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Home > Operation Research calculators > Heuristic method-1 example
7. Heuristic method-1 example ( Enter your problem )
( Enter your problem )
Algorithm and examples
  1. Algorithm & Example-1
  2. Example-2
  3. Unbalanced supply and demand example
 		Other related methods
  1. north-west corner method
  2. least cost method
  3. vogel's approximation method
  4. Row minima method
  5. Column minima method
  6. Russell's approximation method
  7. Heuristic method-1
  8. Heuristic method-2
  9. modi method (optimal solution)
  10. stepping stone method (optimal solution)
2. Example-2
(Previous example)		8. Heuristic method-2
(Next method)

3. Unbalanced supply and demand example


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-1
D1D2D3Supply
S148876
S216241682
S38162477
Demand7210241


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`48876
`S_2`16241682
`S_3`8162477
Demand7210241


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
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`488076
`S_2`162416082
`S_3`81624077
Demand721024120


Table-1
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`488076`4=4-0`20`80=4xx20`
`S_2`162416082`16=16-0`56`896=16xx56`
`S_3`81624077`8=8-0`48`384=8xx48`
Demand721024120
Column
Penalty (P)		`4=8-4``8=16-8``8=16-8``0=0-0`
Total (T)2848480
P`xx`T`112=4xx28``384=8xx48``384=8xx48``0=0xx0`


The lowest PT = 0, occurs in column `D_(dummy)`.

The minimum `c_(ij)` in this column is `c_14` = 0.

The maximum allocation in this cell is min(76,20) = 20.
It satisfy demand of `D_(dummy)` and adjust the supply of `S_1` from 76 to 56 (76 - 20 = 56).

Table-2
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`4880(20)56`4=8-4`20`80=4xx20`
`S_2`162416082`0=16-16`56`0=0xx56`
`S_3`81624077`8=16-8`48`384=8xx48`
Demand72102410
Column
Penalty (P)		`4=8-4``8=16-8``8=16-8`--
Total (T)2848480
P`xx`T`112=4xx28``384=8xx48``384=8xx48`--


The lowest PT = 0, occurs in row `S_2`.

The minimum `c_(ij)` in this row is `c_21` = 16.

The maximum allocation in this cell is min(82,72) = 72.
It satisfy demand of `D_1` and adjust the supply of `S_2` from 82 to 10 (82 - 72 = 10).

Table-3
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`4880(20)56`0=8-8`20`0=0xx20`
`S_2`16(72)2416010`8=24-16`56`448=8xx56`
`S_3`81624077`8=24-16`48`384=8xx48`
Demand0102410
Column
Penalty (P)		--`8=16-8``8=16-8`--
Total (T)2848480
P`xx`T--`384=8xx48``384=8xx48`--


The lowest PT = 0, occurs in row `S_1`.

The minimum `c_(ij)` in this row is `c_12` = 8.

The maximum allocation in this cell is min(56,102) = 56.
It satisfy supply of `S_1` and adjust the demand of `D_2` from 102 to 46 (102 - 56 = 46).

Table-4
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`48(56)80(20)0--20--
`S_2`16(72)2416010`8=24-16`56`448=8xx56`
`S_3`81624077`8=24-16`48`384=8xx48`
Demand046410
Column
Penalty (P)		--`8=24-16``8=24-16`--
Total (T)2848480
P`xx`T--`384=8xx48``384=8xx48`--


The lowest PT = 384, occurs in row `S_3`.

The minimum `c_(ij)` in this row is `c_32` = 16.

The maximum allocation in this cell is min(77,46) = 46.
It satisfy demand of `D_2` and adjust the supply of `S_3` from 77 to 31 (77 - 46 = 31).

Table-5
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`48(56)80(20)0--20--
`S_2`16(72)2416010`16`56`896=16xx56`
`S_3`816(46)24031`24`48`1152=24xx48`
Demand00410
Column
Penalty (P)		----`8=24-16`--
Total (T)2848480
P`xx`T----`384=8xx48`--


The lowest PT = 384, occurs in column `D_3`.

The minimum `c_(ij)` in this column is `c_23` = 16.

The maximum allocation in this cell is min(10,41) = 10.
It satisfy supply of `S_2` and adjust the demand of `D_3` from 41 to 31 (41 - 10 = 31).

Table-6
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`48(56)80(20)0--20--
`S_2`16(72)2416(10)00--56--
`S_3`816(46)24031`24`48`1152=24xx48`
Demand00310
Column
Penalty (P)		----`24`--
Total (T)2848480
P`xx`T----`1152=24xx48`--


The lowest PT = 1152, occurs in row `S_3`.

The minimum `c_(ij)` in this row is `c_33` = 24.

The maximum allocation in this cell is min(31,31) = 31.
It satisfy supply of `S_3` and demand of `D_3`.


Initial feasible solution is
`D_1``D_2``D_3``D_(dummy)`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`48(56)80(20)76 4 |  4 |  0 | -- | -- | -- | 20 80 |  80 |  0 | -- | -- | -- |
`S_2`16(72)2416(10)08216 |  0 |  8 |  8 | 16 | -- | 56896 |  0 |  448 |  448 | 896 | -- |
`S_3`816(46)24(31)077 8 |  8 |  8 |  8 | 24 | 24 | 48 384 |  384 |  384 |  384 | 1152 | 1152 |
Demand721024120
Column
Penalty (P)		4
4
--
--
--
--
8
8
8
8
--
--
8
8
8
8
8
24
0
--
--
--
--
--
Total (T)2848480
P`xx`T112
112
--
--
--
--
384
384
384
384
--
--
384
384
384
384
384
1152
0
--
--
--
--
--


The minimum total transportation cost `= 8 xx 56 + 0 xx 20 + 16 xx 72 + 16 xx 10 + 16 xx 46 + 24 xx 31 = 3240`

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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2. Example-2
(Previous example)		8. Heuristic method-2
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