transportation problem using Heuristic method-2 Unbalanced supply and demand example

3. Unbalanced supply and demand example

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Home > Operation Research calculators > Heuristic method-1 example

8. Heuristic method-2 example ( Enter your problem )

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Algorithm and examples

Algorithm & Example-1
Example-2
Unbalanced supply and demand example

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2. Example-2
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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 Heuristic method-2
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

Table-1

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8	8	0	76	$8=8-0$
$S_2$	16	24	16	0	82	$24=24-0$
$S_3$	8	16	24	0	77	$24=24-0$

Demand	72	102	41	20
Column Penalty	$12=16-4$	$16=24-8$	$16=24-8$	$0=0-0$

The maximum penalty, 24, occurs in row

$S_2$ .

The minimum

$c_(ij)$ in this row is

$c_24$ = 0.

The maximum allocation in this cell is min(82,20) = 20.
It satisfy demand of

$D_(dummy)$ and adjust the supply of

$S_2$ from 82 to 62 (82 - 20 = 62).

Table-2

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8	8	0	76	$4=8-4$
$S_2$	16	24	16	0(20)	62	$8=24-16$
$S_3$	8	16	24	0	77	$16=24-8$

Demand	72	102	41	0
Column Penalty	$12=16-4$	$16=24-8$	$16=24-8$	--

The maximum penalty, 16, occurs in column

$D_2$ .

The minimum

$c_(ij)$ in this column is

$c_12$ = 8.

The maximum allocation in this cell is min(76,102) = 76.
It satisfy supply of

$S_1$ and adjust the demand of

$D_2$ from 102 to 26 (102 - 76 = 26).

Table-3

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8(76)	8	0	0	--
$S_2$	16	24	16	0(20)	62	$8=24-16$
$S_3$	8	16	24	0	77	$16=24-8$

Demand	72	26	41	0
Column Penalty	$8=16-8$	$8=24-16$	$8=24-16$	--

The maximum penalty, 16, occurs in row

$S_3$ .

The minimum

$c_(ij)$ in this row is

$c_31$ = 8.

The maximum allocation in this cell is min(77,72) = 72.
It satisfy demand of

$D_1$ and adjust the supply of

$S_3$ from 77 to 5 (77 - 72 = 5).

Table-4

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8(76)	8	0	0	--
$S_2$	16	24	16	0(20)	62	$8=24-16$
$S_3$	8(72)	16	24	0	5	$8=24-16$

Demand	0	26	41	0
Column Penalty	--	$8=24-16$	$8=24-16$	--

The maximum penalty, 8, occurs in row

$S_2$ .

The minimum

$c_(ij)$ in this row is

$c_23$ = 16.

The maximum allocation in this cell is min(62,41) = 41.
It satisfy demand of

$D_3$ and adjust the supply of

$S_2$ from 62 to 21 (62 - 41 = 21).

Table-5

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8(76)	8	0	0	--
$S_2$	16	24	16(41)	0(20)	21	$0=24-24$
$S_3$	8(72)	16	24	0	5	$0=16-16$

Demand	0	26	0	0
Column Penalty	--	$8=24-16$	--	--

The maximum penalty, 8, occurs in column

$D_2$ .

The minimum

$c_(ij)$ in this column is

$c_32$ = 16.

The maximum allocation in this cell is min(5,26) = 5.
It satisfy supply of

$S_3$ and adjust the demand of

$D_2$ from 26 to 21 (26 - 5 = 21).

Table-6

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8(76)	8	0	0	--
$S_2$	16	24	16(41)	0(20)	21	$0=24-24$
$S_3$	8(72)	16(5)	24	0	0	--

Demand	0	21	0	0
Column Penalty	--	$0=24-24$	--	--

The maximum penalty, 0, occurs in row

$S_2$ .

The minimum

$c_(ij)$ in this row is

$c_22$ = 24.

The maximum allocation in this cell is min(21,21) = 21.
It satisfy supply of

$S_2$ and demand of

$D_2$ .

Initial feasible solution is

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	Row Penalty
$S_1$	4	8(76)	8	0	76	8 \| 4 \| -- \| -- \| -- \| -- \|
$S_2$	16	24(21)	16(41)	0(20)	82	24 \| 8 \| 8 \| 8 \| 0 \| 0 \|
$S_3$	8(72)	16(5)	24	0	77	24 \| 16 \| 16 \| 8 \| 0 \| -- \|

Demand	72	102	41	20
Column Penalty	12 12 8 -- -- --	16 16 8 8 8 0	16 16 8 8 -- --	0 -- -- -- -- --

The minimum total transportation cost

$= 8 xx 76 + 24 xx 21 + 16 xx 41 + 0 xx 20 + 8 xx 72 + 16 xx 5 = 2424$

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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