Queuing Analysis Gregory (Grisha) Chockler, Zinovi Rabinovich, Ittai - PowerPoint PPT Presentation

Queuing Analysis Gregory (Grisha) Chockler, Zinovi Rabinovich, Ittai Abraham Operating Systems Course, Spring 2003 Hebrew University of Jerusalem, Israel Queuing Analysis – p.1

Plan • Review of basic probability. • Markov processes and Poisson processes. • Models in Queuing Theory. • Analysis of the M/M/1 model. • Analysis of the M/M/1/b model. • Proof of Poisson process definition equivalence. Queuing Analysis – p.2

Random Variables • A discrete random variable X can get some countable set of values S = { x 1 , x 2 , x 3 . . . , } . • A function f : R → [0 .. 1] such that � x ∈ S f ( x ) = 1 , and ∀ x / ∈ S , f ( x ) = 0 defines the probability function of X : Pr[ X = x ] = f ( x ) . • A continuous random variable X can get values in some intervals of numbers. � ∞ • A function f : R → R + such that −∞ f ( x ) dx = 1 , defines the probability density function for X . � ∞ Thus Pr[ X ≥ x ] = x f ( x ) dx . Queuing Analysis – p.3

Conditional Probability, Independence and Sums • The conditional probability that A occurs given that B occurs is defined to be Pr[ A | B ] = Pr[ A ∩ B ] Pr[ B ] . • Bayes’ theorem: Pr[ A | B ] = Pr[ B | A ] Pr[ A ] . Pr[ B ] • Events A and B are independent if Pr[ A ∩ B ] = Pr[ A ] Pr[ B ] . • X, Y are independent if X ∈ x and Y ∈ y are independent events for all x, y . • If X, Y are independent, with probability functions f X , f Y then for the sum Z = X + Y we have f Z ( z ) = � x ∈ X f X ( x ) f Y ( z − x ) and � ∞ −∞ f X ( x ) f Y ( z − x ) dx . f Z ( z ) = Queuing Analysis – p.4

Expectation • The expectation of a random variable X with probability function f ( x ) is defined as � ∞ x | f ( x ) > 0 f ( x ) x , and E ( x ) = −∞ f ( x ) xdx . E ( X ) = � • Linearity of expectation: E ( aX + bY ) = aE ( X ) + bE ( Y ) . • Example: exponential distribution with � θe − θx x ≥ 0 parameter θ : f ( x ) = 0 x < 0 • Recall: udv , take v = x , u = − θe − θx . � � vdu = vu − � ∞ � ∞ 0 − e − θx = • E ( x ) = 0 θe − θx xdx = − xe − θx | ∞ 0 − � ∞ 0 e − θx = − 1 θ e − θx | ∞ 0 = 1 θ . 0 + Queuing Analysis – p.5

Variance • Variance is defined as: = E ( X 2 ) − E ( X ) 2 (if it ( X − E ( X )) 2 � � var ( X ) = E converges). • Standard deviation: σ X = � var ( X ) , coefficient σ X of variation: cv ( X ) = E ( x ) . • Covariance: cov ( X, Y ) = E ( XY ) − E ( X ) E ( Y ) , cov ( X, X ) = var ( X ) , if X and Y are independent then cov ( X, Y ) = 0 . • var ( aX ) = a 2 var ( X ) , and var ( X + Y ) = var ( X ) + var ( Y ) + 2 cov ( X, Y ) . Queuing Analysis – p.6

Markov Processes • A stochastic process is a series of random variables X 1 , X 2 , . . . . • A markov process is a stochastic process in which the probability of X t is determined by only by X t − 1 . • Formally, Pr( X t = i t | X t − 1 = i t − 1 ) = Pr( X t = i t | X t − 1 = i t − 1 , X t − 2 = i t − 2 , . . . , X 1 = i 1 ) . • A markov process can be viewed as a weighted directed graph in which ∀ v ∈ V , � ( u → v ) ∈ E d ( u, v ) = 1 . Queuing Analysis – p.7

Example of a Markov Process 1/2� 1� A� B� 1/2� 1/2� 1/2� 1/3� 1/3� 1/3� 1/2� 1/2� C� E� D� 1/2� Queuing Analysis – p.8

Stationary Distribution • Under certain conditions, markov process ( V, E, d ) has a stationary distribution π . lim t →∞ X t = π • If X t ∼ π then X t +1 ∼ π . ( � v ∈ V π v = 1 ) • Formally, Pr( X t +1 = v ) = � ( u → v ) Pr( X t = u ) d ( u, v ) , π v = � ( u → v ) π u d ( u, v ) . • Balance property : given a cut ( U, ¯ U ) the flow of probability π is balanced. • Formally, � � π u d ( u, v ) = π v d ( v, u ) ( u → v ) | u ∈ U,v ∈ ¯ ( v → u ) | u ∈ U,v ∈ ¯ U U . Queuing Analysis – p.9

Example of a Stationary Distribution 1/2� A� 1� B� 1/4� 1/4� 1/2� 1/2� 1/2� 1/3� 1/3� 1/3� 1/2� 1/2� C� E� D� 1/6� 1/6� 1/6� 1/2� 1/6(½)+1/6(½)+1/6(½)=¼(1/3)+¼(1/3)+¼(1/3)� Queuing Analysis – p.10

Poisson Process • Consider a stochastic process { N ( T ) | t ≥ 0 } , N (0) = 0 , N ( t ) counts the number of occurrences of an event, thus N ( t ) ∈ N . • N ( t ) = n means that there have been n occurrences by time t . If N ( t ) = n now, from there it can only go to state n + 1 . This happens as soon as the next event takes place. • Let T n denote the time the process spends in state n . • Poisson process: If T n are exponentially distributed with the same mean, say 1 θ , for all n and all the T n ’s are independent. Queuing Analysis – p.11

Poisson Process - Equivalent Definitions • Poisson process 1: All T n ∼ exp ( θ ) , and all the T n ’s are independent. • Poisson process 2: N ( t ) ∼ Poisson ( θt ) . Pr[ N ( t ) = k ] = ( θt ) k k ! e − θt , and the number of events in non-overlapping intervals are independent. • Poisson process 3: N (0) = 0 , for any k as t → 0 , • Pr [ N ( k + t ) = N ( k )] = 1 − θt + o ( t ) . • Pr [ N ( k + t ) = 1 + N ( k )] = θt + o ( t ) . • Pr [ N ( k + t ) ≥ 2 + N ( k )] = o ( t ) • Increments in non-overlapping intervals are independent. (Recall f ( x ) = o ( g ( x )) means f ( x ) g ( x ) = 0 ). lim x → 0 Queuing Analysis – p.12

Queuing Models Things we need to decide about our model: • Job arrival, Job service time. • No. of processor, their qualities. • Queue size, dispatch discipline, No. of queues. • Kandell’s notation: B/D/n/k/dd, B=arrival, D=service time, n=processors, k=queue length, dd=dispach discipline. Queuing Analysis – p.13

The M/M/1 model • Single FIFO queue, single processor. • Jobs arrive at rate λ , thus job births form a poisson process with parameter λ . • The service time of a job is exponentially distributed with parameter µ , thus job deaths form a poisson process with parameter µ . Queuing Analysis – p.14

Analysis of M/M/1 • Examine time scale at at small intervals δ, 2 δ, 3 δ, . . . . • Analyze as δ → 0 so omit o ( δ ) factors. • � � π u d ( u, v ) = π v d ( v, u ) ( u → v ) | u ∈ U,v ∈ ¯ ( v → u ) | u ∈ U,v ∈ ¯ U U Queuing Analysis – p.15

Analysis of M/M/1 (part 2) • Let p be the stationary distribution. For any i , p i is the probability of having i jobs in the system at the steady state. • By the balance property, at the steady state p j ( µδ + o ( δ )) = p j − 1 ( λδ + o ( δ )) • Taking δ → 0 : λδ + o ( δ ) λδ + o ( δ ) δ p 0 = λ p 1 = lim δ → 0 µδ + o ( δ ) p 0 = lim δ → 0 µ p 0 µδ + o ( δ ) δ • Same way for every j > 0 , p j = λ µ p j − 1 . • Denote the traffic intensity ρ = λ µ . • Since p j = ρp j − 1 so p j = ρ j p 0 . Queuing Analysis – p.16

Analysis (part 3) i =0 a i = a k +1 − 1 i =0 a i = • Recall � k a − 1 , � ∞ 1 1 − a , i =0 ia i = � ∞ a (1 − a ) 2 for a < 1 . • Now 1 = � ∞ i =0 p i . Using p j = ρ j p 0 we have 1 = � ∞ p 0 i =0 ρ i p 0 = 1 − ρ so p 0 = 1 − ρ . • The expected number of jobs in the system ∞ ∞ ρ i (1 − ρ ) ρ i = (1 − ρ ) iρ i = � � E [ n ] = 1 − ρ i =0 i =0 • The expected number of waiting jobs in the system i =0 i (1 − ρ ) ρ i +1 = ρ 2 is E [ w ] = � ∞ 1 − ρ . Queuing Analysis – p.17

Analysis (part 4) • Response time r = time waiting + time served. • Little’s law: E [ r ] = E [ n ] (targil). λ • So λ ρ 1 µ E [ r ] = (1 − ρ ) λ = = (1 − λ µ − λ µ ) λ Queuing Analysis – p.18

The M/M/1/b model • Same as M/M/1 but now queue has bounded size b . • When the queue is full, arriving jobs are dropped. • Makes the model a bit more realistic. Queuing Analysis – p.19

Analysis of M/M/1/b • Same as M/M/1 we have p j = ρ j p 0 . i =0 a i = a k +1 − 1 • Recall � k a − 1 . ρ b +1 − 1 • 1 = � b i =0 ρ i p 0 = p 0 ρ − 1 . 1 − ρ • So p 0 = 1 − ρ b +1 i =0 ia i − 1 = ( k +1) a k − a k +1 − 1 • Recall � k ( a − 1) 2 . a − 1 • The expected number of jobs: E [ n ] = � b 1 − ρ 1 − ρ b +1 ρ i i = i =0 ( b +1) ρ b 1 − ρ − ρ ( b +1) ρ b +1 � − ρ b +1 − 1 � 1 − ρ ρ 1 − ρ b +1 ρ = • ρ − 1 ( ρ − 1) 2 1 − ρ b +1 Queuing Analysis – p.20

Poisson Process - Equivalent Definitions • Poisson process 1: All T n ∼ exp ( θ ) , and all the T n ’s are independent. • Poisson process 2: N ( t ) ∼ Poisson ( θt ) . Pr[ N ( t ) = k ] = ( θt ) k k ! e − θt , and the number of events in non-overlapping intervals are independent. • Poisson process 3: N (0) = 0 , for any k as t → 0 , • Pr [ N ( k + t ) = N ( k )] = 1 − θt + o ( t ) . • Pr [ N ( k + t ) = 1 + N ( k )] = θt + o ( t ) . • Pr [ N ( k + t ) ≥ 2 + N ( k )] = o ( t ) • Increments in non-overlapping intervals are independent. (Recall f ( x ) = o ( g ( x )) means f ( x ) g ( x ) = 0 ). lim x → 0 Queuing Analysis – p.21