Transcript Linear Programming - Winthrop University: College of

Chapter 13 Supplement
Linear Programming
Operations Management - 5th Edition
Roberta Russell & Bernard W. Taylor, III
Copyright 2006 John Wiley & Sons, Inc.
Beni Asllani
University of Tennessee at Chattanooga
Lecture Outline





Model Formulation
Graphical Solution Method
Linear Programming Model
Solution
Solving Linear Programming Problems
with Excel
 Sensitivity Analysis
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-2
Objective of Linear
Programming
 Maximize the use of resources to
achieve competitive priorities, subject to
constraints (restrictions).
 Optimum solution is a base case which
must be adjusted to reflect business
realities.
Uses of Linear Programming
 Production Scheduling
Maximize contribution margin
 Minimize cost
Different objectives yield different schedules.

 Determine product or service mix


Maximize contribution margin
Maximize revenue
Uses of Linear Programming (2)
 Scheduling labor in services

Minimize cost, while providing adequate
staff
 Production-location problem

Allocate products and customers to plants
to minimize total cost of production and
distribution
Uses of Linear Programming (3)
 Distribution


Minimize cost
Maximize on-time delivery
 Facility location

Minimize transportation cost.
 Emergency response systems

Minimize average response time.
Uses of Linear Programming (4)
 Investment/ capital budgeting

Determine amount to invest in different
alternatives, given constraints (risk,
diversity, available capital) and investment
objectives
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-7
LP Model Formulation
 Decision variables

mathematical symbols representing levels of activity of an
operation
 Objective function



a linear relationship reflecting the objective of an operation
most frequent objective of business firms is to maximize profit
most frequent objective of individual operational units (such as
a production or packaging department) is to minimize cost
 Constraint

a linear relationship representing a restriction on decision
making
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-8
LP Model Formulation (cont.)
Max/min
z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variables
bi = constraint levels
cj = objective function coefficients
aij = constraint coefficients
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-9
LP Model: Example
RESOURCE REQUIREMENTS
PRODUCT
Bowl
Mug
Labor
(hr/unit)
1
2
Clay
(lb/unit)
4
3
Revenue
($/unit)
40
50
There are 40 hours of labor and 120 pounds of clay
available each day
Decision variables
x1 = number of bowls to produce
x2 = number of mugs to produce
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-10
LP Formulation: Example
Maximize Z = $40 x1 + 50 x2
Subject to
x1 + 2x2 40 hr (labor constraint)
4x1 + 3x2 120 lb (clay constraint)
x1 , x2 0
Solution is x1 = 24 bowls
Revenue = $1,360
Copyright 2006 John Wiley & Sons, Inc.
x2 = 8 mugs
Supplement 13-11
Linear Programming
Terminology
 Feasible solution: Any solution which satisfies
all constraints.
 Feasible region: Set of points which satisfy all
constraints.
 Optimal solution(s): A point which
 Satisfies all constraints
 Maximizes or minimizes the value of the
objective function.
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-12
To Draw the Feasible Region
 Convert each constraint into an equation.
 Draw the corresponding line.
 The set of points bounded by these lines
is the feasible region.
Graphical Solution Method
1. Plot model constraint on a set of coordinates
in a plane
2. Draw the feasible region
3. Plot objective function to find the point on
boundary of this space that maximizes (or
minimizes) value of objective function
4. The optimal solution will be located at a
vertex (corner point) of the feasible region
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-14
Graphical Solution: Example
x2
50 –
40 –
4 x1 + 3 x2 120 lb
30 –
Area common to
both constraints
20 –
10 –
0–
x1 + 2 x2 40 hr
|
10
|
20
|
30
|
40
|
50
|
60
x1
Computing Optimal Values
x1 + 2x2 = 40
4x1 + 3x2 = 120
x2
40 –
4 x1 + 3 x2 120 lb
4x1 + 8x2 = 160
-4x1 - 3x2 = -120
30 –
20 –
x1 + 2 x2 40 hr
5x2 =
x2 =
40
8
x1 + 2(8) =
x1
=
40
24
10 –
8
0–
|
10
| 24 |
20
30
| x1
40
Z = $40(24) + $50(8) = $1,360
Extreme Corner Points
x1 = 0 bowls
x2 = 20 mugs
Z = $1,000
x2
40 –
x1 = 24 bowls
x2 = 8 mugs
Z = $1,360
30 –
20 – A
10 –
0–
B
|
10
|
20
| C|
30 40
x1
x1 = 30 bowls
x2 = 0 mugs
Z = $1,200
Types of Constraints
 Less than or equal to (<)
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Amount available
Is at most
Cannot exceed
Cannot buy (or sell, or use) more than
Is limited to (or is limited by)
 Slack = amount of a resource that is not
used
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-18
Types of Constraints (2)
 Greater than or equal to (>)
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Is at least
Must be (or must have) at least
Cannot be less than
Have a contract to sell (or buy) at least
 Amount by which a right-hand side
quantity exceeds the required amount
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-19
Types of Constraints (3)
 Equals (=)
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

Equals
Is exactly (must be exactly)
Must use all (that is available)
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-20
Minimization Problem
CHEMICAL CONTRIBUTION
Brand
Nitrogen (lb/bag)
Phosphate (lb/bag)
2
4
4
3
Gro-plus
Crop-fast
Minimize Z = $6x1 + $3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x 1, x 2  0
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-21
Graphical Solution
x2
14 –
x1 = 0 bags of Gro-plus
12 – x = 8 bags of Crop-fast
2
Z = $24
10 –
8–A
Z = 6x1 + 3x2
6–
4–
B
2–
0–
|
2
|
4
|
6
|
8
C
|
10
|
12
|
14
x1
Copyright 2006 John Wiley & Sons, Inc.
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use of the information herein.
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-23