The very basic situation of mathematics has unitary functional objective either maximization of profit or minimization of loss or cost, which are subject to constraints that includes resources, procedures, time etc. The function and objectives in linear programming are in linear form.
Linear Objective Function (Maximize profit or Minimize cost):
$$Z=C1X1+C2X2+C3X3+\dots +CnXn$$
1
Linear Constraints:
$$a11X1+a12X2+a13X3+\dots \dots \dots +anXn \left(\ge \text{o}\text{r} \le \right) RHS1$$
2
$$a21X1+a22X2+a23X3+\dots \dots \dots +anXn \left(\ge \text{o}\text{r} \le \right) RHS2$$
3
$$a31Xm1+a32Xm2+a33Xm3+\dots \dots \dots +anXn \left(\ge \text{o}\text{r} \le \right) RHS3$$
4
Non-Negative Constraint:
Any problem whose mathematical formulation fit this general model is linear program problem. The technique of linear programming is used widely to solve problems and thousands of variables can be accommodated. However, the use of variables is limited and represent constraints. It is important that objective function and constraints are related linearly viewed by Gupta & Hira [14]
For the purpose of application of linear programming, it is important to define the objective of the firm. And in most cases, it is either maximizing the profit or minimizing the cost. Therefore, with the available resources and existing outputs, and objective function is drawn.
In the above model, 2 to 4 are Constraints. These expressions specify elements of problem such as lack of available resources, or challenges such as meeting certain demand.
The xi variables in the above given model are decision variables; that is, they are the variables whose value is determined when the Linear Programming model is solved. Their values provided the answers that are being sought in the Linear Programming analysis. Such model needs data to establish the values of the decision variables.
The input data constants are often referred to as Parameters. The “a” and “c” in the above given model are all parameters of the model. While, condition 5 is called non-negativity conditions [7].
Following assumptions are made, while drawing on the objective function.
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That the available resources needed for production are limited.
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Optimum allocation of available resources will help in achieving optimum results i-e maximizing profit and minimizing cost.
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The qualities of the raw material that is used, is assumed to be standard, and not inferior.
3.1. Problem Formulation of PASVEP Chemicals Production Firm
PASVEP chemicals Production Company produce three different types of chemical products i.e. Varnish Paint, PVC Solution and Acrylic and import one product i.e. Enamel Paint. These products produce from the mixture of different types of raw materials and different phases of the machine.
The accounts department of the company, provided that to produce 36 kg of Varnish Paint requires 12 liters of kerosene oil, 24 kg of varnish resin, 200 gm of Petrosin and 10 gm of Cobalt, while to produce 80 kg of PVC Solution requires 65 kg of cyclohexanone, 11 kg of PVC resin, 10 kg of Ethyl acetate and 500 gm of color powder. Furthermore, to produce 100 kg of Acrylic requires 100 liters of filter water and 10 kg of Polyvinyl. The monthly availability of the raw material is given in the below table:
Table 1
Monthly availability of Raw Material
Raw Materials | Monthly Availability |
Kerosene Oil | 100 liter |
Cyclohexanone | 950 kg |
PVC resin | 198 kg |
Ethyl acetate or MEK | 180 kg |
Color powder | 10 kg |
Filter Water | 600 liter |
Polyvinyl | 60 kg |
Varnish resin | 74 kg |
Petrosin | 2 kg |
Cobalt | 2 kg |
The product Varnish Paint is produces and came to final shape through different machines processes, i.e. Drilling/Mixing, Filling and Packing/scaling. To produce 36 kg of varnish paint requires 40 minutes for Drilling/Mixing. After Drilling/Mixing process, the same quantity requires 50 minutes for Filling and at the end, the same quantity requires 20 minutes for Packing and scaling. To produce 80 kg of PVC Solution it requires 70 minutes for Drilling/Mixing and the next stage is filling, which requires 90 minutes for filling and 30 minutes for packing and scaling. Furthermore, to produce 100 kg of Acrylic requires 100 minutes for Drilling/Mixing and 70 minutes for filing and at the end it requires 40 minutes for packing and scaling.
To estimate the limits of time availability of all three machines are as follows:
5 hours (per day) * 60 minutes = 300 minutes (per day), we get, 300 minutes * 22 days = 6600 Minutes available per month.
The company also imports enamel Paint and mixes some chemicals into it and sale the high quality Enamel Paint. For 16 kg of High Quality Enamel Paint requires 14 kg of enamel paint with monthly availability of 368 kg, 2 liter of Kerosene oil with monthly availability of 100 Liters and 2 kg of varnish with monthly availability of 46 kg. On machine, Drilling and mixing require 20 minutes, while in next stage, filling requires 40 minutes and at the end Packing/scaling requires 20 minutes to get 16 kg of High-Quality Enamel Paint.
The Profit generated from the sale of Varnish Paint, PVC Solution, Acrylic and High-Quality Enamel Paint per kg is Rs. 235, 540, 61 and 300, respectively.
3.2. Formulation of Linear Programming Model
For the purpose of formulation of linear programming model, we let the following;
\(X1\) : Quantity of Varnish Paint.
\(X2\) : Quantity of PVC solution.
\(X3\) : Quantity of Acrylic.
\(X4\) : Quantity of High Enamel Paint.
The mathematical form of the problem is given below, to determine which product the company should produce more so as to maximize the total profit.
The Linear Objective Function:
$$Maximize=235X1+540x2+61x3+300x4$$
6
Subject to:
Raw Materials Constraints:
Kerosene Oil:
Varnish Paint:
Petrosin:
Cobalt:
Cyclohexanone:
PVC Resin:
Ethyl Acetate:
Color Powder:
Filter Water:
Polyvinyl:
Enamel Paint:
Varnish:
Machine Time Constraints:
Drilling/Mixing: \(40X1+70X2+100X3+20X4\le 6600\) (19)
Filling: \(50X1+90X2+70X3+40X4\le 6600\) (20)
Scaling/Packing: \(20X1+30X2+40X3+20X4\le 6600\) (21)
Non-negative Constraints:
$$X1, X2, X3, X4 \ge 0$$
22
In general, a linear programming model has linear objective function having equations that has inequalities. The selected problem, therefore, has all the characteristics that are needed by linear programming model, making this case a linear programming problem.