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:
A reduction in the blending cost is the aim of these models. If common constraints are taken into account, it includes 3 part.
Workings of the model can be explained with the help of an 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:
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.
Component I utilised in the find blend J be represented as XIJ.
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 (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)]
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,
x11+ x12+ x13+ x21+ x22+ x23+ x31+ x32 + x33 + x41+ x42+ x43= 6000,000 litres
X21≤ 0·4 (x11 +x21+x31+x41)
-0.4 x11 +0.6 x21-0.4x31-0.4 x41≤ 0
x32 ≥0.20 (x12+ x22+ x32 + x42)
–0.2 x12– 2·0 x22+ 0.8 x32–0.20 x42 ≥ 0
x13= 0.35 (x13+ x23+ x33 + x43)
0.65 x13– 0.35 x23– 0.35 x33 – 0.35 x43= 0
x21+ x41 ≥0.6 (x11+ x21+ x31+ x41 )
-0.6 x11+ 0·4 x21-0·6x31 +0·4 x41≥ 0
x21+ x22+ x23 ≤ 1600,000
x31+ x32 + x33 ≤1200,000
x11 +x21 +x31 +x41 ≥ 2500,000
The complete linear programming model is,
Z= 10 x11+ 8 x12+ 2 x13+ 5 x21+ 3 x22-3x23+ 8 x31+ 6x32 – 0 x33+ 7 x41+ 5 x42– x43
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– 0·20 x22+0·8 x32-0·20 x42≥ 0
+0.65 x13– 0.35x23– 0.35x33– 0.35x43=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