Homework SourseConsider the following linear program Min 8x12y st 1x3y9 2x2
Consider the following linear program:
Min 8x+12y
s.t.
1x+3y?9
2x+2y?10
6x+2y?
X, y?0
Assume that the objective function coefficient for X changes from 8 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution
X=
Y=
Optimal solution=
Assume that the objective function coefficient for X remains 8, but the objective function coefficient for Y changes from 12 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution.
X=
Y=
Optimal solution=
Solution
Given linear program:
Min 8x+12y ---------------(1)
s.t.
1x+3y?9 ---------------(2)
2x+2y?10 ---------------(3)
6x+2y? ---------------(4)
X, y?0 ---------------(5)
Now solve from (2) 1x+3y=9
points are we take x=0 then y=3 (0,3)
take y=0 then x=9 (9,0)
from (3) 2x+2y=10
if y=0 then x=5 so (5,0)
if x=0 then y=5 so (0,5)
from (4) 6x+2y=0
if x=1 then y=-3 (1,-3)
if y=3 then x=-1 (-1,3)
From all these point if we draw graph we see below,
the intersection point of (2),(3),(4) are (0, 9), (2, 3), (3,2), and (9, 0).
min8x+12y
(0, 9)=108; (2, 3)=52; (3,2)=48; and (9, 0)=72
so the optimal solution is (3,2)
if we assume that the objective function coefficient for X changes from 8 to 6.
then function is change 8x+12y?0 to 6x+12y?0
then points from (0, 9), (2, 3), (3,2), and (9, 0).
(0, 9)=108; (2, 3)=48; (3,2)=42; and (9, 0)=54
so the optimal solution is:(3,2)
So the optimal solution changed
Optimal solution=(3,2)
(b)
Assume that the objective function coefficient for X remains 8, but the objective function coefficient for Y changes from 12 to 6.
then 8x+6y=0
then points from (0, 9), (2, 3), (3,2), and (9, 0).
(0, 9)=54; (2, 3)=34; (3,2)=36 and (9, 0)=72
Optimal solution=(2,3)
No change in the optimal solution.
