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 {𝑁𝑡 , 𝑡 ≥0}, where each random variable 𝑁𝑡, for a fixed 𝑡, denotes the cumulative number calls coming into the center by time 𝑡. Since calls record a count, the state space is the set of whole number 𝑆 = {0,1,2, … }.

Definition 1

Let the continuous parameter stochastic process {𝑁𝑡 , 𝑡 ≥ 0} with 𝑁0 = 0 have a state space equal to the nonnegative integers . It is called 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 with rate 𝜆 if

a) 𝑃 𝑁𝑡 = 𝑘 =𝑒−𝜆𝑡 𝜆𝑡 𝑘

𝑘!for all nonnegative 𝑘 and 𝑡 ≥ 0.

b) The event {𝑁𝑠+𝑢 − 𝑁𝑠 = 𝑖} is independent of the event {𝑁𝑡 = 𝑗} if 𝑡 < 𝑠.

c) 𝑃{𝑁𝑠+𝑢 − 𝑁𝑠 = 𝑖} only depends on the value of 𝑢.

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 𝐸 𝑁𝑡 = 𝜆𝑡. In others words, 𝜆 gives the mean arrival rate per unit time for an arrival process that is described by a Poisson process with rate 𝜆.

Because Poisson process has independent and stationary increments, we also have, for nonnegative 𝑘

𝑃 𝑁𝑠+𝑢 − 𝑁𝑠 = 𝑘 =𝑒−𝜆𝑢 𝜆𝑢 𝑘

𝑘!for 𝑢 ≥ 0.

Example 3

Assume that the arrivals to the call center described in can be modeled according to a Poisson process with rate 25/ℎ𝑟. The number of calls that are expected within an eight-hour shift is 𝐸 𝑁𝑡=8 = 25 × 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

𝑃 𝑁0.1 = 0 =𝑒−25 0.1

𝑜!= 0.08208

Properties of Poisson Process

Property 1. A Poisson process with rate λ has exponentially distributed

inter-arrival times with the mean time between arrivals being 1/λ . 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 𝑃 𝑁𝑡 = 0 = 𝑒−𝜆𝑡. Let 𝑇 denote the time that the first arrival occurs. The event {𝑇 > 𝑡} is equivalent to the event {𝑁𝑡 = 0}; therefore,

𝑃 𝑇 ≤ 0 = 1 − 𝑃 𝑇 > 0 = 1 − 𝑒−𝜆𝑡 for 𝑡 ≥ 0

which is the CDF for the exponential distribution.

Property 2. Let 𝑇𝑛 denote the 𝑛𝑡ℎ arrival time for a Poisson process with rate λ . The random variable 𝑇𝑛 has an Erlang distribution with pdf given by

𝑓 𝑡 =𝜆 𝜆 𝑡 𝑘−1𝑒−𝜆 𝑡

𝑘 − 1 !𝑓𝑜𝑟 𝑡 ≥ 0 .

𝑓(𝑡)~𝑔𝑎𝑚𝑚𝑎(𝑘, 𝜆)

Property 3. Superposition of Poisson Processes: Let {𝑁𝑡; 𝑡 ≥ 0} and {𝑀𝑡; 𝑡 ≥ 0} be two independent Poisson process with rates 𝜆1 and 𝜆2. Form a third process by 𝑌𝑡 = 𝑁𝑡 +𝑀𝑡, for each 𝑡 ≥ 0. The process {𝑌𝑡; 𝑡 ≥0} is a Poisson process with rate 𝜆1 + 𝜆2.

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 𝜆1 = 8/min and cars passing Point B

form a Poisson process with rate 𝜆2 = 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 𝑁 = {𝑁𝑡; 𝑡 ≥ 0} be a Poisson process with rate 𝜆 and let {𝑋1, 𝑋2, … } denote an i.i.d sequence of Bernoulli random variables independent of the Poisson process such that 𝑃 𝑋𝑛 = 1 = 𝑝. Let 𝑀 = {𝑀𝑡; 𝑡 ≥ 0} be a new process formed as follows; for each positive 𝑛, consider the 𝑛𝑡ℎ arrival to the 𝑁 process to also be an arrival to the 𝑀 process if 𝑋𝑛 = 1; otherwise, the 𝑛𝑡ℎ arrival to the 𝑁 process is not part of the 𝑀 process. The resulting 𝑀 process is a Poisson process with rate 𝜆𝑝.

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

• The study of waiting line phenomena (a queue is a waiting line)

• 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.

• 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

• 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 Kendall’s 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, …, ∞ 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

Little’s Law

• Consider a queueing system for which steady state occurs. Let 𝐿 = 𝐸[𝑁] denote the mean long-run number in the system, 𝑊 = 𝐸[𝑇] denote the mean long-run waiting time within the system, and 𝜆𝑒 the mean arrival rate of jobs into the system. Also let 𝐿𝑞 = 𝐸[𝑁𝑞] and 𝑊𝑞 = 𝐸[𝑇𝑞]denote the analogous quantities restricted to the queue. Then

𝐿 = 𝜆𝑒𝑊𝐿𝑞 = 𝜆𝑒𝑊𝑞

• We also have the following results;𝑁 𝑡 = 𝑁𝑞 𝑡 + 𝑁𝑠 𝑡 ⇒ 𝑁 = 𝑁𝑞 +𝑁𝑠

𝐿 = 𝐿𝑞 + 𝐿𝑠

And𝑇 = 𝑇𝑞 + 𝑇𝑠𝑊 = 𝑊𝑞 +𝑊𝑠

Examples:M/M/1/∞/FIFO systemG/G/cM/G/1M/M/1/KM/E₄/3/20/∞/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 𝜆, the mean service time 𝐸[𝑆], and the variance of service time 𝜎𝑆

2.• 𝑊: 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.

• 𝐿: 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.

• 𝜌: 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.

• 𝐵: the long run average duration of a server busy period.

• M/G/1 queueing system is a single server queueing system in which the customer arrival process is Poisson with rate 𝜆 and the service time, 𝑆, for each customer follows a general distribution with 𝑃𝐷𝐹 𝑓𝑆(𝑠), mean 𝐸[𝑆], and variance 𝜎𝑆

2.

• Suppose we have been watching the queueing system for a very very long time, say 𝑇∗ minutes where 𝑇∗ 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 𝑇∗, we would obtain 𝜌, the long run fraction of time the server is busy. During the long period of time 𝑇∗, we would expect there have been 𝜆𝑇∗customers arriving to the queueing system, each of who takes on average 𝐸[𝑆] minutes to be served. This means

𝜌 =number of minutes server is busy

𝑇∗=

𝜆𝑇∗ × 𝐸[𝑆]

𝑇∗= 𝜆𝐸[𝑆]

• To compute 𝐵, 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.

• The length of an idle period is, on average, 1/𝜆 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 𝑚𝑒𝑚𝑜𝑟𝑦𝑙𝑒𝑠𝑠, 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 𝑁/𝜆. The fraction of time the server is busy, 𝜌, can now be computed by

𝜌 =𝑁𝐵

(𝑁𝐵 +𝑁𝜆)=

𝐵

𝐵 +1𝜆

.

Solving for 𝐵 and using 𝜌 = 𝜆𝐸[𝑆], we have

𝐵 =𝜌/𝜆

1 − 𝜌=

𝐸[𝑆]

1 − 𝜆𝐸[𝑆].

Let 𝑇 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 𝑊 = 𝐸[𝑇]. Note that we can decompose 𝑇 into the following three random variables:

• 𝑇1, the remaining service time of the customer currently in service.

• 𝑇2, the time required to serve the customers waiting ahead of me in the queue.

• 𝑇3, my service time.

• Clearly, 𝑊 = 𝐸[𝑇1] + 𝐸[𝑇2] + 𝐸[𝑇3]. Since the expected service time for each customer is 𝐸[𝑆], we have

𝐸 𝑇3 = 𝐸[𝑆]To obtain 𝐸[𝑇2], we first compute the conditional expectation of T2 given that there are already 𝑛 customers in the system when I, a randomly arriving customer, arrive in the system. Since one customer is being served and 𝑛 − 1 customers are waiting in the queue,

𝐸 𝑇2 𝑛 = 𝑛 − 1 𝐸 𝑆 , 𝑛 ≥ 10, 𝑛 = 0

𝐸 𝑇2 =

𝑛

𝐸 𝑇2 𝑛 𝑃 𝑁 = 𝑛 =

𝑛≥1

𝑛 − 1 𝐸 𝑆 𝑃(𝑁 = 𝑛)

= 𝐸 𝑆

𝑛≥1

𝑛𝑃 𝑁 = 𝑛 − 𝐸[𝑆]

𝑛≥1

𝑃(𝑁 = 𝑛)

Note that 𝑛≥1𝑛𝑃(𝑁 = 𝑛) = 𝐿 and 𝑛≥1𝑃(𝑁 = 𝑛) = 𝜌. So we have

𝐸 𝑇2 = 𝐸 𝑆 𝐿 − 𝐸 𝑆 𝜌.

• The expected remaining service time for the customer in service when I randomly arrive in the system is given by

𝐸 𝑇1 𝑛 =

𝜎𝑆2 + 𝐸 𝑆 2

2𝐸[𝑆], 𝑛 ≥ 1

0, 𝑛 = 0

𝐸 𝑇1 =

𝑛

𝐸 𝑇1 𝑛 𝑃 𝑁 = 𝑛 =

𝑛

𝜎𝑆2

2𝐸 𝑠+𝐸 𝑆

2𝑃(𝑁 = 𝑛)

=𝜎𝑆2

2𝐸 𝑠+𝐸 𝑆

2𝜌

Hence 𝑊 = 𝐸 𝑇1 + 𝐸 𝑇2 + 𝐸[𝑇3]

=𝜎𝑆2

2𝐸 𝑠+𝐸 𝑆

2𝜌 + 𝐸 𝑆 𝐿 − 𝐸 𝑆 𝜌 + E[S]

We now have one linear relationship between 𝐿 and 𝑊. Combining this with Little’s Law, 𝐿 = 𝜆𝑊, we have two equations for two unknowns, L and W. A little algebra gives us

𝐿 = 𝜌 +𝜌2 + 𝜆2𝜎𝑆

2

2(1 − 𝜌).

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 dealer’s car. The dealer estimates that the dealer cost for providing the service is $10 per day for as long as the customer’s car is in the shop. (Thus, if the customer’s car was in the shop for 1.5 days, the dealer’s 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 customer’s 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 𝜆 =0.5/𝑑𝑎𝑦 and 𝜇 = 0.625/𝑑𝑎𝑦, yielding a 𝜌 = 0.8. (Note the mean rate is the reciprocal of the mean time.)Using the M/M/1 equations, we have 𝐿 = 4 𝑎𝑛𝑑 𝑊 = 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 𝑝𝑒𝑟 𝑑𝑎𝑦. In other words, cost is equal to $10 ×𝑊 × 𝜆 . But by Little’s formula, this is equivalent to $10 × 𝐿. 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 𝑐𝐿.

References:

▪ Feldman, R.M and Valdez-Flores,C.Applied Probablity and Stochastic Processes.Second Ed.,Springer Heidelbrg Dordrecht London New York,2010.

▪ http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-203j-logistical-and-transportation-planning-methods-fall-2004/lecture-notes/class12mg1q.pdf