1.1, 1.2.
The solutions to Exercise 1.1 are all a result of using the graphical method to identify optimal solutions.
In the diagrams below, the feasible region is shaded in gray. The optimal contour plot for the objective is
drawn in a thick blue line and the direction of improvement is drawn as a dashed arrow. The solutions to
Exercise 1.2 are below each corresponding diagram.
(a)
y
(0, 10)
Optimal solution is (0, 10).
2x + y ≤ 10
2x − y ≤ 6
x≥0
(4, 2)
(0, 0)
y≥0
(3, 0)
x
Suppose we change the objective function coefficient of x so that the objective function is ax + 2y. As a
result, the slope of the objective function contour plots is −a/2. Therefore, from the diagram, we see that
we need
a
⇔
a≤4
− ≥ −2
2
for (0, 10) to remain optimal.
Now suppose we change the objective function coefficient of y so that the objective function is 3x + by.
Note that if b < 0, the objective function contour plots improve in the “downward” direction, and so (0, 10)
cannot be optimal. Therefore, if (0, 10) is to remain optimal, we must at least have b ≥ 0. Since the slope
of the objective function contour plots is −3/b, from the diagram, we see that we need
3
3
(b ≥ 0) and
− ≥ −2
⇔
b≥
b
2
for (0, 10) to remain optimal.
1
(b)
y
−2x + y ≤ 0
(3, 6)
Line segment from (3, 6) to (6, 3) is optimal.
x+y ≤9
x≥0
(6, 3)
x − 2y ≤ 0
(0, 0)
x
y≥0
For the entire line segment between (3, 6) and (6, 3) to remain optimal, neither objective function coefficient can change by itself.
We can take this a little further. How about keeping just (3, 6) optimal? Suppose we change the objective
function coefficient of x so that the objective function is ax + y. The slope of the objective function contour
plots is −a. Therefore, from the diagram, we see that we need
−1 ≤ −a ≤ 2
⇔
−2 ≤ a ≤ 1
for (3, 6) to remain optimal.
Similarly, suppose we change the objective function coefficient of y so that the objective function is x+by.
As with part (a), if (3, 6) is optimal, then we must at least have b ≥ 0. Since the slope of the objective
function contour plots is −1/b, from the diagram, we see that we need
1
(b ≥ 0) and
−1 ≤ − ≤ 2
⇔
b≥1
b
for (3, 6) to remain optimal.
Now, how about keeping just (6, 3) optimal? Suppose we change the objective function coefficient of x
so that the objective function is ax + y. Note that if (6, 3) is optimal, then we must at least have a ≥ 0;
otherwise, if a < 0, then the objective function contour plots improve in the “leftward” direction. Since the
slope of the objective function contour plots is −a, from the diagram, we see that we need
1
⇔
a≥1
(a ≥ 0) and
(−a ≤ −1) or
−a ≥
2
for (6, 3) to remain optimal.
Similarly, suppose we change the objective function coefficient of y so that the objective function is x+by.
Since the slope of the objective function contour plots is −1/b, from the diagram, we see that we need
1
1
1
− ≤ −1
or
− ≥
⇔
−2 ≤ b ≤ 1
b
b
2
for (6, 3) to remain optimal.
2
(c)
x2
x1 ≥ 0
2x1 + x2 ≤ 350
Line segment from (0, 150)
to (100, 100) is optimal.
(0, 150)
(100, 100)
(150, 50)
x1 + 2x2 ≤ 300
(175, 0)
(0, 0)
x1
x2 ≥ 0
x1 + x2 ≤ 200
As in part (b), for the entire line segment between (0, 150) and (100, 100) to remain optimal, neither
objective function coefficient can change by itself.
Again, we can take this a little further. How about keeping just (0, 150) optimal? Using the same
approach as in parts (a) and (b), we find that if we change the objective function coefficient on x1 so that
the objective function is c1 x1 + 50x2 , then we need c1 ≤ 25 for (0, 150) to remain optimal. If we change the
objective function coefficient on x2 so that the objective function is 25x1 + c2 x2 , then we need c2 ≥ 50 for
(0, 150) to remain optimal.
Using the same idea, we find that we need 25 ≤ c1 ≤ 50 or 25 ≤ c2 ≤ 50 for (100, 100) to remain optimal.
3
(d)
x2
x1 + x2 ≥ 3
x1 ≥ 0
(2, 1) is optimal
x2 ≥ 0
(3, 0)
(4, 0)
x1 − x2 ≥ 1
x1
x1 + 2x2 ≤ 4
Suppose we change the objective function coefficient of x1 so that the objective function is c1 x1 + 3x2 .
Now the slope of the objective function contours is −c1 /3. From the diagram, we see that we need
−
c1
1
≥−
3
2
⇔
c1 ≤
3
2
to keep (2, 1) optimal.
Now suppose we change the objective function coefficient of x2 so that the objective function is x1 + c2 x2 .
If (2, 1) is optimal, we must have at least c2 ≥ 0. Since the slope of the objective function contours is −1/c2 ,
from the diagram, we see that we need
1
1
(c2 ≥ 0) and
− ≥−
⇔
c2 ≥ 2
c2
2
for (2, 1) to remain optimal.
4