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Blending Models

Blending models are formulated to determine an optimal combination of different ingredients to be blended in a final product. Such models have found wide applications specially in blending of petroleum products, feed-mixes for agricultural use, fertilisers, teas, coffees etc. The objective with such models usually is to minimise the cost of the blend. Typical constraints include lot size requirement for each blend, technological requirements and limited availability of the ingredients.

Example. A smail refinery is about to blend four petroleum products into three final blends of gasoline. Although, the blending formulas are not precise, there are some restrictions which must be adhered to in the blending process. These are:

  1. Component 2 should constitute no more than 40 percent of the volume of blend 1.
  2. Component 3 should constitute atleast 20 percent of the value of blend 2.
  3. Component I should be exactly 35 percent of blend 3.
  4. Components 2 and 4 together should constitute atleast 60 percent of the volume of blend 1.

There is a limited availability of components 2 and 3 of 1600 K. litres and [200 K. litres. The production manager wants to blend a total of 6000 k. litres. Of this total at least 2500 k. litres of final blend I should be produced. The wholesale price per litre from the sale of each final blend equals Rs. 20, Rs. 18 and Rs. 12 respectively. The input components cost Rs. 10, Rs. 15, Rs. 12 and Rs. 13 per litre respectively. Find out the number of litres of each component to be used in each final blend so as to maximise the total· profit contribution from the production run.

Sol. Let xijrepresent number of litres of component i used in the find blend j.

Total profit contribution = Total revenue from all 3 blends - total cost of the 4 components

= 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)

Hence, the objective function is

Z= [20 (x11 +x21 +x31+x41) + 18 (x12 +x22+x32+x42) + 12 (x13+x23+x33 +x43)] – [10 (x11+ x12+ x13) + 15 (x21+ x22+ x23) + 12 (x31+ x32+ x33) + 13 (x41+ x42+ x43)]

or Z = 10 x11+ 8 x12+ 2 x13+ 5 x21+ 3 x22- 3x23 + 8 x31+ 6 x32+ 0 x33+ 7 x41+ 5 x41-x43.

The constraints are

(a) Total production run must be 6000 K. litres or

x11+ x12+ x13+ x21+ x22+ x23+ x31+ x32 + x33 + x41+ x42+ x43= 6000,000 litres

(b) Recipe restrictions are

Amount of components 2 used s 40% of amount of final blend 1 in blend 1

X21≤ 0·4 (x11 +x21+x31+x41)

or-0.4 x11 +0.6 x21-0.4x31-0.4 x41≤ 0

Amount of component 3 used in blend 2  20% amount of final blend 2

x32 ≥0.20 (x12+ x22+ x32 + x42)

or–0.2 x12- 2· 0 x22+ 0.8 x32–0.20 x42≥ 0.

Component 1should be exactly 35% of blend 3

x13= 0.35 (x13+ x23+ x33 + x43)

or.65 x13- .35 x23- .35 x33 – .35 x43= 0.

Component 2 and 4 together should constitute at least 60 percent of the value of blend 1 or

x21+ x41 ≥0.6 (x11+ x21+ x31+ x41 )

or-0.6 x11+ 0·4 x21-0·6x31 +0·4 x41≥ 0

Limitation of availability of components 2 and 3 can be represented as

x21+ x22+ x23 ≤  1600,000

x31+ x32 + x33 ≤1200,000

Also, there is a constraint of minimum production requirement of final blend 1, that is

x11 +x21 +x31 +x41 ≥ 2500,000

The complete LP model is

Maximise Z= 10 x11+ 8 x12+ 2 x13+ 5 x21+ 3 x22-3x23+ 8 x31+ 6x32 -0· x33+ 7 x41+ 5 x42- x43

Subject to

x11+ x12+.x13+ x21+ x22+ x23+ x31+ x32+ x33+ x41 + X42+ x43= 6000,000

- 0 . 4 x11+ 0.6 .x21 – 0.4 x31 – 0.4 -x41 ≤ 0

-0.2x12- ·20 x22+0·8 x32-0·20 x42≥ 0

.65 x13- .35x23- .35x33- .35x43=0

-0·6 x11+0·4x21-0·6 x31+0·4x41 ≥0

x21+ x22+ x23 ≤1600,000

x31+ x32+ x33 ≤1200,000

x11 +x21 +x31+x41≤ 2500,000

x11, x12,x13,x21, x22, x23, x31, x32, x33, x41, x42, x43 ≥0