Queueing Theory (Delay Models)

Introduction

Total delay of the i-th customer in the system T i = W i + τ i N(t) : the number of customers in the system N q (t) : the number of customers in the queue N s (t) : the number of customers in the service N : the avg number of customers in the queue τ : the service time

T : the total delay in the system λ: the customer arrival rate [#/sec]

Little ’ s Theorem E[N] = λE[T] Number of customer in the system at t N(t)=A(t)-D(t) where D(t) : the number of customer departures up to time t A(t) : the number of customer arrivals up to time t

Poisson Process The interarrival probability density function mean: 1/λ, variance: 1/λ 2 for every t, δ≥0 where

Poisson Process Characteristics of the Poisson process Interarrival times are independent and exponentially distributed If t n denotes the n-th arrival time and the interval τ n = t n+1 - t n, the probability distribution is

Sum of Poisson Random Variables X i, i =1,2, …,n, are independent RVs X i follows Poisson distribution with parameter i Partial sum defined as: S n follows Poisson distribution with parameter

Sum of Poisson Random Variables

Sampling a Poisson Variable X follows Poisson distribution with parameter Each of the X arrivals is of type i with probability p i, i =1,2, …,n, independently of other arrivals; p 1 + p 2 + … + p n = 1 X i denotes the number of type i arrivals X 1, X 2, … X n are independent X i follows Poisson distribution with parameter i  p i

Sampling a Poisson Variable (cont.)

Merging & Splitting Poisson Processes A 1, …, A k independent Poisson processes with rates 1, …, k Merged in a single process A= A 1 + … + A k A is Poisson process with rate = 1 + … + k A: Poisson processes with rate Split into processes A 1 and A 2 independently, with probabilities p and 1-p respectively A 1 is Poisson with rate 1 = p A 2 is Poisson with rate 2 = (1-p)      p (1-p) p 1-p

Poisson Variable mean variance Memoryless property (if exponentially distributed)

Review of Markov chain theory Discrete time Markov chains discrete time stochastic process {X n |n=0,1,2,..} taking values from the set of nonnegative integers Markov chain if where

Markov chain Markov chain formulation Consider a discrete time MC where N k is the number of customers at time k and N(t) is the number of customers at time t probabilities where the arrival and departure processes are independent

The transition probability matrix n-step transition probabilities Review of Markov chain theory

Chapman-Kolmogorov equations detailed balance equations for birth-death systems (in steady state) Review of Markov chain theory

Example -1  1  2 /2  1  2 /2 1-  2 /2  2 /2 01 P 0 =(2-  2 )/(2-  2 (1-  1 )) the throuput= P 0 *P(s=0| P 0 )*0+ P 0 *P(s=1| P 0 )*1+ P 0 *P(s=2| P 0 )*2+ P 1 *P(s=0| P 1 )*0+ P 1 *P(s=1| P 1 )*1+ P 1 *P(s=2| P 1 )*2

Continuous time Markov chains {X(t)| t≥0} taking nonnegative integer values υ i : the transition rate from state i q ij : the transition rate from state i to j q ij = υ i P ij the steady state occupancy probability of state j Analog of detailed balance equations for DTMC

Queueing Theory21 Birth-And-Death Process

Queueing Theory22 Birth-And-Death Process(cont.) l Equation Expressing This: State Rate In = Rate Out 0  1 P 1 = 0 P P 0 +  2 P 2 = ( 1 +  1 ) P P 1 +  3 P 3 = ( 2 +  2 ) P N-1 N-2 P N-2 +  N P N = ( N-1 +  N-1 ) P N-1 N N-1 P N-1 +  N+1 P N+1 = ( N +  N ) P N

Queueing Theory23 Birth-And-Death Process(cont.) l Finding Steady State Process: State 0:P 1 = ( 0 /  1 ) P 0 1:P 2 = ( 1 /  2 ) P 1 + (  1 P P 0 ) /  2 = ( 1 /  2 ) P 1 + (  1 P 1 -  1 P 1 ) /  2 = ( 1 /  2 ) P 1 =

Queueing Theory24 Birth-And-Death Process(cont.) l Finding Steady State Process(cont.): State n-1:P n = ( n-1 /  n ) P n-1 + (  n-1 P n-1 - n-2 P n-2 ) /  n = ( n-1 /  n ) P n-1 + (  n-1 P n-1 -  n-1 P n-1 ) /  n = ( n-1 /  n ) P n-1

Queueing Theory25 Birth-And-Death Process(cont.) l Finding Steady State Process(cont.): N:P n+1 = ( n /  n+1 ) P n + (  n P n - n-1 P n-1 ) /  n+1 = ( n /  n+1 ) P n To Simplify: Let C = ( n-1 n ) / (  n  n  1 ) Then P n = C n P 0, N = 1, 2,....

M/M/1 queueing system Arrival statistics: stochastic process taking nonnegative integer values is called a Poisson process with rate λ if A(t) is a counting process representing the total number of arrivals from 0 to t arrivals are independent probability distribution function

P[1 arrival and no departure in δ]= where the arrival and departure processes are independent M/M/1 queueing system

Global balance equation M/M/1 queueing system

from Then Average number of customers in the system M/M/1 queueing system

Average delay per customer (waiting time + service time) by Little ’ s theorem Average waiting time Average number of customer in queue Server utilization M/M/1 queueing system

example 1/λ=4 ms, 1/μ=3 ms M/M/1 queueing system

תרגיל קצב הגע למערכת הוא *n קצב השרות הוא  *n א. שרטט את דיאגרמת המצבים ב. מצא את הסתברויות הסטציונרות ג. מצא את מספר הצרכנים הממוצע במערכת במצב היציב ד. מצא את זמן ההשהייה הממוצע במערכת באמצעות משפט LITTLE

פתרון א ב

פתרון ג ד

תרגיל צומת ברשת משתמש בשיטת הניתוב הבאה : כאשר חבילה מגיע אליו ללא תלות ביעדה הוא מפנה אותה לקו יצאה אםם התור לקו זה הוא ריק ולא נשלחת ברגע זה שום חבילה דרך קו זה. אחרת חבילה זו מופנת חבילה זו דרך כו אחר כלשהו. נניח כי שמופע ההודעות לצומת הוא פואסוני עם אורך החבילות מתפלג אקספוננצילי אם  וכיבולת הקו היא C. א. איזה חלק מהחבילות מגעות דרך קו ההעדיף ? ב. אם הוחלט להצמיד תור לקו המהיר, מה אורכו המינימלי של התור כך שההסתברות שחבילה תשודרנה בקו זה תהיה לפחות 0.9 בהנחה ש ? ( האם המערכת במצב יציב ?)

פתרון א Message Length: Transmission Rate: Transmission Time: Service Rate:

פתרון ב

Queueing Theory39 M/M/1 Example I Traffic to a message switching center for one of the outgoing communication lines arrive in a random pattern at an average rate of 240 messages per minute. The line has a transmission rate of 800 characters per second. The message length distribution (including control characters) is approximately exponential with an average length of 176 characters. Calculate the following principal statistical measures of system performance, assuming that a very large number of message buffers are provided:

Queueing Theory40 M/M/1 Example I (cont.) l (a) Average number of messages in the system l (b) Average number of messages in the queue waiting to be transmitted. l (c) Average time a message spends in the system. l (d) Average time a message waits for transmission l (e) Probability that 10 or more messages are waiting to be transmitted.

Queueing Theory41 M/M/1 Example I (cont.) 1. E[s] = Average Message Length / Line Speed = {176 char/message} / {800 char/sec} = 0.22 sec/message or   = 1 / 0.22 {message / sec} = 4.55 message / sec  = 240 message / min = 4 message / sec   = E[s] = /  = 0.88

Queueing Theory42 M/M/1 Example I (cont.) l (a) N=  / (1 -  ) = 7.33 (messages) l (b) N q =   / (1 -  ) = 6.45 (messages) l (c) W = E[s] / (1 -  ) = 1.83 (sec) l (d) W q =    E[s] / (1 -  ) = 1.61 (sec) l (e) P [11 or more messages in the system] =   = 0.245

Queueing Theory43 M/M/1 Example II A branch office of a large engineering firm has one on-line terminal that is connected to a central computer system during the normal eight-hour working day. Engineers, who work throughout the city, drive to the branch office to use the terminal to make routine calculations. Statistics collected over a period of time indicate that the arrival pattern of people at the branch office to use the terminal has a Poisson (random) distribution, with a mean of 10 people coming to use the terminal each day. The distribution of time spent by an engineer at a terminal is exponential, with a

Queueing Theory44 M/M/1 Example II (cont.) mean of 30 minutes. The branch office receives complains from the staff about the terminal service. It is reported that individuals often wait over an hour to use the terminal and it rarely takes less than an hour and a half in the office to complete a few calculations. The manager is puzzled because the statistics show that the terminal is in use only 5 hours out of 8, on the average. This level of utilization would not seem to justify the acquisition of another terminal. What insight can queueing theory provide?

Queueing Theory45 M/M/1 Example II (cont.) l {10 person / day}  {1 day / 8hr}  {1hr / 60 min} = 10 person / 480 min = 1 person / 48 min ==>  = 1 / 48 (person / min) l 30 minutes : 1 person = 1 (min) : 1/30 (person) ==>  = 1 / 30 (person / min) l  = /  = {1/48} / {1/30} = 30 / 48 = 5 / 8

Queueing Theory46 M/M/1 Example II (cont.) l Arrival Rate = 1 / 48 (customer / min) l Server Utilization  = /  = 5 / 8 = l Probability of 2 or more customers in system P[N  2] =   = l Mean steady-state number in the system L = E[N] =  / (1 -  ) = l S.D. of number of customers in the system  N = sqrt(  ) / (1 -  ) = 2.108

Queueing Theory47 M/M/1 Example II (cont.) l Mean time a customer spends in the system W = E[w] = E[s] / (1 -  ) = 80 (min) l S.D. of time a customer spends in the system  w = E[w] = 80 (min) l Mean steady-state number of customers in queueN q =   / (1 -  ) = 1.04 l Mean steady-state queue length of nonempty QsE[N q | N q > 0] = 1 / (1 -  ) = 2.67 l Mean time in queue W q = E[q] =  E[s] / (1 -  ) = 50 (min)

Queueing Theory48 M/M/1 Example II (cont.) l Mean time in queue for those who must wait E[q | q > 0] = E[w] = 80 (min) l 90th percentile of the time in queue  q (90) = E[w] ln (10  ) = 80 * = (min)

M/M/m, M/M/m/m, M/M/∞ M/M/m (infinite buffer) detailed balance equations in steady state

M/M/m where From

M/M/m The probability that all servers are busy - Erlang C formula expected number of customers waiting in queue

M/M/m average waiting time of a customer in queue average delay per customer average number of customer in the system by Little ’ s theorem

Queueing Theory53 M/M/s Case Example I Find p 0

Queueing Theory54 M/M/s Case Example I (cont.) = (  43% of time, system is empty) as compared to m = 1: P 0 = 0.20

Queueing Theory55 M/M/s Case Example I (cont.) Find W W q = L q / = / (1/10) = 1.52 (min) W = W q + 1 /  = / (1/8) = 9.52 min) What proportion of time is both repairman busy? (long run) P(N  2) = 1 - P 0 - P 1 = = 0.228(Good or Bad?)

M/M/∞ M/M/∞: The infinite server case The detailed balance equations Then

M/M/m/m M/M/m/m : The m server loss system when m servers are busy, next arrival will be lost circuit switched network model

M/M/m/m The blocking probability (Erlang-B formula)

Moment Generating Function

Discrete Random Variables

Continuous Random Variables