Stochastic Process and Queuing systems incomplete

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Stochastic Process and Queuing systems(incomplete
)

• Summarized by
• Neetesh Purohit
• Lecturer, IIIT,
• Allahabad, UP, India
• http//profile.iiita.ac.in/np/
• 0532 2922236 (O), 0532 2922347 (R)

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Source

• Probability and statistics with reliability,
  Queuing and Computer Science Applications, IInd
  Edition, authored by K. S. Trivedi, John Wiley.
• Communication Systems, IV Edition, A. B.
  Carlson and others, McGraw Hills.
• Notes of Prof. Anirban Mahanti, available at
• http//pages.cpsc.ucalgary.ca/mahanti/teaching/F0
  5/CPSC531/

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Stochastic Process X(s,t)

• It represents a family of random variables
• Involves states, s
• (associated type of random variable )
• as well as time, t
• (either sampling is done at discrete instant of
  time or continuously the system is being
  monitored).
• Four possible types
• DSDT, DSCT, CSDT, CSCT

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Basic Queuing Model
Service discipline
Arrival process
Queue
Servers
Customer population
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• Define stochastic processes W, X, Y, Z over the
  queuing system such that
• W is DSDT
• X is DSCT
• Y is CSDT
• Z is CSCT
• Write down and show your notebook.

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• W number of jobs in the system at the time of
  departure of the kth customer.
• X(t) number of jobs in the system at time t.
• Y time that the kth customer has to wait in the
  system before receiving service.
• Z(t) cumulative service requirement of all jobs
  in the system at time t.

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Other types of stochastic processes

• Strictly stationary
• Widesense stationary
• Ergodic, etc

• Markov Process
• Birth-Death Process
• Poisson Process, etc

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Markov Process

• If future state probabilities independent of the
  past states and depends only on the present, then
  process called Markov.
• Markov property makes analysis easier, since past
  history need not be remembered.
• Markov Chain Discrete-state Markov process

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Markov Chains

• Interested in Markov chains with stationary state
  transition probabilities. I.e.,

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Birth-Death Process

• Birth-Death process A Markov chain that only
  allows transitions to neighboring states
• Represent states by integers. A process in state
  n can transition to state n-1 or n1
• E.g., of jobs in queue in a single server
  system can be represented by birth-death process
• Birth gt arrival of a job gt 1 state transition
• Death gt departure gt -1 state transition
• Batched arrivals cannot be modeled!

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• Using stochastic flow balance, the steady state
  probability of being in state n can be computed.

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Relationships between Processes
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Important parameters of a queuing system model
Service discipline
Arrival process
Queue
Servers
Customer population
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Population size

• Potential customers who can enter the queue
• Real systems have finite population
• However, if population is large, assume infinite
  for ease of analysis.

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Arrival process

• Customers arrive at t1,t2,,tj
• Interarrival time ?j tj-tj-1
• Assume interarrival times ?j are IID (independent
  and identically distributed) random variables
• E.g., Poisson process, Erlang, hyperexponential,
  general.

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Service time distribution

• Assume IID random variables
• E.g., exponential, Erlang, hyperexponential, and
  general

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Service disciplines

• FIFO (or FCFS) most common
• Random
• Round Robin
• Priority disciplines

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• Number of servers
• One/many (identical) servers
• System capacity
• Waiting space number in service
• Infinite assume if cap. large

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The Kendal Notation

• A/S/m/B/K/SD
• A is interarrival time distribution
• S is service time distribution
• m is number of servers
• B is number of buffers (system capacity)
• K is population size
• SD is service discipline

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Some common abbreviations

• M (Markov) denotes exponential (and thus
  memoryless distr.)
• D (Deterministic) values are constant
• Ek (Erlang) Erlang distribution with k phases
• G (General) denotes distribution not specified
  results are valid for all distributions
• Bulk arrivals denoted using superscripts. E.g.
  Mx

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Examples

• M/M/3/20/1500/FCFS is a single queue system with
• Interarrivals are exponentially distributed
• Service times exponentially distributed
• Three servers
• System capacity 20 queue size is 20 3 17
• Population size is 1500
• Service discipline is FCFS
• M/M/3 Typically, assume infinite system
  capacity, infinite population, and FIFO service.
  In such cases, last three parameters dropped.

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Random Variables for Queues
r
All variable except ? and ? are random

• nq of jobs waiting in queue
• ns of jobs recv. service
• n of jobs in system
• n nq ns is the queue length
• r response time
• Waiting for service plus time receiving service
• w waiting time
• Start of service minus arrival time

• ? interarrival time
• ? mean arrival rate 1/E?
• May depend upon of jobs in system
• s service time per job
• ? mean service rate per server 1/Es total
  service rate m?

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Rules for All Queues

• Stability Condition
• For stability, mean arrival rate should be less
  than mean service rate
• ? lt m?
• Does not apply to finite population systems and
  finite capacity systems
• Queues for finite population systems cannot grow
  indefinitely
• By definition, queues for finite buffer systems
  cannot grow indefinitely.

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• Number in System versus Number in Queue
• Number of jobs in system equals those waiting and
  servicing
• n nq ns
• En EnqEns
• That is, mean number of jobs in system equals
  mean number in queue plus mean number being
  serviced
• If service rate of servers independent of number
  of jobs in the queue, then
• Varn VarnqVarns
• Cov(nq,ns) 0

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• Number versus Time
• If jobs not lost due to buffer becoming full, the
  mean jobs is related to response time as
• mean of jobs in system arrival rate x mean
  resp. time
• mean of jobs in queue arrival rate x mean
  waiting time
• The above are obtained by applying Littles Law
• Littles law can be applied for finite buffer
  systems by considering the effective arrival rate

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• Time in System versus Time in Queue
• Time spent by a job in system is sum of waiting
  time and service time. That is, r w s
• Mean response time equals sum of the mean waiting
  time and the mean service time. Er Ew
  Es
• If service rate independent of of jobs in queue
• Cov(w,s) 0
• Varr Varw Vars

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Analysis of existing systems
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Operational Laws

• Operational gt directly measured
• Operational assumptions (e.g. number of arrivals
  number of completions) verifiable by
  measurements
• Operational quantities can be directly measured
  during a finite observation period

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Notation

• T observation period
• A number of arrivals
• ? arrival rate A/T
• C number of completions
• X throughput C/T
• B busy time
• U utilization B/T

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Notation

• S mean service timeB/C
• R mean time in system (residence time)
• Z think time of a terminal user
• D service demand
• V number of visits
• N mean number in system

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Utilization Law

• Utilization is the fraction of time a system is
  busy.
• U is always between 0 and 1.

Utilization of a system (or resource) is equal to
the throughput times the average service time
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Utilization Law Example 1

• A disk is serving 50 requests/sec each request
  requires 0.005 seconds of service.
• 1) What is the Utilization?
• U 50x0.005 .25 (25)
• 2) Maximum possible service rate?
• U 1 (0.005)X
• X 200 requests/sec

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Utilization Law Example 2

• A router forwards 100 packets/second onto a link.
  The transmission time (i.e., time to put packets
  onto the link), on average, is 1 ms. What is the
  link utilization?
• Link throughput X 100 packets/sec
• Service time S 0.001 sec
• U XS 0.1 (10)
• Link capacity?

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Littles Law

• Assume Job Flow Balance
• - Number of arrivals equal number of
  completions
• New jobs not generated in the system
• Jobs not lost (forever) in the system
• If jobs lost in the system (e.g., due to finite
  capacity), law applies with adjusted arrival rate

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• Mean in system arrival rate x mean response
  time
• N ? R X
• Average number in system equals product of the
  throughput of the system and the average
  residence time of a customer.

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Littles Law Example 1

• During a busy hour, approximately 20 clients
  arrive at a barbershop. Each client, on average,
  spends 15 minutes in the shop. How many clients
  are simultaneously present in the shop?
• From Littles law N ? R
• 20 customers/hour x (15/60)
• 5 customers

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Littles Law Example 2

• Littles law can be applied to any subsystem
• A disk is serving 50 requests/sec each request
  requires 5 ms of service average time to satisfy
  a request is 20 ms
• Consider subsystem 1 X 50, S 0.005 U .25.
  In this case, N U and S R. Why?
• Consider subsystem 2 X 50, R 0.02 N XR
• N 1 (queued at server) Avg. waiting for
  service ?

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Response Time Law

• N avg. of terminals
• Z avg. think time
• R avg. sys. Resp. time
• Using Littles law on the system (see the fig. in
  the bottom RHS)
• N X(RZ)
• R (N/X) - Z

RZ
A Model of a Time Sharing System
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• Littles law applicable at both Box 1 and Box 2
• Littles law on Box 2
• NS XR
• Littles law on Box 1
• NT XZ
• Note that
• N NT NS

A Model of a Time Sharing System
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Response Time Law Example 1

• 64 interactive users avg. think time of 15 s
  system throughput is 4 interactions/sec. What is
  R?
• R (N/X) Z 1 Sec

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Response Time Law Example 2

• A Web server was measured for a one hour period.
  During this period, 50 clients were observed. The
  average CPU time demand per request was measured
  to be 5ms, with CPU utilization being 0.25. The
  average response time for requests was 0.25 sec.
  What was the average client think time (in
  seconds)?

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• N 50 Scpu 0.005s Ucpu 0.25 R 0.25s
• Using utilization law
• X Ucpu/Scpu 50 req./s
• Using the response time law
• Z (N/X) R 0.75s
• At any given time, how many clients are
  thinking?

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Forced Flow Law

• Each system-level request may require multiple
  visits to a system resource.
• E.g., A database transaction may require several
  disk accesses
• Visit ratio (Visit count) ratio of number of
  subsystem completions to the number of system
  completions Vk Ck/C

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• If Vk visits are made to resource K in time T and
  a total of X system-level completions are
  observed in time T, then
• Xk VkX
• Uk XkSk
• X VkSk
• XDk
• Sk mean service time per visit to res. K
• Dk total service demand at resource K

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Forced Flow Law Example 1

• A database server is monitored for 15 minutes.
  During this time, the servers CPU was busy for
  12 minutes. It was observed that each transaction
  required on average 2 visits to the CPU, and the
  service requirement per visit was 1 ms. What is
  the system throughput (in transaction per
  second)?

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• T 15 min
• BC 12 min
• VC 2 SC 1 ms
• UC B/T 0.8
• UC X VCSC gt
• X 0.8/(2x0.001) 400 transactions/sec

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Forced Flow Law Example 2

• Consider an interactive system with the following
  characteristics
• no. of terminals 18
• avg. think time 10s
• visits to a specific disk per interaction 20
• utilization of this disk 0.3
• avg. service req. per visit to disk 20 ms
• Determine 1) disk throughput ? 2) system
  throughput ? 3) response time ?

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Analytical analysis
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M/M/1 Queue

• One server, one queue, FIFO service
• exponentially distributed interarrival and
  service times
• infinite population, infinite capacity
• Can model as a birth-death process

Notation pn steady-state probability of being
in state n
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• Using stochastic flow balance equations
• pn (?/?)n p0 ?np0, n0,1,,?
• ? is traffic intensity (lt1 for stability)
• Probabilities sum to one. Therefore,
• ? pn 1, n 0, 1, , ?
• p0(?0 ?1 ?2 ) 1
• p0 (1-?)
• pn ?n(1-?)
• Important performance measures follow

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• Utilization prob. of one or more jobs in system
• U 1- p0 ?
• Mean jobs in System
• En ?npn , n 0, 1, , ?
• En ?/(1-?)
• Mean response time (Littles law)
• number in system arrival rate x response time
• En ?Er
• Er (1/?)(?/(1-?)) (1/?)(1/(1-?))

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• Mean of jobs in queue
• Enq ?(n-1)pn , n 1, , ?
• Enq (?2)/(1-?)
• Can also be obtained using En Enq Ens
• Mean waiting time in queue (Littles Law)
• Number in queue arrival rate x mean waiting
  time
• Enq ?Ew
• Ew (1/?)((?2)/(1-?)) ?((1/µ)/(1-?))

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• Prob. of finding n or more jobs in system
• P( in system n) ?pj , j n,n1, , ?
• ?(1-?)?j ?n
• Waiting time and response time distributions
• Waiting times in queue exponentially distributed
• P0 lt w t 1 - ?e-?t(1-?)
• Response times exponentially distributed
• P0 lt r t 1 - e-?t(1-?)

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M/M/1 Queue Example

• Packets arrive at 100 packets/second at a router.
  The router takes 1 ms to transmit the incoming
  packets to an outgoing link. Using an M/M/1
  model, answer the following
• What is utilization?
• Probability of n packets in router?
• Mean time spent in the router?
• Probability of buffer overflow if router could
  buffer only 5 packets?
• Buffer requirement to limit packet loss to 10-6?

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• Arrival rate
• ? 100 pps
• Service rate
• ? 1/.001 1000pps
• Traffic intensity
• ? 0.1
• Mean packet residence time at router
• r (1/?)(1/(1-?))
• 1.01 ms
• Prob. of buffer overflow
• P( 6) ?6 10-12
• To limit loss to less than 10-6
• ?n 10-6
• n gt log(10-6)/log(0.1) gt 3

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M/M/c Queue

• c servers, one queue, FIFO service, exponentially
  distributed interarrival and service times
• infinite population, infinite capacity
• Model as a birth-death process with K states

State transition diagram for M/M/C Queue with C3
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• Mean arrival rate ?
• Mean service rate c?
• Traffic Intensity (avg. utilization) ? ?/(c?)
• ? lt 1 for stability
• Flow balance equations yield
• pn ((c?)n/n!)p0 n 1,,c-1
• pn ((c?)n/(c!cn-c))p0 n c
• Using law of total probability

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• Newly arrivals wait if all servers are busy,
  i.e., c or more jobs are in the system
• P( c jobs) pc pc1 pc2
• ? (c?)c/c!(1-?) p0
• ? Known as Erlangs C formula
• Mean of jobs in system
• En ?npn n 0, 1, , ?
• p0(c?)c/c!(1-?)2 c?
• c? ??/(1-?)

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• Mean of jobs in queue
• Enq ?(n-c)pn n c, , ?
• p0?(c?)c/c!(1-?)2
• ??/(1-?)
• Mean response time (Littles law)
• En ?Er
• Er 1/? ?/c?(1-?)
• Mean waiting time
• Ew Enq/ ? ??/(1-?)/? ?/c?(1-?)