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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)
1. Algorithm & Example-1
(Previous example)		3. Unbalanced supply and demand example
(Next example)

2. Example-2


Find Solution using Heuristic method-1
D1D2D3D4Supply
S111131714250
S216181410300
S321241310400
Demand200225275250


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`11131714250
`S_2`16181410300
`S_3`21241310400
Demand200225275250


Table-1
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11131714250`2=13-11`55`110=2xx55`
`S_2`16181410300`4=14-10`58`232=4xx58`
`S_3`21241310400`3=13-10`68`204=3xx68`
Demand200225275250
Column
Penalty (P)		`5=16-11``5=18-13``1=14-13``0=10-10`
Total (T)48554434
P`xx`T`240=5xx48``275=5xx55``44=1xx44``0=0xx34`


The lowest PT = 0, occurs in column `D_4`.

The minimum `c_(ij)` in this column is `c_24` = 10.

The maximum allocation in this cell is min(300,250) = 250.
It satisfy demand of `D_4` and adjust the supply of `S_2` from 300 to 50 (300 - 250 = 50).

Table-2
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11131714250`2=13-11`55`110=2xx55`
`S_2`16181410(250)50`2=16-14`58`116=2xx58`
`S_3`21241310400`8=21-13`68`544=8xx68`
Demand2002252750
Column
Penalty (P)		`5=16-11``5=18-13``1=14-13`--
Total (T)48554434
P`xx`T`240=5xx48``275=5xx55``44=1xx44`--


The lowest PT = 44, occurs in column `D_3`.

The minimum `c_(ij)` in this column is `c_33` = 13.

The maximum allocation in this cell is min(400,275) = 275.
It satisfy demand of `D_3` and adjust the supply of `S_3` from 400 to 125 (400 - 275 = 125).

Table-3
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11131714250`2=13-11`55`110=2xx55`
`S_2`16181410(250)50`2=18-16`58`116=2xx58`
`S_3`212413(275)10125`3=24-21`68`204=3xx68`
Demand20022500
Column
Penalty (P)		`5=16-11``5=18-13`----
Total (T)48554434
P`xx`T`240=5xx48``275=5xx55`----


The lowest PT = 110, occurs in row `S_1`.

The minimum `c_(ij)` in this row is `c_11` = 11.

The maximum allocation in this cell is min(250,200) = 200.
It satisfy demand of `D_1` and adjust the supply of `S_1` from 250 to 50 (250 - 200 = 50).

Table-4
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11(200)13171450`13`55`715=13xx55`
`S_2`16181410(250)50`18`58`1044=18xx58`
`S_3`212413(275)10125`24`68`1632=24xx68`
Demand022500
Column
Penalty (P)		--`5=18-13`----
Total (T)48554434
P`xx`T--`275=5xx55`----


The lowest PT = 275, occurs in column `D_2`.

The minimum `c_(ij)` in this column is `c_12` = 13.

The maximum allocation in this cell is min(50,225) = 50.
It satisfy supply of `S_1` and adjust the demand of `D_2` from 225 to 175 (225 - 50 = 175).

Table-5
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11(200)13(50)17140--55--
`S_2`16181410(250)50`18`58`1044=18xx58`
`S_3`212413(275)10125`24`68`1632=24xx68`
Demand017500
Column
Penalty (P)		--`6=24-18`----
Total (T)48554434
P`xx`T--`330=6xx55`----


The lowest PT = 330, occurs in column `D_2`.

The minimum `c_(ij)` in this column is `c_22` = 18.

The maximum allocation in this cell is min(50,175) = 50.
It satisfy supply of `S_2` and adjust the demand of `D_2` from 175 to 125 (175 - 50 = 125).

Table-6
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11(200)13(50)17140--55--
`S_2`1618(50)1410(250)0--58--
`S_3`212413(275)10125`24`68`1632=24xx68`
Demand012500
Column
Penalty (P)		--`24`----
Total (T)48554434
P`xx`T--`1320=24xx55`----


The lowest PT = 1320, occurs in column `D_2`.

The minimum `c_(ij)` in this column is `c_32` = 24.

The maximum allocation in this cell is min(125,125) = 125.
It satisfy supply of `S_3` and demand of `D_2`.


Initial feasible solution is
`D_1``D_2``D_3``D_4`SupplyRow Penalty (P)Total (T)P`xx`T
`S_1`11(200)13(50)1714250 2 |  2 |  2 | 13 | -- | -- | 55 110 |  110 |  110 | 715 | -- | -- |
`S_2`1618(50)1410(250)300 4 |  2 |  2 | 18 | 18 | -- | 58 232 |  116 |  116 | 1044 | 1044 | -- |
`S_3`2124(125)13(275)10400 3 |  8 |  3 | 24 | 24 | 24 | 68 204 |  544 |  204 | 1632 | 1632 | 1632 |
Demand200225275250
Column
Penalty (P)		5
5
5
--
--
--
5
5
5
5
6
24
1
1
--
--
--
--
0
--
--
--
--
--
Total (T)48554434
P`xx`T240
240
240
--
--
--
275
275
275
275
330
1320
44
44
--
--
--
--
0
--
--
--
--
--


The minimum total transportation cost `= 11 xx 200 + 13 xx 50 + 18 xx 50 + 10 xx 250 + 24 xx 125 + 13 xx 275 = 12825`

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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1. Algorithm & Example-1
(Previous example)		3. Unbalanced supply and demand example
(Next example)


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