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:

- p.taylor@ms.unimelb.edu.au 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!