1 Welcome to 620-261 Introduction to Operations Research.
Transcript of 1 Welcome to 620-261 Introduction to Operations Research.
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Welcome to620-261
Introduction toOperations Research
Welcome to620-261
Introduction toOperations Research
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620-261: Introduction to Operations
Research
620-261: Introduction to Operations
Research Lecturer: Peter Taylor Heads Office Richard Berry Building Tel: 8344 7887 E-mail:
- [email protected] Course due to: Moshe Sniedovich
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ScheduleSchedule
Lectures:
Mon, Wed, Friday 3:15 PM
Tutorial:
Check Notice Board and Web site
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Office HoursOffice Hours
Monday 2-3 PM Wednesday 2-3PM Friday 2-3PM These may have to vary sometimes – see my
assistant Lisa Mifsud
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AssessmentAssessment
Assignments: 10% Final Exam: 90%
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Group ProjectsGroup Projects
You are encouraged to study with friends, but you are expected to compose your own reports.
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CommunicationCommunication
You are expected to respond to questions asked (by the lecturer) during the lectures
Suggestions, comments, complaints:– Directly to lecturerDirectly to lecturer– via Student Representativevia Student Representative
Don’t wait till you are asked to complain!
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Lecture NotesLecture Notes
On Sale (Book Room) If out-of-print, let me know
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Thou Shall NotThou Shall Not
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Thou Shall NotThou Shall Not
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Student Representative(SSLC)
Student Representative(SSLC)
Pizza!!!!! Two meetings Questionnaire
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Web SiteWeb Site
http://www.ms.unimelb.edu.au/~dmk/620-261
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Reference MaterialReference Material
Lecture Notes Bibliography (10 copies of Winston in Maths
library, reserved Shelves) Hand-outs
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Computer LiteracyComputer Literacy
_ Applied Mathematics is computational._ I don’t expect any specific knowledge, but I
do expect an open attitude to things computational.
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PerspectivePerspective
Universe Applied maths
OR620-261
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What is OR ?What is OR ?
Controversial question! Surf the WWW for answers Roughly:
– .... Applications of quantitative scientific methods .... Applications of quantitative scientific methods to decision making and support in business, to decision making and support in business, industrial and military organisations, with the industrial and military organisations, with the objective of improving the quality of managerial objective of improving the quality of managerial decisions ..... decisions .....
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Basic CharacteristicsBasic Characteristics
Applies scientific methods Adopts a systems approach Utilises a team concept Relies on computer technologies
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OR StreamOR Stream 620-261: Introduction to Operations
Research 620-262: Decision Making 620-361: Operations Research Methods and
Algorithms 620-362: Applied Operations Research Probability and Statistics are useful other
subjects to study
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and more and more
Honours MSc PhD
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JobsJobs
_ There is a shortage of people with OR skills_ Graduates with these skills get good jobs
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Reading .....Reading .....
Appendix A Appendix B Appendix E Chapters 1,2,3,4 Web
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The OR Problem Solving Schema
Solution
Formulation
Realization
Modelling
Analysis
Implementation
Monitoring
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In Practice
Solution
Formulation
Realization
Modelling
Analysis
Implementation
Monitoring
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Important CommentImportant Comment
In 620-261: Formulation and Modelling Analysis and Solution
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Chapter 2:Optimization Problems
Chapter 2:Optimization Problems
General formulation
f Objective function
x Decision variable
Decision Space
opt Optimality criterion
z* Optimal return/cost
z*:optx
f ( x )
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Observe the distinction between f and f(x). Note that f is assumed to be a real valued
function on .
z*:optx
f ( x )
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ExampleExample
z*:max x n2 xn
3 n 1
10
s. t.
x nn1
10
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xn {0,1} , n 1, .. .,10
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f (x): xn2 xn
3 n1
10
: x10 : xnn1
10
5, xn {0,1} , n 1,...,10
opt max
z* unknown
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We let* denote the set of optimal decisions associated with the optimization problem. That is * denotes the subset of whose elements are an optimal solution to the optimization problem. Formally,
*:={x*: x*, f(x*)=opt {f(x): x }}.
By construction * is a subset of , namely optimality entails feasibility.
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RemarksRemarks
The set of feasible solution, , is usually defined by a system of constraints.
Thus, an optimization problem has three ingredients:– Objective functionObjective function– ConstraintsConstraints– Optimality CriterionOptimality Criterion
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Classification of Optimal Solutions
Classification of Optimal Solutions
Consider the case where opt=min. Then by definition:
x* * iff f(x*) f(x) x If opt=max:
x* * iff f(x*) ≥ f(x) x Solutions of this type are called global optimal
solutions.
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f(x)
Global max
Global min
X
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Question:
How do we solve optimization problems of this type?
Answer:
There are no general purpose solution methods. The methods used are very much problem-dependent.
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SuggestionSuggestion
Try to think about optimization problems in terms of the format:
Z*:= opt f(x)
s.t.----------------------------------------------------------
----------------------------- constraints----------------------------- constraints
----------------------------------------------------------
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Thus ......... Modelling = Thus .........
Modelling = opt = ? f(x) = ? Constraints = ?
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TipTip
You may find it useful to adopt the following approach:
Step 1: Identify and formulate the decision variables.
Step 2: Formulate the objective function and optimality criterion.
Step 3: Formulate the constraints.
But do not be dogmatic about it !!!!!
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Example 2.4.2
False Coin ProblemExample 2.4.2
False Coin Problem
N coins N-1 have the same weight (“good”) 1 is heavier (“false”) Find the best weighing scheme using a
balance beam.
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ObservationsObservations It does not make sense to put a different
number of coins on each side of the scale. The result of any non-trivial weighing must
fall into exactly one of the following cases:– False coin is on the False coin is on the left-handleft-hand side side– False coin is on the False coin is on the right-handright-hand side side– False coin is False coin is not on the scalenot on the scale
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The scheme should tell us what to do at each “trial”, i.e. how many coins to place on each side of the scale, depending on how many coins are still to be inspected.
The term “Best” needs some clarification:
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Best = ???Best = ??? “Best” = “fewest number of weighings”
is not well defined because a priori we don’t know how many weighings will be needed by a given scheme.
This is so because we do not know where the false coin will be placed.
The bottom line: who decides where the false coin will be as we implement the weighing scheme ?
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We need help!!!We need help!!!
Many of the difficulties are nicely resolved if we assume that
Mother Nature Always Plays Against Us! Of course, if you are an optimist you may
prefer to assume that Mother Nature Always Plays in Our Favour!
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AssumptionAssumption
Mother Nature Always Plays Against Us !
Observe that this assumption resolves the question of where the false coin will be.
Nature will always select the largest of (nL,nR,no)
nL nR
no
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SolutionSolution
Let
n := Number of weighings required to identify the false coin.
xj := Number of coins placed on each side of the scale in the j-th weighing (j=1,2,3,...,n)
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Thus, our objective function is
f(x1,x2,...,xn):= n
and
opt=min.
To complete the formulation of the problem we have to determine .
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ConstraintsConstraintsLet
sj := Number of coins left for inspection after the j-th weighing (j = 0,1,2,...,n)
Then clearly,
s0 := N (All coins are yet to be inspected)
sn := 1 (Only false coin is left for inspection)
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xj {0,1,2,...,[sj-1/2)]}
where
[z]:= Integer part of z.
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Dynamics = ????Dynamics = ???? We have to specify the dynamics of the
process: how the {sj} are related to the {xj}.
This is not difficult because we assume that Nature Plays Against Us:
sj = max {xj , sj-1-2xj}
xj xj
sj-1-2xj
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xj xj
sj-1-2xj
j-th weighing:
(j-1) weighing:
sj-1
coins left
sj = max {xj , sj-1-2xj}
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Complete FormulationComplete Formulationz* :min n
s.t.
so N
sn 1
s j max{x j ,sj 1 2x j}, j 1,2,..,n
x j sj 1
2x j {0,1,2,...} (Erase N)
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Complete FormulationComplete Formulation
z * :min n
s.t.
so N
sn 1
sj max{x j , s j 1 2x j}, j 1, 2,..,n
x sj 1
2x j{0,1, 2,...}
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Examples of OR ProblemsExamples of OR Problems Example 2.4.6 Towers of Hanoi Task: Move the discs from left to right Rules:
– One disc at a timeOne disc at a time– No large disc on a small oneNo large disc on a small one
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Example 2.4.4 Travelling Salesman Problem
Visit N cities, starting the tour and terminating it in the home city such that:
Each city (except the home city) is visited exactly once
The tour is as short as possible.
Question: What is the optimal tour?
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Remark:
There are (N-1)! distinct tours. This means that for 11 cities there are 3,628,800 possible tours and for N=21 cities there are 2x1018 possible tours !!!!!!!!!!!!!!!
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If we try to enumerate all the feasible tours for N=21 using a super fast computer capable of enumerating 1,000,000,000 tours per second, we will complete the enumeration of all the feasible tours in approximately 800 years.
This phenomenon is known as
The Curse of Dimensionality!