SPM - Lecture - Part2 - Queueing (1)

7/25/2019 SPM - Lecture - Part2 - Queueing (1)

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Copyright : Nahrstedt, Angrave, Abdelzaher, Caccamo 1

Queueing Systems


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Copyright : Nahrstedt, Angrave, Abdelzaher

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Content of This Lecture

Goals: Introduction to Principles for Reasoning

about Process Management/Scheduling

Things covered in this lecture:

Introduction to Queuing Theory


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3

Queueing Model

Random Arrivals modeled as Poisson process

Service times follow exponential distribution


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Discussion

If a bus arrives at a bus stop every 15minutes, how long do you have to waitat the bus stop assuming you start to

wait at a random time?


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Discussion

The mean value is (0+15)/2 = 7.5 minutes

What assumption have you made about thedistribution of your arrival time?


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Discussion

The mean value is (0+15)/2 = 7.5 minutes

What assumption have you made about thedistribution of your arrival time?

The above mean assumes that your arrival time tothe bus station is uniformly distributed within [0, 15]


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Queuing Theory (M/M/1 queue)

ARRIVAL RATE (Poisson process)

SERVICE RATE Input Queue

Server

the distribution of inter-arrival times between two consecutive arrivals isexponential (arrivals are modeled as Poisson process)

service time is exponentially distributed with parameter


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M/M/1 queue

The M/M/1 queue assumes that arrivals are a Poisson process and theservice time is exponentially distributed.

Interarrival times of a Poisson process are IID (Independent and Identically

Distributed) exponential random variables with parameter

Arrival rate CPU

Service rate

1t

2

Arrival times:

- independent from each other!

- each interarrival i follows

an exponential distribution


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Appendix: exponentialdistribution

If is the exponential random variable describing the distribution of inter-arrival times between two consecutive arrivals, it follows that:

The probability density function (pdf) is:

tetPtA == 1}{)(

t

etAdt

dta

== )()(

Arrival rate CPU

Service rate

Probability to have the firstarrival within is 1-e-

t

cumulative distribution

function (cdf)

0


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Queueing Theory

Queuing theory assumes that the queue is in a steady state

M/M/1 queue model: Poisson arrival with constant average arrival rate (customers per unit time) Each arrival is independent. Interarrival times are IID (Independent and Identically Distributed) exponential

random variables with parameter What are the odds of seeing the first arrivalbefore time t?

See http://en.wikipedia.org/wiki/Exponential_distribution

for additional details

tetP

= 1}{


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Analysis of Queue Behavior

Poisson arrivals: probability ncustomers arrive within time interval t is

( )

!n

te nt


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Probability ncustomers arrive within time interval t is:

Do you see any connection between previous formulas and the above one?

( )!nte

nt


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Probability ncustomers arrive in time interval t is:

Do you see any connection between previous formulas and the above one?

Consider the waiting time until the first arrival. Clearly that time is morethan t if and only if the number of arrivals before time tis 0.

( )!nte

nt

( ) ( ) t

t

ete

tP

==>

!0

0

( ) ( ) tetPtP =>= 11


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Littles Law in queuing theory

The average number L of customers in a stable system is equal to the averagearrival rate times the average time W a customer spends in the system It does not make any assumption about the specific probability distribution followed by the

interarrival times between customers

Wq= mean time a customer spends in the queue

= arrival rate

Lq = Wq number of customers in queue

W = mean time a customer spends in the entire system (queue+server)

L = W number of customers in the system

In words average number of customers is arrival rate times average waiting time


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Analysis of M/M/1 queuemodel

1


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Analysis of M/M/1 queuemodel

Quiz: how can we derive the average time W in the system, and the average timeWq in the queue?

Use Littles theorem

Time in the system is:

Time in the queue is:

Number of customers in the queue is:

= 1W

=qW

=

1

2

qL

Try to derive them usingLittles Law!


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Hamburger Problem

7 Hamburgers arrive on average every time unit

8 Hamburgers are processed by Joe on average every unit

1. Av. time hamburger waiting to be eaten? (Do they get cold?) Ans = ????

2. Av number of hamburgers waiting in queue to be eaten? Ans = ????

Queue

78


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Hamburger Problem

7 Hamburgers arrive on average every time unit

8 Hamburgers are processed by Joe on average every unit

1) How long is a hamburger waiting to be eaten? (Do they get cold?) Ans = 7/8

time units2) How many hamburgers are waiting in queue to be serviced? Ans = 49/8

Queue

78


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Example: How busy is theserver?

=2=3


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Example: How busy is theserver?

=2=3

66%


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How long is an eater in thesystem?

=2

=3


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How long is an eater in thesystem?

=2=3

= 1W = 1/(3-2)= 1


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How long is someone in thequeue?

=2=3


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How long is someone in thequeue?

=2=3

66.

23

66. =

=

=

qW


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How many people in queue?

=2=3


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How many people in queue?

=2=3

33.166.1266.

1

2=

=

=

qL


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Interesting Fact

As approaches one, the queue lengthbecomes infinitely large.


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Until Now We Looked at SingleServer, Single Queue

ARRIVAL RATE

SERVICE RATE

Input Queue

Server


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Sum of Independent Poisson Arrivals

ARRIVAL RATE 1

SERVICE RATE Input Queue

Server

ARRIVAL RATE 2

=1+ 2

If two or more arrival processes are independent and Poisson with parameter i,then their sum is also Poisson with parameter equal to the sum of i

C i h N h d A Abd l h


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As long as service times areexponentially distributed...

ARRIVAL RATE

SERVICE RATE 1

Input Queue

Server

Server

SERVICE RATE 2

Combined=1+2

C i ht N h t dt A Abd l h


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Question: McDonalds Problem

A) Separate Queues per Server B) Same Queue for Servers

Quiz: if WA is waiting time for system A, and WB is waiting time for systemB, which queuing system is better (in terms of waiting time)?

Copyright : Nahrstedt Angrave Abdelzaher


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X = lamda; u = mikro; p = epsilon

Scenario A; x=2 & u=4

P = 2/4 = 0.5

Wq = p/(u-x) = 0.5/(4-2) = 0.5/2 =0.25 seconds


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Copyright : Nahrstedt Angrave Abdelzaher


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X = lamda; u = mikro; p = epsilon Scenario B; x=2 & u=4

Xt = x1 + x2+x3 = 6

Ut = U1+U2+U3 = 12

P = 6/(3*4) = 6/12 = 0.5 Wq = p/(Ut-Xt) = 0.5/(12-6) = 0.5/6 = 0.0833

seconds


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SPM - Lecture - Part2 - Queueing (1)

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