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Unformatted text preview: Probability and Statistics with Reliability, Queuing and Computer Science Applications Second edition by K.S. Trivedi Publisher-John Wiley & Sons Chapter 8 (Part 3) :Continuous Time Markov Chains “Pure” Performance Modeling Dept. of Electrical & Computer engineering Duke University Email: kst@ee.duke.edu URL: www.ee.duke.edu/~kst Copyright © 2003 by K.S. Trivedi 1 Performance Modeling using Elementary Queuing Theory • Birth-Death Process – – – – – – M/M/1 queue M/M/m queue M/M/1/n queue Cyclic Queuing Model Response Time in a Client-Server System Wireless Handoff Model • Non-Birth-Death Process – M/M/2 Heterogeneous queue – MMPP Arrival Process – MMPP/M/1 queue Copyright © 2003 by K.S. Trivedi 2 Continuous Time Birth-Death Process • A CTMC with I={0,1,2,…} forms a B-D process, if ?i, i=0,1,2,.. and µi, i=1,2,.. exist, such that the transition rates are given by given Birth rate ?i ≥ 0 and Death rate µi ≥ 0 Copyright © 2003 by K.S. Trivedi 3 Continuous Time Birth-Death Process (contd.) The state diagram of the birth-death process is given as The generator matrix Q can be shown to be as Copyright © 2003 by K.S. Trivedi 4 Continuous Time Birth-Death Process (contd.) In Steady-state, Copyright © 2003 by K.S. Trivedi 5 Steady State Balance Equations These are called balance eqns. Re-arranging above, v ?i and µi should be > 0 for irreducibility condition, which means every state can be reached from every other state. Copyright © 2003 by K.S. Trivedi 6 Steady State Balance Equations (contd.) • Note that limiting state probability vector [p0, p1, . . .] is now completely determined and are non zero provided converges which also means that all states of the CTMC are recurrent nonnull. Copyright © 2003 by K.S. Trivedi 7 M/M/1 Queue • Arrivals process is Poisson, i.e., interarrival times are all i.i.d, EXP(?). Poisson arrival Process with rate ? • Service times are i.i.d, EXP(µ). • N(t) is a birth-death process, with ?k=?; µk=µ. • Define, ?=?/µ (traffic intensity, in Erlangs) Copyright © 2003 by K.S. Trivedi 8 M/M/1 queue (contd.) • From the balance equations, we get • Uo = 1 – πo is known as server utilization and is interpreted as the proportion of time the server is busy • Expected # of customers, Copyright © 2003 by K.S. Trivedi 9 M/M/1 queue (contd.) • This measure (E[N(t)]) can be viewed as a weighted average, . • By choosing suitable weights for the states of a CTMC, we can get most measures of interest and the resulting model is known as an MRM(Markov Reward Model). Above weighted average is the expected reward rate in the steady state • Other measures: – Average queue length (E[Q]) – Average response time – Average waiting time (avg. time spent in the queue before service begins) etc. Copyright © 2003 by K.S. Trivedi 10 M/M/1 queue: Little’s formula • Let the random variable R denote the response time (defined as the time elapsed from the instant of job arrival until its completion) Little’s law states E[R] = E[N]/? • Here • Response time (R) = waiting time (W) + service time (S ) E[W] = E[R] – E[S] = 1/(µ(1-?)) - 1/ µ . Copyright © 2003 by K.S. Trivedi 11 Reward rate assignment for Measures for M/M/1 queue Mean no. in the system E[N] rj=j Mean Response time E[R] rj=j/? Throughput E[T] r0=0 rj=µ ρ 1− ρ 1/ µ 1− ρ ? j>0 Copyright © 2003 by K.S. Trivedi 12 Response time distribution (Tagged job approach) • With FCFS scheduling and steady-state conditions – If there are already n jobs in the system, the next job (n+1)st will experience a response time =R= S*+S’1+S2+..+Sn – S* : service time for the (n+1)st job; S’1: residual service time for job currently undergoing service (#1). – Because of the memoryless property, these times are EXP(µ). – Hence, for N=n, the conditional LST of R is, Copyright © 2003 by K.S. Trivedi 13 Response time distribution (Tagged job approach) • With FCFS scheduling and steady-state conditions – Note for a Poisson arrival process, it is known that, the state of the system as seen by an arriving customer is statistically the same as the state of the system at an arbitrary time. This Property is known as PASTA –Poisson Arrival See Time Average. Hence unconditioning using – Then, taking inverse transform, Copyright © 2003 by K.S. Trivedi 14 M/M/m queue • m-servers service the queue. 1 Poisson arrivals (rate ? ) µ . . . m Copyright © 2003 by K.S. Trivedi 15 M/M/m Queue Solution Copyright © 2003 by K.S. Trivedi 16 M/M/m Queue performance measures • Average number in system E[N]: rk = k Copyright © 2003 by K.S. Trivedi 17 M/M/m Queue performance measures • Server utilization: rv M - number of busy servers. For number of customers 0 <= k <= m, the number of busy servers = k. Beyond that the number of busy servers = m. • A customer may have to join the queue if # customers already in the system >= m Copyright © 2003 by K.S. Trivedi 18 Separate or common queue? • M/M/2. • 2 cases – Case 1: Two independent queues Two separate Poisson streams à 2 separate M/M/1 queues λ/2 µ λ/2 µ – Case 2: M/M/2 case λ/2 Two separate Poisson streams λ/2 µ 1 λ Combined Poisson steams Copyright © 2003 by K.S. Trivedi µ 2 19 Comparative performance • Case 1: For each M/M/1 queue, • Case 2: Common queue M/M/2 Copyright © 2003 by K.S. Trivedi 20 M/M/1/n Queue • Finite queue size, finite buffer space à finite state space. Max. capacity n Poisson Stream λ µ Arrivals exceeding capacity are rejected 0 1 µ λ λ λ λ 2 µ • Steady State Solution: n µ µ v Note that given irreducibility (?i and µi should be >0) , being a finite CTMC all states are recurrent non-null and hence there is no additional condition for stability. Copyright © 2003 by K.S. Trivedi 21 M/M/1/n Queue Performance Measures • Expected # of jobs in the system: – rk = k, • Loss probability – rn = 1, rk = 0, k=0,1,..,n-1 • Throughput – rk = µ , k=1,2, ..,n; r0 = 0 (or, rk = λ , k=0,1,2, ..,n1; rn = 0) • Mean response time E[R]=E[N]/ (λ(1−πn)) Copyright © 2003 by K.S. Trivedi 22 M/M/1/n: Response time distribution • Response time distribution: Job may be rejected (or accepted) – Unconditional – Conditional (conditioned on the job being accepted): • Reward assignment: for the kth state, response time experienced by the tagged task is sum of k+1 service times, each of which is EXP(µ), i.e., (k+1)-stage Erlang. Copyright © 2003 by K.S. Trivedi 23 M/M/1/n: Response time distribution – Unconditional (defective) distribution: – There is a defect (or mass at infinity) equal to πn , the probability that the job is rejected - Conditioned on job being accepted, R has a non-defective distribution: Copyright © 2003 by K.S. Trivedi 24 Measures for M/M/1/n system Copyright © 2003 by K.S. Trivedi 25 Model made in SHARPE GUI Copyright © 2003 by K.S. Trivedi 26 Analysis frame showing output asked Steady State prob. in all state asked Throughput, Mean no. in system asked Copyright © 2003 by K.S. Trivedi 27 Reward Rate Assignment Copyright © 2003 by K.S. Trivedi 28 Output Generated by SHARPE Steady State prob. for all states Mean no. in the system Throughput Copyright © 2003 by K.S. Trivedi 29 Cyclic Queuing model -Example • Consider the cyclic queuing model shown below. • Lengths of successive CPU execution bursts are independent exponentially distributed rv with mean 1/µ and that successive I/O burst times are independent exponentially distributed variables with mean 1/? • At the end of a CPU burst, a program requests an I/O operation with probability q1 or it completes execution with probability q0 q0 + q1 = 1. Copyright © 2003 by K.S. Trivedi 30 Cyclic Queuing model –Example (contd.) • At the end of a program completion, another statistically identical program enters the system, leaving the number of programs in the system at a constant level n (known as the degree of multiprogramming). Copyright © 2003 by K.S. Trivedi 31 Cyclic Queuing model –Example (contd.) • Let the number of programs in the CPU queue including any being served at the CPU denote the state of the system, i, where 0 = i = n. Then state diagram becomes • Steady State Probability is given by where ρ = λ/(µq1) Copyright © 2003 by K.S. Trivedi 32 Cyclic Queuing model –Example (contd.) • System throughput is defined as • E[T] is proportional to the CPU utilization, U0 for fixed values of µ and q0. • Let the random variable B0 denote the total CPU time requirement of a tagged program. Then B0 is EXP(µq0 ). • the average number of visits V0 to the CPU is V0 = 1/q0 , and thus E[B0] = V0E[S0] =1/(µq0 ), where E[S0] = 1/µ is the average CPU time per burst. • If B1 represents the total I/O service time per program, then Copyright © 2003 by K.S. Trivedi 33 Cyclic Queuing model – Example (contd.) • Defining parameter ? as • CPU utilization as a function of the degree of multiprogramming • Thus ? indicates the relative measure of the CPU versus I/O requirements of a program. If the CPU requirement E[B0] is less than the I/O requirement E[B1] (i.e., ? < 1), the program is said to be I/Obound; if ? > 1, then program is said to be CPUbound; and otherwise it is called balanced. Copyright © 2003 by K.S. Trivedi 34 Model made in SHARPE GUI Copyright © 2003 by K.S. Trivedi 35 Analysis Frame Copyright © 2003 by K.S. Trivedi SS prob. and Utilization asked 36 Reward Rate assigned Copyright © 2003 by K.S. Trivedi 37 Output generated by SHARPE SS prob. and Utilization calculated Copyright © 2003 by K.S. Trivedi 38 Response Time in a Client-Server System • Model can be given as • Consider a client–server system with M clients in which individual think times are exponentially distributed with mean 1/? seconds and the service rate is µ Copyright © 2003 by K.S. Trivedi 39 Response Time in a Client-Server System • State Diagram can be drawn as Μλ 0 (Μ−1)λ µ 2λ …... 2 1 µ (M-2)λ µ µ λ M-1 M µ • Assuming service time per request, B0, is exponentially distributed with mean E[B0] = 1/µ seconds. • Then the steady-state probability that there are n requests executing or waiting on the CPU is given by Copyright © 2003 by K.S. Trivedi 40 Response Time in a Client-Server System (contd.) • Probability that the CPU is idle is • The CPU utilization U0 is 1-p0, and the average rate of request completion is E[T] = µ(1- p 0) = U0/E[B0]. • If E[R] is average response time, then on the average a request is generated by a given client in E[R] + (1/?) seconds. • Thus the average request generation rate of the client subsystem is M/[E[R] + (1/?)]. Copyright © 2003 by K.S. Trivedi 41 Response Time in a Client-Server System (contd.) • In the steady state, the request generation and completion rates must be equal. •E[R] is plotted as a function of the number of clients, M, assuming 1/? = 15 s and 1/µ = 1 s. Copyright © 2003 by K.S. Trivedi 42 Response Time in a Client-Server System (contd.) • When the number of clients M = 1, there is no queuing and the response time E[R] equals the average service time E[B0]. • As the number of clients increases, there is increased congestion as the server utilization U0 approaches 1. In the limit M ? 8 ,E[R] is a linear function [ME[B0] - (1/?)] of M. • In this limit, the installation of an additional client increases every other client’s response time by the new client’s service time E[B0]. • The number of clients, M*, for which the heavy-load asymptote E[R] = ME[B0] - (1/?) intersects with the lightload asymptote E[R] = E[B0] is therefore called the saturation number and is given by Copyright © 2003 by K.S. Trivedi 43 Handoffs in wireless cellular networks • Handoff: When an MS moves across a cell boundary, the channel in the old BS is released and an idle channel is required in the new BS • Hard handoff: the old radio link is broken before the new radio link is established (AMPS, GSM, DECT, D-AMPS, and PHS) Copyright © 2003 by K.S. Trivedi 44 Wireless Cellular System Traffic in a cell New Calls Common Channel Pool Call completion Handoff Calls Handoff out From neighboring cells To neighboring cells A Cell Copyright © 2003 by K.S. Trivedi 45 Performance Measures: Loss formulas or probabilities • When a new call (NC) is attempted in an cell covered by a base station (BS), the NC is connected if an idle channel is available in the cell. Otherwise, the call is blocked • If an idle channel exists in the target cell, the handoff call (HC) continues nearly transparently to the user. Otherwise, the HC is dropped • Loss Formulas – New call blocking probability, Pb : Percentage of new calls rejected – Handoff call dropping probability, Pd : Percentage of calls forcefully terminated while crossing cells Copyright © 2003 by K.S. Trivedi 46 Guard Channel Scheme Handoff dropping less desirable than new call blocking! Handoff call has Higher Priority: Guard Channel Scheme GCS: g channels are reserved for handoff calls. g trade-off between Pb Copyright © 2003 by K.S. Trivedi & Pd 47 Assumptions • • • • Poisson arrival stream of new calls ?1 Poisson stream of handoff arrivals ?2 Limited number of channels: n Exponentially distributed completion time of ongoing calls µ1 • Exponentially distributed cell departure time of ongoing calls µ2 Copyright © 2003 by K.S. Trivedi 48 Markov chain model of wireless hard handoff C(t): the number of busy channels at time t Steady-state probability Copyright © 2003 by K.S. Trivedi 49 Loss formulas for wireless network with hard handoff Dropping probability for handoff: Blocking probability of new calls: q Notation: if we set g=0, the above expressions reduces to the classical Erlang-B loss formula Copyright © 2003 by K.S. Trivedi 50 Computational aspects • Overflow and underflow might occur if n is large • Numerically stable methods of computation are required – Recursive computation of dropping probability for wireless networks – Recursive computation of the blocking probability – For loss formula calculator, see webpage: http://www.ee.duke.edu/~kst/wireless.html Copyright © 2003 by K.S. Trivedi 51 Non-birth-death examples • We consider two example performance models which give rise to non-birth-death CTMC models – M/M/2 heterogeneous queue And – A queue with non-Poisson arrival stream; in fact with MMPP (Markov modulated Poisson process) arrivals Copyright © 2003 by K.S. Trivedi 52 M/M/2 Queue with Heterogeneous Servers (µ1 > µ2). • Variant of M/M/2 with homogenous rate. • The state of the system is defined to be the tuple (n1, n2) where n1 = 0 denotes the number of jobs in the queue including any at the faster server, and n2 ∈{0, 1} denotes the number of jobs at the slower server • When both servers are idle, the faster server is scheduled for service before the slower one. • The block diagram of the system is given Copyright © 2003 by K.S. Trivedi 53 M/M/2 Queue with Heterogeneous Servers (contd.) • State Diagram is given by • Solving balance eqns. we have Copyright © 2003 by K.S. Trivedi 54 M/M/2 Queue with Heterogeneous Servers (contd.). • The average number of jobs in the system may now be computed by assigning the reward rate equal rn1, n2 = n1 + n2 as the number of customers in the system in state (n1, n2). • Therefore, the average number of jobs is given by Copyright © 2003 by K.S. Trivedi 55 Example- M/M/2 server • Assume that the job stream modeled as a Poisson process with rate ? jobs/minute, and the service times µ2 jobs/minute. • Thus, ?2 = ?/µ2, and the average response time is given by E[R2] = (1/µ2)/(1-?2). • Suppose this response time is considered intolerable by the users and an extra server is added to the system. Let the service rate µ1 of the new server be equal to αµ2 for some α > 1. • Now for both machines we have Copyright © 2003 by K.S. Trivedi 56 Example- M/M/2 server (contd.) • Average no. of customer in the system is given by • Now only using the faster server, response time becomes • The condition under which E[R1] = E[R] can be simplified to Copyright © 2003 by K.S. Trivedi 57 Example- M/M/2 server (contd.) • Thus, for example, if ? = 0.2 and µ2 = 0.25 so that ?2 = 0.8, then if the new server is more than 3 times faster than the old one, the inequality shown above is satisfied, and surprisingly, it is better to disconnect the slower machine altogether. • Average response times (in minutes) of the three configurations for different values of α with ? = 0.2 and µ2 = 0.25. Copyright © 2003 by K.S. Trivedi 58 M/M/2 Queue with Heterogeneous Servers (contd.) • Also study perceived response time and perceived queue length idea (Problem 4 on p. 483) • Study the optimization problem (Problem 2 on p. 483) and see the non-intuitive result Copyright © 2003 by K.S. Trivedi 59 Poisson Process • Consider a pure birth process (i.e. all death rates are 0 or µk = 0, k = 0,1,2,…) • Constant birth rate (i.e. λk = λ for k = 0,1,2,…) • N(t) is the number of arrivals (births) in time t The state diagram can be shown as λ 0 λ 1 λ 2 λ …... λ k Copyright © 2003 by K.S. Trivedi λ k+1 ….. 60 Poisson Process (contd.) • Solving the above differential equations with the initial conditions that • We get, Copyright © 2003 by K.S. Trivedi 61 Markov Modulated Poisson Process • An MMPP is a doubly stochastic Poisson process whose arrival rate is “modulated” by an irreducible continuous time Markov chain. • Let Q = [qij ]m×m be the generator matrix of the CTMC with m states. Each state i is assigned a Poisson arrival rate, ?i , i= 1, 2, . . .,m. • The Poisson arrival rate is determined by the state of the CTMC; thus, when the Markov chain is in state i, arrivals occur according to a Poisson process of rate ?i . • Let ? denote the arrival rate vector ? = [?1, ?2, . . . , ?m]T. • Let e = [1, 1, . . . , 1]T . Copyright © 2003 by K.S. Trivedi 62 Markov Modulated Poisson Process (contd.) • The reason for using this in traffic modeling is that MMPP has the capability of capturing some of the most important correlations between interarrival times and still remains analytically tractable. • MMPP is a special case of the Markovian arrival process (MAP). Copyright © 2003 by K.S. Trivedi 63 MMPP-the counting process • Lets see the associated counting process of MMPP. • Let N(t) be the number of arrivals in (0, t] and J(t) the state of the modulating CTMC. • The bivariate process {J(t),N(t), t ≥ 0} is the counting process whose state space is {1, 2, . . .,m} × {0, 1, . ..}. • Transitions between states with the same number of arrivals—say, (i, n) and (j, n)—are the same as transitions between state i and j of the CTMC with rate qij and qji, respectively. • An arrival may occur in any of the modulating CTMC states, resulting in the counter increasing by one. • So a transition from state (i, n) to state (i, n+1) with rate ?i takes place. • The state diagram of the bivariate process for a three-state MMPP is shown. Copyright © 2003 by K.S. Trivedi 64 MMPP-the counting process (contd.) • The counting process is also a homogeneous CTMC. Let π be the steady-state vector of the MMPP, which is the solution to • So we can see that • The result shows that steady-state expected number of arrivals in an interval of length t is the product of time duration t and the average arrival rate, πλ. where πλ is the sum of rates weighted by steady-state probabilities of the modulating CTMC. Copyright © 2003 by K.S. Trivedi 65 MMPP/M/1 Queue • We now consider an MMPP/M/1 queue in which the arrival process is an MMPP characterized by the generator matrix Q of the modulating CTMC and the arrival rate vector ?. • The service time is exponentially distributed with mean 1/µ. The buffer size of the queue is assumed to be infinite. • The state diagram for an MMPP/M/1 queue is shown next, in which the MMPP arrival process has three states. Copyright © 2003 by K.S. Trivedi 66 MMPP/M/1 Queue (contd.) • The structure of the Markov chain suggests that its steady-state solution may have the same form as that of an M/M/1 queue. • Let pi,n be the steady state probability that the MMPP modulating process is in state i and the system has n jobs. • It can be shown that • where ?i = ?i /µ is the traffic intensity conditioned on the MMPP being in state i. Copyright © 2003 by K.S. Trivedi 67 Other arrival processes • MAP, BMAP, PH type arrival processes have been studied [LUCA 1990, NEUT 1978, FISC 1993] • Also periodic NHPP type arrivals have been studied [RIND 1995] Copyright © 2003 by K.S. Trivedi 68 ...
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