Answer to Question #126640 in Operations Research for eyob tilahun

Question

1.two machines (I and II) produce two grades of tires, A and B.in one hour of operation, machine i produces 20 unit of grade A and 10 units of grade B tires, while machine II produce 30 units of grade A and 40 units of grade B tires. the machines are required to meet a production schedule of at least 1400 units of grade A and 1200 units of grade B tires. the cost of operating machine I is $50 per hour and the cost of operating machine II is $80 per hour. based on the given information:-
Required:-
A. formulate the linear programming model
B. determine the optimal solution if the objective is to minimize the cost of operating the machines using graphic method to solve the problem.

Answer 1

A. formulate the linear programming model

• Identifying Decision Variables...............................................(1)

Let x be the total number of units produced by machine I

Let y be the total number of units produced by machine II

Now, Let the total cost of production be represented by z

• Creating Our Objective Function............................................(2)

The total cost of production spent to produce is product A and B multiplied by its cost of production per hour which is $50 and $80 respectively

Cost of Production: $minZ=50x+80y$

Which means we have to minimize z

• Writing Constraints................................................................(3)

From the constraints, we derive

$20x+30y \le1400$

$10x+40y \le1200$

• Non-negative Restrictions..................................................... (4)

So, we have two constraints,

$x \ge 0$ and

$y \ge 0$

We have formulated our linear program.

B. determine the optimal solution if the objective is to minimize the cost of operating the machines using graphic method to solve the problem.

Since we know that $x, y ≥ 0$ . We will consider only the first quadrant.

To plot for the graph for the above equations, first I will simplify all the equations.

$20x + 30y ≤ 1400$ can be simplified to $2x + 3y ≤ 140$ by dividing by 10.

$10x + 40y ≤ 1200$ can be simplified to $x + 4y ≤ 120$ by dividing by 10.

Plot the first 2 lines on a graph in the first quadrant (like shown below)

The optimal feasible solution is achieved at the point of intersection where the units produced & cost of production constraints are active. This means the point at which the equations $2x + 3y ≤ 140$ and$\, x + 4y ≤ 120$ intersect gives us the optimal solution.

The values for X and Y which gives the optimal solution is at $(20,40)$ .

The minimize the cost of operating the machines,

$$minZ=50(20)+80(40)=4200$$