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T2 - 1
Optimisation Methods
T2 - 2
Today
 General introduction to the subject
T2 - 3
 Optimization is derived from the Latin
word “optimus”, the best.
 Optimization characterizes the activities
involved to find “the best”.
 People have been “optimizing” forever,
but the roots for modern day optimization
can be traced to the Second World War.
Introduction
T2 - 4
Operational Research
 Operational Research originated from the
activities performed by multidisciplinary
teams formed in the British armed forces
involved in solving complex strategic and
tactical problems in World War II.
 Waddington describes the main objectives of
the Operational Research Section in the
British armed forces as
“The prediction of the effects of
new weapons and tactics.”
T2 - 5
Motivation for Operational Research
 Many problems associated with the
Allied military effort were simply too
complicated to expect adequate
solutions from a single individual, or
even a single discipline.
Due to the diversity of the membership,
one of the earliest groups in Britain
became known as “Blacket's circus”.
They were very successful. Any idea
why?
T2 - 6
OR
 All persons selected were talented men + wartime
pressure + synergism generated from the
interactions of different disciplines
 Due to their success, other allied nations adopted
the same approach.
 Because the work assigned to these groups were
in the nature of military operations their work
became known as operational research in the
United Kingdom and as operations research in
the United States.
 The abbreviation OR is commonly used for both
operational research and operations research.
 Wartime examples: radar deployment, anti-
aircraft fire control, fleet convoy sizing, submarine
detection.
T2 - 7
Operational Research
Operational Research was defined by the Operational
Research Society of Great Britain as follows:
 "Operational research is the application of the methods of
science to complex problems arising in the direction and
management of large systems of men, machines, materials
and money in industry, business, government, and defense.
 The distinctive approach is to develop a scientific model of
the system, incorporating measurements of factors such as
chance and risk, with which to predict and compare the
outcomes of alternative decisions, strategies or controls.
 The purpose is to help management determine its policy
and actions scientifically."
T2 - 8
Current
Definition
 “Operational Research ("OR"), also
known as Operations Research or
Management Science ("OR/MS")
looks at an organisation's operations
and uses mathematical or computer
models, or other analytical
approaches, to find better ways of
doing them.”
T2 - 9
Operations Research
• Although the US effort started later than the British, it produced more
fundamental advances in the mathematical techniques for analyzing
military problems.
 The Operations Research Society of America
had the following definition:
“Operations research is concerned with
scientifically deciding how to best design
and operate man-machine systems,
usually under conditions requiring the
allocation of scarce resources”.
T2 - 10
IFORS
 “Operational Research can be described as a
scientific approach to the solution of problems in the
management of complex systems.
 In a rapidly changing environment an understanding
is sought which will facilitate the choice of more
effective solutions which, typically, may involve
complicated interaction among people, materials, and
money.”
IFORS - International Federation of Operational
Research Societies (www.ifors.org):
T2 - 11
After the War...
 Many of the scientists in the OR groups turned their
activities to applying their approach to civilian problems.
 Some returned to universities to develop a sound
foundation for the hastily developed techniques, others
concentrated on developing new techniques.
 First civilian organizations interested were large profit
making corporations. For example, petroleum
companies were the first to use linear programming on a
large scale for production planning.
 First only big business could afford it.
 Applications in the service industries did not start until
the mid 1960s.
One concurrent technological development has been
critical for OR...
T2 - 12
Electronic Computers
It is generally accepted that without computers, OR and
optimization would not be what they are today.
 Earlier mathematical models (such as calculus, Lagrange
multipliers) relied on sophistication of technique to solve
the problem classes for which they were suited.
 Methods of mathematical optimization (e.g., Linear
Programming) rely far less on mathematical sophistication
than they do on an unusual “adaptibility to the mode of
solution inherent in the modern digital computer”.
 Particularly striking is the simplicity of these methods of
mathematics coupled with their iterative processes (i.e.,
the repeated performance of a relatively simple set of
operations)
T2 - 13
The Computer and Linear Programming
Consider the case of Linear Programming (more in later
lectures):
 The first large scale computer became a practical reality in
1946 at the University of Pennsylvania.
 This was just one year before the development of simplex.
 The simplex method for linear programming consists only of a
few steps and these steps require only the most basic
mathematical operations which a computer is well suited to
handle.
 However, these steps must be repeated over and over before
one finally obtains an answer.
 The first successful computer solution of a LP problem was in
January 1952 on the National Bureau of Standards SEAC
computer.
T2 - 14
By the way...
You will often hear the phrase “programming” as in:
mathematical programming,
linear programming,
nonlinear programming,
mixed integer programming, etc.
T2 - 15
By the way ... (cont.)
This has (in principle) nothing to do with modern
day computer programming.
In the early days, a set of values
which represented a solution to a
problem was referred to as a
“program”.
However, nowadays you program (software) to
find a program!
T2 - 16
Types of Optimization
Do not forget:
Optimization methods fall in the category of
“decision support systems/methods”
 Note: There are MANY different
optimization methods/algorithms
 However, they are can be grouped by
fundamental principles of:
 model, or
 solution method/algorithm
T2 - 17
Learning Objectives
 Explain linear programming
 Formulate linear programming
problems
 Solve linear programming problems
using graphical methods
 Corner point
 Iso-profit line
T2 - 18
What Is
Linear Programming?
 Mathematical technique
 Not computer programming
 Allocates scarce resources to
achieve an objective
 Pioneered by George Dantzig in
World War 2
 Developed workable solution in 1947
 Called Simplex Method
T2 - 19
Linear Programming
Applications
 Find product mix given machine &
labor hours to maximize profit
 Schedule production given demand
to minimize costs
 Allocate police given limited patrol
cars to minimize response time
 Plan menus given minimum daily diet
requirements to minimize cost
T2 - 20
Linear Programming
Requirements
 One objective
 Maximize or minimize
 Constraints (e.g., limited resources)
 Alternative courses of action
(decision variables)
 Divisible (fractional values)
 Non-negative
 Linear relationships
T2 - 21
Solving Linear
Programming Problems
 Formulate problem
 Define decision variables
 Write objective function
 Write constraint functions
 Find solution
 Graphical methods
 Simplex method
 Karmarkar’s algorithm © 1984-1994 T/Maker Co.
T2 - 22
Maximization Problems
T2 - 23
Maximization Example
You’re an analyst for a
division of Levi, which
makes baseball &
western hats. Each hat
must be run on each of
2 machines. Machine 1
has 40 hr. available per
week; machine 2, 120
hr. How many of each
hat should be made?
Rs.4 profit
1 hr. machine 1
4 hr. machine 2
Rs.5 profit
2 hr. machine 1
3 hr. machine 2
© 1984-1994 T/Maker Co.
© 1984-1994
T/Maker Co.
T2 - 24
Problem Setup
Hours/Unit
Machine Ball
(X1)
Western
(X2)
Capacity
(Hrs/Wk)
Machine 1 1 hr 2 hr 40 hr
Machine 2 4 hr 3 hr 120 hr
Unit Profit $4 $5
T2 - 25
Formulation Solution
 Decision variables
 X1 = no. baseball
hats produced
 X2 = no. western
hats produced
 Objective function
 Maximize profits =
Z = 4X1 + 5X2
T2 - 26
 Decision variables
 X1 = no. baseball
hats produced
 X2 = no. western
hats produced
 Objective function
 Maximize profits =
Z = 4X1 + 5X2
Formulation Solution
 Constraint functions
 Machine 1 capacity:
1X1 + 2X2 40
 Machine 2 capacity:
4X1 + 3X2 120
 Non-negativity
X1 0
X2 0
T2 - 27
Formulation Solution
Summary
Max Z = 4X1 + 5X2
subject to:
1X1 + 2X2 40
4X1 + 3X2 120
X1, X2  0
T2 - 28
Graphical Solution
Method Steps
 Draw graph with vertical & horizontal
axes (1st quadrant only)
 Plot constraints as lines, then as planes
 Use (X1,0), (0,X2) for line
 Find feasible region
 Find optimal solution
 Corner point method
 Iso-profit line method
T2 - 29
Graphical Solution
T2 - 30
Graphical Solution
0
10
20
30
40
0 10 20 30 40
X1
X2
Draw graph with
vertical & horizontal
axes (1st quadrant only)
T2 - 31
Graphical Solution
0
10
20
30
40
0 10 20 30 40
X1
X2
Write 1X1 + 2X2  40 as
equality: 1X1 + 2X2 = 40.
Pick 2 points (X1,0), (0,X2)
to draw line: (40,0), (0,20)
(0, 20)
(40, 0)
T2 - 32
Graphical Solution
0
10
20
30
40
0 10 20 30 40
X1
X2
(0, 20)
(40, 0)
Write 1X1 + 2X2  40 as
equality: 1X1 + 2X2 = 40.
Pick 2 points (X1,0), (0,X2)
to draw line: (40,0), (0,20)
T2 - 33
0
10
20
30
40
0 10 20 30 40
X1
X2
Graphical Solution
Inequality is a plane.
To find which side of
line inequality applies,
test any 2 points in
1X1 + 2X2 40
T2 - 34
0
10
20
30
40
0 10 20 30 40
X1
X2
Graphical Solution
Try (10, 30) in
1X1 + 2X2 40:
(1)(10) + (2)(30) = 70.
Violates constraint.
(10, 30)
T2 - 35
0
10
20
30
40
0 10 20 30 40
X1
X2
Graphical Solution
(10, 30)
(0, 0)
Try (0, 0) in
1X1 + 2X2 40:
(1)(0) + (2)(0) = 0.
Satisfies constraint.
X
T2 - 36
Graphical Solution
Inequality is satisfied
by all points BELOW
the line.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 37
Graphical Solution
(0, 40)
(30, 0)
Write 4X1 + 3X2  120 as
equality: 4X1 + 3X2 = 120.
Pick 2 points (X1,0), (0,X2)
to draw line: (30,0), (0,40)
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 38
Graphical Solution
(0, 40)
(30, 0)
Write 4X1 + 3X2  120 as
equality: 4X1 + 3X2 = 120.
Pick 2 points (X1,0), (0,X2)
to draw line: (30,0), (0,40)
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 39
Graphical Solution
Inequality is a plane.
To find which side of
line inequality applies,
test any 2 points in
4X1 + 3X2 120
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 40
Graphical Solution
(10, 40)
Try (10, 40) in
4X1 + 3X2 120:
(4)(10) + (3)(40) = 160.
Violates constraint.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 41
Graphical Solution
(10, 40)
Try (0, 0) in
4X1 + 3X2 120:
(4)(0) + (3)(0) = 0.
Satisfies constraint.
(0, 0)
X
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 42
Graphical Solution
Inequality is satisfied
by all points BELOW
the line.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 43
Graphical Solution
Feasible region is
intersection of all
planes. It satisfies
all constraints.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 44
Graphical Solution
Feasible region is
intersection of all
planes. It satisfies
all constraints.
Feasible region
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 45
Optimal Solution:
Corner Point Method
A
B
C
D
Optimal solution
is at corner point
of feasible region.
Feasible region
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 46
Optimal Solution:
Corner Point Method
A (0, 0) (4)(0) + (5)(0) = 0
B (0, 20) (4)(0) + (5)(20) = 100
C (24, 8) (4)(24) + (5)(8) = 136
D (30, 0) (4)(30) + (5)(0) = 120
Point C has highest profit. Produce 24 baseball
hats (X1) & 8 western hats (X2).
Point Coordinates Profit: Z = 4X1 + 5X2
T2 - 47
Finding Point C
Coordinates
Point C is the intersection of 2 lines. To solve
simultaneously, multiply 1st equation by - 4:
Add to 2nd equation:
Substitute into 2nd equation to get :
       
  
   


4 1 2 40 4 8 160
4 3 120
4 8 160
8
24
1 2 1 2
1 2
1 2
2
1
1
X X X X
X X
X X
X
X
X
a f
T2 - 48
Optimal Solution:
Iso-Profit Line Method
T2 - 49
Optimal Solution:
Iso-Profit Line Method
Feasible region
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 50
Optimal Solution:
Iso-Profit Line Method
Choose an arbitrary
value for Z = 4X1 + 5X2:
20 = 4X1 + 5X2.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 51
Optimal Solution:
Iso-Profit Line Method
Plot 20 = 4X1 + 5X2.
Pick 2 points (X1,0), (0,X2):
(5,0), (0,4). Connect points.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 52
Optimal Solution:
Iso-Profit Line Method
Plot 20 = 4X1 + 5X2.
Pick 2 points (X1,0), (0,X2):
(5,0), (0,4). Connect points.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 53
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 54
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 55
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 56
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 57
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 58
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Higher
Profit
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 59
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 60
Optimal Solution:
Iso-Profit Line Method
Draw parallel lines moving
outward. Optimal solution
is point where line just
touches feasible region.
Optimum:
‘Just touches’
0
10
20
30
40
0 10 20 30 40
X1
X2
T2 - 61
Linear Programming
Thinking Challenge
You’re an analyst for RCA,
which makes VCR’s &
TV’s. Each product must
be worked on in each of
2 departments. The
electronics dept. has 240
hr. available per week; the
assembly dept., 100 hr.
How many of each
product should be made?
Rs.7 profit
4 hr. Electronics
2 hr. Assembly
Rs.5 profit
3 hr. Electronics
1 hr. Assembly
© 1995 Corel Corp.
© 1995 Corel Corp.
Alone Group Class
T2 - 64
Problem Setup*
Hours Required to
Produce 1 Unit
Dept. VCR
X1
TV
X2
Available
Hrs./Week
Electronic 4 3 240
Assembly 2 1 100
Profit/unit $7 $5
T2 - 65
Formulation Solution*
Max Z = 7X1 + 5X2
subject to:
4X1 + 3X2 240
2X1 + 1X2 100
X1, X2  0
T2 - 66
Graphical Solution*
T2 - 67
Graphical Solution*
0
20
40
60
80
100
120
0 10 20 30 40 50 60
Number VCR’s
No.
TV's
70 80
T2 - 68
Graphical Solution*
0
20
40
60
80
100
120
0 10 20 30 40 50 60
Number VCR’s
No.
TV's
Electronics
Assembly
70 80
T2 - 69
Graphical Solution*
0
20
40
60
80
100
120
0 10 20 30 40 50 60
Number VCR’s
No.
TV's
Electronics
Assembly
70 80
T2 - 70
Graphical Solution*
Feasible
Region
0
20
40
60
80
100
120
0 10 20 30 40 50 60
Number VCR’s
No.
TV's
Electronics
Assembly
70 80
T2 - 71
Optimal Solution:
Corner Point Method*
Feasible
Region
0
20
40
60
80
100
120
0 10 20 30 40 50 60
Number VCR’s
A
B
C
D
No.
TV's
Electronics
Assembly
70 80
T2 - 72
Optimal Solution:
Corner Point Method*
A (0, 80) (7)(0) + (5)(80) = 400
B (30, 40) (7)(30) + (5)(40) = 410
C (50, 0) (7)(50) + (5)(0) = 350
D (0, 0) (7)(0) + (5)(0) = 0
Point B is the optimum solution.
Point Coordinates Profit: Z = 4X1 + 5X2
T2 - 73
Point B Coordinates*
To solve simultaneously,
multiple 2nd equation by - 2:
Add to 1st Equation:
Substitute into 1st equation to get :
       
  
   


2 2 1 100 4 2 200
4 3 240
4 2 200
40
30
1 2 1 2
1 2
1 2
2
1
1
X X X X
X X
X X
X
X
X
a f
T2 - 74
Optimal Solution:
Iso-Profit Line Method*
0
20
40
60
80
100
120
0 10 20 30 40 50 60 40 80
Number VCR’s
No.
TV's
Electronics
Assembly Higher Profit
T2 - 75
Minimization Problems
T2 - 76
Minimization Example
You’re an analyst for a
division of Kodak, which
makes BW & color chemicals.
At least 30 tons of BW & at
least 20 tons of color must be
made each month. The total
chemicals made must be at
least 60 tons. How many of
each chemical should be
made to minimize costs?
BW: Rs.2,500
mfg. cost per
month
Color: Rs. 3,000
mfg. cost per month
© 1995 Corel Corp.
T2 - 77
Formulation Solution
 Decision variables
 X1 = tons of BW chemical produced
 X2 = tons of color chemical produced
 Objective
 Minimize Z = 2500X1 + 3000X2
 Constraints
 X1 30 (BW); X2 20 (Color)
 X1 + X2 60 (Total tonnage)
 X1  0; X2  0 (Non-negativity)
T2 - 78
Graphical Solution
T2 - 79
Graphical Solution
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
Draw axes.
- 40 - 20
T2 - 80
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
- 40 - 20
Everything is possible
without constraints.
T2 - 81
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
- 40 - 20
Find values
satisfying X1  0.
T2 - 82
Graphical Solution
Feasible
Region
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
0
Find values
satisfying X1  0.
T2 - 83
Graphical Solution
Feasible
Region
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
0
Find values
satisfying X2  0.
T2 - 84
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
Find values
satisfying X2  0.
T2 - 85
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW Draw X1  30 as
equality: X1 = 30.
T2 - 86
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW Find values
satisfying X1  30.
T2 - 87
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW Find values
satisfying X1  30.
T2 - 88
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Draw X2  20 as
equality: X2 = 20.
T2 - 89
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Find values
for X2  20.
T2 - 90
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Find values
for X2  20.
T2 - 91
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Draw X1 + X2  60
as equality:
X1 + X2 = 60.
T2 - 92
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Find values for
X1 + X2  60.
T2 - 93
Graphical Solution
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Find values for
X1 + X2  60.
T2 - 94
Optimal Solution:
Corner Point Method
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Find corner
points.
T2 - 95
Optimal Solution:
Corner Point Method
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Find corner
points.
A
B
T2 - 96
Optimal Solution:
Corner Point Method
A (40, 20) (2500)(40) + (3000)(20) = 160,000
B (30, 30) (2500)(30) + (3000)(30) = 165,000
Point A has lowest cost. Produce 40 tons BW
chemicals (X1) & 20 tons color chemicals (X2).
Point Coord. Z = 2500X1 + 3000X2
T2 - 97
Optimal Solution:
Iso-Profit Line Method
T2 - 98
Optimal Solution:
Iso-Profit Line Method
Feasible
Region
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
T2 - 99
Optimal Solution:
Iso-Profit Line Method
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Lower
Cost
T2 - 100
Optimal Solution:
Iso-Profit Line Method
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Lower
Cost
T2 - 101
Optimal Solution:
Iso-Profit Line Method
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Lower
Cost
T2 - 102
Optimal Solution:
Iso-Profit Line Method
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Lower
Cost
T2 - 103
Optimal Solution:
Iso-Profit Line Method
0
20
40
60
80
0
Tons,
Color
Chemical
20 40 60 80
Tons, BW Chemical
BW
Color
Total
Lower
Cost
T2 - 104
Conclusion
 Explained linear programming
 Formulated linear programming
problems
 Solved linear programming problems
using graphical methods
 Corner point
 Iso-profit line
T2 - 105
This Class...
 What was the most important thing
you learned in class today?
 What do you still have questions
about?
 How can today’s class be improved?
Please take a moment to answer the
following questions in writing:
End of Chapter
Any blank slides that follow are
blank intentionally.

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SG 1 Introduction to OR.ppsx for engineering

  • 2. T2 - 2 Today  General introduction to the subject
  • 3. T2 - 3  Optimization is derived from the Latin word “optimus”, the best.  Optimization characterizes the activities involved to find “the best”.  People have been “optimizing” forever, but the roots for modern day optimization can be traced to the Second World War. Introduction
  • 4. T2 - 4 Operational Research  Operational Research originated from the activities performed by multidisciplinary teams formed in the British armed forces involved in solving complex strategic and tactical problems in World War II.  Waddington describes the main objectives of the Operational Research Section in the British armed forces as “The prediction of the effects of new weapons and tactics.”
  • 5. T2 - 5 Motivation for Operational Research  Many problems associated with the Allied military effort were simply too complicated to expect adequate solutions from a single individual, or even a single discipline. Due to the diversity of the membership, one of the earliest groups in Britain became known as “Blacket's circus”. They were very successful. Any idea why?
  • 6. T2 - 6 OR  All persons selected were talented men + wartime pressure + synergism generated from the interactions of different disciplines  Due to their success, other allied nations adopted the same approach.  Because the work assigned to these groups were in the nature of military operations their work became known as operational research in the United Kingdom and as operations research in the United States.  The abbreviation OR is commonly used for both operational research and operations research.  Wartime examples: radar deployment, anti- aircraft fire control, fleet convoy sizing, submarine detection.
  • 7. T2 - 7 Operational Research Operational Research was defined by the Operational Research Society of Great Britain as follows:  "Operational research is the application of the methods of science to complex problems arising in the direction and management of large systems of men, machines, materials and money in industry, business, government, and defense.  The distinctive approach is to develop a scientific model of the system, incorporating measurements of factors such as chance and risk, with which to predict and compare the outcomes of alternative decisions, strategies or controls.  The purpose is to help management determine its policy and actions scientifically."
  • 8. T2 - 8 Current Definition  “Operational Research ("OR"), also known as Operations Research or Management Science ("OR/MS") looks at an organisation's operations and uses mathematical or computer models, or other analytical approaches, to find better ways of doing them.”
  • 9. T2 - 9 Operations Research • Although the US effort started later than the British, it produced more fundamental advances in the mathematical techniques for analyzing military problems.  The Operations Research Society of America had the following definition: “Operations research is concerned with scientifically deciding how to best design and operate man-machine systems, usually under conditions requiring the allocation of scarce resources”.
  • 10. T2 - 10 IFORS  “Operational Research can be described as a scientific approach to the solution of problems in the management of complex systems.  In a rapidly changing environment an understanding is sought which will facilitate the choice of more effective solutions which, typically, may involve complicated interaction among people, materials, and money.” IFORS - International Federation of Operational Research Societies (www.ifors.org):
  • 11. T2 - 11 After the War...  Many of the scientists in the OR groups turned their activities to applying their approach to civilian problems.  Some returned to universities to develop a sound foundation for the hastily developed techniques, others concentrated on developing new techniques.  First civilian organizations interested were large profit making corporations. For example, petroleum companies were the first to use linear programming on a large scale for production planning.  First only big business could afford it.  Applications in the service industries did not start until the mid 1960s. One concurrent technological development has been critical for OR...
  • 12. T2 - 12 Electronic Computers It is generally accepted that without computers, OR and optimization would not be what they are today.  Earlier mathematical models (such as calculus, Lagrange multipliers) relied on sophistication of technique to solve the problem classes for which they were suited.  Methods of mathematical optimization (e.g., Linear Programming) rely far less on mathematical sophistication than they do on an unusual “adaptibility to the mode of solution inherent in the modern digital computer”.  Particularly striking is the simplicity of these methods of mathematics coupled with their iterative processes (i.e., the repeated performance of a relatively simple set of operations)
  • 13. T2 - 13 The Computer and Linear Programming Consider the case of Linear Programming (more in later lectures):  The first large scale computer became a practical reality in 1946 at the University of Pennsylvania.  This was just one year before the development of simplex.  The simplex method for linear programming consists only of a few steps and these steps require only the most basic mathematical operations which a computer is well suited to handle.  However, these steps must be repeated over and over before one finally obtains an answer.  The first successful computer solution of a LP problem was in January 1952 on the National Bureau of Standards SEAC computer.
  • 14. T2 - 14 By the way... You will often hear the phrase “programming” as in: mathematical programming, linear programming, nonlinear programming, mixed integer programming, etc.
  • 15. T2 - 15 By the way ... (cont.) This has (in principle) nothing to do with modern day computer programming. In the early days, a set of values which represented a solution to a problem was referred to as a “program”. However, nowadays you program (software) to find a program!
  • 16. T2 - 16 Types of Optimization Do not forget: Optimization methods fall in the category of “decision support systems/methods”  Note: There are MANY different optimization methods/algorithms  However, they are can be grouped by fundamental principles of:  model, or  solution method/algorithm
  • 17. T2 - 17 Learning Objectives  Explain linear programming  Formulate linear programming problems  Solve linear programming problems using graphical methods  Corner point  Iso-profit line
  • 18. T2 - 18 What Is Linear Programming?  Mathematical technique  Not computer programming  Allocates scarce resources to achieve an objective  Pioneered by George Dantzig in World War 2  Developed workable solution in 1947  Called Simplex Method
  • 19. T2 - 19 Linear Programming Applications  Find product mix given machine & labor hours to maximize profit  Schedule production given demand to minimize costs  Allocate police given limited patrol cars to minimize response time  Plan menus given minimum daily diet requirements to minimize cost
  • 20. T2 - 20 Linear Programming Requirements  One objective  Maximize or minimize  Constraints (e.g., limited resources)  Alternative courses of action (decision variables)  Divisible (fractional values)  Non-negative  Linear relationships
  • 21. T2 - 21 Solving Linear Programming Problems  Formulate problem  Define decision variables  Write objective function  Write constraint functions  Find solution  Graphical methods  Simplex method  Karmarkar’s algorithm © 1984-1994 T/Maker Co.
  • 23. T2 - 23 Maximization Example You’re an analyst for a division of Levi, which makes baseball & western hats. Each hat must be run on each of 2 machines. Machine 1 has 40 hr. available per week; machine 2, 120 hr. How many of each hat should be made? Rs.4 profit 1 hr. machine 1 4 hr. machine 2 Rs.5 profit 2 hr. machine 1 3 hr. machine 2 © 1984-1994 T/Maker Co. © 1984-1994 T/Maker Co.
  • 24. T2 - 24 Problem Setup Hours/Unit Machine Ball (X1) Western (X2) Capacity (Hrs/Wk) Machine 1 1 hr 2 hr 40 hr Machine 2 4 hr 3 hr 120 hr Unit Profit $4 $5
  • 25. T2 - 25 Formulation Solution  Decision variables  X1 = no. baseball hats produced  X2 = no. western hats produced  Objective function  Maximize profits = Z = 4X1 + 5X2
  • 26. T2 - 26  Decision variables  X1 = no. baseball hats produced  X2 = no. western hats produced  Objective function  Maximize profits = Z = 4X1 + 5X2 Formulation Solution  Constraint functions  Machine 1 capacity: 1X1 + 2X2 40  Machine 2 capacity: 4X1 + 3X2 120  Non-negativity X1 0 X2 0
  • 27. T2 - 27 Formulation Solution Summary Max Z = 4X1 + 5X2 subject to: 1X1 + 2X2 40 4X1 + 3X2 120 X1, X2  0
  • 28. T2 - 28 Graphical Solution Method Steps  Draw graph with vertical & horizontal axes (1st quadrant only)  Plot constraints as lines, then as planes  Use (X1,0), (0,X2) for line  Find feasible region  Find optimal solution  Corner point method  Iso-profit line method
  • 29. T2 - 29 Graphical Solution
  • 30. T2 - 30 Graphical Solution 0 10 20 30 40 0 10 20 30 40 X1 X2 Draw graph with vertical & horizontal axes (1st quadrant only)
  • 31. T2 - 31 Graphical Solution 0 10 20 30 40 0 10 20 30 40 X1 X2 Write 1X1 + 2X2  40 as equality: 1X1 + 2X2 = 40. Pick 2 points (X1,0), (0,X2) to draw line: (40,0), (0,20) (0, 20) (40, 0)
  • 32. T2 - 32 Graphical Solution 0 10 20 30 40 0 10 20 30 40 X1 X2 (0, 20) (40, 0) Write 1X1 + 2X2  40 as equality: 1X1 + 2X2 = 40. Pick 2 points (X1,0), (0,X2) to draw line: (40,0), (0,20)
  • 33. T2 - 33 0 10 20 30 40 0 10 20 30 40 X1 X2 Graphical Solution Inequality is a plane. To find which side of line inequality applies, test any 2 points in 1X1 + 2X2 40
  • 34. T2 - 34 0 10 20 30 40 0 10 20 30 40 X1 X2 Graphical Solution Try (10, 30) in 1X1 + 2X2 40: (1)(10) + (2)(30) = 70. Violates constraint. (10, 30)
  • 35. T2 - 35 0 10 20 30 40 0 10 20 30 40 X1 X2 Graphical Solution (10, 30) (0, 0) Try (0, 0) in 1X1 + 2X2 40: (1)(0) + (2)(0) = 0. Satisfies constraint. X
  • 36. T2 - 36 Graphical Solution Inequality is satisfied by all points BELOW the line. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 37. T2 - 37 Graphical Solution (0, 40) (30, 0) Write 4X1 + 3X2  120 as equality: 4X1 + 3X2 = 120. Pick 2 points (X1,0), (0,X2) to draw line: (30,0), (0,40) 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 38. T2 - 38 Graphical Solution (0, 40) (30, 0) Write 4X1 + 3X2  120 as equality: 4X1 + 3X2 = 120. Pick 2 points (X1,0), (0,X2) to draw line: (30,0), (0,40) 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 39. T2 - 39 Graphical Solution Inequality is a plane. To find which side of line inequality applies, test any 2 points in 4X1 + 3X2 120 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 40. T2 - 40 Graphical Solution (10, 40) Try (10, 40) in 4X1 + 3X2 120: (4)(10) + (3)(40) = 160. Violates constraint. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 41. T2 - 41 Graphical Solution (10, 40) Try (0, 0) in 4X1 + 3X2 120: (4)(0) + (3)(0) = 0. Satisfies constraint. (0, 0) X 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 42. T2 - 42 Graphical Solution Inequality is satisfied by all points BELOW the line. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 43. T2 - 43 Graphical Solution Feasible region is intersection of all planes. It satisfies all constraints. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 44. T2 - 44 Graphical Solution Feasible region is intersection of all planes. It satisfies all constraints. Feasible region 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 45. T2 - 45 Optimal Solution: Corner Point Method A B C D Optimal solution is at corner point of feasible region. Feasible region 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 46. T2 - 46 Optimal Solution: Corner Point Method A (0, 0) (4)(0) + (5)(0) = 0 B (0, 20) (4)(0) + (5)(20) = 100 C (24, 8) (4)(24) + (5)(8) = 136 D (30, 0) (4)(30) + (5)(0) = 120 Point C has highest profit. Produce 24 baseball hats (X1) & 8 western hats (X2). Point Coordinates Profit: Z = 4X1 + 5X2
  • 47. T2 - 47 Finding Point C Coordinates Point C is the intersection of 2 lines. To solve simultaneously, multiply 1st equation by - 4: Add to 2nd equation: Substitute into 2nd equation to get :                  4 1 2 40 4 8 160 4 3 120 4 8 160 8 24 1 2 1 2 1 2 1 2 2 1 1 X X X X X X X X X X X a f
  • 48. T2 - 48 Optimal Solution: Iso-Profit Line Method
  • 49. T2 - 49 Optimal Solution: Iso-Profit Line Method Feasible region 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 50. T2 - 50 Optimal Solution: Iso-Profit Line Method Choose an arbitrary value for Z = 4X1 + 5X2: 20 = 4X1 + 5X2. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 51. T2 - 51 Optimal Solution: Iso-Profit Line Method Plot 20 = 4X1 + 5X2. Pick 2 points (X1,0), (0,X2): (5,0), (0,4). Connect points. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 52. T2 - 52 Optimal Solution: Iso-Profit Line Method Plot 20 = 4X1 + 5X2. Pick 2 points (X1,0), (0,X2): (5,0), (0,4). Connect points. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 53. T2 - 53 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 54. T2 - 54 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 55. T2 - 55 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 56. T2 - 56 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 57. T2 - 57 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 58. T2 - 58 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Higher Profit 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 59. T2 - 59 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 60. T2 - 60 Optimal Solution: Iso-Profit Line Method Draw parallel lines moving outward. Optimal solution is point where line just touches feasible region. Optimum: ‘Just touches’ 0 10 20 30 40 0 10 20 30 40 X1 X2
  • 61. T2 - 61 Linear Programming Thinking Challenge You’re an analyst for RCA, which makes VCR’s & TV’s. Each product must be worked on in each of 2 departments. The electronics dept. has 240 hr. available per week; the assembly dept., 100 hr. How many of each product should be made? Rs.7 profit 4 hr. Electronics 2 hr. Assembly Rs.5 profit 3 hr. Electronics 1 hr. Assembly © 1995 Corel Corp. © 1995 Corel Corp. Alone Group Class
  • 62. T2 - 64 Problem Setup* Hours Required to Produce 1 Unit Dept. VCR X1 TV X2 Available Hrs./Week Electronic 4 3 240 Assembly 2 1 100 Profit/unit $7 $5
  • 63. T2 - 65 Formulation Solution* Max Z = 7X1 + 5X2 subject to: 4X1 + 3X2 240 2X1 + 1X2 100 X1, X2  0
  • 64. T2 - 66 Graphical Solution*
  • 65. T2 - 67 Graphical Solution* 0 20 40 60 80 100 120 0 10 20 30 40 50 60 Number VCR’s No. TV's 70 80
  • 66. T2 - 68 Graphical Solution* 0 20 40 60 80 100 120 0 10 20 30 40 50 60 Number VCR’s No. TV's Electronics Assembly 70 80
  • 67. T2 - 69 Graphical Solution* 0 20 40 60 80 100 120 0 10 20 30 40 50 60 Number VCR’s No. TV's Electronics Assembly 70 80
  • 68. T2 - 70 Graphical Solution* Feasible Region 0 20 40 60 80 100 120 0 10 20 30 40 50 60 Number VCR’s No. TV's Electronics Assembly 70 80
  • 69. T2 - 71 Optimal Solution: Corner Point Method* Feasible Region 0 20 40 60 80 100 120 0 10 20 30 40 50 60 Number VCR’s A B C D No. TV's Electronics Assembly 70 80
  • 70. T2 - 72 Optimal Solution: Corner Point Method* A (0, 80) (7)(0) + (5)(80) = 400 B (30, 40) (7)(30) + (5)(40) = 410 C (50, 0) (7)(50) + (5)(0) = 350 D (0, 0) (7)(0) + (5)(0) = 0 Point B is the optimum solution. Point Coordinates Profit: Z = 4X1 + 5X2
  • 71. T2 - 73 Point B Coordinates* To solve simultaneously, multiple 2nd equation by - 2: Add to 1st Equation: Substitute into 1st equation to get :                  2 2 1 100 4 2 200 4 3 240 4 2 200 40 30 1 2 1 2 1 2 1 2 2 1 1 X X X X X X X X X X X a f
  • 72. T2 - 74 Optimal Solution: Iso-Profit Line Method* 0 20 40 60 80 100 120 0 10 20 30 40 50 60 40 80 Number VCR’s No. TV's Electronics Assembly Higher Profit
  • 74. T2 - 76 Minimization Example You’re an analyst for a division of Kodak, which makes BW & color chemicals. At least 30 tons of BW & at least 20 tons of color must be made each month. The total chemicals made must be at least 60 tons. How many of each chemical should be made to minimize costs? BW: Rs.2,500 mfg. cost per month Color: Rs. 3,000 mfg. cost per month © 1995 Corel Corp.
  • 75. T2 - 77 Formulation Solution  Decision variables  X1 = tons of BW chemical produced  X2 = tons of color chemical produced  Objective  Minimize Z = 2500X1 + 3000X2  Constraints  X1 30 (BW); X2 20 (Color)  X1 + X2 60 (Total tonnage)  X1  0; X2  0 (Non-negativity)
  • 76. T2 - 78 Graphical Solution
  • 77. T2 - 79 Graphical Solution 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical Draw axes. - 40 - 20
  • 78. T2 - 80 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical - 40 - 20 Everything is possible without constraints.
  • 79. T2 - 81 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical - 40 - 20 Find values satisfying X1  0.
  • 80. T2 - 82 Graphical Solution Feasible Region 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical 0 Find values satisfying X1  0.
  • 81. T2 - 83 Graphical Solution Feasible Region 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical 0 Find values satisfying X2  0.
  • 82. T2 - 84 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical Find values satisfying X2  0.
  • 83. T2 - 85 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Draw X1  30 as equality: X1 = 30.
  • 84. T2 - 86 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Find values satisfying X1  30.
  • 85. T2 - 87 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Find values satisfying X1  30.
  • 86. T2 - 88 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Draw X2  20 as equality: X2 = 20.
  • 87. T2 - 89 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Find values for X2  20.
  • 88. T2 - 90 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Find values for X2  20.
  • 89. T2 - 91 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Draw X1 + X2  60 as equality: X1 + X2 = 60.
  • 90. T2 - 92 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Find values for X1 + X2  60.
  • 91. T2 - 93 Graphical Solution Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Find values for X1 + X2  60.
  • 92. T2 - 94 Optimal Solution: Corner Point Method Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Find corner points.
  • 93. T2 - 95 Optimal Solution: Corner Point Method Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Find corner points. A B
  • 94. T2 - 96 Optimal Solution: Corner Point Method A (40, 20) (2500)(40) + (3000)(20) = 160,000 B (30, 30) (2500)(30) + (3000)(30) = 165,000 Point A has lowest cost. Produce 40 tons BW chemicals (X1) & 20 tons color chemicals (X2). Point Coord. Z = 2500X1 + 3000X2
  • 95. T2 - 97 Optimal Solution: Iso-Profit Line Method
  • 96. T2 - 98 Optimal Solution: Iso-Profit Line Method Feasible Region 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total
  • 97. T2 - 99 Optimal Solution: Iso-Profit Line Method 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Lower Cost
  • 98. T2 - 100 Optimal Solution: Iso-Profit Line Method 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Lower Cost
  • 99. T2 - 101 Optimal Solution: Iso-Profit Line Method 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Lower Cost
  • 100. T2 - 102 Optimal Solution: Iso-Profit Line Method 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Lower Cost
  • 101. T2 - 103 Optimal Solution: Iso-Profit Line Method 0 20 40 60 80 0 Tons, Color Chemical 20 40 60 80 Tons, BW Chemical BW Color Total Lower Cost
  • 102. T2 - 104 Conclusion  Explained linear programming  Formulated linear programming problems  Solved linear programming problems using graphical methods  Corner point  Iso-profit line
  • 103. T2 - 105 This Class...  What was the most important thing you learned in class today?  What do you still have questions about?  How can today’s class be improved? Please take a moment to answer the following questions in writing:
  • 104. End of Chapter Any blank slides that follow are blank intentionally.