$= 4 xx 56 + 0 xx 20 + 24 xx 41 + 16 xx 41 + 8 xx 16 + 16 xx 61 = 2968$

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6

$:.$ This solution is non-degenerate

This material is intended as a summary. Use your textbook for detail explanation.
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