Question

Suppose you’ve down-selected to two versions of your system (called “System A” and “System B”). Estimated profit for System A is $0.65 per unit and for System B is $0.75 per unit. You seek to maximize total profit.3 However, to manufacture these systems at your company, you’ll need to compete for internal resources. To manufacture a single unit of System A requires 3.0 minutes of machining, 2.0 minutes at a soldering station, and 4.0 minutes for final software testing. System B requires 2.5 minutes of machining and 6 minutes of software testing. Capacity of the machining department is 1,100 minutes per week, capacity of the soldering station is 500 minutes per week, and capacity of final software testing is 1,500 minutes per week.

(a) Write this problem in the form of a constrained linear optimization program.

(b) How many units of System A and B should you produce, and what is the total profit?

Answer 1

(a) Let us suppose that x units of system A and y units of system B are produced .

Profit on system A is 0.65 per unit and profit on system B is 0.75 per unit .

Therefore , total profit for x units system A and y units of system B is -

Resource ( limit )	System A(x)	System B (y)
Machining (1,100 min)	3 min	2.5 min
Soldering station ( 500 min )	2 min	-
Final software testing ( 1500min )	4 min	6 min

Thus , from above table , we get that -

For machining -

For soldering -

And for software testing -

Thus , we have the following optimization model for the given problem -

(b) Now , let us solve the above LPP problem -

We will solve it using Graphical method , for which , we will first change the above inequalities into equations , so that we have -

   .....(2)

.......(3)

   .....(4)

And then we will plot these equations of straight lines onto a graph as shown below -

Here the shaded region shows the solution region which is obtained by shading the common region enclosed by all the three lines . Clearly , the shaded region is the polygon OABC , with corner points O(0,0) ; A(0,250) , B(250,250/3) and C(250,0) .

The point B is on the intersection of the lines (3) and (4) , therefore , the coordinates of point B is found out by solving equations (3) and (4) simultaneously .

Which gives x = 250 and y = 250/3.

Now consider the following table , we have -

Corner point (x,y)
O(0,0)
A(0,250)
B(250,250/3)
C(250,0)

From above , it is clear that In order to obtain maximum profit , 250 units of system A and 250/3 i.e., about 83 units of system B should be produced in order to get the maximum profit .

And in producing 250 units of system A and 83 units of system B , maximum profit we get is equal to -