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Welcome to 620-261 Introduction to Operations Research
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620-261: Introduction to Operations Research
Lecturer: Peter Taylor Heads Office Richard Berry Building Tel: - Course due to: Moshe Sniedovich
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Schedule Lectures: Mon, Wed, Friday 3:15 PM Tutorial:
Check Notice Board and Web site
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Office 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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Assessment Assignments: 10% Final Exam: 90%
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Group Projects You are encouraged to study with friends, but you are expected to compose your own reports.
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Communication You are expected to respond to questions asked (by the lecturer) during the lectures Suggestions, comments, complaints: Directly to lecturer via Student Representative Don’t wait till you are asked to complain!
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Lecture Notes On Sale (Book Room) If out-of-print, let me know
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Thou Shall Not
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Thou Shall Not
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Student Representative (SSLC)
Pizza!!!!! Two meetings Questionnaire
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Web Site
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Reference Material Lecture Notes
Bibliography (10 copies of Winston in Maths library, reserved Shelves) Hand-outs
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Computer 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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Perspective Universe Applied maths OR
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What is OR ? Controversial question! Surf the WWW for answers Roughly:
.... Applications of quantitative scientific methods to decision making and support in business, industrial and military organisations, with the objective of improving the quality of managerial decisions .....
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Basic Characteristics
Applies scientific methods Adopts a systems approach Utilises a team concept Relies on computer technologies
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OR Stream 620-261: Introduction to Operations Research
: Decision Making : Operations Research Methods and Algorithms : Applied Operations Research Probability and Statistics are useful other subjects to study
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and more Honours MSc PhD
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Jobs There is a shortage of people with OR skills
Graduates with these skills get good jobs
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Reading ..... Appendix A Appendix B Appendix E Chapters 1,2,3,4 Web
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The OR Problem Solving Schema
Formulation Monitoring Realization Modelling Implementation Solution Analysis
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In Practice Formulation Monitoring Realization Modelling
Implementation Solution Analysis
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Important Comment In 620-261: Formulation and Modelling
Analysis and Solution
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Chapter 2: Optimization Problems
General formulation f Objective function x Decision variable Decision Space opt Optimality criterion z* Optimal return/cost
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Observe the distinction between f and f(x).
Note that f is assumed to be a real valued function on .
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Example
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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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Remarks The set of feasible solution, , is usually defined by a system of constraints. Thus, an optimization problem has three ingredients: Objective function Constraints Optimality Criterion
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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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Suggestion Try to think about optimization problems in terms of the format: Z*:= opt f(x) s.t. constraints
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Thus Modelling = opt = ? f(x) = ? Constraints = ?
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Tip But do not be dogmatic about it !!!!!
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 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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Observations 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 left-hand side False coin is on the right-hand side False coin is not on the scale
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The scheme should tell us what to do at each “trial”, i. e
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” = “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!!! 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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Assumption 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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Solution 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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Constraints Let 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 = ???? 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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(j-1) weighing: coins left xj xj j-th weighing: sj-1-2xj
sj = max {xj , sj-1-2xj}
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Complete Formulation (Erase N)
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Complete Formulation
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Examples of OR Problems
Example Towers of Hanoi Task: Move the discs from left to right Rules: One disc at a time No 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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The Curse of Dimensionality!
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!
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