Course Notes for CS 1538 Introduction to Simulation By John C. Ramirez Department of Computer Science University of Pittsburgh • These notes are intended for use by students in CS1538 at the University of Pittsburgh and no one else • These notes are provided free of charge and may not be sold in any shape or form • These notes are NOT a substitute for material covered during course lectures. If you miss a lecture, you should definitely obtain both these notes and notes written by a student who attended the lecture. • Material from these notes is obtained from various sources, including, but not limited to, the following: Discrete-Event System Simulation, Fourth Edition by Banks, Carson, Nelson and Nicol (Prentice Hall) • Also (same title and authors) Third Edition Object-Oriented Discrete-Event Simulation with Java by Garrido (Kluwer Academic/Plenum Publishers) Simulation Modeling and Analysis, Third Edition by Law and Kelton (McGraw Hill) A First Course in Monte Carlo by George S. Fishman (Thomson & Brooks/ Cole 2 Goals of Course • To understand the basics of computer simulation, including: Simulation concepts and terminology When it is useful Why it is useful How to approach a simulation How to develop / run a simulation How to interpret / analyze the results 3 Goals of Course • To understand and utilize some of the mathematics required in simulations Statistical models and probability distributions • How various models are defined • Which models are correct for which situations Simple queuing theory • Characteristics • Performance measures • Markovian models 4 Goals of Course Random number theory • Generating and testing pseudo-random numbers • Generating pseudo-random values within various distributions Analysis / generation of input data • How is input data generated? • Is the data correct and appropriate for the simulation? Analysis / measurement of output data • What does the output data mean and what can be derived from it? • How confident are we in our results? 5 Goals of Course • To implement some simulation tools and some simulation projects What enhancements do typical programming languages need to facilitate simulation? Programming will be done in Java • Review if you are rusty • Find / keep a good Java reference • There are special-purpose simulation languages, but we will probably not be using them 6 Introduction to Simulation • What is simulation? Banks, et al: • "A simulation is the imitation of the operation of a real-world process or system over time". It "involves the generation of an artificial history of a system, and the observation of that artificial history to draw inferences … " Law & Kelton: • "In a simulation we use a computer to evaluate a model (of a system) numerically, and data are gathered in order to estimate the desired true characteristics of the model" 7 Introduction to Simulation More specifically (but still superficially) • We develop a model of some real-world system that (we hope) represents the essential characteristics of that system – Does not need to exactly represent the system – just the relevant parts • We use a program (usually) to test / analyze that model – Carefully choosing input and output • We use the results of the program to make some deductions about the real-world system 8 Introduction to Simulation • Why (or when) do we use simulation? • This is fairly intuitive Consider arbitrary large system X • Could be a computer system, a highway, a factory, a space probe, etc. We'd like to evaluate X under different conditions • Option 1: Build system X and generate the conditions, then examine the results – This is not always feasible for many reasons: > X may be difficult to build > X may be expensive to build 9 Introduction to Simulation > We may not want to build X unless it is "worthwhile" > The conditions that we are testing may be difficult or expensive to generate for the real system • For example: – A company needs to increase its production and needs to decide whether it should build a new plant or it should try to increase production in the plants it already has > Which option is more cost-effective for the company? – Clearly, building the new plant would be very expensive and would not be desirable to do unless it is the more cost-effective solution – But how can we know this unless we have built the new plant? 10 Introduction to Simulation • Option 2: Model system X, simulate the conditions and use the simulation results to decide – Continuing with the same example: – Model both possibilities for increasing production and simulate them both > We then choose the solution that is most economically feasible • Clearly, this is itself not a trivial task – Simulations are often large, complex and difficult to develop – Just developing the correct system model can be a daunting task – However, if a new plant costs hundreds of millions or even billions of dollars, spending on the order of thousands (or even hundreds of thousands) of dollars on a simulation could be a bargain 11 Introduction to Simulation When is simulation NOT a good idea? – See Section 1.2 of Banks text • Don't use a simulation when the problem can be solved in a "simpler" or more exact way – Some things that we think may have to be simulated can be solved analytically – Ex: Given N rolls of a fair pair of dice, what are the relative expected frequencies of each of the possible values {2, 3, 4, … 12} ? > We could certainly simulate this, "rolling" the dice N times and counting > However, based on the probability of each possible result, we can derive a more exact answer analytically 12 Introduction to Simulation > How many ways do we have of obtaining each outcome? 2:1, 3:2, 4:3, 5:4, 6:5, 7:6, 8:5, 9:4, 10:3, 11:2, 12:1 Total of 36 possible outcomes For N "rolls", the expected frequency of value i is N * (Pi) = N * (outcomes yielding i / total outcomes) > For example, for 900 rolls, the expected number of 9s generated would be 900 * (4 / 36) = 100 > Note that the expected value may not be a whole number (nor should it necessarily be) > Given 500 rolls, the expected number of 9s is 500 * (4 / 36) 55.55 – Note: You should be familiar with the general approach above from CS 0441 > We will be looking at some more complex analytical models later on 13 Introduction to Simulation • Don't use a simulation if it is easier or cheaper to experiment directly on a real system – Ex: A 24 hour supermarket manager wants to know how to best handle the cash register during the "midnight shift": > Have one cashier at all times > Have two cashiers at all times > Have one cashier at all times, and a second cashier available (but only working as cashier if the line gets too long) – Each of these can be done during operating hours > An extra employee can be used to keep track of queue data (and would not be too expensive) > Differences are (likely) not that drastic so that customers will be alienated 14 Introduction to Simulation • Don't use a simulation if the system is too complex to model correctly / accurately – This is often not obvious – Can depend on cost and alternatives as well – Ex: Simulation of damage to the space shuttle – results were disputed but what was the alternative? 15 Some Definitions • System "A group of objects that are joined together in some regular interaction or interdependence toward the accomplishment of some purpose" (Banks et al) • Note that this is a very general definition • We will represent this system in our simulation using variables (objects) and operations The state of a system is the variables (and their values) at one instance in time 16 Some Definitions • Discrete vs. Continuous Systems Discrete System • State variables change at discrete points in time – Ex: Number of students in CS 1538 > When a registration or add is completed, number of students increases, and when a drop is completed, number of students decreases Continuous System • State variables change continuously over time – Ex: Volume of CO2 in the atmosphere > CO2 is being generated via people (breathing), industries and natural events and is being consumed by plants 17 Some Definitions – Models of continuous systems typically use differential equations to indicate rate of change of state variables – Note that if we make the time increment and the unit of measurement small enough, we may be able to convert a continuous system into a discrete one > However, this may not be feasible to do > Why? – Also note that systems are not necessarily exclusively discrete or exclusively continuous • We will be primarily concerned with Discrete Systems in this course 18 Some Definitions • System Components Entities • Objects of interest within a system – Typically "active" in some way – Ex: Customers, Employees, Devices, Machines, etc • Contain attributes to store information about them – Ex: For Customer: items purchased, total bill • May perform activities while in the system – Ex: For Customer: shopping, paying bill – In normal cases it is really just the period of time required to perform the activity • Note how nicely this meshes with object-oriented programming 19 Some Definitions Events • Instantaneous occurrences that may change the state of a system – Note that the event itself does not take any time – Ex: A customer arrives at a store – Note that they "may" change the state of the system > Example of when they would not? • Endogenous event – Events occurring within the system – Ex: Customer moves from shopping to the check-out • Exogenous event – Events relating / connecting the system to the outside – Ex: Customer enters or leaves the store 20 Some Definitions • System Model A representation of the system to be used / studied in place of the actual system • Allows us to study a system without actually building it (which, as we discussed previously, could be very expensive and time-consuming to do) Physical Model • A physical representation of the system (often scaled down) that is actually constructed – Tests are then run on the model and the results used to make decisions about the system – Ex: Development of the "bouncing bomb" in WWII > http://www.bbc.co.uk/dna/ww2/A2163827 > http://www.computing.dundee.ac.uk/staff/irmurray/bigbounc.asp 21 Some Definitions Mathematical Model • Representing the system using logical and mathematical relationships • Simulations are run using the mathematical model, and, assuming it is valid, the results are interpreted for the system in question • Simple ex: d = vot + ½ at2 – This equation can be used to predict the distance traveled by an object at time t – However, will acceleration always be the same? • More often this model is fairly complex and defined • by the entities and events So this is the model we will be using 22 Some Definitions • Analytical evaluation – If the model is not too complex we can sometimes solve it in a closed form using analytical methods – One type of analytical evaluation is the Markov process (or Markov chain) – Nice simple example at: http://en.wikipedia.org/wiki/Examples_of_Markov_chains – We will see this more in Section 6.4 – Often problems that are too complex, even if they can be modeled analytically, are too computation intensive to be practical • Simulation evaluation – More often we need to simulate the behavior of the model 23 Some Definitions Deterministic Model • Inputs to the simulation are known values – No random variables are used – Ex: Customer arrivals to a store are monitored over a period of days and the arrival times are used as input to the simulation Stochastic Model • One or more random variables are used in the simulation – Results can only be interpreted as estimates (or educated guesses) of the true behavior of the system – Quality of the simulation depends heavily on the correctness of the random data distribution > Different situations may require different distributions 24 Some Definitions – Ex: Customers arrive at a store with exponentially distributed interarrival times having a mean of 5 minutes • In most cases we do not know all of the input data in advance, and at least some random data is required – Thus, our simulations will typically use the stochastic model 25 Some Definitions Static Model • Models a system at a single point in time, rather than over a period of time • Sometimes called Monte Carlo simulations – We'll briefly discuss these shortly Dynamic Model • Models a system over time • Our simulations will typically use this model • In summary our models will typically be: discrete, mathematical, stochastic and dynamic 26 The Clock • Since we are using the dynamic model, we need to represent the passage of time We need to use a clock Three fundamental approaches to time progression • Next-event time advance – Clock initialized to zero – As the times of future events are determined, they are put into the future event list (FEL) – Clock is advanced to the time of the next most imminent event, the event is executed and removed from the list – See example in Section 3.1.1 27 The Clock Ex: People (P) using a MAC machine • Event A == arrival of a customer at MAC machine • Event C == completion of a transaction by a customer Clock FEL Event Action 0 (A2,t1), (C1,t2) A1 P1 arrives, is served; Events A2 and C1 generated, placed in FEL t1 (C1,t2), (A3,t3) A2 P2 arrives, waits; Event A3 generated, placed in FEL t2 (A3,t3), (C2,t4) C1 P1 completes; P2 is served; Event C2 generated, placed in FEL t3 (A4,t5), (C2,t4) A3 P3 arrives, waits; Event A4 generated, placed in FEL (note: t5<t4) t5 (C2,t4), (A5,t6) A4 P4 arrives, waits; Event A5 generated, placed in FEL t4 (A5,t6), (C3,t7) C2 P2 completes; P3 is served; Event C3 generated, place in FEL 28 The Clock • Fixed-increment time advance (activity scanning) – Clock initialized to zero – Clock is incremented by a fixed amount (ex. 1) – With each increment, list of events is checked to see which should occur (could be none) – Clock is typically easier to implement in this way – However, execution is less efficient > Potentially many scans for each event • Process-interaction approach – Entities are associated with processes – Processes interact as entities progress through system – Could delay while waiting for a resource, or during an interaction with another process > Can be implemented with multithreading or multiprocessing 29 Simple Example • Let's consider a very simple example: Single-Channel Queue (Example 2.1 in text) • Small grocery store with a single checkout counter • Customers arrive at the checkout at random between 1 and 8 minute apart (uniform) • Service times at the counter vary from 1 to 6 minutes – P(1) = 0.1, P(2) = 0.2, P(3) = 0.3, P(4) = 0.25 P(5) = 0.1, P(6) = 0.05 • Start with first customer arriving at time 0 • Run for a given number of customers (text uses 100) • Calculate some results that may be useful 30 Simple Example The entities are the customers The system is discrete since states are changed at specific points in time • ex: a customer arrives or leaves The model is mathematical (since we don't have real customers) The model is stochastic since we are generating random arrivals and random service times The model is dynamic since we are progressing in time 31 Simple Example What results are we interested in? • In this simple case we may want to know – What fraction of customers have to wait in line – What is the average amount of time that they wait – What is the fraction of time the cashier is idle (or busy) • We probably want to do several runs and get • cumulative results over the runs (ex: averages) There are more complex statistics that may be relevant – We will discuss some of these later 32 Simple Example We can program this example, but in this simple case we could also use a table or spreadsheet to obtain our results • Let's first look at an "Excel novice" approach to this – See sim1.xls • Although some of the spreadsheet formulas require • some thought, this is fairly simple to do Note that each row in the spreadsheet depends only on some local data (generated in that row) and the data in the previous row – We do not need a "memory" of all rows • Authors have a much nicer spreadsheet with macros – See http://www.bcnn.net 33 Programming a Simple Example • If we do program it, how would we do it? Using Java, it is logical to do it in an object- oriented way Let's think about what is involved • We need to represent our entities – As text indicates, for this simple example we do not have to explicitly represent them – However, we can do it if we want to – and have our Customers and CheckOut as simple Java objects • We need to represent our events – We need to store events in our Future Event List (FEL) and we have two different kinds of events (arrival of a customer, finish of a checkout) 34 Programming a Simple Example > We need to distinguish between the different event types (since different actions are taken for different events) > We need to order our events based on the simulation clock time that they will occur – Thus we probably need to explicitly represent the events in some way > Use classes and inheritance to represent the different events > This enables events to share characteristics but also to be distinguished from each other > So we need a event time instance variable and a method to compare event times > Look at SimEvent.java, ArrivalEvent.java, CompletionEvent.java 35 Priority Queue to Represent the FEL • We need to represent the FEL itself – Since we are inserting items and then removing them based on priority (earliest next time of an event is removed first), we should use a priority queue (PQ) with the following operations: > add (Object e) – add a new Object to the PQ > remove() – remove and return the Object with the min (best) priority value > peek() – return the Object with the min (best) priority value without removing it – It's also a good idea to have some helper methods > size() – how many items are in the PQ > isEmpty() – is the PQ empty – There are variations of these ops depending on the implementation, but the idea is the same 36 Priority Queue to Represent the FEL – How to efficiently implement a Priority Queue? > > > > How about an unsorted array or linked list? add is easy but remove is hard – why? – discuss How about a sorted array or linked list? removeMin is easy but add is hard – why? – discuss – Neither implementation is adequate in terms of efficiency > Note that the premise of a PQ is that everything that is inserted is eventually removed > Thus, with N adds you have N removes > Discuss / show on board overall time required for both implementations > You may have seen this already in CS 1501 – Thus we need a better approach > Implementation of choice is the Heap 37 Heap Implementation of a Priority Queue – Idea of a Heap: > Store data in a partially ordered complete binary tree such that the following rule holds for EACH node, V: Priority(V) betterthan Priority(LChild(V)) Priority(V) betterthan Priority(RChild(V)) > This is called the HEAP PROPERTY > Note that betterthan here often means smaller > Note also that there is no ordering of siblings – this is why the overall ordering is only a partial ordering – ex: 10 30 20 40 90 70 45 85 80 38 35 Heap Implementation of a Priority Queue – How to do our operations? > peek() is easy – return the root > add() and remove() are not so obvious > Let's look at them separately – add(Object e) > We want to maintain the heap property > However, we don't know where in advance the new object will end up > We also don't want a lot of rearranging or searching if we can avoid it – remember time is key > Solution: Add new object at the next open leaf in the last level of the tree, then push the node UP the tree until it is in the proper location > This operation is called upHeap > See example on board 39 Heap Implementation of a Priority Queue – remove() > Clearly, the min node is the root > However, removing it will disrupt the tree greatly > How can we solve this problem? • Remember BST delete? – Did not actually delete the root, but rather the _______________ (fill in blank) • We will do a similar thing with our Heap – Copy the last leaf to the root and delete (easily) the leaf node – Then re-establish the heap property by a downHeap – See example on board 40 Heap Implementation of a Priority Queue – Run-Time? > Since our tree is complete, it is balanced and thus for N nodes has a height of ~ lgN > Thus upHeapand downHeap require no more than ~lgN time to complete > Thus, if we have N adds and N removeMins, our total run-time will be NlgN > This is a SIGNIFICANT improvement of the simpler implementations, especially for a long simulation > Ex: Compare N2 with NlgN for N = 1M (= 220) – Note: > For our simple example, a heap is probably not necessary, since we have few items in our FEL at any given time > However, for more complex simulations, with many different event types, a heap is definitely preferable 41 Implementing a Heap – How to Implement a Heap? > We could use a linked binary tree, similar to that used for BST Will work, but we have overhead associated with dynamic memory allocation and access > But note that we are maintaining a complete binary tree for our heap > It turns out that we can easily represent a complete binary tree using an array We simply must map the tree locations onto the array indexes in a reasonable / consistent way – Idea: > Number nodes row-wise starting at 0 (some implementations start at 1) > Use these numbers as index values in the array 42 Implementing a Heap > Now, for node at index i Parent(i) = floor((i-1)/2) LChild(i) = 2i+1 RChild(i) = 2i+2 > See example on board – Now we have the benefit of a tree structure with the speed of an array implementation • So now should we write the code? – No! Luckily, in JDK 1.5 a heap-based PriorityQueue class has been provided! – It's still a good idea to understand the implementation, however – Look at API 43 Queue for Waiting Customers • We need to represent the queue (or line) of customers waiting at the checkout – This is a FIFO queue and can simply be implemented in various ways > We can use a circular array > We can use a linked-list – You should be already familiar with queue implementations from CS 0445 – In JDK 1.5 Queue is an interface which is implemented by the LinkedList class > See API > Q: Would a similar approach using an ArrayList also be good? 44 Programming a Simple Example • We need to represent the clock – This is fairly easy – we can do it with an integer > In some cases it might be better to use a double • We need to implement some activities – These are actually better defined as the time required for activities to execute – Typically interarrival times or service times, either specified exactly (with deterministic model) or by probability distributions (with stochastic model) > In our case, we have the interarrival times of customers and the time required for checkout, specified by the distributions shown on p. 28 of the text > We will discuss various distributions in more detail later 45 Programming a Simple Example Let's put this all together: GrocerySim.java • This is a fairly object-oriented implementation, using newer JDK 1.5 features Note that there is also a Java version from authors in Chapter 4 • Look over this one as well • Does not utilize JDK 1.5 and not quite as object- • oriented The author also switches distributions in this implementation – Uses an exponential distribution for arrivals – Uses a normal distribution for service times > We will look at these later 46 One More Example • Newspaper Seller's Problem Example 2.3 in text Simple inventory problem • Each day new inventory is produced and used, but is not carried over to successive days • Thus, time is more or less removed from this problem Used where goods are only useful for a short time • Ex: newspaper, fresh food In this case, our goal is to try to optimize our profit 47 Newspaper Seller's Problem Specifics of the Newspaper Seller's Problem • Seller buys N newspapers per day for 0.33 each • Seller sells newspapers for 0.50 each • Unused papers are "scrapped' for 0.05 each • If seller runs out, lost revenue is 0.17 for each not sold paper – Text says this is controversial, which is true – How to predict how many would have been sold? > Perhaps seller goes home when he/she runs out > May be a goal to run out every day – easier than returning the papers for scrap • See sim2.xls 48 Newspaper Seller's Problem In fact we do we really need to simulate this problem at all? • The data is simple and highly mathematical • Time is not involved Let's try to come up with an analytical solution to this problem • We have two distributions, the second of which • utilizes the result of the first Let's calculate the expected values for random variables using these distributions – For a given discrete random variable X, the expected value, E(X) = Sum [xi p(xi)] all i 49 (more soon in Chapter 5) Newspaper Seller's Problem • Let our random variable, X, be the number of newspapers sold – Let's first consider the expected value for each of the demands of good, fair and poor Demand Probability Distribution Demand Good Fair Poor 40 0.03 0.10 0.44 50 0.05 0.18 0.22 60 0.15 0.40 0.16 70 0.20 0.20 0.12 80 0.35 0.08 0.06 90 0.15 0.04 0.00 100 0.07 0.00 0.00 50 Newspaper Seller's Problem • Egood(X) = (40)(0.03) + (50)(0.05) + (60)(0.15) + • • (70)(0.20) + (80)(0.35) + (90)(0.15) + (100)(0.07) = 75.2 Efair(X) = (40)(0.10) + (50)(0.18) + (60)(0.40) + (70)(0.20) + (80)(0.08) + (90)(0.04) + (100)(0.00) = 61 Epoor(X) = (40)(0.44) + (50)(0.22) + (60)(0.16) + (70)(0.12) + (80)(0.06) + (90)(0.00) + (100)(0.00) = 51.4 Now we need to use the second distribution (of good, fair and poor days) to determine the overall expected value 51 Newspaper Seller's Problem • E(X) = (Egood(X))(0.35) + (Efair(X))(0.45) + (Epoor(X))(0.20) = 64.05 Now we utilize the expected number of newspapers sold to find results for each of the potential number that we stock • Let sales = expected value calculated above • Let stock = number vendor purchases • Let left = stock – sales (only if stock > sales, else 0) • Let lost = sales – stock (only if sales > stock, else 0) • Profit = (Min(sales,stock))(0.5) – (stock)(0.33) + (left)(0.05) – (lost)(0.17) 52 Newspaper Seller's Problem Stock Profit 40 2.71 50 6.11 60 9.51 70 9.2 80 6.4 90 3.6 100 0.82 53 Expected profit values for given stock anoumts Note that this table shows that 60 is the best choice (more or less agreeing with the simulation results) Newspaper Seller's Problem Is this analytical solution correct? • Not entirely • We are using an expected value to derive another expected value – oversimplifying the actual analysis • The variance from the expected value will cause our actual results to differ • Note that the simulation results are almost identical to the analytical for small and large inventories • In the middle there is more variation and this is where using the expected value is inadequate • However, as a basis for choosing the best number of papers to stock, it still works 54 Other Simulation Examples There are other examples in Chapters 2 and 3 • Read over them carefully • We may look at some of these types of simulations later on in the term 55 Simulation Software • Simulations can be written in any good • programming language However, many things that need to be done in simulations can be built into languages to make them easier Random values from various probability distributions Tools for modeling Tools for generating and analyzing output Graphical tools for displaying results 56 Simulation Software Look at the various described languages Our simple queueing example (Example 2.1) is shown using many of the languages • Even if you don't completely understand all of the code, look it over to note some differences We may look at one of these packages later in the term if we have time 57 Probability and Statistics in Simulation • Why do we need probability and statistics in simulation? Needed to validate the simulation model Needed to determine / choose the input probability distributions • Needed to generate random samples / values from these distributions Needed to analyze the output data / results Needed to design correct / efficient simulation experiments 58 Experiments and Sample Space • Experiment A process which could result in several different outcomes • Sample Space The set of possible outcomes of a given experiment • Example: Experiment: Rolling a single die Sample Space: {1, 2, 3, 4, 5, 6} • Another example? 59 Random Variables • Random Variable A function that assigns a real number to each point in a sample space Example 5.2: • Let X be the value that results when a single die is • rolled Possible values of X are 1, 2, 3, 4, 5, 6 • Discrete Random Variable A random variable for which the number of possible values is finite or countably infinite • Example 5.2 above is discrete – 6 possible values 60 Random Variables and Probability Distribution • Countably infinite means the values can be mapped to the set of integers – Ex: Flip a coin an arbitrary number of times. Let X be the number of times the coin comes up heads • Probability Distribution For each possible value, xi, for discrete random variable X, there is a probability of occurrence, P(X = xi) = p(xi) p(xi) is the probability mass function (pmf) of X, and obeys the following rules: 1) p(xi) >= 0 for all i 2) p ( x i ) = 1 all i 61 Random Variables and Probability Distribution The set of pairs (xi, p(xi)) is the probability distribution of X Examples: • For Example 5.2 (assuming a fair die): – Probability Distribution: > {(1, 1/6), (2, 1/6), (3, 1/6), (4, 1/6), (5, 1/5), (6, 1/6)} • From Example 2.1 for Service Times – Probability Distribution: > {(1, 0.1), (2, 0.2), (3, 0.3), (4, 0.25), (5, 0.1), (6, 0.05)} • From Example 2.3 for Type of Newsday – Probability Distribution: > {(0, 0.35), (1, 0.45), (2, 0.20)} > Note in this case we are assigning the values 0, 1, 2 to the outcomes somewhat arbitrarily 62 Cumulative Distribution • Cumulative Distribution Function The pmf gives probabilities for individual values xi of random variable X The cumulative distribution function (cdf), F(x), gives the probability that the value of random variable X is <= x, or F(x) = P(X <= x) For a discrete random variable, this can be calculated simply by addition: F(x) = p ( xi ) xi x 63 Cumulative Distribution Properties of cdf, F: 1) F is non-decreasing 2) lim F ( x ) 1 3) lim F ( x ) 0 and x x P(a < X b) = F(b) – F(a) for all a < b Ex: Probability that a roll of two dice will result in a value > 7? • Discuss Ex: Probability that 10 flips of a fair coin will yield between 6 and 8 (inclusive) heads? • Discuss 64 Expected Value • Expected Value (for discrete random variables) E(X ) x i p ( xi ) all i Also called the mean Ex: Expected value for roll of 2 fair dice? E(X) = (2)(1/36) + (3)(2/36) + (4)(3/36) + (5)(4/36) + (6)(5/36) + (7)(6/36) + (8)(5/36) + (9)(4/36) + (10)(3/36) + (11)(2/36) + (12)(1/36) =7 • Note that in this case the expected value is an actual value, but not necessarily 65 Expected Value and Variance If each value has the same "probability", we often add the values together and divide by the number of values to get the mean (average) • Ex: Average score on an exam • Variance V ( X ) E [( X E [ X ]) ] 2 E ( X ) [ E ( X )] 2 2 ( Equation 5 . 10 ) We won't prove the identity, but it is useful 66 Expected Value and Variance • In the original definition, we need to subtract the mean from each of the X values before squaring – So we need each X value to calculate the mean AND AFTER the mean has been calculated > Must look at them twice • In the right side of the equation (Equation 5.10 in the text), we need to calculate the mean of X and the mean of the squares of X – We can do this as we process the individual X values and need to look at them only one time • Ex: What is the variance of the following group of exam scores: { 75, 90, 40, 95, 80 } – Since each value occurs once, we can consider this to have a uniform distribution 67 Expected Value and Variance • V(X) using original definition: E(X) = (75+90+40+95+80)/5 = 76 V(X) = E[(X – E[X])2] = [(75-76)2 + (90-76)2 + (40-76)2 + (95-76)2 + (80-76)2]/5 = (1 + 196 + 1296 + 361 + 16)/5 = 374 • V(X) using Equation 5.10 E(X) = (75+90+40+95+80)/5 = 76 E(X2) = (5625+8100+1600+9025+6400)/5 = 6150 V(X) = 6150 – (76)2 = 374 – Note that in this case we can add each number to one sum and its square to another, so we can calculate our overall answer with one a single "look" at each number 68 Discrete Distributions • Discrete Distributions of interest: Bernoulli Trials and the Bernoulli Distribution • Consider an experiment with the following properties – n independent trials are performed – each trial has two possible results – success or failure – the probability of success, p and failure, q (= 1 – p) is constant from trial to trial – for random variable X, X = 1 for a success and X = 0 for a failure • Probability Distribution: P(X = 1) = p P(X = 0) = 1 – p = q or 0 for all other values of X 69 Bernoulli Distribution • Expected Value – E(X) = (0)(q) + (1)(p) = p • Variance – V(X) = [02q + 12p] – p2 = p(1 – p) A single Bernoulli trial is not that interesting • Typically, multiple trials are performed, from which we can derive other distributions: – Binomial Distribution – Geometric Distribution 70 Binomial Distribution Binomial Distribution • Given n Bernoulli trials, let random variable X denote the number of successes in those trials • Note that the order of the successes is not important, just the number of successes – Thus, we can achieve the same number of successes in various different ways n x n x p q , p ( x ) x 0, x 0 ,1, , n otherwise – Since the trials are independent, we can multiply the probabilities for each trial to get the overall probability for the sequence 71 Binomial Distribution • Recall that the number of combinations of n items taken x at a time is n n! x x! ( n x )! • E(X) = np – Discuss • V(X) = npq • Consider an example: – Exercise 5.1 – Read – Do solution on board 72 Binomial Distribution • Consider again coin-flip ex. on slide 63 • Generally speaking binomial distributions can be used to determine the probability of a given number of defective items in a batch, or the probability of a given number of people having a certain characteristic – Ex: The trait of having a klinkled flooje occurs on average in 10% of Kreptoplomians (krep-tō-plō'-mēəns). Given a group ot 20 Kreptoplomians, what is the probability that 3 of them have klinkled floojes? P(3) = (20 C 3)(0.1)3(0.9)17 = (1140)(0.001)(0.1668) = 0.1902 73 Geometric Distribution Geometric Distribution • Given a sequence of Bernoulli trials, let X represent the number of trials required until the first success q x 1 p , x 1, 2 , p(x) otherwise 0 , – i.e. we have x – 1 failures, followed by a success – Note that the maximum probability for this is at X = 1, regardless of p and q • E(X) = 1/p • V(X) = q/p2 – We will omit the proofs of the above, since they are fairly complex (involving series solutions) 74 Geometric Distribution • Ex: What is the probability that the first Kreptoplomian found to have a klinkled flooje will be the 5th Kreptoplomian overall? (0.9)4(0.1) = 0.0656 • Ex: The probability that a certain computer will fail during any 1-hour period is 0.001 – What is the probability that the computer will survive at least 3 hours? > Here p = 0.001 and q = (1 – p) = 0.999 – Using a geometric distribution, we want to solve P(X >= 4) = 1 – P(1) – P(2) – P(3) = 1 – (0.001) – (0.999)(0.001) – (0.999)2(0.001) = 0.997 75 Geometric Distribution • The Geometric Distribution is memoryless – Consider the following two scenarios where p = probability that a component will fail in the next hour. Assume the current hour is hour 0. 1) What is the probability that the component will fail by the end of hour 3? 2) What is the probability that the component will fail by the end of hour 6, given that it has not failed by the end of hour 3 ? > For 1) the solution is P(1) + P(2) + P(3) > For 2), since the component did NOT fail by the end of hour 3, and since the probability is for the next hour (whatever that hour may be), the solution is the same – We can prove this property with fairly simple algebra > First we need one additional definition 76 Geometric Distribution • The conditional probability of an event, A, given that another event, B, has occurred is defined to be: P(A | B) P(A B) P(B) • Applying this to the geometric distribution we get P ( X s t | X s) P ([ X s t ] [ X s ]) P( X s) – Clearly, if X > s+t, then X > t, so we get P ( X s t | X s) 77 P( X s t) P ( X s) Geometric Distribution – Consider that P(X > s) = q q q j 1 p j s 1 s q j 1 q (1 q ) j s 1 s 1 q j 1 q j j s 1 s 1 q s2 q s2 q s3 q – We can use similar logic to determine that P(X > s + t) = qs+t – Now our conditional probability becomes P ( X s t | X s) q st q q P( X t) t s – and thus we have shown that the geometric distribution is memoryless > We will see shortly that the exponential distribution is also memoryless 78 s Poisson Distribution Poisson Distribution • Often used to model arrival processes with constant arrival rates • Gives (probability of) the number of events that occur in a given period • Formula looks quite complicated (and NOT discrete), but it is discrete and using it is not that difficult e x , x 0 ,1, p ( x ) x! 0 , otherwise • Where is the mean arrival rate. Note that must • be positive E(X) = V(X) = 79 Poisson Distribution x F ( x) i0 e i i! • Note: The Poisson Distribution is actually the convergence of the Binomial Distribution as the number of trials, n, approaches infinity – If we think of n as the number of subintervals of a given unit of time – As n , the subintervals get smaller and smaller – We will skip the detailed math here – One nice feature of this is that we can use a Poisson Distribution to approximate a Binomial Distribution when n is large 80 Poisson Distribution • Example – Number of people logging onto a computer per minute is Poisson Distributed with mean 0.7 – What is the probability that 5 or more people will log onto the computer in the next 10 minutes? – Solution? >Must first convert the mean to the 10 minute period – if mean is 0.7 in 1 minute, it will be (0.7)(10) = 7 in a ten minute period >Now we can plug in the formula >P(X >= 5) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 1 – F(4) (where F is the cdf) = 1 – 0.173 (from Table A.4 in the text) = 0.827 81 Continuous Random Variables • Continuous Random Variable Random variable X is continuous if its sample space is a range or collection of ranges of real values More formally, there exists non-negative function f(x), called the probability density function, such that for any set of real numbers, S a) f(x) >= 0 for all x in the range space b) f ( x ) dx 1 – the total area under f(x) is 1 range space c) f(x) = 0 for all x not in the range space – Note that f(x) does NOT give the probability that X = x – Unlike the pmf for discrete random variables 82 Continuous Random Variables • The probability that X lies in a given interval [a,b] is P (a X b) b f ( x ) dx a – We see this visually as the "area under the curve" – Note that for continuous random variables, P(X = x) = 0 for any x (see from formula above) – Rather we always look at the probability of x within a given range (although the range could be very small) • The cumulative density function (cdf), F(x) is simply the integral from - to x or F ( x) x f ( t ) dt – This gives us the probability up to x 83 Continuous Random Variables • Ex: Consider the uniform distribution on the range [a,b] (see text p. 166) 1 if a x b f ( x) b a 0 otherwise F ( x) x f ( y ) dy x 1 dy xa if a x b ba ba – Look at plots on board for example range [0,1] a a > What about F(x) when x < a or x > b? Expected Value for a continuous random variable E(X ) xf ( x ) dx – Compare to the discrete expected value 84 Continuous Random Variables Variance for continuous random variables • Defined in same way as for discrete variables – Calculating it will clearly be different, however Ex: Uniform Distribution b x a ba E(X ) dx ( b a )( b a ) 2 (b a ) b x b a x 2 ba b a 2 a 2 (b a ) 2 2 (b a ) ba 2 V ( X ) E ( X ) E ( X ) 2 2 2 ( Eq 5 . 10 ) dx E ( X ) 2 85 b x 3 a 3(b a ) [ E ( X )] 2 Continuous Random Variables b a V (X ) [ E ( X )] a 3(b a ) 3(b a ) 2 (b a ) b x b a 3 3 2 2 3 2 2 2 ( b a )( b a ) b a (b a ) (b a ) 2 3(b a ) 2 (b a ) 3(b a ) 4 (b a ) b a 3 3 4 (b a ) 3 3 12 ( b a ) 2 12 ( b a ) 2 2 12 ( b a ) 4 b 4 a 3 b 6 ab 3 a b 3 ab 6 a b 3 a 3 3 2 2 2 2 12 ( b a ) b a 3 ab 3 a b 3 2 3(b a ) (b a ) 3 3 3 4 b 4 a 3[ b 2 ab a ]( b a ) 3 3 3 2 12 ( b a ) 2 86 (b a ) 3 12 ( b a ) (b a ) 12 2 3 2 Uniform Distribution • Generally speaking we will not be calculating these values from scratch (good news, in all likelihood!) – However, it is good to see how it can be done for at least one (simple) distribution The Uniform Distribution will be useful primarily in the generation of other distributions • Ex: Most random number generators on computers minimally will provide a uniform value from [0,1) – We will later see how that can be used to generate other random variates (See Chapter 8) 87 Exponential Distribution Given a value > 0, the pdf of the Exponential Distribution is • • e x , x 0 f (x) otherwise 0 , Since the exponent is negative, the pdf will decrease as x increases – see shape on board and p. 168 is the rate: number of occurrences per time unit – Ex: arrivals per hour; failures per day • Note that 1/ can thus be considered to be the time between events or the duration of events – Ex: 10 arrivals per hour 1/10 hr (6min) average between arrivals – Ex: 20 customers served per hour 1/20 hr (3min) average service time 88 Exponential Distribution • Some more definitions E(X ) 1 V (X ) 1 2 0 , x 0 F ( x ) x t x e dt 1 e , 0 x0 – See shape of cdf on board and p. 168 • Ex: Assume the hard drives manufactured by Herb's Hard Drives have a mean lifetime of 4 years, exponentially distributed a) What is the probability that one of Herb's Hard Drives will fail within the first two years? b) What is the probability that one of Herb's Hard Drives will last at least 8 years (or twice the mean)? 89 Exponential Distribution • Like the geometric distribution, the exponential distribution is memoryless – Implies that P(X > s+t | X > s) = P(X > t) – The proof of this is similar in nature to that for the geometric distribution > See pp. 169 in text • Ex: Exercise 5.19 in the text – Component has exponential time-to-failure with mean 10,000 hrs a) Component has already been in operation for its mean life. What is the prob. that it will fail by 15,000 hrs? b) Component is still ok at 15,000 hrs. What is the prob. that it will operate for another 5,000 hrs The first thing we need to do here is be sure we understand the problem correctly 90 Exponential Distribution – First, let's determine our distribution > X is the time to failure, and is exponentially distributed with mean 10,000 hours > This gives us a cumulative distribution x F ( x ) 1 e 10000 , x0 > Here the failure rate = = 1/10000 – a) is pretty clear – we want the probability that it lasts at most 15,000 given that it has lasted 10,000 > We want P(X <= 15000 | X > 10000) > Due to the memoryless property of the exponential distribution, this reduces to P(X <= 5000) which is 1 – e-1/2 = 0.3935 91 Exponential Distribution – b) is trickier (and actually not phrased well) > Do they mean the prob. that it will last exactly 5000 more hours? If so, the probability is 0, since continuous distributions have 0 prob. for a specific value > Do they mean it will last at most 5000 more hours? If so, the answer is the same as a) due to the memoryless property > Do they mean it will last at least 5000 more hours? If so, we want 1 – F(5000) = 0.6065 92 Gamma Distribution Gamma Distribution • More general than exponential • Based on the gamma function, which is a continuous generalization of the the factorial function ( ) ( 1) ( 1) ( 1)! when base case (1) = 1 – is called the shape parameter – The gamma function also has , the scale parameter – This leads the the following pdf: 1 x ( x ) e , f (x) ( ) 0 , otherwise 93 x0 Gamma Distribution E ( x) 1 V (X ) 1 F (x) 0 , x ( ) 1 ( t ) 2 1 e t dt , x0 x0 – These formulas look complicated (and they are) – However, they allow for more flexibility in the distributions > Allow more different curves (See Fig. 5.10a) – Note that if =1, the equations simplify to the exponential distribution with rate > Look at the formulas to see this 94 Erlang Distribution When is an arbitrary positive integer, k, the Gamma Distribution is also called the Erlang Distribution of order k • In general terms, it represents the sum of k independent, exponentially distributed variables, each with rate k (= ) X = X1 + X2 + … + Xk leading to E(X ) V (X ) 1 k 1 k 1 (k ) 1 F (x) 0 , 2 k 1 e 1 (k ) k x i0 x0 2 1 k (k x ) i! 95 1 1 (k ) 2 i , x0 1 k 2 Erlang Distribution • This allows us to determine probabilities for sequences of exponentially distributed events – Note that the rates for all events in the sequence must be the same • Ex: Exercise 5.21 in text – Time to serve a customer at a bank is exponentially distributed with mean 50 sec a)Probability that two customers in a row will each require less than 1 minute for their transaction b)Probability that the two customers together will require less than 2 minutes for their transactions – It is important to recognize the difference between these two problems 96 Erlang Distribution – For a) we are looking at the probability that each of two independent events will be < 1 minute > In this case, the probability overall is the product of the two probabilities, each of which is exponential = rate = 1/mean = 1/50 We want P(X < 60) = F(60) = 1 – e-(1/50)(60) = 0.6988 > Thus the total probability is (0.6988)2 = 0.4883 – For b) we are looking at the probability that two events together will be < 2 minutes > In this case the probability overall is an Erlang distribution with k = 2 and k = 1/50, and we want to determine P(X < 120) = F(120) > Substituting into our equation for F, we get 97 Erlang Distribution 2 1 F (120 ) 1 e (120 / 50 ) (120 / 50 ) i i! i0 1 ( 0 . 0907 ) ( 0 . 2177 ) 0 . 6916 – Note that these results are fairly intuitive > Requiring both to be < 1 minute is more restrictive a condition than requiring the sum to be < 2 minutes, and would seem to have a lower probability – How about if we add another part: Probability that the next 3 customers will have a cumulative time of more than 2.5 minutes? > Now we want P(X > 150) = 1 – F(150) > But F has changed since we now have 3 events > Let's do this one on the board 98 Normal Distribution A very common distribution is the Normal Distribution • It has some nice properties lim x f ( x ) 0 lim x f ( x) f ( x) f ( x) max( f ( x )) f ( ) – Discuss these • The pdf for the normal distribution is also quite complex – We won't even show it here – However, we can use tables and another nice property to allow solution for arbitrary normal distributions 99 Normal Distribution • Define (z) to be the normal distribution with mean () 0 and variance (2) 1 = N(0, 1) – We call this the standard normal distribution – This can be calculated using numerical methods, and its cdf is typically provided in tables in statistics (and simulation) textbooks (see Table A.3 in text) – We use the notation Z ~ N(0, 1) to mean that Z is a random variable with a standard normal distribution • Naturally, most normal distributions of interest will not be the standard normal distribution – However, Eqs 5.42 & 5.43 in the text relate any normal distribution to the standard normal distribution in the following way 100 Normal Distribution – Given an arbitrary normal distribution, X ~ N(, 2), let Z = (X – )/ – Through Eq. 5.43, we know that x F ( x) > where (z) is the cumulative density function for the standard normal distribution – Thus we can use the tabulated values of the standard normal distribution to determine probabilities for arbitrary normal distributions – Ex:Student GPA's are approximately normally distributed with =2.4 and =0.8. What fraction of students will possess a GPA in excess of 3.0? Example is from From Mathematical Statistics with Applications, Second Edition by Mendenhall, Scheaffer and Wackerly 101 Normal Distribution – Let Z = (X – 2.4)/0.8 – We want the area under the normal curve with mean 2.4 and standard deviation 0.8 where x > 3.0 This will be 1 – F(3.0) F(3.0) = [(3.0 – 2.4)/0.8] = (0.75) – Looking up (0.75) in Table A.3 we find 0.77337 – Recall that we want 1 – F(3.0), which gives us our final answer of 1 – 0.77337 = 0.2266 • The idea in general is that we are moving from the mean in units of standard deviations – The relationship of the mean to standard deviation is the same for all normal distributions, which is why we can use the method indicated 102 Other Distributions There are a LOT of probability distributions • More in the text that we did not discuss • Many others not in the text For simulation, the idea for using them is: • How well does the distribution of choice model the actual distribution of events / times that are relevant to our model • The more possibilities and variations, the more closely we can model our actual behavior • However, we need to be able to determine if a distribution fits observed data – We will look at this in Chapter 9 103 Poisson Arrival Process Before we finish Ch. 5, let's revisit the Poisson Distribution e x , x 0 ,1, p ( x ) x! 0 , otherwise x F ( x) i0 e i i! • In this case, indicates the mean value, or number of arrivals (total) – Does not factor in arrivals over time – However, this can be done, and in this case we say the arrivals follow a Poisson Process > In this case we are counting the number of arrivals over time – However, some rules must be followed 104 Poisson Arrival Process 1) Arrivals occur one at a time 2) The number of arrivals in a given time period depends only on the length of that period and not on the starting point – i.e. the rate does not change over time 3) The number of arrivals in a given time period does not affect the number of arrivals in a subsequent period – i.e. the number of arrivals in given periods are independent of each other – Discuss if these are realistic for actual "arrivals" • We can alter the Poisson distribution to include time – Only difference is that t is substituted for e t ( t ) n , n 0 ,1, P [ N (t ) n ] n! 0 , otherwise 105 Poisson Arrival Process • Just like the Poisson Distribution, V = E = = t • In fact, if you look at the example from slide 81, we are in fact using the Poisson Arrival Process there The Poisson Arrival Process has some nice properties • The three required from the previous slide (obviously) – These imply that the arrivals follow an exponential distribution • Random Splitting – Consider a Poisson Process N(t) with rate t – Assume that arrivals can be divided into two groups, A and B with probability p and (1-p), respectively > Show on board 106 Poisson Arrival Process – In this case N(t) = NA(t) + NB(t) – NA is a Poisson Process with rate p and NB is a Poisson Process with rate (1-p) – Splitting can be used in situations where arrivals are subdivided to different queues in some way > Ex: At immigration US citizens vs. non-US citizens • Pooled Process – Consider two Poisson Processes N1(t) and N2(t), with rates 1 and 2 – The sum of the two, N1,2(t) is also a Poisson Process with rate 1 + 2 – Pooling can be used in situations where multiple arrivals processes feed a single queue > Ex: Cars arrive in New York City from many bridges and tunnels, each at a different rate 107 Poisson Arrival Process • Ex: Exercise 5.28 in text: – An average of 30 customers per hour arrive at the Sticky Donut Ship in accordance with a Poisson process. What is the probability that more than 5 minutes will elapse before both of the next two customers walk through the door? > As usual, the first thing is to identify what it is that we are trying to solve. > Discuss (and see Notes) – [I added this part] If (on average) 75% of Sticky Donut Shop's customers get their orders to go, what is the probability that 3 or more new customers will sit in the dining room in the next 10 minutes? > Discuss 108 Brief Intro. to Monte Carlo Simulation • We discussed previously that our simulations will typically follow the dynamic model Progress over time • Stochastic simulations using the static model are often called Monte Carlo Simulations Idea is to determine some quantity / value using random numbers that could be very difficult to do by other means • Ex: Evaluating an integral that has no closed analytical form 109 Brief Intro. to Monte Carlo Simulation • Before any formal definitions, let's consider a simple example Let's assume we don't know the formula for the area of a circle, but we do know the formula for the area of a square We'd like to somehow find the area of a circle of a given radius (let's say 1) 2 110 Brief Intro. to Monte Carlo Simulation Let's generate a (large) number of random points known to be within the square • We then test to see if each point is also within the circle – Since we know the circle has a radius of 1, we can put its center at the origin and any random point a distance <= 1 from the origin is within the circle • The ratio of points in the circle to total points • • generated should approximate the ratio of the area of circle to the area of the square We can then calculate the area of the circle by multiplying the area of the square by that ratio See Circle.java 111 Brief Intro. to Monte Carlo Simulation • Some informal theory behind M.C. Empirical probability • Consider a random experiment with possible outcome C • Run the experiment N times, counting the number of C outcomes, NC • The relative frequency of occurrence of C is the ratio NC/N • As N , NC/N converges to the probability of C, or p ( C ) lim N NC N 112 Brief Intro. to Monte Carlo Simulation Axiomatic probability • Set theoretic approach that determines probabilities of events based on the number of ways they can occur out of the total number of possible outcomes • Gives the "true" probability of a given event, whereas empirical probability only gives an estimate (since we cannot actually have N be infinity) • However, for complex situations this could be quite difficult to do When axiomatic probability is not practical, empirical probability (via Monte Carlo sims) can often be a good substitute • Can also be used to verify axiomatic results 113 Let's Make a Deal • Ex: Famous Let's Make a Deal problem Player is given choice of 3 curtains • One has a grand prize • Other two are duds After player chooses a curtain, Monty shows one of the other two, which has a dud • Now player has option to keep the same curtain or to switch to the remaining curtain What should player do? At first thought, it seems like it should not matter • However, it DOES matter – player should always switch 114 Let's Make a Deal We can look at this axiomatically • Initially there is a 1/3 probability that the player's choice is correct and 2/3 that it is incorrect • Revealing an incorrect curtain does not change that probability, so if the user does not switch his/her chance of winning is still 1/3 • However, now what do we know? – There is a 2/3 chance that the winning curtain is NOT the one originally picked – Of that 2/3, there is a 0 chance that it is the curtain already revealed – Therefore, there is a 2/3 chance that the remaining curtain is the winner, so we should switch to it 115 Let's Make a Deal In case we are still skeptical, we can verify this result with a Monte Carlo Simulation • See MontyMonte.java Note that the larger our number of trials, the better our result agrees with the axiomatic result 116 Monte Carlo Integration • Let's apply this idea to another common problem – evaluating an integral Many integrals have no closed form and can also be very difficult to evaluate with "traditional" numerical methods How can we utilize Monte Carlo simulation to evaluate these? Let's look at this in a somewhat simplified way (i.e. we will be light on the theory) 117 Monte Carlo Integration Consider function f(x) that is defined and continuous on the range [a,b] • The first mean value theorem for integral calculus states that there exists some number c, with a < c < b such that: 1 ba b f ( x ) dx f ( c ) or a b f ( x ) dx ( b a ) f ( c ) a – The idea is that there is some point within the range (a,b) that is the "average" height of the curve – So the area of the rectangle with length (b-a) and height f(c) is the same as the area under the curve 118 Monte Carlo Integration So now all we have to do is determine f(c) and we can evaluate the integral We can estimate f(c) using Monte Carlo methods • Choose N random values x1, … , xN in [a,b] • Calculate the average (or expected) value, ḟ(x) in that range: f ( x) • 1 N f ( xi ) f (c ) N i 1 Now we can estimate the integral value as b f ( x ) dx ( b a ) f ( x ) a 119 Monte Carlo Integration There is some error in this, but as N the error approaches 0 • It is inversely proportional to the square root of N • Thus we may need a fairly large N to get satisfactory results Let's look at a few simple examples • In practice, these would be solved either analytically or through other numerical methods • Monte Carlo methods are most useful for multiple integrals that are not analytically solvable • See Monte.java 120 Simple Queueing Theory • Many simulations involve use of one or more queues People waiting in line to be served Jobs in a process or print queue Cars at a toll booth Orders to be shipped from a company • Queueing Theory can get quite complex We are interested in a few of the more important results / guidelines 121 Simple Queueing Theory • First, we should define queue characteristics in a consistent way Standard Queueing Notation: A/B/c/N/K • where A is the interarrival time distribution • where B is the service-time distribution – A and B can follow any distribution (ex: the ones we discussed in Chapter 5): > D Deterministic Distribution is not random (ex: real data that has been measured / calculated) 122 Simple Queueing Theory > M Exponential (Markov) This is probably the most common and most studied of the random distributions – more on this below > Ek Erlang of order k > G General So why is an exponential distribution called Markov? • Relates to Markov processes (continuous time) and • Markov chains (discrete time) Let's consider a Markov chain for now (it is easier to conceptualize) 123 Simple Queueing Theory • A set of random variables (or states) X1, X2, … forms a Markov Chain if the probability that we transition from state Xi to state Xi+1 does not depend on any of the previous states X1, … Xi-1 – In other words, the past history of the chain does not affect its future – The idea is the same for a continuous time Markov process • Let's consider now a random variable Y that describes how long it will be in one state before transitioning to a different state – For example, in a queueing system, this could model how long before another arrival into the system (which changes the system state) 124 Simple Queueing Theory – This time should not depend on how long the process has been in the current state > i.e. it must be memoryless – As we discussed in Chapter 5, the exponential distribution is the only continuous distribution that is memoryless – Thus, when arrivals or services times are exponentially distributed, they are often called Markovian – We will touch on a bit more of this theory later – Now back to our description of the terminology… A/B/c/N/K 125 Simple Queueing Theory • where c is the number of parallel servers – A single queue could feed into multiple servers • where N is the system capacity – Could be "infinite" if the queue can grow arbitrarily large (or at least larger than is ever necessary) > Ex: a queue to go up the Eiffel Tower – Space could be limited in the system > Ex: a bank or any building > This can affect the effective arrival rate, since some arrivals may have to be discarded • where K is the size of the calling population – How large is the pool of customers for the system? – It could be some relatively small, fixed size > Ex: The computers in a lab that may require service 126 Simple Queueing Theory – It could be very large (effectively infinite) > Ex: The cars coming upon a toll booth – The size of the calling population has an important effect on the arrival rate > If the calling population is infinite, customers that are removed from the population and enter the queueing system do not affect the arrival rate of future customers (-1 = ) > If the calling population is finite, removal of customers from the population (and putting them into and later out of the system) must affect future arrival rates Ex: In a computer lab with 10 computers, each has a 10% chance of going down in a given day. If a computer goes down, the repair takes a mean of 2 days, exponentially distributed 127 Simple Queueing Theory In the first day, the expected number of failures is 10*0.1=1 However, once a failure occurs, the faulty computer is out of the calling population, so the expected number of failures in the next day is 9 * 0.1 = 0.9 Clearly this changes again if another computer fails • What about the system you are testing in Programming Assignment 1? – We cannot actually classify it with this notation, due to the single initial queue branching into multiple queues for the toll booths – However, if we consider only the toll booth queues, we could make each queue M/M/1/10/ > However, since a single queue buffers all of the individual queues, the arrival rate into the queues will change if traffic backs up onto the single queue 128 Long-Run Measures of Performance • What are some important queueing measurements? L = long-run average number of customers in the system LQ = long-run average number in queue w = long-run average time spent in system wq = long-run average time spent in queue = server utilization (fraction of time server is busy) 129 Time-Average Number in System • Let's discuss these in more detail Time-Average Number in the System, L • Given a queueing system operating for some period of time, T, we'd like to determine the time-weighted average number of people in the system, L • We'd also like to know the time-weighted average number of people in the queue, L Q – Note that for a single queue with a single server, if the system is always busy, L L 1 Q > However, it is not the case when the server is idle part of the time – The "hats" indicate that the values are "estimators" rather than analytically derived long-term values 130 Time-Average Number in System • We can calculate L for an interval [0,T] in a fairly straightforward manner using a sum: L 1 T iT i i0 – Note that each Ti here represents the total time that the system contained exactly i customers > These may not be contiguous – i is shown going to infinity, but in reality most queuing systems (especially stable queuing systems) will have all Ti = 0 for i > some value > In other words, there is some maximum number in the system that is never exceeded – See GrocerySimB.java 131 Time-Average Number in System Let's think of this value in another way: • Consider the number of customers in the system at any time, t – L(t) = number of customers in system at time t • This value changes as customers enter and leave the • • system We can graph this with t as the x-axis and L(t) as the y-axis Consider now the area under this plot from [0, T] – It represents the sum of all of the customers in the system over all times from [0, T], which can be determined with an integral Area T L ( t ) dt 0 132 Time-Average Number in System • Now to get the time-average we just divide by T, or L 1 T iT i i0 1 T T L ( t ) dt 0 – See on board • For many stable systems, as T (or, practically speaking, as T gets very large) L approaches L, the long-run time-average number of customers in the system – However, initial conditions may determine how large T must be before the long-run average is reached The same principles can be applied to L Q , the time-average number in the queue, and LQ, the long-run time average number in the queue 133 Average Time in System Per Customer Average Time in System Per Customer, w • This is also a straightforward calculation during our simulations w 1 N W i N i 1 – where N is the number of arrivals in the period [0,T] – where each Wi is the time customer i spends in the system during the period [0,T] • If the system is stable, as N , ŵ w – w is the long-run average system time • We can do similar calculations for the queue alone to get the values ŵQ and wQ – We can think of these values as the observed delay and the long-run average delay per customer 134 Arrival Rates, Service Rates and Stability We have seen a few times now • "for stable systems…" • What does this mean and what are the implications? For simple queueing systems such as those we have been examining, stability can be determined in a fairly easy way • The arrival rate, , must be less than the service rate – i.e. customers must arrive with less frequency than they can be served • Consider a simple single queue system with a single server (G/G/1//) – Define the service rate to be – This system is stable if < 135 Arrival Rates, Service Rates and Stability – If > , then, over a period of time there is a net rate of increase in the system of – > This will lead to increase in the number in the system (L(t)) without bound as t increases – Note if == some systems (ex: deterministic) may be stable while others may not be • If we have a system with multiple servers (ex: G/G/c//) then it will be stable if the net service rate of all servers together is greater than the arrival rate – If all servers have the same rate , then the system is stable if < c • See GrocerySimB.java 136 Arrival Rates, Service Rates and Stability • Note that if our system capacity or calling population (or both) are fixed, our system can be stable even if the arrival rate exceeds the service rate – Ex: G/G/c/k/ > With a fixed system capacity, in a sense the system is unstable until it "fills" up to k. At this point, excess arrivals are not allowed into the system, so it is stable from that point on > The idea here is that the net arrival rate decreases once the system has filled – Ex: G/G/c//k > With a fixed calling population, we are in effect restricting the arrival rate > As arrivals occur, the arrival rate decreases and the system again becomes stable 137 Conservation Law An important law in queueing theory states L = w • where L is the long-run number in the system, is the arrival rate and w is the long-run time in the system – Discuss intuitively what this means • Often called "Little's Equation" • This holds for most queueing systems • Text shows the derivation – Read it but we will not cover it in detail here 138 Server Utilization Server Utilization • What fraction of the time is the server busy? • Clearly it is related to the other measures we have discussed (we will see the relationship shortly) • As with our other measures we can calculate this for a given system (G/G/1//) 1 T T i i 1 – We assume that if at least 1 customer is in the system, the server will be busy (which is why we start at T1 rather than T0) • However, we can also calculate the server utilization based on the arrival and service rates 139 Server Utilization G/G/1// systems • Consider again the arrival rate and the service rate • Consider only the server (without the queue) • With a single server, it can be either busy or idle • If it is busy, there is 1 customer in the "server system", otherwise there are 0 customers in the "server system" (excluding the queue) – Thus we can define Ls = = average number of customers in the "server system" • Using the conservation equation this gives us – Ls = sws > where s is the rate of customers coming into the server and ws is the average time spent in the server 140 Server Utilization – For the system to be stable, s = , since we cannot serve faster than customers arrive and if we serve more slowly the line will grow indefinitely – The average time spent in the server is simply 1/ (i.e. 1/(rate of the server)) – Putting these together gives us – = Ls = sws = (1/) = / – Note that this indicates that a stable queueing system must have a server utilization of less than 1 > The closer we get to one, the less idle time for the server, but the longer the lines will be (probably) > Actual line length depends a lot not just on the rates, but also on the variance – we will discuss shortly – See GrocerySimB.java 141 Server Utilization G/G/c// systems • Applying the same techniques we did for the single server, recalling that for a stable system with c servers: < c • we end up with the the final result = /c 142 Markov Systems in Steady-State • Consider stable M/G/c queueing systems • Exponential interarrival times • Arbitrary service times • 1 or more servers As long as they are kept relatively simple, we can calculate some of the long-run steady state performance measures for these analytically May enable us to avoid a simulation for simple systems May give us a good starting point even if the actual system is more complicated 143 Markov Systems in Steady-State First, what do we mean by steady-state? • The probability that the system is in a particular state is not time-dependent • For example, consider a stable queueing system S – What is the probability that fewer than 10 people are in the queue? – If we start with an empty queue, this probability is initially 1 – However, as the system runs over time, the length of the queue will approach its long-run average length LQ > It no longer depends on the initial state > The actual length will still vary, but only due to variations in the arrival and service times 144 Markov Systems in Steady-State Let's look at it another way • Consider a stable M/G/1 queueing system with a given long-run average customer delay wQ • We start the system with a certain number of customers, s, in the queue • For each customer processed, we re-calculate the mean delay up to that point – In other words (ŵQ)j = average customer delay up to and including customer j • As j increases, (ŵQ)j wQ , despite the different • starting values of s – When this occurs the steady-state has been reached See graph from Law text on board 145 Markov Systems in Steady-State We will not concentrate on the derivation of the formulas, but it is useful to know them and how to use them M/G/1 Queue (see p. 222 of text for complete list) Note that is as we previously defined it (and is equal to the average number in the server) 2 2 2 Note that 2 is the variance in (1 ) the service times L 2 (1 ) Note that (intuitively) the 2 2 2 long-run queue length is equal (1 ) LQ to the long-run number in the system minus the average number in the server 2 (1 ) 146 Markov Systems in Steady-State Let's look at these a bit more closely • Consider first if 2 = 0 – i.e. the service times are all the same (= mean) > For example a deterministic distribution – In this case the equations for L and LQ greatly simplified to (1 0 ) 2 LQ 2 2 (1 ) 2 2 2 (1 ) – In this case LQ is dependent solely upon the server utilization, – Note as 0 (low server utilization) LQ 0 – Note as 1 (high server utilization) LQ 147 Markov Systems in Steady-State • However, as 2 increases with a fixed utilization, LQ also increases – Other measures such as w and wQ increase with 2 as well – This indicates that all other factors being equal, a system with a lower variance will tend to have better performance > In fact (See Ex. 6.9) in some cases a lower will give shorter lines than a higher , if it also has a lower 2 Able: 1/ = 24min, 2 = 400min2 Baker: 1/ = 25min, 2 = 4min2 Able: LQ = 2.711, P0 = 0.20 Baker: LQ = 2.097, P0 = 0.167 > Note that Able has a longer long-run queue length despite his faster rate > However, he also has a higher P0, indicating that more people experience no delay 148 Markov Systems in Steady-State We can generalize this idea, comparing various distributions using the coefficient of variation, cv: (cv)2 = V(X)/[E(X)]2 – i.e. it is the ratio of the variance to the square of the expected value • Distributions that have a larger cv have a larger LQ for a given server utilization, ρ – Ex: Consider an exponential service distribution: V(X) = 1/μ2 and E(X) = 1/μ, so (cv)2 = (1/μ)2/[(1/μ)]2 = 1 – See Figure 6.12 for chart of other distributions 149 Markov Systems in Steady-State Consider again the special case of an exponential service distribution (M/M/1), then • The mean service time is 1/ and the variance on the service time is 1/2 This simplifies our equations to (1 2 L 1 ) 2 2 2 (1 ) 1 2 2 (1 2 ) LQ 2 (1 ) 150 (1 ) (1 ) 2 (1 ) 2 (1 ) (1 ) Markov Systems in Steady-State • The M/M/1 queue gives us a closed form expression for a measure that the M/G/1 queue does not: – Pi , the long-run probability that there will be exactly i customers in the system Pn (1 ) L n – Note: We can use this value to show L for this system i Pi 0[1 ] 1[1 ] 2[1 ] 3[1 ] 0 1 2 3 i0 0 2 2 3 3 2 2 3 3 (1 ) ( 2 3 4 1 1 ) > The last equality is based on the solution to an infinite geometric series when the base is < 1 > We know ρ < 1 since system must be stable 151 Markov Systems in Steady-State The text has formulas for a number of different queue possibilities: • M/G/1 already discussed • M/M/1 already discussed • M/M/c Exponential arrivals and service with • multiple servers M/G/ "Infinite" servers (or service capacity >> arrivals, or "how many servers are needed"?) M/M/c/N/ Limited capacity system M/M/c/K/K Finite population • • Let's look at an example where these formulas would be used 152 Markov Systems in Steady-State Exercise 6.6 – Patients arrive for a physical exam according to a Poisson process at the rate of 1/hr – The physical exam requires 3 stages, each one independently and exponentially distributed with a service time of 15min – A patient must go through all 3 stages before the next patient is admitted to the facility – Determine the average number of delayed patients, LQ, for this system – Hint: The variance of the sum of independent random variables is the sum of the variance > Discuss and develop the solution 153 Markov Systems in Steady-State Exercise 6.23 – Copy shop has a self-service copier – Room in the shop for only 4 people (including the person using the machine) > Others must line up outside the shop > This is not desirable to the owners – Customers arrive at rate of 24/hr – Average use time is 2 minutes – What impact will adding another copier have on this system • How to solve this? – Determine the systems in question – Calculate for each system the probability that 5 or more people will be in the system at steady-state 154 Markov Systems in Steady State Exercise 6.21: – A large consumer shopping mall is to be constructed – During busy periods the expected rate of car arrivals is 1000 per hour – Customers spend 3 hrs on average shopping – Designers would like there to be enough spots in the lot 99.9% of the time – How many spaces should they have? • We will model this as an M/G/ queue, where the spaces are the servers – We want to know how many servers, c, are necessary so that the probability of having c+1 or more customers in the system is < 0.001 155 Markov Systems in Steady State Text also has some examples showing the M/M/c/N Queue and the M/M/c/K/K Queue • Idea is similar but the formulas and applications are different 156 Markov Systems in Steady State • Before we finish with this topic… Let's just talk a LITTLE bit about how these formulas are derived (being light on theory) Let's look at the simplest case, an M/M/1 queue with arrival rate and service rate Consider the state of the system to be the number of customers in the system We can then form a state-transition diagram for this system • A transition from state k to state k+1 occurs with • probability A transition from state k+1 to state k occurs with probability 157 Markov Systems in Steady State 0 1 … 2 k-1 k k+1 From this we can obtain k 1 Pk P0 i0 P0 k ( Eq . 138 . 1) We also know that P k 1 k 0 • Since the sum of the probabilities in the distribution • must equal 1 This will allow us to solve for P0 158 Markov Systems in Steady State P k 1 •All that is needed for this derivation is fairly simple algebra k 0 k 0 k P0 1 P0 1 P0 •Before completing the derivation, we must note an important requirement: to be stable, the system utilization / < 1 must be true 1 k 1 1 k 1 k 1 •This allows the series to converge k 159 Markov Systems in Steady State P0 1 / 1 1 ( / ) 1 P0 1 1 1 ( / ) 1 1 ( / ) / 1 ( / ) 1 ( / ) • Which is the solution for P0 from the M/G/1 Queue in Table 6.3 Utilizing Eq. 138.1, we can substitute back to get k k k Pk P0 1 (1 ) • Which is the formula indicated in Table 6.4 • The other values can also be derived in a similar manner 160 Markov Systems in Steady State A lot of theory has been left out of this process If you want to read more about queueing theory, there are a number of books on the topic: • Queueing Systems Volume 1: Theory by L. Kleinrock (Wiley and Sons, 1975) – Old but still relevant and still available – well known text on the topic • Many newer textbooks as well – Go to Amazon.com select "Books" and search "Queueing Theory" One final note: spelling! • Both "Queueing Theory" and "Queuing Theory" seem to be acceptable (and both spellings are in dictionary) – Search both to cover all options 161 Random Numbers • Stochastic simulations require random data If we want true random data, we CANNOT use an algorithm to derive it • We must obtain it from some process that itself has random behavior – http://en.wikipedia.org/wiki/Hardware_random_number_generator • Ex: thermal noise – http://noosphere.princeton.edu/reg.html • Ex: atmospheric noise – http://random.org/essay.html • Ex: radioactive decay – http://www.fourmilab.ch/hotbits/ – http://www.enginova.com/radioactive_random_number_gen era.htm 162 Pseudo-Random Numbers However, these require either purchase of hardware or a service For stochastic simulations we also need very large amounts of random data quickly • Possibly more than can be obtained from a true random source in a reasonable amount of time Often in testing and running our simulations, we also want to reuse the same "random" values • If we are debugging it helps to be able to see the • execution repeatedly with the same data If we are comparing systems, we may want to use the same data on both systems – Ex: Project 1 163 Pseudo-Random Numbers • Thus, more often than not, simulations rely on pseudo-random numbers These numbers are generated deterministically (i.e. can be reproduced) However, they have many (most, we hope) of the properties of true random numbers: • Numbers are distributed uniformly on [0,1] – Assuming a generator from [0,1), which is the most common • Numbers should show no correlation with each other – Must appear to be independent – There are no discernable patterns in the numbers 164 Pseudo-Random Numbers Let's look at a decidedly non-random distribution in the range [0,1) • Ex: Assume 99 numbers, X1…X99 are generated X1 = 0.01; Xi = Xi-1 + 0.01 – This will generate the sequence of numbers 0.01, 0.02, 0.03, … , 0.99 – Clearly these numbers are uniformly distributed throughout the range of values – However, also as clearly they are not independent and thus would not be considered to be "random" – Often, uniformity and independence are not as obvious, and must be determined mathematically > We will discuss how to test for each shortly 165 Linear Congruential Generators • Linear Congruential Generators These are perhaps the best known pseudo- random number generators • Easy and fairly efficient – Depending upon parameter choices • Simple to reproduce sequences • Give good results (for the most part, and when used properly) – Again depending upon parameter choices Based on mathematical operations of multiplication and modulus 166 Linear Congruential Generators Standard Equation: X0 = seed value Xi+1 = (aXi + c) mod m for i = 1, 2, … – where a = multiplier c = increment m = modulus • For c == 0 it is called a multiplicative congruential • generator For c != 0 it is called a mixed congruential generator – Both can achieve good results • Initially proposed by Lehmer and studied extensively by Knuth – See references at end of the chapter 167 Linear Congruential Generators Note that the generator as shown will produce integers If we want numbers in the range [0,1) we will have to convert the integer to a float in some way • This can be done in a fairly straightforward way by • • dividing by m However, sophisticated generators can do the conversion in a better (more efficient) way Clearly, however, the more possible integers, the denser the values will be in the range [0,1) 168 Linear Congruential Generators For now, consider three properties • The density of the distribution – How many different values in the range can be generated? • The period of the generator – How many numbers will be generated before the generator cycles? > Since the values are deterministic this will inevitably happen > Clearly, a large period is desirable, especially if a lot of numbers will be needed > A large period also implies a denser distribution • The ease of calculation – We'd like the numbers to be generated reasonably quickly, with few complex operations 169 Linear Congruential Generators Consider a multiplicative linear congruential generator (i.e. c == 0) • Xi+1 = (aXi) mod m If m is prime, this will produce a maximum period of (m–1) if ak – 1 is not divisible by m for all k < (m–1) – The period and density here are good, but with m as a prime, the mod calculation is somewhat time-consuming • If m = 2b for some b, this will produce a maximum period of 2b-2 of X0 is odd and either a = 3 + 8k or a = 5 + 8k for some k – This allows more efficient calculation of the numbers, but gives a smaller period – The importance of a good seed also makes this less practical (programmer must understand algorithm and seed requirements to use generator effectively) 170 Linear Congruential Generators Consider a mixed linear congruential generator (i.e. c != 0) • • • • • • Xi+1 = (aXi + c) mod m If m = 2b for some b, this will produce a maximum period of 2b if c and m are relatively prime and a = (1+4k) for some k This allows a good period, an easy mod calculation with the relatively small overhead of an addition This is the generator used by Java in the JDK Let's take a look at that in more detail See JDKRandom.java See TestRandom.java 171 Quality of Linear Congruential Generators The previous criteria for m, a and c can guarantee a full period of m (or m-1 for multiplicative congruential generators) • However, this does not guarantee that the generator • • • will be good We must also check for uniformity and independence in the values generated General criteria for good values of m, a and c are still unsure If you are creating a new LCG, a good rule of thumb is: – Choose m, a and c to guarantee a good period – Test the resulting generator for uniformity and independence 172 Chi Square Testing for Uniformity • Idea: Consider discrete random variable X, which has possible values x1, x2, …, xk and probabilities p1, p2, …, pk (which sum to 1) Assume n random values for X are chosen Then the expected number of times each value will occur Ei = npi Now assume n random values of distribution Y thought to be the same as X are chosen • How can we tell if distribution Y is the same as X? • We can at least get a good idea if the occurrences more or less match those of the Ei 173 Chi Square Testing for Uniformity Ex: Consider our dice-throwing example from earlier this term (see Slides 12-13) E2 = (n/36), E3 = (2n/36), E4 = (3n/36), E5 = (4n/36) E6 = (5n/36), E7 = (6n/36), E8 = (5n/36), E9 = (4n/36) E10 = (3n/36), E11 = (2n/36), E12 = (n/36) • Now consider a sequence of n = 360 rolls with the following distribution N2 = 8, N3 = 19, N4 = 35, N5 = 38, N6 = 53, N7 = 64 N8 = 46, N9 = 40, N10 = 27, N11 = 17, N12 = 13 – Question: Are these dice "fair"? • We need a way of comparing empirical results with theoretical predictions – We can use the Chi-Square (2) test for this 174 Chi Square Test Consider the following formula: V k Yi E i 2 i 1 Ei • Where Yi is the observed number of occurrences of • • value xi and Ei is the expected number of occurrences of value xi This is showing the square of the differences the observed values and the expected values (divided by the expected values to normalize it) Now consider two hypotheses: – NULL Hypothesis, H0 : Y matches distribution of X – Hypothesis H1: Y does not match distribution of X 175 Chi Square Test • If V is too large, we reject the null hypothesis (i.e. the • distributions do not match) If V is small(ish) we do not reject the null hypothesis (i.e. the distributions may match) Staying with our dice example: V = (8-10)2/10 + (19-20)2/20 + (35-30)2/30 + (38-40)2/40 + (53-50)2/50 + (64-60)2/60 + (46-50)2/50 + (40-40)2/40 + (27-30)2/30 + (17-20)2/20 + (13-10)2/10 = 4/10 + 1/20 + 25/30 + 4/40 + 9/50 + 16/60 + 16/50 + 0/40 + 9/30 + 9/20 + 9/10 = 228/60 • Ok, so what does this number mean? • Should we accept the null hypothesis or reject it? 176 Chi Square Test • To answer this question we need to look at the test a • bit more closely The test is called the Chi Square Test because for n large enough (standard rule is that each Yi should be at least 5) the value for V gives a Chi Square distribution under the null hypothesis – Since V is normalized it is not dependent on the original distributions – only the differences between them and the "degrees of freedom" (k – 1, where k is the number of possible values of the random variable). – Thus a standard chart can be used – see p. 584 in text to see how it looks – The idea is that, given the null hypothesis, it is more likely that V will fall into the "middle" part than either end 177 Chi Square Test – The closer we get to the ends (the "tails") the less likely it is that V should fall here, given the null hypothesis – Thus if V is too large (or in some cases small), we reject the null hypothesis • How to determine values for "too large" or "too small"? – We must set a level of significance, = Pr(reject H0 | H0 true) > Or, is the probability that we reject the null hypothesis even though it is true (i.e. the probability that we reject it by mistake) > Clearly, the lower the value for the less likely we are to reject a distribution by mistake > Ex: if = 0.1, we are saying we are willing to reject the null hypothesis even though there is a 10% chance that it is true 178 Chi Square Test – Mapping this onto our Chi Square graph, we locate the portion point of the graph that corresponds to our level of significance, > For example, if = 0.1, which values of V would cause us to reject the null hypothesis? > This corresponds to the area in the Chi Square graph where there is less than 0.1 chance that V would legitimately be mapped > See handout – Depending upon the application, we can use the full on one "tail" of the graph (one-sided test) > Usually the right tail – in this case V can be 0 but should not be too large – We may also want to split between the left and right "tails" of the graph > If it is either too big or too small we reject the null hypothesis 179 Chi Square Test Consider again our die-tossing experiment • We have 10 degrees of freedom (since we have 11 possible values) • Our value for V = 228/60 = 3.8 • Assume our level of significance is 0.05, using a twotailed test (so 0.025 on each side) – In other words, if the expected values are varied from by too much, OR if they are matched too closely, we will assume that the sequence of throws is not random • Looking at the charts in the handout, we see: – For the right tail, with critical value 0.025, the value of V for 10 degrees of freedom is 20.483 > Since our value of V is much less than this, we do not reject the null hypothesis on this basis 180 Chi Square Test – For the left tail, with critical value 0.025, we actually want the critical value for (1 – 0.025) = 0.975 > The value of V here for 10 degrees of freedom is 3.247 > Since our value of V is GREATER than this (but not by a lot) we do not reject the null hypothesis on this basis either • Note that under repeated testing, there is a reasonable chance that we WILL occasionally reject a null hypothesis (perhaps) improperly – Ex: if = 0.1 and we do 100 trials, we are likely to reject 10 of them even if the distribution is valid – We need to take this into account during testing 181 Chi Square Test So how do we apply this test to our Uniform distribution? • After all, the values shown are discrete, and the uniform distribution is continuous! We simply divide U[0,1) into subranges that represent each discrete value • Ex: For 10 subranges we can have {[0,0.1), [0.1,0.2), • …, [0.9,1.0)} Then we count the values in each subrange and perform the Chi Square test as indicated – Note that for a uniform distribution all of the Ei will be the same = n/(# of subranges) • We probably want the two-tail test here (why?) 182 Chi Square Test Look at handout RandTest1.java • Here some linear congruential generators are tested with the Chi Square Test • Note the comments and why some of the generators are poor in some situations The Chi Square test is not perfect • Doesn't work well for small number of trials • As shown does not directly test independence – Thus we need other tests as well before deciding upon quality of a random number generator • However, Chi Square can be applied in conjunction with other tests – we may see this later 183 Kolmogorov-Smirnov Test • Another test for uniformity is the Kolmogorov-Smirnov test Has a different approach than Chi-Square • Does not count number of values in each subrange of U[0,1) • Instead it compares the empirical distribution (i.e. the numbers generated) to the cumulative distribution function (cdf) for a uniform distribution – Recall from Chapter 5 and Slide 84: F(x) = x where 0 <= x <= 1 – Think about what this means > Discuss 184 Kolmogorov-Smirnov Test – Now consider the empirical distribution generated by our random number generator (assume N values are gen'ed) > Call it SN(x) – Of the N total values generated by our random number generator, say that k of them are <= x > Then SN(x) in this case is k/N – For any value x, we can determine SN(x) by counting the number of values <= x and dividing by N – For example, consider the following 10 values: (0.275, 0.547, 0.171, 0.133, 0.865, 0.112, 0.806, 0.155, 0.572, 0.222) SN(0.25) = 5/10 since 5 values above are <= 0.25 SN(0.5) = 6/10 since 6 values above are <= 0.5 SN(0.75) = 8/10 since 8 values above are <= 0.75 – K-S test checks to see how much SN(x) differs from F(x) > If it is too much we reject the null hypothesis 185 Kolmogorov-Smirnov Test More specifically: • Sort the empirical data • Calculate the maximum "SN(x) above F(x)" value, D+ for the empirical data • Calculate the maximum "SN(x) below F(x)" value, Dfor the empirical data – Idea is that if the empirical data matches the uniform distribution, it should not ever be too much above or below the cdf – See Figure 7.2 • Test max(D+, D-) for the desired level of significance against the table of critical values for the given degree of freedom (see Table A.8) – If it is too high, reject the null hypothesis 186 Kolmogorov-Smirnov Test • Ex: For previous data, call it R (0.275, 0.547, 0.171, 0.133, 0.865, 0.112, 0.806, 0.155, 0.572, 0.222) • Sorted, and compared to F(x) (actually, i/N), we get (0.112, 0.133, 0.155, 0.171, 0.222, 0.275, 0.547, 0.572, 0.806, 0.865) (0.100, 0.200, 0.300, 0.400, 0.500, 0.600, 0.700, 0.800, 0.900, 1.0) • i/N – Ri ( ----- , 0.067, 0.145, 0.229, 0.278, 0.325, 0.153, 0.228, 0.094, 0.135) • Ri – (i-1)/N (0.112, 0.033, ----- , ----- , ----- , ----- , ----- , ----- , 0.006, ----- ) • D+ = 0.325; D- = 0.112 • D = 0.325 • Assume = 0.05 – N = 10, critical value = 0.41 (from Table A.8) – So the null hypothesis is not rejected 187 Chi-Square vs. Kolmogorov-Smirnov There is debate about which test is more effective • Each seems to work better under certain • circumstances It is probably best to run both tests on the data to thoroughly test for uniformity 188 Tests for Independence • There are many tests for independence There are a lot of ways data can correlate From one point of view it could appear independent, while from another it could appear very non-independent For best results, many different independence tests should be run on the data • Only if it "passes" them all should the generator be accepted We will briefly look at a few of these tests • Most not taken from the text – Many taken from previous (3rd) edition 189 Runs Tests A run can be considered a sequence of events whose outcome is "the same" • For example, a sequence of increasing values – Old game show "Card Sharks" was based on this > 2, 5, 8, J, Q, A In random data the number and length of various runs should not be too great (or too small) • If so, it indicates some non-independence of the data Many types of runs can be tested • We'll look at a few 190 Runs Tests Runs up and Runs down • In a random sequence of N values in U[0,1), some properties of the runs up and runs down are Max = N – 1 (every other number switches) Min = 1 (monotone increasing or decreasing) = (2N – 1)/2 2 = (16N – 29)/90 (don't worry about why this is true) • If N > 20, this distribution is about normal • Assume we observed r actual runs in our data – If our data is random, r should not be too far from the mean > If it is we reject it 191 Runs Tests – We normalize using our standard formula: Z0 = (r - )/ – And look up the result in our normal table > If is our level of significance, our value r should not be less than /2 standard deviations from either side of the normal curve > If it is we reject the null hypothesis – See example in handout and show on board Runs above and below the mean • Runs up and down don't indicate any absolute organization of the data (or lack thereof) • We may also want to test the number of runs (i.e consecutive values) above and below the mean – Too many or too few is not a good sign 192 Runs Tests • Calculation for mean and variance is more complex, but we handle this test in more or less the same way as the runs up and runs down test – Don't want the our number to be too far from the mean – Use the standard normal distribution to test – See handout for formulas We can also test the lengths of runs for both of our previous options • The math in this case is more complex, but the idea is similar – Given a sequence of random numbers, we can expect a certain number of runs of each length 1, 2, 3, … > Clearly the probability should get lower as the lengths increase 193 Runs Tests – If the run lengths in our empirical tests differ too much from the expected distribution, we can reject the generator – Idea is to determine the empirical distribution of runs of various lengths, then compare it with the expected distribution given random data > The distributions are compared using the Chi-Square test – See handout for details 194 Other Tests Other tests for independence include • Autocorrelation tests – Tests relationship of items m locations apart > See simple ex on board • Gap tests – Tests the gap between repetitions of the same digit > For U[0,1) we can discretize it and test for the gap between values in the same bin > See simple ex on board • Poker test – Tests frequencies of digits within numbers – Looks at possible poker hands and their frequencies vs. the expected values > See simple ex on board 195 Other Tests • Serial Test – Let's start with an example here: – Given a sequence of random values in U[0,1): X1, X2, X3, X4, … – Consider the values in non-overlapping pairs (X1, X2), (X3, X4), … – Now let's consider the distribution of the pairs of values in a plane > We can approach this in a similar way to what we did with the Chi-Square frequency test > Now, however, rather than dividing U[0,1) into (for example) 10 bins we instead divide U[0,1)2 into (for example) 100 bins, one for each pair > We then check the empirical results with the expected values for each bin (using Chi-Square) 196 Other Tests – We can generalize this to an arbitrary number of dimensions – However, note that the number of bins increases dramatically as we increase dimensions > Recall the Chi-Square is only effective if the number of values in each bin is ~5 or more > For large dimensions we have to generate an incredibly large number of values to obtain accurate results – However, if our generator "passes" in 2 and 3 dimensions, for many purposes it should be ok > As long as it passes other tests as well 197 Other Tests There are many other empirical tests that we can run on our generators There are also theoretical ways to test generators • In this case we do not generate any actual values • with a generator Rather, we use the generation algorithm to determine how well (or poorly) it "can perform" Two famous theoretical tests are the spectral test and the lattice test – Both are similar in what they are trying to do • Measure the theoretical randomness / distribution of sequences of values generated 198 Spectral Test & Lattice Test These tests are (more or less) theoretical extensions of the serial test Let's look at a simple example with sets of pairs • (Xn, Xn+1) for n = 1 up to period – 1. • If we consider these sets in two-dimensional space, we get points in a plane – These points tend to fall on a set of parallel lines > Show on board – The more lines required to touch all of the points, the better the distribution (for d = 2) – If few lines are required, it indicates that the generator does not disperse the pairs well in 2 dimensions – See handout 199 Spectral Test & Lattice Test • We can extend this test to an arbitrary number of dimensions – Generally, for dimension d we will look at (d-1)-tuples and see how many d-1 dimension hyperplanes are needed to capture all of the points > Actually we are looking at "how far apart" the hyperplanes are, but it is more or less the same idea • See handout from Knuth – Note that the process is highly mathematical – However, if a generator does not pass this test in at least the first few dimensions, it is probably not a good generator to use 200 Other Random Number Generators • Other random number generation techniques have been tried, with varying degrees of success Middle Square method • Original method proposed by von Neumann Start with a four-digit positive integer Z0 and square it to obtain an integer with up to eight digits; if necessary, append zeros to the left to make it exactly eight digits. Take the middle four digits of this eight-digit number as the next four-digit number, Z1. Place a decimal point at the left of Z1 to obtain the first U[0,1) random number, U1. Then let Z2 be the middle four digits of Z12 and let U2 be Z2 with a decimal point at the left, and so on. – Try an example • Interesting idea and at first glance looks good, but has been shown to be quite poor 201 Other Random Number Generators Combined LCGs • In this case multiple LCGs are combined (in a specific way) to produce generators with longer periods – See Section 7.3.2 • These must be handled very carefully – Oddly enough, the best random generators tend to be those that follow strict mathematical formulas Additive Generators • These generators add previously generated values to generate new values • They typically differ on how many values are added and which previous values are chosen to add – These choices have a large effect on the quality! 202 Additive Generators Famous initial attempt: Fibonacci Generator: • Initialize X0 and X1 using an LCG Xi = (Xi-1 + Xi-2) mod 232 – The name is fairly obvious • Unfortunately, this generator does not do too well in empirical tests Researchers determined better choices for terms to add Ex: The generator used for Unix BSD random() • Initialize X0, X1, … X31 using an LCG Xi = Xi-31 + Xi-3 – Has a HUGE period – good for long simulations – See handout for details 203 Random Variate Generation • Generating Random Variates Once we have obtained / created and verified a quality random number generator for U[0,1), we can use that to obtain random values in other distributions • Ex: Exponential, Normal, etc. A variety of techniques exist for generating random variates • Some are more efficient than others • Some are easier to implement than others We will briefly look at just a few 204 Random Variate Generation • Inverse Transform Technique Distributions that have a closed mathematical formula can be generated in this way Idea is that some function F-1 will map values from U[0,1) into the desired distribution An obvious example is the exponential distribution that you used in Project 1 Let's look this one in more detail • Recall what the distribution is 205 Inverse Transform Technique Recall the cdf, as shown below 0 , x 0 F ( x) x 1 e , x0 Now consider a uniform distribution over U[0,1) – we will call this R We then set F(x) = R and solve for x • This will give us the transformation needed to convert from R to F(x) 1 e e x x R 1 R x ln( 1 R ) x ( 1 / ) ln( 1 R ) ( 1 / ) ln( R ) 206 Inverse Transform Technique The last equality seems incorrect, but note that if R is a random distribution on U[0,1), then so is 1-R, so we can use the slightly simpler form • Let's look at another example: The Uniform Distribution over an arbitrary range • We could have it continuous, such as U(1,2) or • U(0,4) We could have it discrete, such as – Uniform distribution over integers 1-10 – Uniform distribution over integers 100-200 207 Uniform Distributions on Different Ranges • Continuous uniform distributions – Given R U[0,1), we may need to do 2 different things to get an arbitrary uniform distribution > Expand (or compress) the range to be larger (or smaller) > Shift it to get a different starting point – The text shows a derivation of the formula, but we can do it intuitively, based on the two goals above > To expand the range we need to multiply R by the length of the range desired > To get the correct starting point, we need to add (to 0) the starting point we want – Consider the desired range [a,b]. Our transformations yield X = a + (b – a)R 208 Uniform Distributions on Different Ranges • Discrete uniform distributions – The formula in this case is quite similar to that for the continuous case – However, care must be paid to the range and any error due to truncation – Consider a discrete uniform range [m, n] for integers m, m+1, … , n-1, n – If we use X = m + (n – m)R > The minimum value is correct, since the minimum result for X is m (since R can be 0) > Now we need to be sure of two things: 1) The maximum value is correct 2) The probabilities of all values are the same 209 Uniform Distributions on Different Ranges – First let's consider the maximum value > Although the continuous range (b – a) worked as we expected, the discrete range (n – m) does NOT work > Since R U[0,1), we know it can never actually be 1 > Thus, (n – m)R will always be strictly less than n – m and thus m + (n – m)R will always be strictly less than n (it will be n – 1 , since it is discrete) – To solve this problem, we need to increase the range by 1 > (n – m + 1) – So we get overall X = m + (n – m + 1)R – Now are all values uniform (i.e. equal probability)? > Consider breaking U[0,1) up into (n – m + 1) intervals 210 Uniform Distributions on Different Ranges 1 1 2 n m 1 nm 0, n m 1 , n m 1 , n m 1 n m 1 , n m 1 • Each sub-interval is equal in range and thus has • equal probability Given R in each sub-interval, a distinct value in the range [m,n] will result – Ex: Assume R is in the second sub-interval above – We then have m + (n – m + 1)(1/(n – m + 1)) = m + 1 X < m + (n – m + 1)(2/(n – m + 1)) = m + 2 – Since X is an integer, and is strictly less than m + 2, the entire sub-interval maps into m + 1 211 Inverse Transform Technique The inverse transform technique can be applied to other distributions as well • See text for more examples Note that in some situations, the inverse transform technique may NOT be the most efficient way of generating random variates • However for relatively simple cdfs such as the exponential distribution, it can be easily done 212 Convolution Technique Recall that the Erlang distribution can be thought of as the sum of K independent exponential random variables, each with the same mean 1/K • Thus, since we can already calculate the exponential • • distribution using the inverse transform technique, we should be able to easily also generate an Erlang distribution However, for efficiency, we can generate the answer in a slightly different way Let's look at it in some detail 213 Erlang Distribution • From the text we see: k X X i 1 k i 1 ( k ln Ri ) i 1 – where the contents of the parenthesis is an exponential variate with mean 1/k – We could calculate the sum directly, but that would require calculating a log k times > Since logarithms are typically time-consuming, we'd like to calculate fewer if possible – The equation continues in the text k k k 1 1 X X i ( ln R i ) ln R i k k i 1 i 1 i 1 – Now only one logarithm is required > But how did they get that last step? 214 Erlang Distribution • Recall properties of logarithms Given values x1 and x2 logb(x1) + logb(x2) = logb(x1x2) – Ex: log10(1000) + log10(100) = log10(100000) = 5 • Applying the rule multiple times gives us our desired • result We haven't talked too much about efficiency up to this point, but note that for large simulations, very high numbers of variates may be required – If we save even a few cycles in each generation overall it can add up to substantial savings • In fact there are more efficient ways than this as well, which would probably be used in a large simulation 215 Other Techniques Not all distributions can be easily transformed • Some may not have a closed form inverse • We can still use the inverse transform technique if we are willing to approximate the distribution • However, if we want to be more precise we may have to use other means In these cases we can use alternative techniques to generate the variates • There are many different techniques, some of which are quite specific to a particular distribution 216 Normal Variates Recall that the cdf for a normal distribution is complex and cannot be inverted in a closed form • However, with a different approach we can still • generate normal variates with good results The text shows a polar technique that generates two normal values at a time – There are many other techniques as well Let's take a look at these four distributions, using the implementations just discussed • Look at Variates.java 217 Variate Generation Summary • In a professional simulation Software used will have predefined functions for all of these variates • However, it is good to know some of the theory for how they are derived You may perhaps have to derive one You may need to examine the implementation to perhaps improve the efficiency 218 Input Modeling • In the real world, input data distributions are not always obvious or even clearly defined Often we have some sample data but not enough for our entire simulation Thus we'd like to determine the distribution of the sample data, then generate an arbitrary number of values within that distribution for our simulation • However, the input data may not even be of a single • distribution May differ at different times / days 219 Input Modeling • Also, in some cases we may not be able to generate real data, because the means to do so do not yet exist – Ex: If we are building a new network or road system, we don't yet have a system to provide us with the real data Assuming we can get sample input data, determining the distribution is not easy • Usually requires multiple steps, and a combination of computer and "by hand" work Let's try to do this for a small real example • Arrivals at Panera downstairs • We'll monitor Panera for the rest of class 220 Input Modeling • We will stand at the door and record the interarrival times of customers over a 15-20 minute period – We could do this by hand, but we'll use a computer since that is easier and more precise • We'll then look at the data to see if we can fit it to a distribution – We'll do this next class 221 Input Modeling Once we have the data, how do we fit it to a distribution? • The first step typically involves creating one or more • histograms of the data and graphing them, to see the basic "shape" of the distribution This requires some skill and finesse – How "wide" is each group to be? – How can we know if a shape is of a particular distribution? > Sometimes we just "eyeball it" > There are also some analytical techniques for verification • If we have an idea of a possible distribution, then we can try to verify it 222 Input Modeling • Estimate parameters based on the distribution – Ex: lamba for exponential – Ex: beta and theta for gamma – Ex: mu and sigma for normal • We can then try some goodness of fit tests to see if the distribution works – As with the random number testers, we can use the Chi-Square test and the Kolmogorov-Smirnov test Let's try this process with our collected data 223 Panera Example Consider the initial data points acquired • First try to form one or more histograms to “eyeball” the distribution versus known distributions – In this case, the exponential distribution seems to best match the data – Note that we may have to try a few different intervals for the histogram • Once a family of distributions has been determined, we’d like to determine the parameter(s) to the distribution – With exponential, we’d like to estimate the value of the rate, lambda – Since this should be 1/mean, we can calculate it easily using the sample data 224 Panera Example • Next we should try one or more “goodness of fit” tests to see if the proposed distribution does in fact reasonably match our sample data – In this case we will use Chi-Square or KomolgorovSmirnov – Consider in this case Chi-Square as shown in the panera.xls handout – First we will consider an “uneven” distribution based on the histogram shown > We generate the cumulative distribution with the given lambda at the interval endpoints > Next we determine the expected values for each interval > We then run the test on the results, with the following modifications: 225 Panera Example > The degrees of freedom will be n – s – 1, where n is the number of intervals, and s is the number of estimated parameters (in this case s = 1) > Groups with fewer than 5 expected values will be merged into a single group – Note that the p-values described in the Banks text (section 9.4.4) can be easily calculated in Excel using the CHITEST function > However there is an issue with degrees of freedom – see panera.xls – As indicated in the Banks text (section 9.4.2) we should also consider an equal probability Chi Square test > Note that for our data this test does not perform nearly as well as the “uneven” probability test (in fact it fails fairly badly) • We could also use a K-S test to test the distribution – See Example 9.16 in Banks text 226 Panera Example If we find Exponential Distribution unacceptable (based on the even prob. Chi-Square test) it might behoove us to try a different distribution to see if it matches better • Some options are – Gamma – Weibull > Exponential is a special case of both of these – Lognormal > This is actually tried in the panera.xls spreadsheet, but it is not good • Clearly this is not a simple process • Also note that we really need more data here 227 Input Modeling What if we don’t have sample data to use? • We must do the best that we can, relying on – Experts > What do people in the area in question think, based on their knowledge and experience – Engineering specs > Ex: A device is built to have some mean time to failure, based on the production environment. We can use that value as a starting point for mean time to failure of the device in our real environment – Similarity to something we already know > Ex: What is the distribution of people wanting to ride the new “Super Wacky Crazy Loopmeister” ride? We can use as a starting point the measured distribution from last year’s “Bizarro Zany Upsidedowner” ride 228 Is Data a Single Distribution? • We already discussed the fact that often input data (ex: interarrival times) will not remain constant over time There are two issues to deal with here: 1) How to determine if the data is a single distribution or not – One simple way to check this is with a linear – – regression through a scatter plot of the points If the distribution is consistent over time the line should be horizontal If it has a positive slope it means the mean is gradually increasing (similar idea for negative slope) 229 Is Data a Single Distribution? 2) How to formulate the distribution(s) if it is determined to be of multiple distributions • There are various ways of "combining" the • distributions for a simulation One simple approach for Poisson processes is shown in Section 9.5 of the Banks text – Idea is to subdivide time intervals such that each interval achieves single distribution as discussed in 1) – Then use the appropriate distribution in the appropriate time period • See Banks section 9.5 for more information 230 Is Data Independent? • Single distributions such as exponential require independent arrivals However, real data may have some dependencies • Perhaps each arrival affects the next – Ex: A small restaurant has a large window to a busy, trendy street. Initially, an empty restaurant is not appealing, and arrivals are few. As more people fill the restaurant more want to go, and the arrivals increase. However, once it gets too full, people avoid it because it is too crowded. • We can test for this using lag-h autocorrelation tests, as discussed in Section 9.7 of the banks text – If this occurs we can model the distribution as a time series input model – There are various models that can be used to for timeseries inputs – see Section 9.7.3 in the Banks text – See CoVar.java and Covariance.xls 231 Multivariate Inputs Sometimes two or more seemingly independent distributions may in fact be related • Ex: Customers in a grocery store spend T1 time shopping and T2 time in the checkout (not counting wait time) – At first glance we might try to formulate two distinct distributions for these times – However, customers who spend more time in the store probably buy more items and thus take longer to checkout – Thus the distributions are not independent – In this case can represent the overall distribution as a distribution of two-value vectors (one value for each Ti) • See Section 9.7.2 in Banks text 232 Verification and Validation • First we must distinguish between the two Verification • Is the process of building / implementing the model correct? – Is the final result an accurate implementation of the model? – Note that verification does not address the issue of whether the model itself is accurate or not Validation • Is the model that is being implemented the correct one (does it accurately model the system in question)? • This is testing the correctness of the model itself 233 Verification The techniques here are similar to general “program correctness” techniques that apply for software engineering and general programming – Ex: Run with some known data and examine output – Ex: Look for obvious problems with special cases and / or obviously bogus results > Negative wait time, etc. – Ex: Document code well, and use clear, well-named variables – Ex: Do a detailed trace, keeping track of variable states as the execution progresses • Seem simple to do but in practice, verification can be extremely difficult and time-consuming – Especially for large, complex simulations 234 Validation These techniques are more specific to simulation • How do we know that the model we have developed • is an accurate representation of the real system? Often an iterative process (calibration) – A system is evaluated against the real system, modified, and evaluated again – The process continues until the model is deemed satisfactory overall • Evaluation is typically done in two ways • Subjective – Experts examining the model and perhaps some test output, determining if it fits the real system 235 Validation – Programmer must be involved here, but in mainly in the role of making the recommended changes • Objective – Tests here deal more specifically with the data produced by the simulation vs. the real data > Clearly, in this case, we must have at least some “real data” to compare the simulation data to – without it the evaluation will be primarily subjective – Idea is that we collect some sample input and sample output for the system to be modeled > Use the sample input to create input distribution(s) (as discussed in Chapter 9) > Run the simulation using the generated input > Compare simulation output with collected output data 236 3-Step Approach Naylor and Finger give a very intuitive 3-step approach to validation: 1) Build a model with high face validity 2) Validate model assumptions 3) Compare model I/O with real system I/O Let’s talk about these more • And where pertinent we will relate them to Assignment 3 – Panera Simulation 237 Face Validity Is the model reasonable to informed parties? • If the model is specified clearly, an expert in the area (not necessarily even a programmer) can review it for obvious problems / inconsistencies • This can be done prior to implementation Does the model behave in an expected way in a given situation? Is the output reasonable? • Note here we are not analyzing the data in detail (we save that for step 3) – Rather we are just giving it a coarse look for blatant incorrectness – Ex: Two queue system and average length of one queue is 0 and the other is high > This does not seem reasonable, so we try to find out why it is occurring 238 Face Validity – Ex: Average wait does not change as arrival rate increases > Note that these above issues could be due to either an incorrect model or an incorrect implementation > Often validation and verification are done hand-in-hand • These must clearly be done during / after implementation – However, they don’t require a lot of calculation and can typically be done quickly (relatively speaking) > Finding the problem at least – fixing it may take more time At this point in our Panera simulation, we cannot completely assess the "reasonableness" of the model • Since it is not completely specified in the assignment – you must fill in some details yourselves 239 Face Validity But let’s look at a few things: • 2 input distributions at the doors – We are not yet sure what these are – However the two together give the input to the restaurant > Yet not all of the customers will actually buy something > You must determine the probability of staying • 2 different types of service 1) The "hot food" line – Single (FIFO) queue to multiple servers (unknown distribution) – Additional queue for pick up (also multiple servers, unknown distribution) – You must determine the relationship between the ordering and the preparing of the food for pick-up – May not be FIFO 240 Face Validity 2) The "coffee and pastry" line – Single (FIFO) queue to multiple servers > The service distribution is likely (but not certainly) different from the "hot food" line > At least the parameters will be different – No separate pick up line – people get their order before the next is processed (I think – verify!) • Most customers will use one line or the other, but perhaps an occasional customer will use both – You will have to check this to find out 241 Validate Model Assumptions • Banks classifies these into two groups: Structural assumptions • Relate to the operation of the system • How data is manipulated once inside • Deals with the data structures and algorithms used – Ex: Single FIFO Queue Data assumptions • Are the distributions used correct / accurate? • This takes the data issues from the face validity and looks at them in more detail • More or less what we did in Chapter 9 242 Validate Input / Output Transformations Here we develop an objective way of testing the model for correctness / accuracy • Identify the input and output variables of interest • Obtain sample data of the real system of interest – If possible • Run the simulation and compare the results to those obtained for the real system – Analyze mathematically to see if the model “fits” the system – If the results are poor / incorrect, we need to find out why and revise the model 243 Validate Input / Output Transformations Consider Project 3 • In order to validate the I/O transformations correctly, we must know the structural assumptions of Panera – ex: How are customers transferred from the "hot food" line to the "waiting for food" line – This could be obtained from the manager > If we cannot obtain this we have to do our best based on observation • However, we can definitely validate the I/O in the ways previously discussed 244 Input / Output Transformations Let’s at least look at the way we might set up these tests for Project 3 We want to consider two different kinds of variables needed by our simulation • Uncontrollable variables, whose values are determined by the external environment – These could be interarrival times, delays, and possibly service times – Denote these by the array X • Decision variables, whose values are controllable within the simulation (i.e. can be changed) – These could be number of servers, maximum queue length, and possibly service times – Denote these by the array D 245 Input / Output Transformations • Note that service times and generally speaking several variables could be uncontrollable variables but they also could be decision variables – This depends on the model and whether or not these can be altered – Ex: Consider the time required to service a customer > This could be uncontrollable if we deem it to be fixed based on observation > This could be a decision variable if we can train the employees in order to speed up service • Given X and D, our simulation will produce some output data, Y, based on the protocols implemented – Call this transformation f, leading to f(X, D) = Y 246 Input / Output Transformations • Let’s try to come up with (some of) the contents of X, D, and Y for Project 3 – They won’t be completely specified until we have completed our input analysis > See Banks Table 10.1 for an example > See notes below for some suggestions X – Consider first X1 = customer arrivals > How do we show differences for different entrances? > How do we show differences for different times / days (if we were taking these into account) – Consider next X2 = decision about what to do once inside > We could go to the hot line or the coffee line or both or neither 247 Input / Output Transformations – Consider next X3 = service at the "coffee" line > What we want here is the service distribution for one server > The number of servers will be a decision variable > Note that this could be VERY complex if we wanted it to be -- each server could have a separate rate and possibly even a separate distribution -- different servers could be on duty on different days – Consider next X4 = service at the "hot food" line > This could be complex or even two separate variables if we consider the "purchase" segment and the "pick up" segments as distinct from one another – Other Xi as well once model is more fully developed 248 Input / Output Transformations D – – – – What are our decision variables in this simulation? D1 = number of servers at coffee area D2 = number of servers at hot foods area D3 = number of people preparing meals in hot foods area – Plus any other decision variables you may have (ex: number of various appliances, etc) – You could potentially also make service times a decision variable (as mentioned before) if you could change it via training 249 Input / Output Transformations Y – Clearly, we can have our program output many different things – We need to identify those that are significant / important to our goals – Determining these can be done by the programmer or anyone with common sense in simple cases – However, in very complex cases it may require experts in the field and / or experts in statistics to correctly identify the output variables that are significant – We can probably come up with at least a few for Project 3 right now 250 Input / Output Transformations – Y1 = how long a customer must wait > We could make this a double to reflect the two different lines – Y2 = average time in the system – Y3 = utilization of the various servers – These first three are fairly obvious, and are specified in the assignment sheet – Any other suggestions? 251 Input / Output Transformations • What about f? – This is determined by the logic of the simulation – For some simulations this could be very complex > You may actually be comparing different algorithms > For large systems these could require a lot of code (ex: flight simulator, damage to space shuttle) – For Project 3 this is fairly straightforward – how do we actually implement the transformations specified in our program > Ex: algorithm for new customer proceeding to one or more lines for food, waiting in line and eventually being served 252 Comparing Generated and Measured Data How do we actually test to see if our output matches the measured data? Here is one approach outlined in Banks: • Consider some output variable Yk, taken over multiple, independent simulation runs – We calculate the average value and standard deviation of the output over those runs • We’d like to see if this “matches” our measured value for the variable – Since the simulation is stochastic, they will not match exactly, but we want to see if the simulated values are consistent with the measured values – To do this we can use the t-test 253 Comparing Generated and Measured Data – This enables us to statistically determine whether the sample data (from the simulation) matches the population (the measured data) – Specifically, the formula for the t-test is t0 > > > > Yk 0 S/ n Where Y = the mean of the simulation results k Where µ0 = the measured (hypothesized) mean Where S = the standard deviation of the simulation results Where n = the number of simulation runs used – More commonly, the t-test is used to compare two sets of (normal) data to see if they are from the same population > In this case, the formula is t0 Ya Yb Va n V Where Va and Vb are the variances of each set of data b n 254 Comparing Generated and Measured Data – The idea is that we look at the difference of the means divided by the variability of the data > Look at some normal curves to see the idea – If we assume that the mean of one group is absolutely fixed, with no variability, its variance is 0 and the formula reduces to the one shown in the text – We use this formula together with a table in a way similar to that for the Chi-Square test: > H0: The populations match > H1: The populations do not match – We determine a level of significance, α, and typically do a two-sided test, with /2 on each side > See Table A.5 on p. 583 of Banks text – If the value is too large the null hypothesis is rejected 255 Comparing Generated and Measured Data For example, assume we know = 28 We have run a simulation that generated the following 10 values: 19.60598106 25.87288018 24.25071858 19.48842905 33.80513566 35.27167241 24.78146019 25.81875958 24.62708968 22.79642059 Y_bar = 25.63 S = 5.22 t0 Y S/ n 25 . 63 28 5 . 22 / 10 1 . 436 Assume = 0.1 Thus, we look up t0.05,9 in the table and find: 1.83 Since the table is symmetric, we can use the absolute value, and |-1.436| < 1.83 so we do not reject the null hypothesis 256 Comparing Generated and Measured Data • Note that up until this point we have been concentrating on the following: P(H0 rejected | H0 is true) = – We call this a Type I error – The smaller the value of α, the lower the chance we reject the hypothesis given that it is true – Thus, if we DO reject the hypothesis, it is with a high level of confidence • However, it is perhaps more important to make sure we don’t accept a hypothesis when it is false P(failure to reject H0 | H1 is true) = β – We call this a Type II error – In other words β is the probability that we (incorrectly) fail to reject a distribution that should be rejected – (1 – β), therefore is the probability that we (correctly) reject a distribution that should be rejected 257 Comparing Generated and Measured Data – We call (1 – β) the power of the test, and clearly a high value for (1 – β) [equal to a low value for β] is desirable – Unfortunately, for a given value of , the only way to increase the power of the test (i.e. reduce Type II error) is to increase the number of points, n – See ttest.xls for more info • The existence of Type II errors is why we are cautious with our Type I error test: – “We fail to reject the null hypothesis” – This is also why we looked at the p-values in Chapter 9 when discussing distribution fits – The better the fit, the higher the p-value and the lower the chance of Type II errors 258 Brief Review • Let’s review a bit: So far we have looked at • Why we need simulation and what it is used for • Different types of simulations and simulation techniques – Ex: Discrete Event Simulation • Probability background necessary for simulation – Some definitions and important distributions • Elementary queuing theory – Sometimes used in place of simulation – Often used to complement simulation 259 Brief Review • Random number theory – The basis for stochastic processes – If we don’t have a good source of pseudo-random numbers, our simulation results will be bogus • Random variate generation – Transform U[0,1) into the distributions we really need • Input modeling – How do we determine the data to use for our simulation input? – If possible, obtain samples of real data, then “fit” it to one or more distributions to generate input – Use goodness-of-fit tests such as Chi-Square and Kolmogorov-Smirnov to check for a match 260 Brief Review • Verify and validate your simulation – Verify to make sure the model has been correctly implemented – Validate to make sure the model is itself correct – Can be done in various ways, some subjective and some objective Where does this leave us? • We need to know how to interpret the results of our simulation • What can we infer from the results and what can we not infer • We will first look at output from a single model • Then we will look at output from alternative designs 261 Output Analysis for a Single Model • Consider a “typical” computer program to solve some problem Programmer / team (say P) gets problem specifications P works to code / debug / test the program • Clearly this can take a considerable amount of time and effort Once confident that the program is correct, P runs the program and obtains the results • For simulations, however, we must be careful with obtaining and interpreting the results 262 Output Analysis for a Single Model Since most simulations are stochastic in nature, their output can vary from run to run due to random chance • The results from any single run may not be useful • Instead we typically need to analyze our results over • many runs Consider the data in ttest.xls from GrocerySimC.java – Look in column A (200 customers) 12.43207958 14.45303307 15.62356045 35.63605098 26.67449414 11.54200814 21.00290498 18.16938983 9.996530427 Note the large variation in the results Clearly, any one of them would be inappropriate as an answer 263 Output Analysis for a Single Model – The values have very great differences – None in itself would necessarily accurately represent the “correct” result – Thus, we need to combine these results in some way First some more background: • Simulations typically attempt to model one of two behaviors for a stochastic process: 1) Transient behavior – Indicated by a simulation with a specific termination event (ex: runs for X minutes, runs until C customers have been processed, runs until inventory is exhausted) 2) Steady-state behavior – Indicated by a simulation that runs over a very long period of simulated time, or with no stated stop event 264 Output Analysis for a Single Model Mathematically, we can distinguish between the two based on the distribution of the output data • Consider output Y1, Y2, … for a simulation with initial conditions I – Ex: The conditions I could represent the number of people in line outside a bank when it opens • Consider distribution Fi(y|I) = P(Yi≤y|I) for i = 1, 2, … – In words, F gives the probability distribution for each output conditional upon the initial conditions – For a stochastic process with transient behavior > Each Fi will be different for different i and for different I > In other words the output values vary from each other and are different for different initial conditions 265 Output Analysis for a Single Model – For a stochastic process with steady state behavior > Eventually a point is reached such that Fi(y|I) F(y) > Or, the output converges to a distribution that is independent of i and I • We’ll concentrate first on transient behavior – This is what you will be doing in Project 3 – We may then briefly cover steady state behavior 266 Transient Behavior Processes • Consider an output value of interest Ex: Queue length, Q Ex: Wait time in queue, W Within a single run of a simulation, the values will autocorrelate, and thus will not be independent • Ex: A delay or long service time for one customer will • tend to delay customers “near” to the delayed customer Because of this we cannot do “classical” statistical analysis on these values within a single simulation run – “Classical” analysis requires data to be IID (Independent, Identically Distributed) 267 Transient Behavior Processes • However, from one run to the next, as long as • different random number streams were used, the results should be independent In more mathematical terms, let Wij represent the wait time for customer i during run j of the simulation – Wij for j constant and i = 1, 2, … are NOT independent – Wij for i constant and j = 1, 2, … ARE independent, and can be analyzed as such – If we get the average wait for each run, we can also analyze that value across the runs • The next step is to determine how to analyze the values – We’d like to have some confidence about the accuracy / validity of our results 268 Confidence Intervals Consider result Yj for j = 1, 2, …, n that we obtain for the n runs of our simulation • Assume that the overall result for Y has a mean µ and • standard deviation σ – however, since we are just generating this data, we don’t know what they in fact are at this point If we calculate the mean of our results, Ῡ, we’d like to know if it is close to the actual mean, µ, of the output distribution – They may not match since we are doing a fixed number of iterations • A confidence interval will tell us that the actual mean is within a certain range with a certain probability – Ex: µ = 5.2 0.32 with confidence 90% > This means that there is a 90% chance that the actual mean of our output distribution is between 4.88 and 5.52 269 Confidence Intervals • Clearly, we’d like – The confidence to be as high as possible (close to 1) – The interval to be as small as possible (close to 0) • The actual confidence and interval values are dependent upon a few factors, including – The variation in the data produced by the simulation – The number of simulation runs performed (i.e. n) • So now the question is – How do we determine the confidence interval? • We will look at it from two points of view: 1) Given n runs and a confidence probability, what is our confidence interval? 2) Given a desired confidence interval, how many runs are necessary to achieve that interval? 270 Confidence Intervals Let’s first look at 1) Given n runs and a confidence probability, what is our confidence interval? • First, we determine the point estimate, Ῡ, for our distribution n – This is simply Y j j 1 n > The mean of the sample points (Equation 11.3) • We next need to determine the sample variance, S2 n Y S 2 2 j nY 2 j 1 n 1 271 Confidence Intervals • This gives us an estimate of the actual variance, σ2, of Ῡ through the following: 2 (Y ) S 2 n – Taking the square root of this gives us the standard error of the point estimator (Y ) S 2 n S = s.e.(Ῡ) n • Before we put these together, we need a bit more theory – Luckily, we have seen this before 272 Confidence Intervals • As long as our estimators are not too biased, the random variable tn Y (Y ) – Is approximately t-distributed with n-1 degrees of freedom – So what does this mean and imply? Y 1 P t / 2 , n 1 t / 2 , n 1 2 S /n P Y t P t / 2 , n 1 S / n Y t / 2 , n 1 2 S / n Y t / 2 , n 1 2 / 2 , n 1 273 S /n 2 2 S /n Confidence Intervals – In words, the formula above will give us the interval about our point estimate of the mean where the actual mean will fall with a (1 - ) probability – We look up the value in the t-distribution table and plug in the other numbers – We can also think of it visually, looking at a tdistribution curve > It is actually quite similar to a normal curve and is also symmetric > The value is divided into /2 on either extreme of the curve – The t-distribution approaches a standard normal distribution as degrees of freedom > If n is large enough we can substitute z for t 274 Confidence Intervals • Let’s look at an example using some data generated by GrocerySimC.java – We’ll try 3 different confidence intervals: > = 0.2 > = 0.1 > = 0.05 – We’ll also try 3 different numbers of runs > n = 10 > n = 100 > n = 1000 – Look at confidence.xls for more information 275 Confidence Intervals Consider now item 2) from slide 270: a situation where a specific amount of precision is required of the output • Ex: The simulation runs should be within 0.4 of the actual mean – i.e. The half-length of the confidence interval must be 0.4 with the desired confidence, • We know the variance lessens as the runs increase – This will reduce the half length of the confidence interval • What we need to know is how many runs to do in order to guarantee the desired interval 276 Confidence Intervals Consider some desired half-length interval • We’d like our sample mean to be within of the actual mean, with a stated probability, (1 - ), or P Y 1 • We will do this in a multistep process: • Choose an n0 as an initial number of runs to try. From n0 we calculate an S0 (initial variance) – We know that the precision increases (and halflength decreases) with increased n – Looking more specifically, the last line of slide 273 shows that the half-length decreases proportionally to the square root of n (if S remains the same) 277 Confidence Intervals • From our previous derivations, we know that the half-length of a confidence interval is – half-length = t / 2 , n 1 S 0 n – for confidence probability 1 - • Which we would like to solve for n such that t / 2 , n 1 S 0 n 2 – However, since we don’t know n yet, we cannot yet calculate t/2, n-1 278 Confidence Intervals – Therefore we need to substitute another similar value > We know that the t-distribution approaches the standard normal distribution as n > If we substitute z/2 we can get a ballpark value for R • This leads us to z / 2 S 0 n 2 • However, this may not be exactly correct, since we • substituted z for t in our formula Since tn > z the value we calculate for n may be a bit small – We can then do an iterative process of updating n and testing it until we get the desired precision – See confidence.xls 279 Quantiles and Percentiles • Confidence intervals can give us good • • estimates of average values However, sometimes we are more interested in the probability that the result for a given run will be <= some value Looking at it a different way, consider a probability p and a value θ such that Pr(Y <= θ) = p We say the value θ is the pth quantile (as a fraction) or percentile (as a percentage) of Y 280 Quantiles and Percentiles These can be useful in some results • Ex: I simulate a number of (independent) days in my grocery store, keeping track of the average time in the system for each – I'd like to determine a time, θ, such that the average time in the system is <= θ on 80% of the days • We typically determine these in the following general way: Determine the desired probability Find the value that satisfies that probability 281 Quantiles and Percentiles • More specifically: Calculate the point estimate, by taking the appropriate proportion of the sample points • Ex: If we have 1000 sample points and we want the 80th percentile, we estimate the 80th percentile as the 0.8(1000)th point (sorted) We then calculate the (1 - ) confidence interval using formulas 11.20 in the text p l p z / 2 p (1 p ) n 1 p u p z / 2 282 p (1 p ) n 1 Quantiles and Percentiles These give us the range of probabilities that we then convert to quantiles • Ex: The lower bound, θl, is the pl(1000)th point in our sample (sorted) and the upper bound, θu, is the pu(1000)th point in our sample (sorted) Ex: Thus the 80th percentile of our sample will be between θl and θu with probability (1 - ) This gets a bit confusing because we are talking about probabilities at two different levels • The quantile itself gives a probability • The confidence interval also gives a probability See quantile.xls 283 Steady State Analysis • If our simulation is evaluating steady state • behavior, we use somewhat different evaluation techniques Recall that for a given measure, if we are determining a long run value, then lim n 1 n Y n i i 1 • For some variable of interest Y, where n is the • number of observations of Y Where θ is independent of any initial conditions of the simulation 284 Steady State Analysis However, it may not be obvious how large n has to be until we approach the long run value Initial conditions CAN affect the value of n needed to get to a steady state • These introduce a bias into the data that can take a lot of time to dissipate The text outlines some ways to reduce this initial bias: 1) Set the initial conditions of the simulation in an intelligent way, as close to the steady state values as possible 285 Steady State Analysis • These conditions could be obtained through • observation of the real system They could also be approximated using analytical techniques (Markov analysis, as discussed in Chapter 6) – Determine long run values and use these as the initial conditions for our simulation – We likely cannot model the exact system but we can get something close enough to reduce the initial bias 2) Run the simulation for a while without tabulating any data • By the time data begins to be collected, the conditions should be close to the steady state 286 Steady State Analysis But how can we determine how long to wait before tabulating data? In other words, if we divide our run into two segments, T0 and TE, how long should we make T0 and TE be? • This is most likely done via experimentation • The text discusses how to break up runs into batches and calculate ensemble averages in order to accomplish this Idea: • Let's say we want to do R runs of our simulation • Let's say each run will be for time T 287 Steady State Analysis • Break up T into k subdivisions (batches) • Given variable of interest, Y, calculate the average value of Y for each batch across all runs to get the ensemble averages Y. j 1 R Y R rj r 1 • Examine the ensemble averages and the cumulative • average Also consider cumulative average when some of the initial batches are not considered – In other words, we start tabulating the data with batch m, with m >= 0 288 Steady State Analysis • We can plot ensemble averages over time and see if • • there is a trend If necessary we can decide to eliminate some of the initial batches if they show too much bias Let's look at how this could be done in yet another version of the GrocerySim program – See GrocerySimD.java 289 Steady State Analysis Once we have done our runs we still need to analyze the data Text discusses various techniques for error estimation and how number of replications and sample size relate to this Due to time limitations we will not go into these details • Just read over the sections but only for the "big picture" 290 Comparing Two Alternative Designs Assuming we have the same output variable of interest for both designs, we typically would like to see how they compare • But how do we know if differences between them • are statistically significant? A common approach is to look at the difference of the values – We can form a confidence interval of this difference, similar to what we did for a single value – If the confidence interval of the difference indicates with good confidence a large enough value, we can determine that the output variables from the two distributions differ statistically > We need to judge separately whether the difference is practically significant 291 Comparing Two Alternative Designs – Ex: -4.5 ≤ Ῡ1 - Ῡ2 ≤ 3.6 with 95% confidence > In this case, since the interval contains 0 and the magnitude on each side does not differ greatly, we cannot make any strong statement about the two designs > Also, the interval is fairly large, indicating a lack of precision in our analysis – Ex: 3.0 ≤ Ῡ1 - Ῡ2 ≤ 4.3 with 95% confidence > In this case, it is clear that Ῡ1 is expected to be greater than Ῡ2 with a high degree of confidence > Also, the interval is smaller, indicating better precision in our analysis > Overall result of which design is better depends on the parameter, and on how significant the output data is to the quality of the system 292 Comparing Two Alternative Designs • How do we obtain each of the Ῡi and then how can • we compare them effectively? Two ways of obtaining the Ῡi – Independent Sampling > Separate random data streams are used for the different simulation variations > We may or may not have the same number of runs for each Ῡi – Correlated Sampling > Identical random data streams are used for the different simulation variations > i.e. the input data is identical for both versions (not just the same distribution, but, rather the same exact data) > Will always have the same number of runs in this case 293 Comparing Two Alternative Designs – Correlated sampling tends to reduce variance in the difference of the results, thereby giving better confidence intervals for the same number of runs > However, it is not always possible or easy to do • For each situation, we will be comparing result Y1 and Y2 Independent Sampling with Equal Variances • Also called two-sample-t approach • Can actually be used in two situations: – Variances for Y1 and Y2 are the same and the number of runs of each may or may not be the same – The number of runs for Y1 and Y2 are the same and the variances may or may not be the same 294 Comparing Two Alternative Designs First recall what it is that we want to calculate: • The confidence interval of the difference of the two results, or (Ῡ.1 - Ῡ.2) ± t/2,v [s.e.(Ῡ.1 - Ῡ.2)] with prob. (1-) – To calculate this we need to determine the standard error for the difference of the two means, and the degrees of freedom, v • Consider Ῡ.1 - Ῡ.2 and, more specifically the variance • of Ῡ.1 - Ῡ.2 Since the runs are independent Var(Ῡ.1 - Ῡ.2) = Var(Ῡ.1) + Var(Ῡ.2) = σ12/R1 + σ22/R2 (See Slide 271) 295 Comparing Two Alternative Designs • The standard deviation of the difference will be the square root of this sum – And the standard error is the sample approximation of the standard deviation • So we calculate the sample variance S2 for each • version (recall that Si2 is an approximation of σi2) We then pool these variances together ( R1 1) S 1 ( R 2 1) S 2 2 Sp 2 2 R1 R 2 2 • Idea is that we are getting the weighted average of the two sample variances (which should both be estimates of the same population variance) – Multiply each sample variance by its degrees of freedom and divide by the sum of d.o.f. for both 296 Comparing Two Alternative Designs – Note that R1 may equal R2 if we decided to use the "same number of runs" (in which case the sample variances may not necessarily estimate the same population variance) – Now we substitute Sp for both σ12 and σ22 in the formula for the variance: Var(Ῡ.1 - Ῡ.2) = Var(Ῡ.1) + Var(Ῡ.2) = σ12/R1 + σ22/R2 ≈ Sp2/R1 + Sp2/R2 – And the standard error (i.e. estimate of the standard deviation) is 1 1 s.e.(Ῡ.1 - Ῡ.2) = Sp R1 R 2 > with v = degrees of freedom = R1+ R2 - 2 297 Comparing Two Alternative Designs • We then plug this into the formula on Slide 295 to determine the confidence interval Note: Recall that we can use this technique if either the variances or the number of runs (or both) are the same • In practice it is difficult to know for sure if the • variances for both versions are the same, since the true variances are often unknown Thus if the number of runs for the two versions is not the same, it is probably not a good idea to use this technique – Rather use the one we will discuss next 298 Comparing Two Alternative Designs Independent Sampling with Unequal Variances • Also called modified two-sample-t approach • Used when variances for Y1 and Y2 are not equal AND R1 != R2 (number of runs do not match) • In this case we use the same formula for the confidence interval as in the previous approach (see Slide 295 – We calculate the estimated variances for each sequence of values, just as in the regular two-samplet approach – However, now we will use different formulas for the standard error (s.e.) and the degrees of freedom, v 299 Comparing Two Alternative Designs • Because the true variances differ, we cannot pool the sample variances, and must calculate the standard error using the following equation s .e .( Y1 Y 2 ) var( Y1 ) var( Y 2 ) S1 2 R1 S2 R2 • The degrees of freedom is then estimated by ( S 1 / R1 S 2 / R 2 ) 2 v 2 2 [( S 1 / R1 ) /( R1 1)] [( S 2 / R 2 ) /( R 2 1)] 2 2 2 2 – This technique is actually very old, developed by Welch in 1938 and discussed [Scheffe, 1970] 300 2 Comparing Two Alternative Designs Correlated Sampling • Also called paired-t approach • In this case, both versions use the identical random data for the identical number of runs • Thus, the output from run Xi for each version is not independent, but, rather it is correlated • This enables us to process the difference of the versions Yr1 – Yr2 as a new, separate random variable for each run r • We can then process this difference result in the same manner that we processed our single result data in Chapter 10 301 Comparing Two Alternative Designs • The advantage in this approach is that it reduces the • variance (which, in turn reduces the confidence interval) This is because Var(Ῡ.1 - Ῡ.2) = Var(Ῡ.1) + Var(Ῡ.2) – 2cov(Ῡ.1, Ῡ.2) = σ12/R + σ22/R – (2ρ12σ1σ2/R) – where ρ12 is the correlation between Ῡ.1 and Ῡ.2 – So a positive correlation will make the variance smaller • See compare.xls for all three versions 302 Conclusions If identical random data and identical runs can be used for both design versions, it should be • A positive correlation between the versions reduces • the variance, thereby improving the confidence interval Note that caution must be taken to ensure that the data used is actually identical – So for each data stream, value Xi in version 1 must be used for the same purpose in version 2 • Also, it is a good idea to use separate random streams for different input variables – See more guidelines on pp. 439-441 in text 303 Conclusions • However, identical data streams do not always guarantee a positive correlation – We also need monotonicity with regard to the output related to the input – An obvious example of this is the relationship between service times and total time in system > Clearly, as the service time increases, assuming no other changes, the total time in system should increase for any version of the system > This is monotonic – A variable that does not have this monotonic behavior may not give a positive correlation between systems, even with identical data > Ex: As input X increases, Y1 increases but Y2 decreases 304 Conclusions In some circumstances, using identical data may not be possible • One version may have different amounts of input • • from the other, or different input parameters In these cases, we need to use independent sampling, as discussed previously If possible, we’d like the number of runs to be the same Generally speaking, we prefer to get results that will have the least variance, thereby giving the best confidence intervals • However, we can always improve our confidence intervals for any system by increasing the runs 305 Comparing Multiple Designs Sometimes we must consider more than two alternatives, say C How do we apply statistical comparisons in these cases? • One approach is to compare all C versions against a “standard” – The standard might be a goal or it might be the value in a real system – Requires C confidence intervals and we can choose the version that is closest to the standard – But we may not have a “standard” to compare against 306 Comparing Multiple Designs • Another approach is to use the confidence interval technique previously described pair-wise over all alternatives – This would required C(C-1)/2 comparisons (think edges in a completely connected graph) – We then must interpret these results to determine the best alternative – this depends on the goals of the simulation 307 Bonferroni Approach Bonferroni Approach to Multiple Comparisons • Used to evaluate several confidence intervals together • Assume that for confidence interval i, the probability that the goal parameter is contained within the interval is (1 – i) • Assume now that we want the probability that all goal parameters will be contained within the confidence intervals • Bonferroni sets a lower bound on this probability – Or an upper bound on the probability of a false conclusion C P ( all S i true ) 1 i 1 to C j 1 308 j 1E Bonferroni Approach • This is not a tight bound, since the probability of • error may in fact be lower It can be used to 1) Compare / evaluate multiple parameters within the same system – What is the probability that all parameters' actual values fall within their confidence intervals 2) Compare multiple different systems to a standard system or control – What is the probability that the difference between each system and the standard is within its c.i. 3) Compare / evaluate multiple different systems to each other – What is the probability that each pairwise difference is within its c.i. 309 Evaluating Multiple Parameters • Recall that in our prev. analysis we looked at a single output variable of interest, Y We determined a confidence interval with a probability (1 - ) that the true value of the variable falls within the interval • Often in a simulation we want to evaluate several output variables (say K) Ex: Average queue length, average wait time, average number in the system 310 Evaluating Multiple Parameters If we consider each of these individually, we can use the techniques discussed previously (see Chapter 11) to come up with the However, if we consider them all together, we must evaluate the probability that ALL true values fall within the calculated confidence intervals Using the Bonferroni Inequality we can determine that K P ( all Y i within c .i .) 1 j 1 311 j 1E Evaluating Multiple Parameters This means that each individual probability must be higher so that the total will be acceptable • Ex: If we want the probability that all K true values • are within the confidence intervals with a probability (1 - ), each individual probability must be (1 - /K) Ex: If we have 5 variables of interest and = 0.1 then each i must be 0.1/5 = 0.02 – The implication of this is that, with a fixed number of runs, this will cause the c.i.s to be very large (the more variables, the larger the c.i.s will be) – Alternatively, if we can vary the runs, we will need more runs to obtain the given probability 312 Comparing Different Systems to a Control • In this case we are looking at differences, • as we discussed prev. in Chapter 12 We want to know if any/all of the (say K) systems differ (statistically) from the control, say S Determine c.i. for difference Yi – S with probability (1 - /K) If a given c.i. does not include 0 we can say that the value differs from the control Again, we must make the indiv. probs. higher to achieve the correct overall prob. 313 Comparing Multiple Systems Pairwise • As mentioned previously, in this case we have K(K-1)/2 pairs for K different systems We look for differences pairwise, just as in the previous technique However, now we have a lot more pairs to consider and therefore the individual probabilities must be a lot higher • This can present a problem for even a moderate K • Ex: = 0.1, K = 5 – Each i must be (0.1)/[(5)(4)/2] = 0.01 • Again, this will create either very large intervals or require a large number of runs 314 Selecting the "Best" System Bonferroni also has an approach for • selecting the best out of multiple system designs using a two stage procedure • selecting a subset of the systems in which the best is likely to occur • See Section 12.2.2 315 Other Evaluation Issues (Briefly) • Metamodeling Often an output parameter is affected by several independent variables In these cases, sometimes it is desirable to know the relationship between the independent variables and the output variable • Programmer may know this in advance but often it is not known This is often not trivial to determine • Linear Regression, Multiple Linear Regression and other tools are used to help with this 316 Other Evaluation Issues (Briefly) • Optimization via Simulation In some cases many different system versions are possible • If an input parameter or design decision is varied slightly, the overall system can produce different results There may be so many different versions, each with different results, that we cannot compare them each “by hand” • We must thus automate the process and have a program analyze the results and attempt to determine the optimal system 317 Other Evaluation Issues (Briefly) • This is itself a difficult process and can often be quite complex to implement – We need to determine how to “evaluate” each system using the output parameters – We need to mechanically try all variations and determine how to compare them to each other – Since outputs have confidence intervals and may be estimates, there is a lot of inherent uncertainty in the process > We need to choose evaluation that is least susceptible to this uncertainty • Heuristics to do this process attempt to “converge” to the overall best solution – Ex: Genetic Algorithms with selection, crossover and mutation 318

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