Answer to Question #175095 in Operations Research for Nomi

Question

A company produces two products that are processed on two assembly lines. Assembly line 1 has 100 available hours, and assembly line 2 has 42 available hours. Each product requires 10 hours of processing time on line 1, while on line 2 product 1 requires 7 hours and product 2 requires 3 hours. The profit for product 1 is $6 per unit, and the profit for product 2 is $4 per unit. Formulate a linear programming model for this problem and solve using the simplex method.

Answer 1

Solution:

Let the number of product 1 and product 2 be $x, y$ respectively.

Objective function, maximize $Z=6x+4y$

subject to the constraints:

$10x+10y\le100 \\ 7x+3y\le42 \\x,y\ge0$

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate

1. As the constraint-1 is of type '≤' we should add slack variable $S_1$

2. As the constraint-2 is of type '≤' we should add slack variable $S_2$

Max $Z=6x+4y+0S_1+0S_2$

subject to $10x+10y+S_1=100$

$7x+3y+S_2=42$

and $x,y,S_1,S_2≥0$

Negative minimum $Z_j-C_j$  is -6 and its column index is 1. So, the entering variable is $x$ .

Minimum ratio is 6 and its row index is 2. So, the leaving basis variable is $S_2$ .

∴ The pivot element is 7.

Entering =$x$ , Departing =$S_2$ , Key Element =7

Then, we have iteration-2 and similarly, iteration-3.

Since all $Z_j-C_j≥0$

Hence, optimal solution is obtained with value of variables as :

x=3,y=7 and Max Z=46