#### Transcript Module B - Faculty

```B
Linear Programming
PowerPoint presentation to accompany
Heizer and Render
Operations Management, 10e
Principles of Operations Management, 8e
PowerPoint slides by Jeff Heyl
© 2011 Pearson Education, Inc. publishing as Prentice Hall
B-1
Outline

Requirements of a Linear Programming Problem

Formulating Linear Programming Problems


Graphical Solution to a Linear Programming Problem

Graphical Representation of Constraints

Iso-Profit Line Solution Method

Corner-Point Solution Method

Solving Minimization Problems

Linear Programming Applications

Production-Mix Example

Diet Problem Example
2
Learning Objectives
1. Formulate linear programming models,
including an objective function and
constraints
2. Graphically solve an LP problem with the
iso-profit line method
3. Graphically solve an LP problem with the
corner-point method
4. Construct and solve a minimization
problem
3
Why Use Linear Programming?
 A mathematical technique to help
plan and make decisions relative to
resources
 Will find the minimum or maximum
value of the objective
 Guarantees the optimal solution to
the model formulated
4
LP Applications
1. Scheduling school buses to minimize
total distance traveled
2. Allocating police patrol units to high
crime areas in order to minimize
response time to 911 calls
3. Scheduling tellers at banks so that
needs are met during each hour of the
day while minimizing the total cost of
labor
5
LP Applications
4. Selecting the product mix in a factory
to make best use of machine- and
labor-hours available while maximizing
the firm’s profit
5. Picking blends of raw materials in feed
mills to produce finished feed
combinations at minimum costs
6. Determining the distribution system
that will minimize total shipping cost
6
Requirements of an
LP Problem
1. LP problems seek to maximize or
minimize some quantity
2. The presence of restrictions, or
constraints
3. There must be alternative courses of
action to choose from
4. The objective and constraints in linear
programming problems must be
expressed in terms of linear equations
or inequalities
7
Formulating LP Problems
The product-mix problem at Shader Electronics
 Two products
1. Shader x-pod, a portable music
player
2. Shader BlueBerry, an internetconnected color telephone
 Determine the mix of products that will
produce the maximum profit
8
Formulating LP Problems
Hours Required
to Produce 1 Unit
Department
Electronic
Assembly
Profit per unit
x-pods
(X1)
BlueBerrys
(X2)
4
2
3
1
\$7
\$5
Available Hours
This Week
240
100
Table B.1
Decision Variables:
X1 = number of x-pods to be produced
X2 = number of BlueBerrys to be produced
9
Formulating LP Problems
Objective Function:
Maximize Profit = \$7X1 + \$5X2
There are three types of constraints
 Upper limits where the amount used is ≤
the amount of a resource
 Lower limits where the amount used is ≥
the amount of the resource
 Equalities where the amount used is =
the amount of the resource
10
Formulating LP Problems
First Constraint:
Electronic
time used
is ≤
Electronic
time available
4X1 + 3X2 ≤ 240 (hours of electronic time)
Second Constraint:
Assembly
time used
is ≤
Assembly
time available
2X1 + 1X2 ≤ 100 (hours of assembly time)
11
Graphical Solution
 Can be used when there are two
decision variables
1. Plot the constraint equations at their
limits by converting each equation
to an equality
2. Identify the feasible solution space
3. Create an iso-profit line based on
the objective function
4. Move this line outwards until the
optimal point is identified
12
Graphical Solution
X2
100 –
Number of BlueBerrys
–
80 –
–
60 –
–
40 –
Electronics (Constraint A)
–
20 – Feasible
–
|–
0
Figure B.3
Assembly (Constraint B)
region
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of x-pods
13
Graphical Solution
X2
Iso-Profit
Line Solution Method
100 –
Number of BlueBerrys
Choose a–possible value for the
objective
80 –function Assembly (Constraint B)
–
\$210 = 7X1 + 5X2
60 –
–
Solve for
40 –the axis intercepts of the function
Electronics (Constraint A)
and plot the
line
–
20 – Feasible
–
|–
0
Figure B.3
region
X2 =
|
|
20
|
42
|
40
X1 = 30
|
|
60
|
|
80
|
|
100
X1
Number of x-pods
14
Graphical Solution
X2
100 –
Number of BlueBerrys
–
80 –
–
60 –
–
40 –
\$210 = \$7X1 + \$5X2
(0, 42)
–
20 –
(30, 0)
–
|–
0
Figure B.4
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of x-pods
15
Graphical Solution
X2
100 –
Number of BlueBerrys
–
\$350 = \$7X1 + \$5X2
80 –
\$280 = \$7X1 + \$5X2
–
60 –
\$210 = \$7X1 + \$5X2
–
40 –
–
\$420 = \$7X1 + \$5X2
20 –
–
|–
0
Figure B.5
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of x-pods
16
Graphical Solution
X2
100 –
Number of BlueBerrys
–
Maximum profit line
80 –
–
60 –
Optimal solution point
(X1 = 30, X2 = 40)
–
40 –
–
\$410 = \$7X1 + \$5X2
20 –
–
|–
0
Figure B.6
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of x-pods
17
Corner-Point Method
X2
Number of BlueBerrys
100 –
2
–
80 –
–
60 –
–
3
40 –
–
20 –
–
1
Figure B.7
|–
0
|
|
20
|
|
40
|
4
|
60
|
|
80
|
|
100
X1
Number of x-pods
18
Corner-Point Method
 The optimal value will always be at a
corner point
 Find the objective function value at each
corner point and choose the one with the
highest profit
Point 1 :
(X1 = 0, X2 = 0)
Profit \$7(0) + \$5(0) = \$0
Point 2 :
(X1 = 0, X2 = 80)
Profit \$7(0) + \$5(80) = \$400
Point 4 :
(X1 = 50, X2 = 0)
Profit \$7(50) + \$5(0) = \$350
19
Corner-Point Method
 The optimal value will always be at a
Solvepoint
for the intersection of two constraints
corner
(electronics
time)
1 + 3X2 ≤ 240
 Find the4Xobjective
function
value
at each
(assembly
time)with the
corner 2X
point
and
choose
the one
1 + 1X
2 ≤ 100
highest profit
Point 1 :
4X1 + 3X2 = 240
- 4X1 - 2X2 = -200
(X1 = 0, X2 = 0)
+ 1X2 =
40
4X1 + 3(40) = 240
4X1 + 120 = 240
Profit \$7(0) + \$5(0) = \$0
X1 = 30
Point 2 :
(X1 = 0, X2 = 80)
Profit \$7(0) + \$5(80) = \$400
Point 4 :
(X1 = 50, X2 = 0)
Profit \$7(50) + \$5(0) = \$350
20
Corner-Point Method
 The optimal value will always be at a
corner point
 Find the objective function value at each
corner point and choose the one with the
highest profit
Point 1 :
(X1 = 0, X2 = 0)
Profit \$7(0) + \$5(0) = \$0
Point 2 :
(X1 = 0, X2 = 80)
Profit \$7(0) + \$5(80) = \$400
Point 4 :
(X1 = 50, X2 = 0)
Profit \$7(50) + \$5(0) = \$350
Point 3 :
(X1 = 30, X2 = 40)
Profit \$7(30) + \$5(40) = \$410
21
Solving Minimization
Problems
 Formulated and solved in much the
same way as maximization problems
 In the graphical approach an iso-cost
line is used
 The objective is to move the iso-cost line
inwards until it reaches the lowest cost
corner point
22
Minimization Example
X1 = number of tons of black-and-white picture
chemical produced
X2 = number of tons of color picture chemical
produced
Minimize total cost = 2,500X1 + 3,000X2
Subject to:
X1
X2
X1 + X2
X1, X2
≥ 30 tons of black-and-white chemical
≥ 20 tons of color chemical
≥ 60 tons total
≥ \$0 nonnegativity requirements
23
Minimization Example
Table B.9
X2
60 –
X1 + X2 = 60
50 –
Feasible
region
40 –
30 –
b
20 –
a
10 –
|–
0
X1 = 30
|
10
|
20
X2 = 20
|
30
|
40
|
50
|
60
X1
24
Minimization Example
Total cost at a = 2,500X1
+ 3,000X2
= 2,500 (40) + 3,000(20)
= \$160,000
Total cost at b = 2,500X1
+ 3,000X2
= 2,500 (30) + 3,000(30)
= \$165,000
Lowest total cost is at point a
25
LP Applications
Production-Mix Example
Department
Product
XJ201
XM897
TR29
BR788
Wiring Drilling
.5
1.5
1.5
1.0
3
1
2
3
Assembly
2
4
1
2
Inspection
.5
1.0
.5
.5
Unit Profit
\$ 9
\$12
\$15
\$11
Department
Capacity
(in hours)
Product
Minimum
Production Level
Wiring
Drilling
Assembly
Inspection
1,500
2,350
2,600
1,200
XJ201
XM897
TR29
BR788
150
100
300
400
26
LP Applications
X1 = number of units of XJ201 produced
X2 = number of units of XM897 produced
X3 = number of units of TR29 produced
X4 = number of units of BR788 produced
Maximize profit = 9X1 + 12X2 + 15X3 + 11X4
subject to
.5X1 + 1.5X2 + 1.5X3 + 1X4
3X1 + 1X2 + 2X3 + 3X4
2X1 + 4X2 + 1X3 + 2X4
.5X1 + 1X2 + .5X3 + .5X4
X1
X2
X3
X4
≤ 1,500 hours of wiring
≤ 2,350 hours of drilling
≤ 2,600 hours of assembly
≤ 1,200 hours of inspection
≥ 150 units of XJ201
≥ 100 units of XM897
≥ 300 units of TR29
≥ 400 units of BR788
27
LP Applications
Diet Problem Example
Feed
Product
A
B
C
D
Stock X
Stock Y
Stock Z
3 oz
2 oz
1 oz
6 oz
2 oz
3 oz
0 oz
8 oz
4 oz
1 oz
2 oz
4 oz
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LP Applications
X1 = number of pounds of stock X purchased per cow each month
X2 = number of pounds of stock Y purchased per cow each month
X3 = number of pounds of stock Z purchased per cow each month
Minimize cost = .02X1 + .04X2 + .025X3
Ingredient A requirement:
Ingredient B requirement:
Ingredient C requirement:
Ingredient D requirement:
Stock Z limitation:
3X1 +
2X1 +
1X1 +
6X1 +
2X2 +
3X2 +
0X2 +
8X2 +
4X3
1X3
2X3
4X3
X3
X1, X2, X3
≥ 64
≥ 80
≥ 16
≥ 128
≤ 80
≥0
Cheapest solution is to purchase 40 pounds of grain X
at a cost of \$0.80 per cow
29
In-Class Problems from the
Lecture Guide Practice Problems
Problem 1:
Chad’s Pottery Barn has enough clay to make 24 small vases or 6
large vases. He has only enough of a special glazing compound to
glaze 16 of the small vases or 8 of the large vases. Let X1 = the
number of small vases and X2 = the number of large vases.
The smaller vases sell for \$3 each, and the larger vases would bring
\$9 each.
(a) Formulate the problem
(b) Solve the problem graphically
30
In-Class Problems from the
Lecture Guide Practice Problems
Problem 2:
A fabric firm has received an order for cloth specified to contain at
least 45 pounds of cotton and 25 pounds of silk. The cloth can be
woven out of any suitable mix of two yarns A and B. They contain the
proportions of cotton and silk (by weight) as shown in the following
table:
A
B
Cotton
30%
60%
Silk
50%
10%
Material A costs \$3 per pound, and B costs \$2 per pound. What
quantities (pounds) of A and B yarns should be used to minimize the
cost of this order?
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