Question

AutoIgnite produces electronic ignition systems for automobiles at a plant in Cleveland, Ohio. Each ignition system is assembled from two components produced at AutoIgnite’s plants in Buffalo, New York, and Dayton, Ohio. The Buffalo plant can produce 2,000 units of component 1, 1,000 units of component 2, or any combination of the two components each day. For instance, 60% of Buffalo’s production time could be used to produce component 1 and 40% of Buffalo’s production time could be used to produce component 2; in this case, the Buffalo plant would be able to produce 0.6(2,000) = 1,200 units of component 1 each day and 0.4(1,000) = 400 units of component 2 each day. The Dayton plant can produce 600 units of component 1, 1,400 units of component 2, or any combination of the two components each day. At the end of each day, the component production at Buffalo and Dayton is sent to Cleveland for assembly of the ignition systems on the following workday.

1. Formulate a linear programming model that can be used to develop a daily production schedule for the Buffalo and Dayton plants that will maximize daily production of ignition systems at Cleveland. If required, round your answer to the nearest whole number.
   

   Let B	=	proportion of Buffalo's time used to produce component 1
   D	=	proportion of Dayton's time used to produce component 1

1. 	B	+	D
   s.t.
   	B	+	D
   	B
   			D
   			B, D

2. Find the optimal solution. If an amount is zero, enter "0". If required, round your answer to the nearest whole number.
   

   Optimal Production Plan	Buffalo	Dayton
   Component 1
   Component 2

Answer 1

Formulate a linear programming model that can be used to develop a daily production schedule for the Buffalo and Dayton plants as :

Objective function,

Z_Max = B+D

S.t ,

B+D ≤ 2600

B≤ 2000

D≤ 600

B,D ≥ 0.

The same can be solved by use of graphical method or simplex model of LPP model .

1. To draw constraint B+D≤2600→(1)

Treat it as B+D=2600

When B=0 then D=?

⇒(0)+D=2600

⇒D=2600

When D=0 then B=?

⇒B+(0)=2600

⇒D=2600

B	0	2600
D	2600	0

Similarly for To draw constraint B≤2000→(2)

Treat it as B=2000

Here line is parallel to Y-axis

B	2000	2000
D	0	1

To draw constraint D≤600→(3)

Treat it as D=600

Here line is parallel to X-axis

B	0	1
D	600	600

The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (x1,x2)	Lines through Extreme Point	Objective function value z=B+D
O(0,0)	4→B≥0 5→D≥0	1(0)+1(0)=0
A(2000,0)	2→B≤2000 5→D≥0	1(2000)+1(0)=2000
B(2000,600)	1→B+D≤2600 2→D≤2000	1(2000)+1(600)=2600
C(0,600)	3→B≤600 4→D≥0	1(0)+1(600)=600

The maximum value of the objective function z=2600 occurs at the extreme point (2000,600).

Hence, the optimal solution to the given LP problem is : B=2000,D=600 and max z=2600.