# An Introduction To Management Science: Quantitative Approach, 15th Edition Solution Manual

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Chapter 2
An Introduction to Linear Programming
Learning Objectives
1.
Obtain an overview of the kinds of problems linear programming has been used to solve.
2.
Learn how to develop linear programming models for simple problems.
3.
Be able to identify the special features of a model that make it a linear programming model.
4.
Learn how to solve two variable linear programming models by the graphical solution procedure.
5.
Understand the importance of extreme points in obtaining the optimal solution.
6.
Know the use and interpretation of slack and surplus variables.
7.
Be able to interpret the computer solution of a linear programming problem.
8.
Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear
programming problems.
9.
Understand the following terms:
problem formulation
constraint function
objective function
solution
optimal solution
nonnegativity constraints
mathematical model
linear program
linear functions
feasible solution
feasible region
slack variable
standard form
redundant constraint
extreme point
surplus variable
alternative optimal solutions
infeasibility
unbounded
2-1
Chapter 2
Solutions:
1.
a, b, and e, are acceptable linear programming relationships.
c is not acceptable because of ๏ญ2B 2
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.
B
8
4
0
4
8
4
8
A
b.
B
8
4
0
A
c.
B
Points on line
are only feasible
points
8
4
0
4
2-2
8
A
An Introduction to Linear Programming
3.
a.
B
(0,9)
A
0
(6,0)
b.
B
(0,60)
A
0
(40,0)
c.
B
Points
on line are only
feasible solutions
(0,20)
A
(40,0)
0
4.
a.
B
(20,0)
(0,-15)
2-3
A
Chapter 2
b.
B
(0,12)
(-10,0)
A
c.
B
(10,25)
Note: Point shown was
used to locate position of
the constraint line
A
0
5.
B
a
300
c
200
100
b
A
0
100
200
2-4
300
An Introduction to Linear Programming
6.
7A + 10B = 420 is labeled (a)
6A + 4B = 420 is labeled (b)
-4A + 7B = 420 is labeled (c)
B
100
80
60
(b)
(c)
40
20
(a)
A
-100
-80
-60
-40
-20
0
20
40
60
80
100
7.
B
100
50
A
0
50
100
150
2-5
200
250
Chapter 2
8.
B
200
133 1/3
(100,200)
A
-200
0
-100
100
200
9.
B
(150,225)
200
100
0
(150,100)
100
-100
-200
2-6
200
300
A
An Introduction to Linear Programming
10.
B
5
4
Optimal Solution
A = 12/7, B = 15/7
3
Value of Objective Function = 2(12/7) + 3(15/7) = 69/7
2
1
A
0
1
2
(1) ร 5
(2) – (3)
4
5
2B
3B
10B
7B
B
=
6
=
15
=
30
= -15
= 15/7
3
A
5A
5A
+
+
+
–
From (1), A = 6 – 2(15/7) = 6 – 30/7 = 12/7
2-7
6
(1)
(2)
(3)
Chapter 2
11.
B
A = 100
Optimal Solution
A = 100, B = 50
Value of Objective Function = 750
100
B = 80
A
0
100
200
12. a.
B
6
5
4
Optimal Solution
A = 3, B = 1.5
Value of Objective Function = 13.5
3
(3,1.5)
2
1
A
(0,0)
1
2
3
2-8
4
(4,0)
5
6
An Introduction to Linear Programming
b.
B
3
Optimal Solution
A = 0, B = 3
Value of Objective Function = 18
2
1
A
(0,0)
1
c.
2
3
4
5
6
7
8
There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).
13. a.
B
8
6
Feasible Region
consists of this line
segment only
4
2
0
A
2
b.
4
The extreme points are (5, 1) and (2, 4).
2-9
6
8
9
10
Chapter 2
c.
B
8
6
Optimal Solution
A = 2, B = 4
4
2
0
A
2
14. a.
4
6
Let F = number of tons of fuel additive
S = number of tons of solvent base
Max
s.t.
40F
+
30S
2/5F
+
1/ S
2
1/ S
5
๏ฃ 200
Material 1
๏ฃ
5
Material 2
3/ S
10
๏ฃ
21
Material 3
3/ F
5
F, S ๏ณ 0
+
2 – 10
8
An Introduction to Linear Programming
b.
S
F
c.
Material 2: 4 tons are used, 1 ton is unused.
d.
No redundant constraints.
15. a.
2 – 11
Chapter 2
b.
Similar to part (a): the same feasible region with a different objective function. The optimal solution
occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160.
c.
The sewing constraint is redundant. Such a change would not change the optimal solution to the
original problem.
16. a.
b.
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 3S + 9D.
Optimal Solution is S = 0 and D = 540.
c.
Department
Cutting and Dyeing
Sewing
Finishing
Inspection and Packaging
Hours Used
1(540) = 540
5
/6(540) = 450
2
/3(540) = 360
1
/4(540) = 135
Max. Available
630
600
708
135
+
17.
Max
s.t.
5A
+ 2B
+
0S1
1A
2A
6A
– 2B
+ 3B
– 1B
+
1S1
+ 0S2
0S3
+ 1S2
1S3
+
=
=
=
420
610
125
A, B, S1, S2, S3 ๏ณ 0
18. a.
Max
s.t.
4A
+ 1B
+ 0S1
10A
3A
2A
+ 2B
+ 2B
+ 2B
+ 1S1
+ 0S2
+ 0S3
+ 1S2
+ 1S3
A, B, S1, S2, S3 ๏ณ 0
2 – 12
= 30
= 12
= 10
Slack
90
150
348
0
An Introduction to Linear Programming
b.
B
14
12
10
8
6
Optimal Solution
A = 18/7, B = 15/7, Value = 87/7
4
2
0
c.
A
2
4
6
8
+ 0S2
+ 0S3
10
S1 = 0, S2 = 0, S3 = 4/7
19. a.
Max
s.t.
3A
+
4B
+ 0S1
-1A
1A
2A
+
+
+
2B
2B
1B
+ 1S1
+ 1S2
+ 1S3
A, B, S1, S2, S3 ๏ณ 0
2 – 13
= 8
= 12
= 16
(1)
(2)
(3)
Chapter 2
b.
B
14
(3)
12
10
(1)
8
6
Optimal Solution
A = 20/3, B = 8/3
Value = 30 2/3
4
2
(2)
0
A
2
c.
4
6
8
10
12
S1 = 8 + A โ 2B = 8 + 20/3 – 16/3 = 28/3
S2 = 12 – A โ 2B = 12 – 20/3 – 16/3 = 0
S3 = 16 โ 2A – B = 16 – 40/3 – 8/3 = 0
20. a.
Max
s.t.
3A
+ 2B
A
3A
A
A
+ B
+ 4B
– S1
+ S2
– S3
–
– S4
B
A, B, S1, S2, S3, S4 ๏ณ 0
2 – 14
=
=
=
=
4
24
2
0
An Introduction to Linear Programming
b.
c.
S1 = (3.43 + 3.43) – 4 = 2.86
S2 = 24 – [3(3.43) + 4(3.43)] = 0
S3 = 3.43 – 2 = 1.43
S4 = 0 – (3.43 – 3.43) = 0
2 – 15
Chapter 2
21. a. and b.
B
90
80
70
Constraint 2
60
50
40
Optimal Solution
Constraint 1
Constraint 3
30
Feasible Region
20
10
2A + 3B = 60
A
0
10
c.
20
30
40
50
60
70
80
90
100
Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2,
B = 10 + A
Substituting for B in constraint 1 we obtain
5A + 5(10 + A)
5A + 50 + 5A
10A
A
= 400
= 400
= 350
= 35
B = 10 + A = 10 + 35 = 45
Optimal solution is A = 35, B = 45
d.
Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding
constraints.
2 – 16
An Introduction to Linear Programming
e.
Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 1(35) + 3(45) = 170.
Because 170 exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associated
with this constraint.
22. a.
C
3500
3000
2500
Inspection and
Packaging
2000
Cutting and
Dyeing
5
1500
Feasible Region
4
1000
Sewing
3
5A + 4C = 4000
500
0
2
1
500
1000
1500
2000
2500
Number of All-Pro Footballs
A
3000
b.
Extreme Point
1
2
3
4
5
Coordinates
(0, 0)
(1700, 0)
(1400, 600)
(800, 1200)
(0, 1680)
Profit
5(0) + 4(0) = 0
5(1700) + 4(0) = 8500
5(1400) + 4(600) = 9400
5(800) + 4(1200) = 8800
5(0) + 4(1680) = 6720
Extreme point 3 generates the highest profit.
c.
Optimal solution is A = 1400, C = 600
d.
The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection
and packaging constraint. Therefore these two constraints are the binding constraints.
e.
New optimal solution is A = 800, C = 1200
Profit = 4(800) + 5(1200) = 9200
2 – 17
Chapter 2
23. a.
Let E = number of units of the EZ-Rider produced
L = number of units of the Lady-Sport produced
Max
s.t.
2400E
+
6E
+
2E
+
1800L
3L ๏ฃ 2100
L ๏ฃ 280
2.5L ๏ฃ 1000
E, L ๏ณ 0
Engine time
Lady-Sport maximum
Assembly and testing
b.
L
700
Number of EZ-Rider Produced
600
Engine
Manufacturing Time
500
400
Frames for Lady-Sport
300
Optimal Solution E = 250, L = 200
Profit = $960,000
200
100
Assembly and Testing
0
E
100
300
200
400
500
Number of Lady-Sport Produced
c.
24. a.
The binding constraints are the manufacturing time and the assembly and testing time.
Let R = number of units of regular model.
C = number of units of catcherโs model.
Max
s.t.
5R
+
8C
1R
+
1/ R
2
1/ R
8
+
3/ C
2
1/ C
3
1/ C
4
+
๏ฃ
900
Cutting and sewing
๏ฃ
300
Finishing
๏ฃ
100
Packing and Shipping
R, C ๏ณ 0
2 – 18
.
An Introduction to Linear Programming
b.
C
1000
F
Catcher’s Model
800
600
C&
400
S
P&
Optimal Solution
(500,150)
S
200
R
0
200
400
600
800
1000
Regular Model
c.
5(500) + 8(150) = $3,700
d.
C&S
1(500) + 3/2(150) = 725
F
1/ (500) + 1/ (150) = 300
2
3
P&S
1/ (500) + 1/ (150) = 100
8
4
e.
Department
C&S
F
P&S
25. a.
Usage
725
300
100
Slack
175 hours
0 hours
0 hours
Let B = percentage of funds invested in the bond fund
S = percentage of funds invested in the stock fund
Max
s.t.
b.
Capacity
900
300
100
0.06 B
+
0.10 S
B
0.06 B
B
+
+
0.10 S
S
๏ณ
๏ณ
=
0.3
0.075
1
Optimal solution: B = 0.3, S = 0.7
Value of optimal solution is 0.088 or 8.8%
2 – 19
Bond fund minimum
Minimum return
Percentage requirement
Chapter 2
26. a.
Let D = amount spent on digital advertising
R = amount spent on radio advertising
Max
s.t.
50D
+ 80R
D
D
+
D
R = 1000 Budget
๏ณ 250 Digital min.
R ๏ณ 250 Radio min.
-2R ๏ณ
0 Digital ๏ณ 2 Radio
D, R ๏ณ 0
b.
27.
Let I = Internet fund investment in thousands
B = Blue Chip fund investment in thousands
Max
s.t.
0.12I
+
0.09B
1I
1I
6I
+
1B
+
4B
I, B ๏ณ 0
๏ฃ
๏ฃ
๏ฃ
50
35
240
Available investment funds
Maximum investment in the internet fund
Maximum risk for a moderate investor
2 – 20
An Introduction to Linear Programming
B
Blue Chip Fund (000s)
60
Risk Constraint
Optimal Solution
I = 20, B = 30
$5,100
50
40
Maximum
Internet Funds
30
20
10
Objective
Function
0.12I + 0.09B
Available Funds
$50,000
0
I
10
30
20
40
50
60
Internet Fund (000s)
Internet fund
Blue Chip fund
Annual return
b.
$20,000
$30,000
$ 5,100
The third constraint for the aggressive investor becomes
6I + 4B ๏ฃ 320
This constraint is redundant; the available funds and the maximum Internet fund investment
constraints define the feasible region. The optimal solution is:
Internet fund
Blue Chip fund
Annual return
$35,000
$15,000
$ 5,550
The aggressive investor places as much funds as possible in the high return but high risk Internet
fund.
c.
The third constraint for the conservative investor becomes
6I + 4B ๏ฃ 160
This constraint becomes a binding constraint. The optimal solution is
Internet fund
Blue Chip fund
Annual return
$0
$40,000
$ 3,600
2 – 21
Chapter 2
The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fund
is still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. The
remaining $10,000 could be invested in low-risk bonds or certificates of deposit.
28. a.
Let W = number of jars of Western Foods Salsa produced
M = number of jars of Mexico City Salsa produced
Max
s.t.
1W
+
1.25M
5W
3W
+
2W
+
W, M ๏ณ 0
7M
1M
2M
๏ฃ
๏ฃ
๏ฃ
4480
2080
1600
Whole tomatoes
Tomato sauce
Tomato paste
Note: units for constraints are ounces
b.
Optimal solution: W = 560, M = 240
Value of optimal solution is 860
29. a.
Let B = proportion of Buffalo’s time used to produce component 1
D = proportion of Dayton’s time used to produce component 1
Buffalo
Dayton
Maximum Daily Production
Component 1
Component 2
2000
1000
600
1400
Number of units of component 1 produced: 2000B + 600D
Number of units of component 2 produced: 1000(1 – B) + 600(1 – D)
For assembly of the ignition systems, the number of units of component 1 produced must equal the
number of units of component 2 produced.
Therefore,
2000B + 600D = 1000(1 – B) + 1400(1 – D)
2000B + 600D = 1000 – 1000B + 1400 – 1400D
3000B + 2000D = 2400
Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can
maximize the number of electronic ignition systems produced by maximizing the number of units of
subassembly 1 produced.
Max 2000B + 600D
In addition, B ๏ฃ 1 and D ๏ฃ 1.
2 – 22
An Introduction to Linear Programming
The linear programming model is:
Max
s.t.
2000B
+ 600D
3000B
B
+ 2000D
= 2400
๏ฃ1
๏ฃ1
๏ณ0
D
B, D
The graphical solution is shown below.
D
1.2
1.0
30
.8
00
B+
20
.6
00
D
=2
40
.4
0
Optimal
Solution
2000B + 600D = 300
.2
B
0
.2
.4
.6
.8
1.0
Optimal Solution: B = .8, D = 0
Optimal Production Plan
Buffalo – Component 1
Buffalo – Component 2
Dayton – Component 1
Dayton – Component 2
.8(2000) = 1600
.2(1000) = 200
0(600) = 0
1(1400) = 1400
Total units of electronic ignition system = 1600 per day.
2 – 23
1.2
Chapter 2
30. a.
Let
E = number of shares of Eastern Cable
C = number of shares of ComSwitch
Max
s.t.
15E
+ 18C
40E
40E
+ 25C
25C
25C
E, C ๏ณ 0
๏ฃ
๏ณ
๏ณ
๏ฃ
50,000
15,000
10,000
25,000
Maximum Investment
Eastern Cable Minimum
ComSwitch Minimum
ComSwitch Maximum
b.
C
Number of Shares of ComSwitch
2000
Minimum Eastern Cable
1500
Maximum Comswitch
1000
Maximum Investment
500
Minimum Conswitch
0
500
1000
1500
Number of Shares of Eastern Cable
E
c.
There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000)
d.
Optimal solution is E = 625, C = 1000
Total return = $27,375
2 – 24
An Introduction to Linear Programming
31.
B
6
Feasible
Region
4
2
A
0
2
4
6
3A + 4B = 13
Optimal Solution
A = 3, B = 1
Objective Function Value = 13
32.
A
B
A
B
A
2 – 25
Chapter 2
Extreme Points
(A = 250, B = 100)
(A = 125, B = 225)
(A = 125, B = 350)
Objective
Function Value
800
925
1300
Surplus
Demand
125
โ
โ
Surplus
Total Production
โ
โ
125
33. a.
xB2
6
4
2
0
xA1
2
4
6
Optimal Solution: A = 3, B = 1, value = 5
b.
(1)
(2)
(3)
(4)
3 + 4(1) = 7
2(3) + 1 = 7
3(3) + 1.5 = 10.5
-2(3) +6(1) = 0
Slack = 21 – 7 = 14
Surplus = 7 – 7 = 0
Slack = 21 – 10.5 = 10.5
Surplus = 0 – 0 = 0
2 – 26
Slack
Processing Time
โ
125
โ
An Introduction to Linear Programming
c.
B
A
Optimal Solution: A = 6, B = 2, value = 34
34. a.
B
x2
4
3
Fe asible
Re gion
2
1
(21/4, 9/4)
(4,1)
x1A
0
1
2
3
4
5
b.
There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4)
c.
The optimal solution is A = 4, B = 1
2 – 27
6
Chapter 2
35. a.
Min
s.t.
6A
+
4B
+
0S1
2A
1A
+
+
1B
1B
1B
–
S1
+
–
0S2
+
0S3
S2
+
S3
=
=
=
A, B, S1, S2, S3 ๏ณ 0
b.
The optimal solution is A = 6, B = 4.
c.
S1 = 4, S2 = 0, S3 = 0.
36. a.
Let
Max
s.t.
T =
P =
number of training programs on teaming
number of training programs on problem solving
10,000T
+
8,000P
+
+
P
P
2P
T
T
3T
๏ณ
๏ณ
๏ณ
๏ฃ
8
10
25
84
T, P ๏ณ 0
2 – 28
Minimum Teaming
Minimum Problem Solving
Minimum Total
Days Available
12
10
4
An Introduction to Linear Programming
b.
P
Minimum Teaming
Number of Problem-Solving Programs
40
30
Minimum
Total
20
Days Available
Minimum Problem Solving
10
0
10
20
Number of Teaming Programs
c.
There are four extreme points: (15,10); (21.33,10); (8,30); (8,17)
d.
The minimum cost solution is T = 8, P = 17
Total cost = $216,000
30
37.
Mild
Extra Sharp
Regular
80%
20%
Zesty
60%
40%
8100
3000
Let R = number of containers of Regular
Z = number of containers of Zesty
Each container holds 12/16 or 0.75 pounds of cheese
Pounds of mild cheese used
=
=
0.80 (0.75) R + 0.60 (0.75) Z
0.60 R + 0.45 Z
Pounds of extra sharp cheese used =
=
0.20 (0.75) R + 0.40 (0.75) Z
0.15 R + 0.30 Z
2 – 29
T
Chapter 2
Cost of Cheese
=
=
=
=
Cost of mild + Cost of extra sharp
1.20 (0.60 R + 0.45 Z) + 1.40 (0.15 R + 0.30 Z)
0.72 R + 0.54 Z + 0.21 R + 0.42 Z
0.93 R + 0.96 Z
Packaging Cost = 0.20 R + 0.20 Z
Total Cost
= (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z)
= 1.13 R + 1.16 Z
Revenue
= 1.95 R + 2.20 Z
Profit Contribution = Revenue – Total Cost
= (1.95 R + 2.20 Z) – (1.13 R + 1.16 Z)
= 0.82 R + 1.04 Z
Max
s.t.
0.82 R
+
1.04 Z
0.60 R
+
0.15 R
+
R, Z ๏ณ 0
0.45 Z
0.30 Z
๏ฃ
๏ฃ
8100
3000
Mild
Extra Sharp
Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280
38. a.
Let
S = yards of the standard grade material per frame
P = yards of the professional grade material per frame
Min
s.t.
7.50S
+ 9.00P
0.10S + 0.30P
0.06S + 0.12P
S +
P
S, P ๏ณ 0
๏ณ 6
๏ฃ 3
= 30
carbon fiber (at least 20% of 30 yards)
kevlar (no more than 10% of 30 yards)
total (30 yards)
2 – 30
An Introduction to Linear Programming
b.
P
Professional Grade (yards)
50
40
total
Extreme Point
S = 10 P = 20
30
Feasible region is the
line segment
20
kevlar
carbon fiber
10
Extreme Point
S = 15 P = 15
S
0
10
20
30
40
50
60
Standard Grade (yards)
c.
Extreme Point
(15, 15)
(10, 20)
Cost
7.50(15) + 9.00(15) = 247.50
7.50(10) + 9.00(20) = 255.00
The optimal solution is S = 15, P = 15
d.
Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is
reduced to 7.50(15) + 8(15) = $232.50.
e.
At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is
reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change
the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).
39. a.
Let S = number of units purchased in the stock fund
M = number of units purchased in the money market fund
Min
s.t.
8S
+
50S
5S
+
+
3M
100M
4M
M
S, M, ๏ณ 0
๏ฃ
๏ณ
๏ณ
1,200,000 Funds available
60,000 Annual income
3,000 Minimum units in money market
2 – 31
Chapter 2
Units of Mo ney Mark et Fu nd
x2
M
20 000
+ 3M = 62,000
8x8S
1 + 3×2 = 62,000
15 000
Optim al Solution
.
10 000
50 00
0
50 00
10 000
15 000
20 000
x1S
Units of Stock Fun d
Optimal Solution: S = 4000, M = 10000, value = 62000
40.
b.
Annual income = 5(4000) + 4(10000) = 60,000
c.
Invest everything in the stock fund.
Let P1 = gallons of product 1
P2 = gallons of product 2
Min
s.t.
1P1
+
1P1
+
1P1
+
1P2
1P2
2P2
๏ณ
๏ณ
๏ณ
P1 , P2 ๏ณ 0
2 – 32
30
20
80
Product 1 minimum
Product 2 minimum
Raw material
An Introduction to Linear Programming
P2
Feasible
Region
+1
1
60
1P
=
P2
55
Number of Gallons of Product 2
80
40
20
Use
8
(30,25)
0
0g
als.
40
20
60
80
Number of Gallons of Product 1
P1
Optimal Solution: P1 = 30, P2 = 25 Cost = $55
41. a.
Let R = number of gallons of regular gasoline produced
P = number of gallons of premium gasoline produced
Max
s.t.
0.30R
+
0.50P
0.30R
1R
+
+
0.60P
1P
1P
๏ฃ
๏ฃ
๏ฃ
18,000
50,000
20,000
R, P ๏ณ 0
2 – 33
Grade A crude oil available
Production capacity
Demand for premium
Chapter 2
b.
P
Gallons of Premium Gasoline
60,000
50,000
Production Capacity
40,000
30,000
Maximum Premium
20,000
Optimal Solution
R = 40,000, P = 10,000
$17,000
10,000
Grade A Crude Oil
0
R
10,000 20,000 30,000 40,000 50,000 60,000
Gallons of Regular Gasoline
Optimal Solution:
40,000 gallons of regular gasoline
10,000 gallons of premium gasoline
Total profit contribution = $17,000
c.
Constraint
1
2
3
d.
Value of Slack
Variable
0
0
10,000
Interpretation
All available grade A crude oil is used
Total production capacity is used
Premium gasoline production is 10,000 gallons less than
the maximum demand
Grade A crude oil and production capacity are the binding constraints.
2 – 34
An Introduction to Linear Programming
42.
B
x2
14
Sa tisfies Constraint #2
12
10
8
Infe asibility
6
4
Sa tisfies Constraint #1
2
0
4
2
6
8
10
12
x1A
43.
Bx
2
4
Unbounded
3
2
1
0
1
2
3
x1A
44. a.
xB2
Objective Function
Optimal Solution
(30/16 , 30/16 )
Value = 60/16
4
2
0
b.
2
New optimal solution is A = 0, B = 3, value = 6.
2 – 35
4
xA1
Chapter 2
45. a.
B
A
A
A
46.
B
A
b.
Feasible region is unbounded.
c.
Optimal Solution: A = 3, B = 0, z = 3.
d.
An unbounded feasible region does not imply the problem is unbounded. This will only be the case
when it is unbounded in the direction of improvement for the objective function.
Let
N = number of sq. ft. for national brands
G = number of sq. ft. for generic brands
Problem Constraints:
N
N
+
G
G
๏ฃ
๏ณ
๏ณ
2 – 36
200
120
20
Space available
National brands
Generic
An Introduction to Linear Programming
Extreme Point
1
2
3
N
120
180
120
G
20
20
80
a.
Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the generic
brand.
b.
Alternative optimal solutions. Any point on the line segment joining extreme point 2 and extreme
point 3 is optimal.
c.
Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic
brand.
2 – 37
Chapter 2
47.
Bx2
P ro
600
s
ce s
ing
500
e
Tim
400
300
Alte rnate optim a
(125,225)
200
100
(250,100)
0
100
200
300
400
x1A
Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100).
Cost
= 3(125) + 3(225) = 1050
Cost
= 3(250) + 3(100) = 1050
or
The solution (A = 250, B = 100) uses all available processing time. However, the solution
(A = 125, B = 225) uses only 2(125) + 1(225) = 475 hours.
.
Thus, (A = 125, B = 225) provides 600 – 475 = 125 hours of slack processing time which may be
used for other products.
2 – 38
An Introduction to Linear Programming
48.
Possible Actions:
i.
Reduce total production to A = 125, B = 350 on 475 gallons.
ii.
Make solution A = 125, B = 375 which would require 2(125) + 1(375) = 625 hours of processing
time. This would involve 25 hours of overtime or extra processing time.
iii. Reduce minimum A production to 100, making A = 100, B = 400 the desired solution.
49. a.
Let
P = number of full-time equivalent pharmacists
T = number of full-time equivalent physicians
The model and the optimal solution are shown below:
MIN 40P+10T
S.T.
1)
2)
3)
P+T >=250
2P-T>=0
P>=90
Optimal Objective Value
5200.00000
Variable
P
T
Value
90.00000
160.00000
2 – 39
Reduced Cost
0.00000
0.00000
Chapter 2
Constraint
1
2
3
Slack/Surplus
0.00000
20.00000
0.00000
Dual Value
10.00000
0.00000
30.00000
The optimal solution requires 90 full-time equivalent pharmacists and 160 full-time equivalent
technicians. The total cost is $5200 per hour.
b.
Pharmacists
Technicians
Current Levels
85
175
Attrition
10
30
Optimal Values
90
160
New Hires Required
15
15
The payroll cost using the current levels of 85 pharmacists and 175 technicians is 40(85) + 10(175) =
$5150 per hour.
The payroll cost using the optimal solution in part (a) is $5200 per hour.
Thus, the payroll cost will go up by $50
50.
Let
M = number of Mount Everest Parkas
R = number of Rocky Mountain Parkas
Max
s.t.
100M
+
150R
30M
45M
0.8M
+
+
–
20R
15R
0.2R
๏ฃ
๏ฃ
๏ณ
7200 Cutting time
7200 Sewing time
0 % requirement
Note: Students often have difficulty formulating constraints such as the % requirement constraint.
We encourage our students to proceed in a systematic step-by-step fashion when formulating these
types of constraints. For example:
M must be at least 20% of total production
M ๏ณ 0.2 (total production)
M ๏ณ 0.2 (M + R)
M ๏ณ 0.2M + 0.2R
0.8M – 0.2R ๏ณ 0
2 – 40
An Introduction to Linear Programming
The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 100(65.45) +
150(261.82) = $45,818. If we think of this situation as an on-going continuous production process,
the fractional values simply represent partially completed products. If this is not the case, we can
approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with
a corresponding profit of $45,650.
51.
Let
C = number sent to current customers
N = number sent to new customers
Note:
Number of current customers that test drive = .25 C
Number of new customers that test drive = .20 N
Number sold = .12 ( .25 C ) + .20 (.20 N )
= .03 C + .04 N
Max
s.t.
.03C
+
.04N
.25 C
.20 N
.25 C
– .40 N
4C +
6N
C, N, ๏ณ 0
30,000
๏ณ
10,000
๏ณ
0
๏ณ
๏ฃ 1,200,000
2 – 41
Current Min
New Min
Current vs. New
Budget
Chapter 2
Current Min.
N
200,000
Current ๏ณ 2 New
Budget
.03
C
+.
04
N
=6
100,000
00
0
Optimal Solution
C = 225,000, N = 50,000
Value = 8,750
New Min.
0
52.
Let
100,000
200,000
S = number of standard size rackets
O = number of oversize size rackets
Max
s.t.
10S
+
15O
0.8S
10S
0.125S
+
+
S, O, ๏ณ 0
0.2O
12O
0.4O
2 – 42
๏ณ
๏ฃ
๏ฃ
0
4800
80
% standard
Time
Alloy
300,000
C
An Introduction to Linear Programming
53. a.
Let
R = time allocated to regular customer service
N = time allocated to new customer service
Max
s.t.
1.2R
+
N
R
25R
-0.6R
+
+
+
N
8N
N
๏ฃ
๏ณ
๏ณ
80
800
0
R, N, ๏ณ 0
b.
Optimal Objective Value
90.00000
Variable
R
N
Value
50.00000
30.00000
Reduced Cost
0.00000
0.00000
Constraint
1
2
3
Slack/Surplus
0.00000
690.00000
0.00000
Dual Value
1.12500
0.00000
-0.12500
Optimal solution: R = 50, N = 30, value = 90
HTS should allocate 50 hours to service for regular customers and 30 hours to calling on new
customers.
54. a.
Let
M1 = number of hours spent on the M-100 machine
M2 = number of hours spent on the M-200 machine
Total Cost
6(40)M1 + 6(50)M2 + 50M1 + 75M2 = 290M1 + 375M2
Total Revenue
25(18)M1 + 40(18)M2 = 450M1 + 720M2
Profit Contribution
(450 – 290)M1 + (720 – 375)M2 = 160M1 + 345M2
2 – 43
Chapter 2
Max
s.t.
160 M1
+
345M2
M1
M2
M1
40 M1
+
M2
50 M2
๏ฃ
๏ฃ
๏ณ
๏ณ
๏ฃ
15
10
5
5
1000
M-100 maximum
M-200 maximum
M-100 minimum
M-200 minimum
Raw material available
M1, M2 ๏ณ 0
b.
Optimal Objective Value
5450.00000
Variable
M1
M2
Value
12.50000
10.00000
Reduced Cost
0.00000
145.00000
Constraint
1
2
3
4
5
Slack/Surplus
2.50000
0.00000
7.50000
5.00000
0.00000
Dual Value
0.00000
145.00000
0.00000
0.00000
4.00000
The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200.
55. Mr. Krtickโs solution cannot be optimal. Every department has unused hours, so there are no binding
constraints. With unused hours in every department, clearly some more product can be made.
56. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible (it had
an optimal solution). Every solution that was feasible is still feasible when we change the constraint to lessthan-or-equal-to, since the new constraint is satisfied at equality (as well as inequality). In summary, we have
relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as
strict inequality).
57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-thanor-equal to constraint as shown below. Constraints 2,3, and 4 form the feasible region and constraint 1 is
redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible.
2 – 44
An Introduction to Linear Programming
Original Problem:
Modified Problem:
58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of
products needed so that everyoneโs top three choices are included. There are only two possible outcomes
relative to the bossโ new constraint. First, suppose the minimum number of products is 15. Then the new constraint makes
the problem infeasible.
2 – 45

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