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
 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
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
> 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
– 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
– 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
– 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
• 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
• 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?
>
>
>
>
add is easy but remove is hard – why? – discuss
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
> 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
> 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
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
• 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
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
– 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
s2
  q
s2
q
s3
   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
st
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) 

i0
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 
xa
if a  x  b
ba
ba
– 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
ba

E(X ) 
dx 
( b  a )( b  a )
2 (b  a )
b
x



b
a
x
2
ba
b a
2
a 2 (b  a )

2
2 (b  a )

ba
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
x0
– 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 ,
x0
> 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
x0
Gamma Distribution
E ( x) 
1
V (X ) 


1 
F (x)  
0 ,



x

( )
1

(  t )
2
 1
e
  t
dt ,
x0
x0
– 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
i0
x0
2
1
k
 
(k x )
i!
95

1

1
(k )
2
i
,
x0

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!
i0
 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
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) 

i0

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
– 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
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

ba
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
i0
– 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
i0

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
• 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
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?
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
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
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
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
• 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
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
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   )
• 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
i0
 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
The text has formulas for a number of different
queue possibilities:
• 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
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
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
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
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
• 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


0
1

…
2

k-1


k

k+1

From this we can obtain
k 1
Pk  P0 
i0

 
 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

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
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
A lot of theory has been left out of this process
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
– http://www.fourmilab.ch/hotbits/
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
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
– 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
• 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
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
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
, x0

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

 
  nm
 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
• 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
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
– 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
• 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
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)
– 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
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
 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
• 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
 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
• 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
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
• 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
• 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
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
 1E
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
 1E
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