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 = {x1, x2, x3 . . . , }.

• 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

fX, fY then for the sum Z = X + Y we have fZ(z) =

x∈X fX(x)fY (z − x) and

fZ(z) = ∞

−∞ fX(x)fY (z − x)dx.

Queuing Analysis – p.4

Expectation

• The expectation of a random variable X with

probability function f(x) is defined as

E(X) =

x|f(x)>0 f(x)x, and E(x) =

∞

−∞ f(x)xdx.

• Linearity of expectation:

E(aX + bY ) = aE(X) + bE(Y ).

• Example: exponential distribution with

parameter θ: f(x) =

• θe−θx

x ≥ 0 x < 0

• Recall:
• vdu = vu −
• udv, take v = x, u = −θe−θx.
• E(x) =

∞

0 θe−θxxdx = −xe−θx|∞ 0 −

∞

0 −e−θx =

0 + ∞

0 e−θx = −1 θe−θx|∞ 0 = 1 θ.

Queuing Analysis – p.5

Variance

• Variance is defined as:

var(X) = E

• (X − E(X))2

= E(X2) − E(X)2 (if it

converges).

• Standard deviation: σX =
• var(X), coefficient
• f variation: cv(X) =

σX 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) = a2var(X), and

var(X + Y ) = var(X) + var(Y ) + 2cov(X, Y ).

Queuing Analysis – p.6

Markov Processes

• A stochastic process is a series of random

variables X1, X2, . . . .

• A markov process is a stochastic process in

which the probability of Xt is determined by only by

Xt−1.

• Formally, Pr(Xt = it|Xt−1 = it−1) = Pr(Xt = it|Xt−1 =

it−1, Xt−2 = it−2, . . . , X1 = i1).

• 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

A B C D E 1/2 1/2 1/2 1/2 1/2 1/2 1 1/2 1/3 1/3 1/3

Queuing Analysis – p.8

Stationary Distribution

• Under certain conditions, markov process (V, E, d)

has a stationary distribution π. limt→∞ Xt = π

• If Xt ∼ π then Xt+1 ∼ π. (

v∈V πv = 1)

• Formally, Pr(Xt+1 = v) =

(u→v) Pr(Xt = u)d(u, v),

πv =

(u→v) πud(u, v).

• Balance property: given a cut (U, ¯

U) the flow of

probability π is balanced.

• Formally,
• (u→v)|u∈U,v∈ ¯

U

πud(u, v) =

• (v→u)|u∈U,v∈ ¯

U

πvd(v, u)

.

Queuing Analysis – p.9

Example of a Stationary Distribution

A 1/4 B 1/4 C 1/6 D 1/6 E 1/6 1/2 1/2 1/2 1/2 1/2 1/2 1 1/2 1/3 1/3 1/3

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

• f an event, thus N(t) ∈ N.
• N(t) = n means that there have been n
• ccurrences 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 Tn denote the time the process spends in state

n.

• Poisson process: If Tn are exponentially

distributed with the same mean, say 1

θ, for all n and

all the Tn’s are independent.

Queuing Analysis – p.11

Poisson Process - Equivalent Definitions

• Poisson process 1: All Tn ∼ exp(θ), and all the

Tn’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

limx→0

f(x) g(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, 2δ, 3δ, . . . .
• Analyze as δ → 0 so omit o(δ) factors.
• (u→v)|u∈U,v∈ ¯

U

πud(u, v) =

• (v→u)|u∈U,v∈ ¯

U

πvd(v, u)

Queuing Analysis – p.15

Analysis of M/M/1 (part 2)

• Let p be the stationary distribution. For any i, pi is

the probability of having i jobs in the system at the steady state.

• By the balance property, at the steady state

pj(µδ + o(δ)) = pj−1(λδ + o(δ))

• Taking δ → 0:

p1 = limδ→0

λδ+o(δ) µδ+o(δ)p0 = limδ→0 λδ+o(δ) δ µδ+o(δ) δ

p0 = λ

µp0

• Same way for every j > 0 , pj = λ

µpj−1.

• Denote the traffic intensity ρ = λ

µ.

• Since pj = ρpj−1 so pj = ρjp0.

Queuing Analysis – p.16

Analysis (part 3)

• Recall k

i=0 ai = ak+1−1 a−1 , ∞ i=0 ai = 1 1−a,

∞

i=0 iai = a (1−a)2 for a < 1.

• Now 1 = ∞

i=0 pi. Using pj = ρjp0 we have

1 = ∞

i=0 ρip0 = p0 1−ρ so p0 = 1 − ρ.

• The expected number of jobs in the system

E[n] =

∞

• i=0

i(1 − ρ)ρi = (1 − ρ)

∞

• i=0

iρi = ρ 1 − ρ

• The expected number of waiting jobs in the system

is E[w] = ∞

i=0 i(1 − ρ)ρi+1 = ρ2 1−ρ.

Queuing Analysis – p.17

Analysis (part 4)

• Response time r = time waiting + time served.
• Little’s law: E[r] = E[n]

λ

(targil).

• So

E[r] = ρ (1 − ρ)λ =

λ µ

(1 − λ

µ)λ

= 1 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 pj = ρjp0.
• Recall k

i=0 ai = ak+1−1 a−1 .

• 1 = b

i=0 ρip0 = p0 ρb+1−1 ρ−1 .

• So p0 =

1−ρ 1−ρb+1

• Recall k

i=0 iai−1 = (k+1)ak a−1

− ak+1−1

(a−1)2 .

• The expected number of jobs:

E[n] = b

i=0 1−ρ 1−ρb+1ρii =

• 1−ρ

1−ρb+1ρ

• (b+1)ρb

ρ−1

− ρb+1−1

(ρ−1)2

• =

ρ 1−ρ − ρ(b+1)ρb+1 1−ρb+1

Queuing Analysis – p.20

Poisson Process - Equivalent Definitions

• Poisson process 1: All Tn ∼ exp(θ), and all the

Tn’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

limx→0

f(x) g(x) = 0).

Queuing Analysis – p.21

Poisson Process - Proof of (1 ⇒ 3)

• Suppose for all n, Tn ∼ exp(θ) are independent.
• Note that Pr[N(k + t) = b + N(k)] = Pr[N(t) = b] due

to independence.

• We need to find

Pr[N(t) = 0], Pr[N(t) = 1], Pr[N(t) > 1].

• Pr[N(t) = 0] = Pr[T0 ≥ t] =

∞

t

θe−θxdx = e−θt.

• Using L

’Hôpital’s rule:

lim

t→0

e−θt − 1 + θt t = lim

t→0 −θe−θt + θ = 0

.

• So Pr[N(t) = 0] = 1 − θt + o(t).

Queuing Analysis – p.22

Poisson Process - Proof of (1 ⇒ 3) (part 2)

• Let Z be the sum of two exponentially distributed

random variables.

• Z = T0 + T1 and for z ≥ 0 we have fZ(z) =

z

0 fT0(x)fT1(z − x)dx =

z

0 θe−θxθe−θ(z−x)dx = θ2ze−θz.

• Pr[N(t) ≥ 2] = Pr[T0 + T1 ≤ t] =

t

0 Fz(x)dx =

θ2 t

0 ze−θz = 1 − (θt + 1)e−θt (integration by parts).

• So Pr[N(t) ≥ 2] = o(t) since limt→0

1−(θt+1)e−θt t

= 0

(L ’Hôpital’s rule).

• Using Pr[N(t) = 0] = 1 − θt + o(t) we get

Pr[N(t) = 1] = θt.

Queuing Analysis – p.23

Poisson Process - Proof of (3 ⇒ 2)

• Recall the binomial distribution

Pr(Bp

n = k) =

n

k

• pk(1 − p)n−k.
• How does N(t) behave ?
• Divide t into small parts of size t/n, when the parts

are small enough then the probability of one event is close to θt/n. So the number of events till time t has a binomial distribution with n trails and

p = θt/n.

• Pr[N(t) = k] equals the limit of the binomial

distribution of k successes where n → ∞, p → 0 while np = θt.

Queuing Analysis – p.24

Poisson Process - Proof of (3 ⇒ 2) (part 2)

So Pr[N(t) = k] = limn→∞

n

k

• (θt

n )k(1 − θt n )n−k

= lim

n→∞

n! k!(n − k)! (θt)k nk 1 − θt n n−k = lim

n→∞

(θt)k k! n! (n − k)!nk

• 1 − θt

n n

θt(θt− kθt n )

= lim

n→∞

(θt)k k! k−1

• i=0

n − i n

• e−1θt− kθt

n

= (θt)k k! e−θt

.

Queuing Analysis – p.25

Poisson Process - Proof of (2 ⇒ 1)

• Given Pr[N(t) = k] = (θt)k

k! e−θt.

• We have Pr[T0 ≥ t] = Pr[N(t) = 0] = eθt. So

T0 ∼ exp(θ).

• For Ti Denote zi =

0≤j<i Tj.

• So generally,

Pr[Ti ≥ t] = Pr[N(zi) = N(zi + t)] = Pr[N(t) = 0] = eθt.

• And Ti ∼ exp(θ).

Queuing Analysis – p.26