Formulation of LP Problem
The formulation of linear programming problem as a mathematical model involves the following basic steps.
Step 1. Find the key decision to be made from the study of the solution. (In this connection looking for variables helps considerably).
Step 2. Identify the variables and assume symbols x1, x2.....for variable quantities noticed in step 1.
Step 3. Express the possible alternatives mathematically in terms of variables. The set of feasible alternatives generally in the given situation is:
{ (x1 x2 ) ; x1 >0,x2>0}
Step 4. Mention the objective quantitatively and express it as a linear of variables.
Step 5. Express the constraints also as linear equalities inequalities in terms of variables.
Example 1. A factory manufactures two articles A and B. To manufacture the article A, a certain machine has to be worked for 1.5 hours and in addition a cratsman has to work for 2 hours. To manufacture the article B, the machine has to be worked for 2-5 hours and in addition the craftsman has to work for 1-5 hours. In a week the factory can avail of 80 hours of machine time and 70 hours of craftsman's time. The profit on each article A is Rs. 5 and that on each article B is Rs. 4 if all the articles produced can be sold away find how many of each kind should be produced can be sold away find how many of each kind should be produced to earn the maximum profit per week. Formulate the problem as LP problem.
Solution. The data of the given problem can be conveniently summarized in the following tabular form:
Decision Article Hours on profit per unit
Variables machine craftsman
X1 A 1.5 2 Rs.5.00
X2 B 2.5 1.5 Rs. 4.00
Hours available 80 70
(per week) maximum maximum
Formulation. We will begin by treating the number of articles of type A and B to be manufactured as unknown quantities or decision variables. Let x1 and x2 be the number of tables and chairs to be manufactured, respectively.
Objective Function. Since total profit consists of profit derived from setting type A articles at Rs. 5 each plus the profit derived from selling type B articles at Rs. 4 each. Thus, 5x1 is the profit earned by selling type A articles and 4x2 is the corresponding profit by selling type B articles. As the factory wants to achieve the greatest possible profit (say Z) it can be stated algebraically by writing profit equation as:
Maximize Z = 5x1 + 4x2
In this form the profit expression provides the objective function.
Constraints. Constraints are limitations or restrictions placed on availability of resources.
1. Machine time for article A + Machine time for article B < Available time of machine
i.e., x1 + 2.5 x2 < 80
(the sign < is read 'less than or equal to')
2. Craftman time for article A + Craftman time for article B < Available craftman time
i.e., 2x1 + 1.5x2 < 70
Further, we cannot have negative production, i.e., either we manufacture or do not manufacture, i.e., x1 > 0 and x2 > 0.
These are called non-negativity restrictions. We can now state the problem also referred to as a linear Programming Problem in full.
Maximize = 5x1 + 4x2
Subject 1.5x1 + 2.5x2 < 80
2x1 + 1.5x2 < 70
X1 > 0, x2 > 0
Example 4. A company is making two products A and B. The cost of producing one unit of product A and B is Rs. 60 and Rs. 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires on labour hour each and total of 500 labour hours are available. The company problem as a linear programming problem.
Solution. Let x1 and x2 be the number of units of product A and B to be manufactured respectively, then the LP model is given by:
Minimise Z = 60x1 + 80x2
Subject to x2 > 200 (agreement constraint)
X1 < 400 (machine hour's constraint for product A)
X1 + x2 < (labour hours constraint)
X1 > 0, x2 > 0
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