Chapter 2 solutions - Chapter 2 An Introduction to Linear...

2 - 1
Chapter 2
An Introduction to Linear Programming
Solutions:
1. a, b, and e, are acceptable linear programming relationships.
c is not acceptable because of 2 2 B 
d is not acceptable because of 3 A
f is not acceptable because of 1AB
c, d, and f could not be found in a linear programming model because they have the above nonlinear
terms.
2. a.
8
4
4 8 0
B
A
b.
B
A
8
4
4 8 0
c.

Chapter 2
2 - 2
B
A
8
4
4 8 0
Points on line
are only feasible
points
6. 7 A + 10 B = 420 is labeled (a)
6 A + 4 B = 420 is labeled (b)
-4 A + 7 B = 420 is labeled (c)
80
40
40 80
B
A
20
60
100
20 60 0 100
-20
-40 -60 -80 -100
(c)
(a)
(b)
7.

An Introduction to Linear Programming
2 - 3
0
B
A
150 50 100 200
250
50
100
10.
0
1
2
3
4
5
B
1
2 3 4 5
6
Optimal Solution
A = 12/7, B = 15/7
Value of Objective Function = 2(12/7) + 3(15/7) = 69/7
A

Chapter 2
2 - 4
A + 2 B = 6 (1)
5 A + 3 B = 15 (2)
(1) × 5 5 A + 10 B = 30 (3)
(2) - (3) - 7 B = -15
B = 15/7
From (1), A = 6 - 2(15/7) = 6 - 30/7 = 12/7
12. a.
(0,0)
1
2
3
4
5
B
1
2 3
4 5
6
Optimal Solution
A = 3, B = 1.5
Value of Objective Function = 13.5
6
(4,0)
(3,1.5)
A

An Introduction to Linear Programming
2 - 5
b.
(0,0)
1
2
3
B
1 2 3 4 5 6
Optimal Solution
A = 0, B = 3
Value of Objective Function = 18
A
7 8 9 10
c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).
13. a.
0
2
4
6
8
B
2 4 6 8
Feasible Region
consists of this line
segment only
A
b. The extreme points are (5, 1) and (2, 4).

Chapter 2
2 - 6
c.
0
2
4
6
8
B
2 4 6 8
Optimal Solution
A
A = 2, B = 4
14. a. Let F = number of tons of fuel additive
S = number of tons of solvent base
Max 40 F + 30 S
s.t.
2/5 F +
1
/
2
S
 200 Material 1
1
/
5
S

5 Material 2
3
/
5
F
+ 3
/
10
S
 21 Material 3
F , S  0

An Introduction to Linear Programming
2 - 7
b.
c. Material 2: 4 tons are used, 1 ton is unused.
d. No redundant constraints.
16. a. A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make
extreme point (0, 540) the optimal solution. For example, one possibility is 3 S + 9 D .
b. Optimal Solution is S = 0 and D = 540.
c.
Department Hours Used Max. Available Slack
Cutting and Dyeing 1(540) = 540 630 90
Sewing
5
/
6
(540) = 450
600 150
Finishing
2
/
3
(540) = 360
708 348
Inspection and Packaging
1
/
4
(540) = 135
135 0
S
F

Chapter 2
2 - 8
17.
Max 5 A + 2 B +
0 S
1
+
0 S
2
+
0 S
3
s.t.
1 A - 2 B +
1 S
1
= 420
2 A + 3 B + 1 S
2
= 610
6 A - 1 B + 1 S
3
= 125
A , B , S
1
, S
2
, S
3
 0
18. a.
Max 4 A + 1 B + 0 S
1
+ 0 S
2
+ 0 S
3
s.t.
10 A + 2 B + 1 S
1
= 30
3 A + 2 B +
1 S
2
= 12
2 A + 2 B + 1 S
3
= 10
A , B , S
1
, S
2
, S
3
 0
b.

An Introduction to Linear Programming
2 - 9
0
2
4
6
8
B
2 4 6 8
Optimal Solution
A
A = 18/7, B = 15/7, Value = 87/7
10
12
14
10
c. S
1
= 0, S
2
= 0, S
3
= 4/7
20. a.