BUSINESS ADMINISTRATION — HONOURS
Paper : A501 (C-11)
(Quantitative Techniques for Management)
Full Marks : 80
- A company is making two products A and B. The cost of producing one unit of product A and B is
60 and80, 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 one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as Linear Programming Problem.
To formulate this problem as a Linear Programming (LP) problem, we need to define the objective function and the constraints.
Objective Function: The objective is to minimize the cost of production. We’ll define this cost as the total cost (in rupees) of producing x units of Product A and y units of Product B.
The cost for Product A is
60 per unit, and for Product B is 80 per unit. Therefore, the objective function is:
Minimize Z = 60x + 80y
- The company must supply at least 200 units of Product B. So, the first constraint is: y ≥ 200
- Product A requires 1 machine hour per unit, and the company has 400 machine hours available. Product B does not have a machine hour constraint. Therefore, the machine hour constraint is: x ≤ 400
- Both Product A and Product B require 1 labor hour per unit, and the company has 500 labor hours available. The labor hour constraint is: x + y ≤ 500
- Non-negativity constraint: x ≥ 0 (the number of units of A cannot be negative) y ≥ 0 (the number of units of B cannot be negative)
So, the complete linear programming problem can be formulated as follows:
Minimize Z = 60x + 80y
- y ≥ 200
- x ≤ 400
- x + y ≤ 500
- x ≥ 0
- y ≥ 0
These constraints ensure that the company supplies at least 200 units of Product B, utilizes the available machine hours efficiently for Product A, and does not exceed the available labor hours. The non-negativity constraints are standard in LP problems to prevent negative production quantities. The objective function aims to minimize the cost of production.
- A small-scale manufacturer has production facilities for producing two different products. Each of the products requires three different operatings : grinding, assembly and testing. Product 1 requires 15, 20 and 10 minutes to grind, assembly and test respectively; whereas product 2 requires 7.5, 40 and 45 minutes for grinding, assembling and testing. The production run calls at least 7.5 hours of grinding time, at least 20 hours of assembling time, and at least 15 hours of test time. If Product 1 costs
and Product 2 costs90 to manufacture; determine the number of units of each product the firm should produce in order to minimize the cost of operations. Solve it graphically
To solve this problem graphically, we’ll first set up the objective function and constraints for the linear programming problem.
Objective Function: The objective is to minimize the cost of production. We’ll define this cost as the total cost (in rupees) of producing x units of Product 1 and y units of Product 2.
The cost for Product 1 is
60 per unit, and for Product 2 is 90 per unit. Therefore, the objective function is:
Minimize Z = 60x + 90y
- Grinding Time Constraint: Product 1 requires 15 minutes of grinding per unit, and Product 2 requires 7.5 minutes. The total grinding time must be at least 7.5 hours (which is 450 minutes): 15x + 7.5y ≥ 450
- Assembly Time Constraint: Product 1 requires 20 minutes of assembly per unit, and Product 2 requires 40 minutes. The total assembly time must be at least 20 hours (which is 1,200 minutes): 20x + 40y ≥ 1200
- Testing Time Constraint: Product 1 requires 10 minutes of testing per unit, and Product 2 requires 45 minutes. The total testing time must be at least 15 hours (which is 900 minutes): 10x + 45y ≥ 900
- Non-negativity constraint: x ≥ 0 (the number of units of Product 1 cannot be negative) y ≥ 0 (the number of units of Product 2 cannot be negative)
Now, let’s plot these constraints on a graph and find the feasible region. Once we have the feasible region, we can determine the optimal solution that minimizes the cost.
The feasible region is the shaded area in the graph. The objective function is to minimize the cost, which is represented by the downward-sloping line. The optimal solution is the point where the objective function line intersects the feasible region.
From the graph, we can see that the optimal solution is at the point where the objective function line intersects the line representing the assembly constraint. At this point, x1 = 10 and x2 = 20. This means that the firm should produce 10 units of Product 1 and 20 units of Product 2 in order to minimize the cost of operations.
The firm should produce 10 units of Product 1 and 20 units of Product 2 in order to minimize the cost of operations.