Poisson Process
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Transcript of Poisson Process
Poisson ProcessPresented by : Johniel E. Babiera
Example 1
We are interested in the dynamics of the arrival of telephone calls to a call center. To describe these arriving calls, consider a collection of random variables {ππ‘ , π‘ β₯0}, where each random variable ππ‘, for a fixed π‘, denotes the cumulative number calls coming into the center by time π‘. Since calls record a count, the state space is the set of whole number π = {0,1,2, β¦ }.
Definition 1
Let the continuous parameter stochastic process {ππ‘ , π‘ β₯ 0} with π0 = 0 have a state space equal to the nonnegative integers . It is called ππππ π ππ ππππππ π with rate π if
a) π ππ‘ = π =πβππ‘ ππ‘ π
π!for all nonnegative π and π‘ β₯ 0.
b) The event {ππ +π’ β ππ = π} is independent of the event {ππ‘ = π} if π‘ < π .
c) π{ππ +π’ β ππ = π} only depends on the value of π’.
Example 2
For example, if the process in example 1 has independent increments, then knowledge of the number of calls that arrived before 11:00AM will not help in predicting the number of calls to arrive between 11:00AM and noon. Likewise, if the process has stationary increments, then the probability that k calls arrive between 10:00AM and 10:15AM will be equal to the probability that k calls arrive between 4:00PM and 4:15PM because the size of both intervals is the same. It turns out that an arrival process with independent and stationary increments in which arrivals can only occur one-at-a-time must be a Poisson process.
We have πΈ ππ‘ = ππ‘. In others words, π gives the mean arrival rate per unit time for an arrival process that is described by a Poisson process with rate π.
Because Poisson process has independent and stationary increments, we also have, for nonnegative π
π ππ +π’ β ππ = π =πβππ’ ππ’ π
π!for π’ β₯ 0.
Example 3
Assume that the arrivals to the call center described in can be modeled according to a Poisson process with rate 25/βπ. The number of calls that are expected within an eight-hour shift is πΈ ππ‘=8 = 25 Γ 8 = 200. Assume we had 20 calls that arrived from 8AM through 9AM. What is the probability that there will be no calls that arrive from 9:00AM through 9:06AM? Because of independent increments, the information regarding the 20 calls is irrelevant. Because of station stationary increments, we only need to know that the length of the
interval is 6 minutes (or 0.1 hours), note that it starts at 9AM; thus, the answer is given as
π π0.1 = 0 =πβ25 0.1
π!= 0.08208
Properties of Poisson Process
Property 1. A Poisson process with rate Ξ» has exponentially distributed
inter-arrival times with the mean time between arrivals being 1/Ξ» . The converse is also true; namely, an arrival process with exponentially distributed inter-arrival times is a Poisson process.
Consider an interval of length t. The probability that no arrival occurs within that
interval is π ππ‘ = 0 = πβππ‘. Let π denote the time that the first arrival occurs. The event {π > π‘} is equivalent to the event {ππ‘ = 0}; therefore,
π π β€ 0 = 1 β π π > 0 = 1 β πβππ‘ for π‘ β₯ 0
which is the CDF for the exponential distribution.
Property 2. Let ππ denote the ππ‘β arrival time for a Poisson process with rate Ξ» . The random variable ππ has an Erlang distribution with pdf given by
π π‘ =π π π‘ πβ1πβπ π‘
π β 1 !πππ π‘ β₯ 0 .
π(π‘)~πππππ(π, π)
Property 3. Superposition of Poisson Processes: Let {ππ‘; π‘ β₯ 0} and {ππ‘; π‘ β₯ 0} be two independent Poisson process with rates π1 and π2. Form a third process by ππ‘ = ππ‘ +ππ‘, for each π‘ β₯ 0. The process {ππ‘; π‘ β₯0} is a Poisson process with rate π1 + π2.
Example 4 (property 3)
Consider the diagram in the right representing a
limited access highway. Assume that the times in
which cars pass Point A form a Poisson process
with rate π1 = 8/min and cars passing Point B
form a Poisson process with rate π2 = 2/min.
After these two streams of cars merge, the times at which cars pass
Point C form a Poisson process with mean rate 10/min.
Property 4. Decomposition of a Poisson Processes: Let π = {ππ‘; π‘ β₯ 0} be a Poisson process with rate π and let {π1, π2, β¦ } denote an i.i.d sequence of Bernoulli random variables independent of the Poisson process such that π ππ = 1 = π. Let π = {ππ‘; π‘ β₯ 0} be a new process formed as follows; for each positive π, consider the ππ‘β arrival to the π process to also be an arrival to the π process if ππ = 1; otherwise, the ππ‘β arrival to the π process is not part of the π process. The resulting π process is a Poisson process with rate ππ.
Example 5 (property 4)
Consider traffic coming to a fork in the road, and assume that the arrival times of cars to the fork form a Poisson process with mean rate 2 per minute. In addition, there is a 30% chance that cars will turn left and a 70% chance that cars will turn right. Under the assumption that all cars act independently, the arrival stream on the left-hand fork form a Poisson process with rate 0.6/min and the stream on the right-hand fork form a Poisson process with rate 1.4/min.
Queuing Theory
Queuing Theory
β’ The study of waiting line phenomena (a queue is a waiting line)
β’ The theory enables mathematical analysis of several related processes, including arriving at the (back of the) queue, waiting in the queue and being served at the front of the queue.
β’ The theory permits the derivation and calculation of several performance measures including the average waiting time in the queue, the expected number waiting or receiving service, and the probability of encountering the system in certain states, such as empty, full, having an available server or having to wait a certain time to be served.
Kendall Notation
β’ General Form
(A/B/c/K/m/Z)A - inter-arrival time distribution
B - service time distribution
c - the number of servers
K - the system capacity (number of customer)
m - the number of source
Z - the queue discipline
Queueing symbols used with Kendallβs notation
Symbols Explanation
M Exponential (Markovian) inter-arrival or service time
D Deterministic inter-arrival or service time
Ek Erlang type k inter-arrival or service time
G General inter-arrival or service time
1, 2, β¦, β Number of parallel servers or capacity
FIFO First In, First Out queue discipline
LIFO Last In, First Out queue discipline
SIRO Service in random order
PRI Priority queue discipline
GD General queue discipline
Littleβs Law
β’ Consider a queueing system for which steady state occurs. Let πΏ = πΈ[π] denote the mean long-run number in the system, π = πΈ[π] denote the mean long-run waiting time within the system, and ππ the mean arrival rate of jobs into the system. Also let πΏπ = πΈ[ππ] and ππ = πΈ[ππ]denote the analogous quantities restricted to the queue. Then
πΏ = ππππΏπ = ππππ
β’ We also have the following results;π π‘ = ππ π‘ + ππ π‘ β π = ππ +ππ
πΏ = πΏπ + πΏπ
Andπ = ππ + ππ π = ππ +ππ
Examples:M/M/1/β/FIFO systemG/G/cM/G/1M/M/1/KM/Eβ/3/20/β/SIRO
M/G/1 queue system
For the M/G/1 queueing system being operated under the FIFO service rule, we derive the expressions of the following quantities in terms of the arrival rate π, the mean service time πΈ[π], and the variance of service time ππ
2.β’ π: the average time a randomly arriving customer will spend in the
system, which is composed of the waiting time in the queue and the service time.
β’ πΏ: the average number of people in the system that a randomly arriving customer finds, which is composed of the number of people in the queue and the person in service.
β’ π: the long run fraction of time the server is busy, which is equivalently the probability that the server is busy at a random point in time.
β’ π΅: the long run average duration of a server busy period.
β’ M/G/1 queueing system is a single server queueing system in which the customer arrival process is Poisson with rate π and the service time, π, for each customer follows a general distribution with ππ·πΉ ππ(π ), mean πΈ[π], and variance ππ
2.
β’ Suppose we have been watching the queueing system for a very very long time, say πβ minutes where πβ is very large. If we recorded the number of minutes the server was busy during this long period of time and then divided it by πβ, we would obtain π, the long run fraction of time the server is busy. During the long period of time πβ, we would expect there have been ππβcustomers arriving to the queueing system, each of who takes on average πΈ[π] minutes to be served. This means
π =number of minutes server is busy
πβ=
ππβ Γ πΈ[π]
πβ= ππΈ[π]
β’ To compute π΅, the long run average length of a server busy period, we again think of observing the system for a long period of time. During this long period of time, there occur a large number, say N, of busy periods. Since every busy period is followed by an idle period, we could say that the number of idle periods is N as well. (The difference between the number of busy periods and the number of idle periods would be at most one, and this is negligible compared to N.). Since the average length of a busy period is B and the number of busy periods is N, the total amount of time the server is busy, over the long period of time we are observing, is NB.
β’ The length of an idle period is, on average, 1/π since an idle period occurs when the server is waiting for a customer to arrive after the queue becomes empty. Since the arrival process is Poisson and therefore ππππππ¦πππ π , the server will wait a negative exponentially distributed amount of time until the next customer arrives. Therefore if there are N idle periods, the total amount of time the server is idle is π/π. The fraction of time the server is busy, π, can now be computed by
π =ππ΅
(ππ΅ +ππ)=
π΅
π΅ +1π
.
Solving for π΅ and using π = ππΈ[π], we have
π΅ =π/π
1 β π=
πΈ[π]
1 β ππΈ[π].
Let π be a random variable denoting the amount of time that a randomly arriving customer, say I, will spend in the system. Our goal is to compute π = πΈ[π]. Note that we can decompose π into the following three random variables:
β’ π1, the remaining service time of the customer currently in service.
β’ π2, the time required to serve the customers waiting ahead of me in the queue.
β’ π3, my service time.
β’ Clearly, π = πΈ[π1] + πΈ[π2] + πΈ[π3]. Since the expected service time for each customer is πΈ[π], we have
πΈ π3 = πΈ[π]To obtain πΈ[π2], we first compute the conditional expectation of T2 given that there are already π customers in the system when I, a randomly arriving customer, arrive in the system. Since one customer is being served and π β 1 customers are waiting in the queue,
πΈ π2 π = π β 1 πΈ π , π β₯ 10, π = 0
πΈ π2 =
π
πΈ π2 π π π = π =
πβ₯1
π β 1 πΈ π π(π = π)
= πΈ π
πβ₯1
ππ π = π β πΈ[π]
πβ₯1
π(π = π)
Note that πβ₯1ππ(π = π) = πΏ and πβ₯1π(π = π) = π. So we have
πΈ π2 = πΈ π πΏ β πΈ π π.
β’ The expected remaining service time for the customer in service when I randomly arrive in the system is given by
πΈ π1 π =
ππ2 + πΈ π 2
2πΈ[π], π β₯ 1
0, π = 0
πΈ π1 =
π
πΈ π1 π π π = π =
π
ππ2
2πΈ π +πΈ π
2π(π = π)
=ππ2
2πΈ π +πΈ π
2π
Hence π = πΈ π1 + πΈ π2 + πΈ[π3]
=ππ2
2πΈ π +πΈ π
2π + πΈ π πΏ β πΈ π π + E[S]
We now have one linear relationship between πΏ and π. Combining this with Littleβs Law, πΏ = ππ, we have two equations for two unknowns, L and W. A little algebra gives us
πΏ = π +π2 + π2ππ
2
2(1 β π).
Example 6
A large car dealer has a policy of providing cars for its customers that have car problems. When a customer brings the car in for repair, that customer has use of a dealerβs car. The dealer estimates that the dealer cost for providing the service is $10 per day for as long as the customerβs car is in the shop. (Thus, if the customerβs car was in the shop for 1.5 days, the dealerβs cost would be $15.) Arrivals to the shop of customers with car problems form a Poisson process with a mean rate of one every other day. There is one mechanic dedicated to those customerβs cars. The time that the mechanic spends on a car can be described by an exponential random variable with a mean of 1.6 days.
We would like to know the expected cost per day of this policy to the car dealer. Assuming infinite capacity, we have the assumptions of the M/M/1 queueing system satisfied, with π =0.5/πππ¦ and π = 0.625/πππ¦, yielding a π = 0.8. (Note the mean rate is the reciprocal of the mean time.)Using the M/M/1 equations, we have πΏ = 4 πππ π = 8 days. Thus, whenever a customer comes in with car problems, it will cost the dealer $80. Since a customer comes in every other day (on the average) the total cost to the dealer for this policy is $40 πππ πππ¦. In other words, cost is equal to $10 Γπ Γ π . But by Littleβs formula, this is equivalent to $10 Γ πΏ. The cost structure illustrated with this example is a very common occurrence for queueing systems. In other words if c is the cost per item per time unit that the item spends in the system, the expected system cost per time unit is ππΏ.
References:
βͺ Feldman, R.M and Valdez-Flores,C.Applied Probablity and Stochastic Processes.Second Ed.,Springer Heidelbrg Dordrecht London New York,2010.
βͺ http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-203j-logistical-and-transportation-planning-methods-fall-2004/lecture-notes/class12mg1q.pdf