Week 7: Arena notes Week 8 notes:
Arena view: ● Some Lousy Generators:
Discrete Probability: DISC(CumulativeP, V) CONT(P,V), POIS(λ), EXPO(λ), NORM(µ; σ), TRIA(a,b,c), UNIF(a,b), WEIB(β; α) 1-random devices like a coin toss (0 and 1 values randomly). Roll a die to generate random numbers between 1 and 6.
Advanced Process Panel: Blocks(Hold, seize, search, delay, release, store, unstore, remove, match, pickup, gather, More sophisticated random devices include Geiger counters and atomic clocks. Naturally, random devices have strong
dropoff, assign attribute, adjust variable, readWrite, signal, search and remove) spreadsheets like(Advanced Set, randomness properties; however, we cannot reproduce their results easily.
Expression, Failure, file, StateSet, Statistic, Storage) 2-Random Number Tables: Published by the RAND corporation, contains one million random digits and one hundred
Basic Process panel: create, dispose, process, decide, batch, clone, separate, assign, record, label, go to label, thousand normal random variates.
adjustable batch. Spreadsheets: Attribute, entity, queue, resource, variable, schedule, set, picture 3-The mid-square method was created by the famous mathematician John von Neumann. The main idea here is we take
an integer, square it, and then use the middle part of that integer as our next random integer, repeating the process as
Advanced Transfer panel: ENTER, LEAVE, PICKSTATION, ROUTE, STATION, ACCESS, CONVEY, EXIT, START, STOP,
many times as we need. To generate the Uniform (0,1) random variable, we would divide each generated integer by the
ACTIVATE, ALLOCATE, FREE, HALT, MOVE, REQUEST, TRANSPORT. SPREADSHEETS (STATION DATA, SEQUENCE,
appropriate power of ten.
CONVEYOR, SEGMENT, TRANSPORTER, DISTANCE, NETWORK, NETWORK LINK, ACTIVITY AREA)
4-Fibonacci Additive Congruential Generators: 𝑋𝑖 = (𝑋𝑖−1 + 𝑋𝑖−2 )𝑚𝑜𝑑 𝑚, 𝑖 = 1,2 … & 𝑅𝑖 = 𝑋𝑖 /𝑚
Trick 1: Pre-assign the service times as an attribute (Sealer Time) in an Assign module immediately after each
● Linear Congruential Generator
customer arrives.
1-LCGs: 𝑋𝑖 = (𝑎𝑋𝑖−1 + 𝑐)𝑚𝑜𝑑 𝑚, 𝑤ℎ𝑒𝑟𝑒 𝑋0 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑒𝑑 𝑎𝑛𝑑 𝑅𝑖 = 𝑋𝑖 /𝑚
Trick 2: While we’re at it, use the Assign to store each customer’s arrival time as an attribute. Use the Arena variable
Example: 𝑋𝑖 = (5𝑋𝑖−1 + 3)𝑚𝑜𝑑 8. 𝑋8 = 𝑋0 = 0. This generator is a full-period generator since it has a cycle length
TNOW to do so.
equal to me
- Sometimes you need to do things that are too complicated for the Process module version of Seize–Delay–Release
Now you can handle things like… 1) Seize–Assign–Delay–Release 2) Non-symmetric multiple Seize’s and Release’s 3)
2- “Desert island” generator: 𝑋𝑖 = 16807𝑋𝑖−1 𝑚𝑜𝑑 231 − 1. This generator has a full-period cycle length > 2 billion
Complicated Seize’s and Release’s that might depend on sets of servers
Let's look at the algorithm for generating each 𝑋𝑖 and 𝑅𝑖 :
Resource spreadsheet in the Basic Process template ➔ Click on Failures column ➔ Add a Failure Name.
𝐾 ← ⌊𝑋𝑖−1 /127773⌋
Go to the Failure spreadsheet in the Advanced Process template, where you’ll see your new failure name. Choose
𝑋𝑖 ← 16807(𝑋𝑖−1 − 127773𝐾) − 2836𝐾
type of failure: Count (failure after a certain # of arrivals) Time (after a certain amount of time) Choose downtime (for
repair): Can be any expression. if 𝑋𝑖 < 0, then set 𝑋𝑖 ← 𝑋𝑖 + 2147483647
-Certain primitive blocks such as the Queue block can’t even connect to a Seize module from the Advanced Process 𝑅𝑖 ← 𝑋𝑖 ∗ 4.656612875E − 10
template or a Process module from the Basic Process template
-Using the Alter block to change the number of resources 3-RANDU Generator (bad) 𝑋𝑖 = 65539𝑋𝑖−1 𝑚𝑜𝑑 231 The random numbers generated by this generator fall entirely on
15 hyperplanes (not iid)
Advanced Transfer Panel Modules ● Various Seize Selection Rules: Cyclical, Random, Preferred Order,
related to movement: Station, Route, Specific Member, Largest Remaining Capacity, Smallest Number ● Tausworthe Generator: 𝐵𝑖 = (𝐵𝑖−𝑟 + 𝐵𝑖−𝑞 )𝑚𝑜𝑑 2 = 𝐵𝑖−𝑟 𝑋𝑂𝑅 𝐵𝑖−𝑞 𝑤ℎ𝑒𝑟𝑒 < 𝑟 < 𝑞. The sequence has a period of
Enter, Leave. Sets and Advanced Sets Busy. 2𝑞 − 1
(e.g., sets of Sequences). ENTER – LEAVE How do we get Unif (0,1) random variables from the 𝐵𝑖 ′s? We can take a sequence of 𝑙 bits and divide them by 2𝑙 to
modules for Station-to-Station ● Station: Somewhere you can go to Route: Tells you where you’re compute a real number between zero and one.
movement. going. Similar to Connect, except it takes time. Enter: General way ● Combined generators:
Resource (e.g., an assistant) to get into a Station. Leave: General way to depart a Station 1- L’Ecuyer’s generator of cycle length 2191 :
-Seize-Release 2- Mersenne Twister: Matsumoto and Nishimura have developed the "Mersenne Twister" generator, which has a period
Transporter (e.g., a car) ● A sequence goes from station to station. Maybe have different of 219937 − 1. This period is beyond sufficient for any modern application; we will often need several billion PRNs, but
-Request-Free service time distribution at each station (depending on customer never more than even 2100 . The period only describes how long it takes for the generator to cycle, but doesn’t have an
-Requires Distance Set type and place in visitation sequence effect on generating the PRNs or the quality of the PRNs.
Conveyor ● Choosing a Good Generator - Theory
- Access-Exit ● Use the Set spreadsheet in the Basic Process template to define
-Requires Segment Set sets. A “vanilla” resource has identical, interchangeable servers. But ● Statistical tests for randomness:
a resource set can have distinct servers, with different schedules, 1- Goodness of fit test - Chi-squared: Are the PRNs approximately Unif (0,1)?? When we design the test, we set the
service speeds, service specialties, etc level of significance α = P (Reject H_0 | H_0 true). Typically, α = 0.05 or 0.1, and is the probability of Type I error.
● Have to be a little careful with Seize-Delay-Release for a resource We can also specify the probability of Type II error, β = P (Accept H_0 | H_0 false). We regard Ho as the status quo -
set. The problem is that you have to make sure that you Release the "that which we currently believe to be true". In this case, Ho refers to the belief that the numbers we are currently
same guy that you originally Seize’d. (If you release a random generating are, in fact, i.i.d Unif (0,1).
(𝑂𝑖 −𝐸𝑖 )2 𝑛
server, some other customer may lose his server!) 𝜒02 = ∑𝑘𝑖=1 where 𝑂𝑖 ∼ Bin(𝑛, 1/𝑘), 𝑖 = 1,2, . . . , 𝑘. And 𝐸𝑖 = 𝐸[𝑂𝑖 ] = 𝑘 , 𝑖 = 1,2, . . . , 𝑘. (1 − 𝛼) quantiles for 𝜒 2
𝐸𝑖
2
distributions with varying degrees of freedom, 𝑛, where we define a quantile, 𝜒𝛼,𝑛 as: 𝑃(𝜒𝑛2 < 𝜒𝛼,𝑛
2
)=1−𝛼
Questions: 2
We reject the null hypothesis 𝐻𝑜 if 𝜒02 > 𝜒𝛼,𝑘−1 2
, and we fail to reject 𝐻0 if 𝜒02 ≤ 𝜒𝛼,𝑘−1 . 𝐸𝑖 = 𝑛/𝑘 should be at least
True: In Arena, a DECIDE module can route customers probabilistically or conditionally to multiple locations. 2
five, and 𝑛 should be at least 30. When k is so large, we cant find tables with values for 𝜒𝛼,𝑘−1 , then we can use the
False: In Arena, a primitive QUEUE block can connect with a PROCESS module approximation below:
True: In Arena, you can pre-assign a service time as an attribute before you actually get to the server that you’ll be 3
2 2 2
using. 𝜒𝛼,𝑘−1 ≈ (𝑘 − 1) [1 − 9(𝑘−1) + 𝑧𝑎 √9(𝑘−1)]
True: In Arena, it’s perfectly OK to have a CREATE module generating one-at-a-time customer arrivals, while also
2- Runs tests for independence:
having a different CREATE module with4-at-a-time customer arrivals elsewhere in the model.
● Runs up and down
True: In Arena, \fake" customers can be used to schedule machine breakdowns, keep track of which time period the 2𝑛−1 16𝑛−29
simulation is currently in, and carry out other duties not associated with actual customers 𝐴 ≈ Nor ( , ) where A denote the total number of ups and downs out of n observations. We reject
3 90
𝐴−𝐸[𝐴]
True: In Arena, the SEIZE, DELAY, RELEASE functionalities are in multiple templates. 𝐻𝑜 𝑖𝑓 𝐴 𝑖𝑠 𝑡𝑜𝑜 𝑏𝑖𝑔 𝑜𝑟 𝑠𝑚𝑎𝑙𝑙. 𝑡ℎ𝑒 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑖𝑠 𝑍0 = and we reject 𝐻𝑜 if |𝑍0 | > 𝑧𝛼/2 here we look for 1-
√Var(𝐴)
True: In Arena, you can schedule failures to occur after a random number of customers have used a resource or after
(alpha/2) in the std normal table
a random amount of time has passed. 2𝑛1 𝑛2 1 2𝑛1 𝑛2(2𝑛1 𝑛2 −𝑛)
False: In Arena, you are only allowed to schedule one type of failure for a particular resource. ● Runs above and below the mean: 𝐵 ≈ Nor ( +2, ) that 𝑛1 refers to the number of observations
𝑛 𝑛2(𝑛−1)
True: In Arena, it is possible for the resource Joey to be a member of three different resource sets. 𝐵−𝐸[𝐵]
greater than or equal to the mean and 𝑛2 = 𝑛 − 𝑛1 . 𝑍0 = where B is the # runs. we reject 𝐻0 if the absolute
√Var(𝐵)
True: For easy display and organizational purposes, our Arena Call Center simulation is divided into several sub
models. value of 𝑍0 is greater than the 𝛼/2 standard normal quantile
True: The customer arrival pattern for our Arena Call Center simulation changes over the day and is therefore
modeled as a nonhomogeneous Poisson process. ● Autocorrelation test (bonus): We define the lag-1 correlation of the 𝑅𝑖 's by 𝜌 ≡ Corr(𝑅𝑖 , 𝑅𝑖+1 ). In other words, the
False: All of the Tech Support servers for our Arena Call Center simulation are equally skilled and have the same lag-1 correlation measures the correlation between one PRN and its immediate successor. Ideally, if the PRNs are
12
concurrent work schedules. ^ ≡ ( ∑𝑛−1
uncorrelated, 𝜌 should be zero. A good estimator for 𝜌 is given by: 𝜌 𝑘=1 𝑅𝑘 𝑅1+𝑘 ) − 3
𝑛−1
-In Arena, what kind of module would you use to change an entity’s picture? ASSIGN In particular, if 𝑛 is large, and 𝐻0 is true: 𝜌
13𝑛−19
^ ≈ Nor (0, (𝑛−1)2 ) Let's transform 𝜌
^ into a standard normal random variable,
-In Arena, where would you find the Expression spreadsheet? Advanced Process panel. 𝜌
^ −𝐸[𝜌
^] 𝜌
^
-Consider the Arena Inventory model that we did in class as a demo. Did this model allow for backlogs? YES 𝑍0 , which we accomplish with the following manipulation: 𝑍0 = = We reject 𝐻0 if the absolute value of
^)
√Var(𝜌 ^)
√Var(𝜌
𝑍0 is greater than the 𝛼/2 standard normal quantile:
● The name of the automatically supplied ARENA attribute that stores a particular entity's sequence is Entity. Week 9 notes:
Sequence ● Inverse transform method: The inverse transform method states that, if 𝑋 is a continuous random variable with cdf
● Where can you set the buffer size of a certain queue? In a Queue block in the Blocks template 𝐹(𝑥), then 𝐹(𝑋) ∼ 𝒰(0,1). In other words, if we plug a random variable, from any distribution, into its own cdf, we get
● There are some great PRN generators out there with incredible cycle lengths 2191 𝑎𝑛𝑑 𝑒𝑣𝑒𝑛 219937 a Unif(0,1) random variable.
● Attributes are properties for individual customers; variables are properties of the entire system. ● Proof: Let 𝑌 = 𝐹(𝑋). Since 𝑌 is a random variable, it has a cdf, which we can denote 𝐺(𝑦). By definition: 𝐺(𝑦) =
● In arena, NORM(a,b) has a variance of b^2 𝑃(𝑌 ≤ 𝑦)
Since 𝑌 = 𝐹(𝑋): ➔ 𝐺(𝑦) = 𝑃(𝐹(𝑋) ≤ 𝑦)
Since 𝑋 is a continuous random variable, its cdf is continuous. Therefore, we can apply the inverse, 𝐹 −1 , to both sides of
On average, A-r algo requires Geom(1/c) trials with expected value = c the inequality: 𝐺(𝑦) = 𝑃(𝐹 −1 (𝐹(𝑋)) ≤ 𝐹 −1 (𝑦))
Polynomial example: Let’s generate a random variable with pdf 𝑓(𝑥) = 60𝑥 3 (1 − 𝑥)2 , 0 ≤ 𝑥 ≤ 1. the maximum of What is 𝐹 −1 (𝐹(𝑋))? Simply, 𝑋: 𝐺(𝑦) = 𝑃(𝑋 ≤ 𝐹 −1 (𝑦))
𝑓(𝑥) occurs at 𝑥 = 0.6: 𝑓(0.6) = 2.0736. With this knowledge, we can generate a basic majorizer, 𝑡(𝑥) = 2.0736. Notice that we have an expression of the form 𝑃(𝑋 ≤ 𝑥), where 𝑥 = 𝐹 −1 (𝑦). We know, by definition, 𝐹(𝑥) = 𝑃(𝑋 ≤
We know that 𝑐 equals the integral of 𝑡(𝑥) from zero to one, so, in this case, 𝑐 = 2.7036. All this to say, 𝑡(𝑥) is a 𝑥), so: 𝐺(𝑦) = 𝐹(𝐹 −1 (𝑦)) = 𝑦
relatively inefficient majorizer. In summary, the cdf of 𝑌 is 𝐺(𝑦) = 𝑦. If we take the derivative of the cdf to get the pdf, we see that 𝑔(𝑦) = 1 Let’s
Like we said, ℎ(𝑥) = 𝑡(𝑥)/𝑐. In this case, ℎ(𝑥) = 1, since 𝑡(𝑥) = 2.7036 = 𝑐. This result means that 𝑌 is a Unif(0,1) remember the pdf for a uniform random variable:
𝑓(𝑥) 60𝑥 3 (1−𝑥)2 1
random variable, since it’s pdf is one. Finally, let’s compute 𝑔(𝑥): 𝑔(𝑥) = = 𝑥 ∈ [𝑎, 𝑏]
𝑡(𝑥) 2.0736 𝑓(𝑥) = {𝑏−𝑎
Let’s look at a simple example. Let’s draw two uniforms, 𝑈 = 0.13 and 𝑌 = 0.25. If we plug and chug, we see that 0 otherwise
𝑔(𝑌) ≈ 0.25. Therefore, 𝑈 ≤ 𝑔(𝑌) and we take 𝑋 ← 0.25 If 𝑎 = 0, 𝑏 = 1, then 𝑓(𝑥) = 1 = 𝑔(𝑦). Therefore, 𝑌 ∼ 𝒰(0,1).
2 −𝑥2 2𝑒
● How Do We Use This Result: Let 𝑈 ∼ 𝒰(0,1). Since 𝐹(𝑋) = 𝑈, then 𝐹 −1 (𝑈) = 𝑋. In other words, if we set 𝐹(𝑋) =
Half-normal example: with the pdf : 𝑓(𝑥) = 𝑒 2 , 𝑥 ≥ 0 . We can use 𝑡(𝑥) = √ 𝜋 𝑒 −𝑥 ≥ 𝑓(𝑥) & 𝑐 = 𝑈, and then solve for 𝑋, we get a random variable from the same distribution as 𝑋, but in terms of 𝑈.
√2𝜋
∞ 2𝑒
∫0 𝑡 (𝑥)𝑑𝑥 = √ 𝜋 ● Inverse Transform Method - Continuous Examples:
𝑥−𝑎
Now, let’s compute ℎ(𝑥): ℎ(𝑥) =
𝑡(𝑥)
= 𝑒 −𝑥 easy exp(1) pdf and 𝑔(𝑥) =
𝑓(𝑥)
= 𝑒 −(𝑥−1)
2 /2
. To generate a half normal, 1- Uniform Example: Consider the 𝒰(𝑎, 𝑏) distribution, which has the following cdf: 𝐹(𝑥) = 𝑏−𝑎 , 𝑎 ≤ 𝑥 ≤ 𝑏. Let's set
𝑐 𝑡(𝑥)
𝐹(𝑋) = 𝑈 and solve for 𝑋:
we simply generate 𝑈 ∼ 𝒰(0,1) and 𝑌 ∼ 𝐸𝑥𝑝(1) and accept 𝑌 if 𝑈 ≤ 𝑔(𝑌). We can use the half-normal result to 𝑋−𝑎
generate a Nor(0,1) random variable. We simply have to “flip back” half of the 𝑋 values over the 𝑦-axis. Given 𝑈 ∼ 𝑈 = 𝑏−𝑎
𝒰(0,1) and 𝑋 from the half-normal distribution, we can see that: (𝑏 − 𝑎)𝑈 =𝑋−𝑎
−𝑋 𝑈 ≤ 1/2 𝑎 + (𝑏 − 𝑎)𝑈 =𝑋
𝑍={ ∼ 𝑁𝑜𝑟(0,1) As always, we can generate a Nor(𝜇, 𝜎 2 ) by applying the transformation 𝜇 + 𝜎𝑍.
𝑋This𝑈>study
1/2 source was downloaded by 100000899606070 from CourseHero.com on 02-22-2026 22:26:36
Intuitively, this GMT
result makes -06:00
perfect sense. If we take a Unif (0,1) random variable, and multiply it by 𝑏 − 𝑎, we end up
𝜆𝑛
Poisson example: The Poisson distribution has the following pmf: 𝑃(𝑋 = 𝑛) = 𝑒 −𝜆 𝑛! , 𝑛 = 0,1, . .. with a Unif (0, 𝑏 − 𝑎) random variable. Then, if we add 𝑎 to each end, we get a Unif (𝑎, 𝑏) random variable.
𝑛
𝑥 = 𝑛 ⟺ 𝑠𝑒𝑒 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑛 𝑃𝑜𝑖𝑠(𝜆)𝑎𝑟𝑟𝑖𝑣𝑎𝑙𝑠 𝑏𝑦 𝑡 = 1. 𝑋 = 𝑛 ⇔ ∑𝑖=1 𝐴𝑖 ≤ 1 < ∑𝑖=1 𝐴𝑖 𝑛+1 2- Exponential example: Consider the Exp(𝜆) distribution, which has the following cdf: 𝐹(𝑥) = 1 − 𝑒 −𝜆𝑥 , 𝑥 ≥ 0 Let’s
set 𝐹(𝑋) = 𝑈 and solve for 𝑋:
https://www.coursehero.com/file/81699636/sim-cheatsheet-part1pdf/
Arena view: ● Some Lousy Generators:
Discrete Probability: DISC(CumulativeP, V) CONT(P,V), POIS(λ), EXPO(λ), NORM(µ; σ), TRIA(a,b,c), UNIF(a,b), WEIB(β; α) 1-random devices like a coin toss (0 and 1 values randomly). Roll a die to generate random numbers between 1 and 6.
Advanced Process Panel: Blocks(Hold, seize, search, delay, release, store, unstore, remove, match, pickup, gather, More sophisticated random devices include Geiger counters and atomic clocks. Naturally, random devices have strong
dropoff, assign attribute, adjust variable, readWrite, signal, search and remove) spreadsheets like(Advanced Set, randomness properties; however, we cannot reproduce their results easily.
Expression, Failure, file, StateSet, Statistic, Storage) 2-Random Number Tables: Published by the RAND corporation, contains one million random digits and one hundred
Basic Process panel: create, dispose, process, decide, batch, clone, separate, assign, record, label, go to label, thousand normal random variates.
adjustable batch. Spreadsheets: Attribute, entity, queue, resource, variable, schedule, set, picture 3-The mid-square method was created by the famous mathematician John von Neumann. The main idea here is we take
an integer, square it, and then use the middle part of that integer as our next random integer, repeating the process as
Advanced Transfer panel: ENTER, LEAVE, PICKSTATION, ROUTE, STATION, ACCESS, CONVEY, EXIT, START, STOP,
many times as we need. To generate the Uniform (0,1) random variable, we would divide each generated integer by the
ACTIVATE, ALLOCATE, FREE, HALT, MOVE, REQUEST, TRANSPORT. SPREADSHEETS (STATION DATA, SEQUENCE,
appropriate power of ten.
CONVEYOR, SEGMENT, TRANSPORTER, DISTANCE, NETWORK, NETWORK LINK, ACTIVITY AREA)
4-Fibonacci Additive Congruential Generators: 𝑋𝑖 = (𝑋𝑖−1 + 𝑋𝑖−2 )𝑚𝑜𝑑 𝑚, 𝑖 = 1,2 … & 𝑅𝑖 = 𝑋𝑖 /𝑚
Trick 1: Pre-assign the service times as an attribute (Sealer Time) in an Assign module immediately after each
● Linear Congruential Generator
customer arrives.
1-LCGs: 𝑋𝑖 = (𝑎𝑋𝑖−1 + 𝑐)𝑚𝑜𝑑 𝑚, 𝑤ℎ𝑒𝑟𝑒 𝑋0 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑒𝑑 𝑎𝑛𝑑 𝑅𝑖 = 𝑋𝑖 /𝑚
Trick 2: While we’re at it, use the Assign to store each customer’s arrival time as an attribute. Use the Arena variable
Example: 𝑋𝑖 = (5𝑋𝑖−1 + 3)𝑚𝑜𝑑 8. 𝑋8 = 𝑋0 = 0. This generator is a full-period generator since it has a cycle length
TNOW to do so.
equal to me
- Sometimes you need to do things that are too complicated for the Process module version of Seize–Delay–Release
Now you can handle things like… 1) Seize–Assign–Delay–Release 2) Non-symmetric multiple Seize’s and Release’s 3)
2- “Desert island” generator: 𝑋𝑖 = 16807𝑋𝑖−1 𝑚𝑜𝑑 231 − 1. This generator has a full-period cycle length > 2 billion
Complicated Seize’s and Release’s that might depend on sets of servers
Let's look at the algorithm for generating each 𝑋𝑖 and 𝑅𝑖 :
Resource spreadsheet in the Basic Process template ➔ Click on Failures column ➔ Add a Failure Name.
𝐾 ← ⌊𝑋𝑖−1 /127773⌋
Go to the Failure spreadsheet in the Advanced Process template, where you’ll see your new failure name. Choose
𝑋𝑖 ← 16807(𝑋𝑖−1 − 127773𝐾) − 2836𝐾
type of failure: Count (failure after a certain # of arrivals) Time (after a certain amount of time) Choose downtime (for
repair): Can be any expression. if 𝑋𝑖 < 0, then set 𝑋𝑖 ← 𝑋𝑖 + 2147483647
-Certain primitive blocks such as the Queue block can’t even connect to a Seize module from the Advanced Process 𝑅𝑖 ← 𝑋𝑖 ∗ 4.656612875E − 10
template or a Process module from the Basic Process template
-Using the Alter block to change the number of resources 3-RANDU Generator (bad) 𝑋𝑖 = 65539𝑋𝑖−1 𝑚𝑜𝑑 231 The random numbers generated by this generator fall entirely on
15 hyperplanes (not iid)
Advanced Transfer Panel Modules ● Various Seize Selection Rules: Cyclical, Random, Preferred Order,
related to movement: Station, Route, Specific Member, Largest Remaining Capacity, Smallest Number ● Tausworthe Generator: 𝐵𝑖 = (𝐵𝑖−𝑟 + 𝐵𝑖−𝑞 )𝑚𝑜𝑑 2 = 𝐵𝑖−𝑟 𝑋𝑂𝑅 𝐵𝑖−𝑞 𝑤ℎ𝑒𝑟𝑒 < 𝑟 < 𝑞. The sequence has a period of
Enter, Leave. Sets and Advanced Sets Busy. 2𝑞 − 1
(e.g., sets of Sequences). ENTER – LEAVE How do we get Unif (0,1) random variables from the 𝐵𝑖 ′s? We can take a sequence of 𝑙 bits and divide them by 2𝑙 to
modules for Station-to-Station ● Station: Somewhere you can go to Route: Tells you where you’re compute a real number between zero and one.
movement. going. Similar to Connect, except it takes time. Enter: General way ● Combined generators:
Resource (e.g., an assistant) to get into a Station. Leave: General way to depart a Station 1- L’Ecuyer’s generator of cycle length 2191 :
-Seize-Release 2- Mersenne Twister: Matsumoto and Nishimura have developed the "Mersenne Twister" generator, which has a period
Transporter (e.g., a car) ● A sequence goes from station to station. Maybe have different of 219937 − 1. This period is beyond sufficient for any modern application; we will often need several billion PRNs, but
-Request-Free service time distribution at each station (depending on customer never more than even 2100 . The period only describes how long it takes for the generator to cycle, but doesn’t have an
-Requires Distance Set type and place in visitation sequence effect on generating the PRNs or the quality of the PRNs.
Conveyor ● Choosing a Good Generator - Theory
- Access-Exit ● Use the Set spreadsheet in the Basic Process template to define
-Requires Segment Set sets. A “vanilla” resource has identical, interchangeable servers. But ● Statistical tests for randomness:
a resource set can have distinct servers, with different schedules, 1- Goodness of fit test - Chi-squared: Are the PRNs approximately Unif (0,1)?? When we design the test, we set the
service speeds, service specialties, etc level of significance α = P (Reject H_0 | H_0 true). Typically, α = 0.05 or 0.1, and is the probability of Type I error.
● Have to be a little careful with Seize-Delay-Release for a resource We can also specify the probability of Type II error, β = P (Accept H_0 | H_0 false). We regard Ho as the status quo -
set. The problem is that you have to make sure that you Release the "that which we currently believe to be true". In this case, Ho refers to the belief that the numbers we are currently
same guy that you originally Seize’d. (If you release a random generating are, in fact, i.i.d Unif (0,1).
(𝑂𝑖 −𝐸𝑖 )2 𝑛
server, some other customer may lose his server!) 𝜒02 = ∑𝑘𝑖=1 where 𝑂𝑖 ∼ Bin(𝑛, 1/𝑘), 𝑖 = 1,2, . . . , 𝑘. And 𝐸𝑖 = 𝐸[𝑂𝑖 ] = 𝑘 , 𝑖 = 1,2, . . . , 𝑘. (1 − 𝛼) quantiles for 𝜒 2
𝐸𝑖
2
distributions with varying degrees of freedom, 𝑛, where we define a quantile, 𝜒𝛼,𝑛 as: 𝑃(𝜒𝑛2 < 𝜒𝛼,𝑛
2
)=1−𝛼
Questions: 2
We reject the null hypothesis 𝐻𝑜 if 𝜒02 > 𝜒𝛼,𝑘−1 2
, and we fail to reject 𝐻0 if 𝜒02 ≤ 𝜒𝛼,𝑘−1 . 𝐸𝑖 = 𝑛/𝑘 should be at least
True: In Arena, a DECIDE module can route customers probabilistically or conditionally to multiple locations. 2
five, and 𝑛 should be at least 30. When k is so large, we cant find tables with values for 𝜒𝛼,𝑘−1 , then we can use the
False: In Arena, a primitive QUEUE block can connect with a PROCESS module approximation below:
True: In Arena, you can pre-assign a service time as an attribute before you actually get to the server that you’ll be 3
2 2 2
using. 𝜒𝛼,𝑘−1 ≈ (𝑘 − 1) [1 − 9(𝑘−1) + 𝑧𝑎 √9(𝑘−1)]
True: In Arena, it’s perfectly OK to have a CREATE module generating one-at-a-time customer arrivals, while also
2- Runs tests for independence:
having a different CREATE module with4-at-a-time customer arrivals elsewhere in the model.
● Runs up and down
True: In Arena, \fake" customers can be used to schedule machine breakdowns, keep track of which time period the 2𝑛−1 16𝑛−29
simulation is currently in, and carry out other duties not associated with actual customers 𝐴 ≈ Nor ( , ) where A denote the total number of ups and downs out of n observations. We reject
3 90
𝐴−𝐸[𝐴]
True: In Arena, the SEIZE, DELAY, RELEASE functionalities are in multiple templates. 𝐻𝑜 𝑖𝑓 𝐴 𝑖𝑠 𝑡𝑜𝑜 𝑏𝑖𝑔 𝑜𝑟 𝑠𝑚𝑎𝑙𝑙. 𝑡ℎ𝑒 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑖𝑠 𝑍0 = and we reject 𝐻𝑜 if |𝑍0 | > 𝑧𝛼/2 here we look for 1-
√Var(𝐴)
True: In Arena, you can schedule failures to occur after a random number of customers have used a resource or after
(alpha/2) in the std normal table
a random amount of time has passed. 2𝑛1 𝑛2 1 2𝑛1 𝑛2(2𝑛1 𝑛2 −𝑛)
False: In Arena, you are only allowed to schedule one type of failure for a particular resource. ● Runs above and below the mean: 𝐵 ≈ Nor ( +2, ) that 𝑛1 refers to the number of observations
𝑛 𝑛2(𝑛−1)
True: In Arena, it is possible for the resource Joey to be a member of three different resource sets. 𝐵−𝐸[𝐵]
greater than or equal to the mean and 𝑛2 = 𝑛 − 𝑛1 . 𝑍0 = where B is the # runs. we reject 𝐻0 if the absolute
√Var(𝐵)
True: For easy display and organizational purposes, our Arena Call Center simulation is divided into several sub
models. value of 𝑍0 is greater than the 𝛼/2 standard normal quantile
True: The customer arrival pattern for our Arena Call Center simulation changes over the day and is therefore
modeled as a nonhomogeneous Poisson process. ● Autocorrelation test (bonus): We define the lag-1 correlation of the 𝑅𝑖 's by 𝜌 ≡ Corr(𝑅𝑖 , 𝑅𝑖+1 ). In other words, the
False: All of the Tech Support servers for our Arena Call Center simulation are equally skilled and have the same lag-1 correlation measures the correlation between one PRN and its immediate successor. Ideally, if the PRNs are
12
concurrent work schedules. ^ ≡ ( ∑𝑛−1
uncorrelated, 𝜌 should be zero. A good estimator for 𝜌 is given by: 𝜌 𝑘=1 𝑅𝑘 𝑅1+𝑘 ) − 3
𝑛−1
-In Arena, what kind of module would you use to change an entity’s picture? ASSIGN In particular, if 𝑛 is large, and 𝐻0 is true: 𝜌
13𝑛−19
^ ≈ Nor (0, (𝑛−1)2 ) Let's transform 𝜌
^ into a standard normal random variable,
-In Arena, where would you find the Expression spreadsheet? Advanced Process panel. 𝜌
^ −𝐸[𝜌
^] 𝜌
^
-Consider the Arena Inventory model that we did in class as a demo. Did this model allow for backlogs? YES 𝑍0 , which we accomplish with the following manipulation: 𝑍0 = = We reject 𝐻0 if the absolute value of
^)
√Var(𝜌 ^)
√Var(𝜌
𝑍0 is greater than the 𝛼/2 standard normal quantile:
● The name of the automatically supplied ARENA attribute that stores a particular entity's sequence is Entity. Week 9 notes:
Sequence ● Inverse transform method: The inverse transform method states that, if 𝑋 is a continuous random variable with cdf
● Where can you set the buffer size of a certain queue? In a Queue block in the Blocks template 𝐹(𝑥), then 𝐹(𝑋) ∼ 𝒰(0,1). In other words, if we plug a random variable, from any distribution, into its own cdf, we get
● There are some great PRN generators out there with incredible cycle lengths 2191 𝑎𝑛𝑑 𝑒𝑣𝑒𝑛 219937 a Unif(0,1) random variable.
● Attributes are properties for individual customers; variables are properties of the entire system. ● Proof: Let 𝑌 = 𝐹(𝑋). Since 𝑌 is a random variable, it has a cdf, which we can denote 𝐺(𝑦). By definition: 𝐺(𝑦) =
● In arena, NORM(a,b) has a variance of b^2 𝑃(𝑌 ≤ 𝑦)
Since 𝑌 = 𝐹(𝑋): ➔ 𝐺(𝑦) = 𝑃(𝐹(𝑋) ≤ 𝑦)
Since 𝑋 is a continuous random variable, its cdf is continuous. Therefore, we can apply the inverse, 𝐹 −1 , to both sides of
On average, A-r algo requires Geom(1/c) trials with expected value = c the inequality: 𝐺(𝑦) = 𝑃(𝐹 −1 (𝐹(𝑋)) ≤ 𝐹 −1 (𝑦))
Polynomial example: Let’s generate a random variable with pdf 𝑓(𝑥) = 60𝑥 3 (1 − 𝑥)2 , 0 ≤ 𝑥 ≤ 1. the maximum of What is 𝐹 −1 (𝐹(𝑋))? Simply, 𝑋: 𝐺(𝑦) = 𝑃(𝑋 ≤ 𝐹 −1 (𝑦))
𝑓(𝑥) occurs at 𝑥 = 0.6: 𝑓(0.6) = 2.0736. With this knowledge, we can generate a basic majorizer, 𝑡(𝑥) = 2.0736. Notice that we have an expression of the form 𝑃(𝑋 ≤ 𝑥), where 𝑥 = 𝐹 −1 (𝑦). We know, by definition, 𝐹(𝑥) = 𝑃(𝑋 ≤
We know that 𝑐 equals the integral of 𝑡(𝑥) from zero to one, so, in this case, 𝑐 = 2.7036. All this to say, 𝑡(𝑥) is a 𝑥), so: 𝐺(𝑦) = 𝐹(𝐹 −1 (𝑦)) = 𝑦
relatively inefficient majorizer. In summary, the cdf of 𝑌 is 𝐺(𝑦) = 𝑦. If we take the derivative of the cdf to get the pdf, we see that 𝑔(𝑦) = 1 Let’s
Like we said, ℎ(𝑥) = 𝑡(𝑥)/𝑐. In this case, ℎ(𝑥) = 1, since 𝑡(𝑥) = 2.7036 = 𝑐. This result means that 𝑌 is a Unif(0,1) remember the pdf for a uniform random variable:
𝑓(𝑥) 60𝑥 3 (1−𝑥)2 1
random variable, since it’s pdf is one. Finally, let’s compute 𝑔(𝑥): 𝑔(𝑥) = = 𝑥 ∈ [𝑎, 𝑏]
𝑡(𝑥) 2.0736 𝑓(𝑥) = {𝑏−𝑎
Let’s look at a simple example. Let’s draw two uniforms, 𝑈 = 0.13 and 𝑌 = 0.25. If we plug and chug, we see that 0 otherwise
𝑔(𝑌) ≈ 0.25. Therefore, 𝑈 ≤ 𝑔(𝑌) and we take 𝑋 ← 0.25 If 𝑎 = 0, 𝑏 = 1, then 𝑓(𝑥) = 1 = 𝑔(𝑦). Therefore, 𝑌 ∼ 𝒰(0,1).
2 −𝑥2 2𝑒
● How Do We Use This Result: Let 𝑈 ∼ 𝒰(0,1). Since 𝐹(𝑋) = 𝑈, then 𝐹 −1 (𝑈) = 𝑋. In other words, if we set 𝐹(𝑋) =
Half-normal example: with the pdf : 𝑓(𝑥) = 𝑒 2 , 𝑥 ≥ 0 . We can use 𝑡(𝑥) = √ 𝜋 𝑒 −𝑥 ≥ 𝑓(𝑥) & 𝑐 = 𝑈, and then solve for 𝑋, we get a random variable from the same distribution as 𝑋, but in terms of 𝑈.
√2𝜋
∞ 2𝑒
∫0 𝑡 (𝑥)𝑑𝑥 = √ 𝜋 ● Inverse Transform Method - Continuous Examples:
𝑥−𝑎
Now, let’s compute ℎ(𝑥): ℎ(𝑥) =
𝑡(𝑥)
= 𝑒 −𝑥 easy exp(1) pdf and 𝑔(𝑥) =
𝑓(𝑥)
= 𝑒 −(𝑥−1)
2 /2
. To generate a half normal, 1- Uniform Example: Consider the 𝒰(𝑎, 𝑏) distribution, which has the following cdf: 𝐹(𝑥) = 𝑏−𝑎 , 𝑎 ≤ 𝑥 ≤ 𝑏. Let's set
𝑐 𝑡(𝑥)
𝐹(𝑋) = 𝑈 and solve for 𝑋:
we simply generate 𝑈 ∼ 𝒰(0,1) and 𝑌 ∼ 𝐸𝑥𝑝(1) and accept 𝑌 if 𝑈 ≤ 𝑔(𝑌). We can use the half-normal result to 𝑋−𝑎
generate a Nor(0,1) random variable. We simply have to “flip back” half of the 𝑋 values over the 𝑦-axis. Given 𝑈 ∼ 𝑈 = 𝑏−𝑎
𝒰(0,1) and 𝑋 from the half-normal distribution, we can see that: (𝑏 − 𝑎)𝑈 =𝑋−𝑎
−𝑋 𝑈 ≤ 1/2 𝑎 + (𝑏 − 𝑎)𝑈 =𝑋
𝑍={ ∼ 𝑁𝑜𝑟(0,1) As always, we can generate a Nor(𝜇, 𝜎 2 ) by applying the transformation 𝜇 + 𝜎𝑍.
𝑋This𝑈>study
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Intuitively, this GMT
result makes -06:00
perfect sense. If we take a Unif (0,1) random variable, and multiply it by 𝑏 − 𝑎, we end up
𝜆𝑛
Poisson example: The Poisson distribution has the following pmf: 𝑃(𝑋 = 𝑛) = 𝑒 −𝜆 𝑛! , 𝑛 = 0,1, . .. with a Unif (0, 𝑏 − 𝑎) random variable. Then, if we add 𝑎 to each end, we get a Unif (𝑎, 𝑏) random variable.
𝑛
𝑥 = 𝑛 ⟺ 𝑠𝑒𝑒 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑛 𝑃𝑜𝑖𝑠(𝜆)𝑎𝑟𝑟𝑖𝑣𝑎𝑙𝑠 𝑏𝑦 𝑡 = 1. 𝑋 = 𝑛 ⇔ ∑𝑖=1 𝐴𝑖 ≤ 1 < ∑𝑖=1 𝐴𝑖 𝑛+1 2- Exponential example: Consider the Exp(𝜆) distribution, which has the following cdf: 𝐹(𝑥) = 1 − 𝑒 −𝜆𝑥 , 𝑥 ≥ 0 Let’s
set 𝐹(𝑋) = 𝑈 and solve for 𝑋:
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