In order to determine the most favourable integration of various components that are required to be mixed to produce a final product is the reason of preparing blending models. This model is widely used in many industrial sectors. Few areas of its blending application are:

- Teas
- Petroleum products
- Coffees
- Feed-mixes to be utilised for agricultural purpose
- Fertilisers and much more.

A reduction in the blending cost is the aim of these models. If common constraints are taken into account, it includes 3 part.

- Demanded size of blends
- Ingredients which are available in finite quantity
- Technological requirements

Workings of the model can be explained with the help of an example.

**Example**

Suppose, there are 3 final blends of gasoline and4 petroleum products which are required to be blended by a small refinery. There are no accurate blending formulas, and in addition to it, there few limitations that are required to be abided in the entire blending process. Few of the limitations are:

- The precise percentage of component 1 should be 35% of blend 3.
- The constituting percentage of component 2 should not exceed 40% of blend 1’s volume.
- On combining component 2 and 4, the blending percentage should be a minimum of 60% of blend 1’s volume.
- 20% of blend 2 should be the constituent of component 3.

Component 2 and 3 is scarcely available in an amount of 1600 Kilolitres and 200 Kilolitres. 6000 Kilolitres in totality is demanded to be blended by the production manager. An estimated blending amount that can be finally produced out of the total quantity is 2500 Kilolitres.After the final blends are sold at a wholesale price, the amount stands to be Rs. 12, Rs. 18, and Rs. 20 for per litre.

The cost of components utilised in the blends are Rs. 10, Rs. 12, Rs. 13 and Rs. 15 per litres. Now the volume of each component is required to be found out in litres for every final blend in the course of production. The motive of this problem is to find the answer by magnifying its total profit contribution.

**Solution**

Component I utilised in the find blend J be represented as X_{IJ}.

Profit contribution in totality = Total earnings from all 3 blends – Costing of all 4 components in totality

= Profit blend 1 + Profit blend 2 + Profit blend 3 – (cost of components I, 2, 3 and 4)

= Rs (20 + 18 + 12) – Rs (10 + 15 + 12 + 13)

The objective function stands at,

Z= [20 (x_{11} +x_{21} +x_{31}+x_{41}) + 18 (x_{12} +x_{22}+x_{32}+x_{42}) + 12 (x_{13}+x_{23}+x_{33} +x_{43})] – [10 (x_{11}+ x_{12}+ x_{13}) + 15 (x_{21}+ x_{22}+ x_{23}) + 12 (x_{31}+ x_{32}+ x_{33}) + 13 (x_{41}+ x_{42}+ x_{43})]

Or

Z = 10 x_{11}+ 8 x_{12}+ 2 x_{13}+ 5 x_{21}+ 3 x_{22}– 3x_{23} + 8 x_{31}+ 6 x_{32}+ 0 x_{33}+ 7 x_{41}+ 5 x_{41}-x_{43}.

The constraints are,

- 6000 Kilolitres should be the total production run.

x_{11}+ x_{12}+ x_{13}+ x_{21}+ x_{22}+ x_{23}+ x_{31}+ x_{32} + x_{33} + x_{41}+ x_{42}+ x_{43}= 6000,000 litres

- Limitations in the recipe are:

- The constituting percentage of component 2 should not exceed 40% of blend 1’s volume.

X21≤ 0·4 (x_{11} +x_{21}+x_{31}+x_{41})

Or

-0.4 x_{11} +0.6 x_{21}-0.4x_{31}-0.4 x_{41}≤ 0

- 20% of blend 2 should be the constituent of component 3.

x_{32} ≥0.20 (x_{12}+ x_{22}+ x_{32} + x_{42})

or

–0.2 x_{12}– 2·0 x_{22}+ 0.8 x_{32}–0.20 x_{42 }≥ 0

- The precise percentage of component 1 should be 35% of blend 3.

x_{13}= 0.35 (x_{13}+ x_{23}+ x_{33} + x_{43})

Or

0.65 x_{13}– 0.35 x_{23}– 0.35 x_{33} – 0.35 x_{43}= 0

- On combining component 2 and 4, the blending percentage should be a minimum of 60% of blend 1’s volume.

x_{21}+ x_{41} ≥0.6 (x_{11}+ x_{21}+ x_{31}+ x_{41} )

Or

-0.6 x_{11}+ 0·4 x_{21}-0·6x_{31} +0·4 x_{41}≥ 0

- Representation of scarcely limited components 2 and 3 can be showed by this equation.

x_{21}+ x_{22}+ x_{23 }≤ 1600,000

x_{31}+ x_{32} + x_{33} ≤1200,000

- Minimum production requirement constraint in final blend 1 can be represented as

x_{11} +x_{21} +x_{31} +x_{41} ≥ 2500,000

The complete linear programming model is,

Maximise

Z= 10 x_{11}+ 8 x_{12}+ 2 x_{13}+ 5 x_{21}+ 3 x_{22}-3x_{23}+ 8 x_{31}+ 6x_{32} – 0 x_{33}+ 7 x_{41}+ 5 x_{42}– x_{43}

Subject to

x_{11}+ x_{12}+ x_{13}+ x_{21}+ x_{22}+ x_{23}+ x_{31}+ x_{32}+ x_{33}+ x_{41} + x_{42}+ x_{43}= 6000,000

-0.4 x_{11}+ 0.6 .x_{21} – 0.4 x_{31} – 0.4 -x_{41} ≤ 0

-0.2x_{12}– 0·20 x_{22}+0·8 x_{32}-0·20 x_{42}≥ 0

+0.65 x_{13}– 0.35x_{23}– 0.35x_{33}– 0.35x_{43}=0

-0·6 x_{11}+0·4x_{21}-0·6 x_{31}+0·4x_{41} ≥0

x_{21}+ x_{22}+ x23 ≤1600,000

x_{31}+ x_{32}+ x_{33} ≤1200,000

x_{11} +x_{21} +x_{31}+x_{41}≤ 2500,000

x_{11}, x_{12},x_{13},x_{21}, x_{22}, x_{23}, x_{31}, x_{32}, x_{33}, x_{41}, x_{42}, x_{43} ≥0