MP CHAPTER 4 SOLUTIONS
SECTION 4.1
1.
max z = 3x
1
+ 2x
2
s.t.
2x
1
+ x
2
+ s
1
= 100
x
1
+ x
2
+ s
2
= 80
x
1
+ s
3
= 40
2.
min z = 50x
1
+ 100x
2
s.t.
7x
1
+ 2x
2
- e
1
= 28
2x
1
+ 12x
2
- e
2
= 243.
3.
min z = 3x
1
+ x
2
s.t.
x
1
- e
1
= 3
x
1
+ x
2
+ s
2
= 4
2x
1
- x
2
= 3
SECTION 4.4
1.
From Figure 2 of Chapter 3 we see that the extreme
points of the feasible region are
Basic Feasible Solution
H = (0, 0)
s
1
= 100, s
2
= 80, s
3
= 40 x
1
= x
2
= x
3
= 0
E = (40, 0)
x
1
= 40, s
1
= 20, s
2
= 40 x
2
= x
3
= s
3
= 0
F = (40, 20) x
1
= 40, x
2
= 20, s
2
= 20 x
3
= s
1
= s
3
= 0
G = (20, 60) x
1
= 20, x
2
= 60, s
3
= 20 x
3
= s
1
= s
2
= 0
D = (0, 80)
s
1
= 20, x
2
= 80, s
3
= 40 s
2
= x
1
= x
3
= 0
2.
From Figure 4 of Chapter 3 we see that the
correspondence is as follows:
Extreme Point
Basic Feasible Solution
E = (3.6, 1.4)
x
1
= 3.6, x
2
= 1.4, e
1
= e
2
= 0
B = (0, 14)
x
2
= 14, e
2
= 144, e
1
= x
1
= 0
C = (12, 0)
x
1
= 12, e
1
= 56, x
2
= e
2
= 0
3.
Basic Variables Basic Feasible Solution Corner
Point
x
1
, x
2
x
1
=150 x
2
=100 s
1
=s
2
=0(150, 100)
x
1
, s
1
x
1
=200, s
1
=150, x
2
=s
2
=0(200, 0)
x
1
, s
2
x
1
=350, s
2
=-300, x
2
=s
1
=0 Infeasible
x
2
, s
1
x
2
=400, s
1
=-450, x
1
=s
2
=0 Infeasible
x
2
, s
2
x
2
=175, s
2
= 225, x
1
=s
1
=0(0, 175)
s
1
, s
2
s
1
=350, s
2
=400, x
1
=x
2
=0 (0, 0)
4. Let
σ
A
= 1/6,
σ
C
= 1/3 and
σ
B
= ½. Then
1