Queueing Theory (Delay Models)

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Queueing Theory (Delay Models)
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Introduction
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• Total delay of the i-th customer in the system
• Ti Wi ti
• N(t) the number of customers in the system
• Nq (t) the number of customers in the queue
• Ns (t) the number of customers in the service
• N the avg number of customers in the queue
• t the service time

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• T the total delay in the system
• ? the customer arrival rate /sec

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Littles Theorem

• EN ?ET
• Number of customer in the system at t
• N(t)A(t)-D(t)
• where
• D(t) the number of customer departures up to
  time t
• A(t) the number of customer arrivals up to time
  t

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Poisson Process

• The interarrival probability density function
• mean 1/?, variance 1/?2
• for every t, d0
• where

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Poisson Process

• Characteristics of the Poisson process
• Interarrival times are independent and
  exponentially distributed
• If t n denotes the n-th arrival time and the
  interval tn t n1- t n , the probability
  distribution is

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Sum of Poisson Random Variables

• Xi , i 1,2,,n, are independent RVs
• Xi follows Poisson distribution with parameter li
• Partial sum defined as
• Sn follows Poisson distribution with parameter l

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Sum of Poisson Random Variables
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Sampling a Poisson Variable

• X follows Poisson distribution with parameter l
• Each of the X arrivals is of type i with
  probability pi, i 1,2,,n, independently of
  other arrivals p1 p2 pn 1
• Xi denotes the number of type i arrivals
• X1 , X2 ,Xn are independent
• Xi follows Poisson distribution with parameter
  li lpi

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Sampling a Poisson Variable (cont.)
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Merging Splitting Poisson Processes
l1
lp
p
l
l1 l2
1-p
l2
l(1-p)

• A Poisson processes with rate l
• Split into processes A1 and A2 independently,
  with probabilities p and 1-p respectively
• A1 is Poisson with rate l1 lpA2 is Poisson with
  rate l2 l(1-p)

• A1,, Ak independent Poisson processes with rates
  l1,, lk
• Merged in a single processA A1 Ak
• A is Poisson process with rate l l1 lk

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Poisson Variable

• mean
• variance
• Memoryless property (if exponentially
  distributed)

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Review of Markov chain theory

• Discrete time Markov chains
• discrete time stochastic process Xn n0,1,2,..
  taking values from the set of nonnegative
  integers
• Markov chain if
• where

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Markov chain

• Markov chain formulation
• Consider a discrete time MC
• where Nk is the number of customers at time k
  and N(t) is the number of customers at time t
• probabilities
• where the arrival and departure processes are
  independent

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Review of Markov chain theory

• The transition probability matrix
• n-step transition probabilities

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Review of Markov chain theory

• Chapman-Kolmogorov equations
• detailed balance equations for birth-death
  systems (in steady state)

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Example
0
1
P0(2-?2)/(2-?2(1-?1))
the throuput P0 P(s0 P0)0 P0 P(s1 P0)1
P0 P(s2 P0 )2 P1 P(s0 P1)0 P1 P(s1
P1)1 P1 P(s2 P1)2
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• Continuous time Markov chains
• X(t) t0 taking nonnegative integer values
• ?i the transition rate from state i
• qij the transition rate from state i to j
• qij ?i Pij
• the steady state occupancy probability of state j
• Analog of detailed balance equations for DTMC

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Birth-And-Death Process
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Birth-And-Death Process(cont.)

• Equation Expressing This
• State Rate In Rate Out
• 0 m1P1 l0P0
• 1 l0P0 m2P2 (l1 m1) P1
• 2 l1P1 m3P3 (l2 m2) P2
• .... ...................
• N-1 lN-2PN-2 mNPN (lN-1 mN-1) PN-1
• N lN-1PN-1 mN1PN1 (lN mN) PN
• .... ...................

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Birth-And-Death Process(cont.)

• Finding Steady State Process
• State
• 0 P1 (l0 / m1) P0
• 1 P2 (l1 / m2) P1 (m1P1 - l0P0) / m2
• (l1 / m2) P1 (m1P1 - m1P1) / m2
• (l1 / m2) P1

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Birth-And-Death Process(cont.)

• Finding Steady State Process(cont.)
• State
• n-1 Pn (ln-1 / mn) Pn-1 (mn-1Pn-1-
  ln-2Pn-2) / mn
• (ln-1 / mn) Pn-1 (mn-1Pn-1- mn-1Pn-1)
  / mn
• (ln-1 / mn) Pn-1

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Birth-And-Death Process(cont.)

• Finding Steady State Process(cont.)
• N Pn1 (ln / mn1) Pn (mnPn - ln-1Pn-1) /
  mn1
• (ln / mn1) Pn
• To Simplify
• Let C (ln-1 ln-2 .... l0) / (mn mn-1
  ......... m1)
• Then Pn Cn P0 , N 1, 2, ....

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M/M/1 queueing system

• Arrival statistics
• stochastic process taking
  nonnegative integer values is called a Poisson
  process with rate ? if
• A(t) is a counting process representing the total
  number of arrivals from 0 to t
• arrivals are independent
• probability distribution function

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M/M/1 queueing system

• P1 arrival and no departure in d
• where the arrival and departure processes are
  independent

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M/M/1 queueing system

• Global balance equation

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M/M/1 queueing system

• from
• Then
• Average number of customers in the system

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M/M/1 queueing system

• Average delay per customer (waiting time
  service time)
• by Littles theorem
• Average waiting time
• Average number of customer in queue
• Server utilization

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M/M/1 queueing system

• example
• 1/?4 ms, 1/µ3 ms

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?????

• ????? ???? ????? ?????? ?? K ?????. ??? ???????
  ???? ???, ??? ?????? ?K ????? ??????. ??? ??????
  ?? ?? ????? ?????? ??? P ?????. ??? ???? ???????
  ???????

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?????

• Arrival rate ?1/K
• Time in the system T?KP
• Applying Littles theorem
• N ?T ?P/K
• Though the process is deterministic, and N(t)
  does not converge to any value, N is well
  defined, interpreted as the time average

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?????

• ??? ???? ?????? ??? ?n
• ??? ????? ??? ?n
• ?. ???? ?? ??????? ??????
• ?. ??? ?? ????????? ??????????
• ?. ??? ?? ???? ??????? ?????? ?????? ???? ?????
• ?. ??? ?? ??? ??????? ?????? ?????? ??????? ????
  LITTLE

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?????

• ?
• ?

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?????

• ?
• ?

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?????

• ???? ???? ????? ????? ?????? ????
• ???? ????? ????? ???? ??? ???? ????? ??? ????
  ???? ??? ????? ?? ???? ??? ?? ??? ??? ??? ?????
  ???? ?? ??? ????? ??? ?? ??. ???? ?????? ????? ??
  ??? ?? ??? ?????. ???? ?? ???? ??????? ????? ???
  ??????? ?? ? ???? ??????? ????? ??????????? ?? ?
  ??????? ??? ??? C.
• ?. ???? ??? ???????? ?????? ??? ??? ????? ?
• ?. ?? ????? ?????? ??? ??? ?????, ?? ?????
  ???????? ?? ???? ?? ????????? ?????? ??????? ???
  ?? ???? ????? 0.9 ????? ? ? (??? ?????? ????
  ?????)

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?????

• ?
• Message Length
• Transmission Rate
• Transmission Time
• Service Rate

 
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?????
 

• ?

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M/M/1 Example I

• Traffic to a message switching center for one of
  the outgoing communication lines arrive in a
  random pattern at an average rate of 240 messages
  per minute. The line has a transmission rate of
  800 characters per second. The message length
  distribution (including control characters) is
  approximately exponential with an average length
  of 176 characters. Calculate the following
  principal statistical measures of system
  performance, assuming that a very large number of
  message buffers are provided

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M/M/1 Example I (cont.)

• (a) Average number of messages in the system
• (b) Average number of messages in the queue
  waiting to be transmitted.
• (c) Average time a message spends in the system.
• (d) Average time a message waits for transmission
• (e) Probability that 10 or more messages are
  waiting to be transmitted.

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M/M/1 Example I (cont.)

1. Es Average Message Length / Line Speed
   176 char/message / 800 char/sec 0.22
   sec/message or
2. m 1 / 0.22 message / sec 4.55 message /
   sec
3. l 240 message / min 4 message / sec
4. r l Es l / m 0.88

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M/M/1 Example I (cont.)

• (a) N r / (1 - r) 7.33 (messages)
• (b) Nq r2 / (1 - r) 6.45 (messages)
• (c) W Es / (1 - r) 1.83 (sec)
• (d) Wq r Es / (1 - r) 1.61 (sec)
• (e) P 11 or more messages in the system
  r11 0.245

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M/M/1 Example II

• A branch office of a large engineering firm has
  one on-line terminal that is connected to a
  central computer system during the normal
  eight-hour working day. Engineers, who work
  throughout the city, drive to the branch office
  to use the terminal to make routine calculations.
  Statistics collected over a period of time
  indicate that the arrival pattern of people at
  the branch office to use the terminal has a
  Poisson (random) distribution, with a mean of 10
  people coming to use the terminal each day. The
  distribution of time spent by an engineer at a
  terminal is exponential, with a

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M/M/1 Example II (cont.)

• mean of 30 minutes. The branch office receives
  complains from the staff about the terminal
  service. It is reported that individuals often
  wait over an hour to use the terminal and it
  rarely takes less than an hour and a half in the
  office to complete a few calculations. The
  manager is puzzled because the statistics show
  that the terminal is in use only 5 hours out of
  8, on the average. This level of utilization
  would not seem to justify the acquisition of
  another terminal. What insight can queueing
  theory provide?

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M/M/1 Example II (cont.)

• 10 person / day1 day / 8hr1hr / 60 min
• 10 person / 480 min
• 1 person / 48 min
• gt l 1 / 48 (person / min)
• 30 minutes 1 person 1 (min) 1/30
  (person) gt m 1 / 30 (person / min)
• r l / m 1/48 / 1/30 30 / 48 5 / 8

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M/M/1 Example II (cont.)

• Arrival Rate l 1 / 48 (customer / min)
• Server Utilization r l / m 5 / 8 0.625
• Probability of 2 or more customers in system PN
  ³ 2 r2 0.391
• Mean steady-state number in the system L EN
  r / (1 - r) 1.667
• S.D. of number of customers in the system sN
  sqrt(r) / (1 - r) 2.108

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M/M/1 Example II (cont.)

• Mean time a customer spends in the system W
  Ew Es / (1 - r) 80 (min)
• S.D. of time a customer spends in the system sw
  Ew 80 (min)
• Mean steady-state number of customers in
  queue Nq r2 / (1 - r) 1.04
• Mean steady-state queue length of nonempty
  Qs ENq Nq gt 0 1 / (1 - r) 2.67
• Mean time in queue Wq Eq rEs / (1 -
  r) 50 (min)

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M/M/1 Example II (cont.)

• Mean time in queue for those who must wait Eq
  q gt 0 Ew 80 (min)
• 90th percentile of the time in queue pq(90)
  Ew ln (10 r) 80 1.8326
  146.6 (min)

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M/M/m, M/M/m/m, M/M/8

• M/M/m (infinite buffer)
• detailed balance equations in steady state

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M/M/m

• where
• From

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M/M/m

• The probability that all servers are busy
• - Erlang C formula
• expected number of customers waiting in queue

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M/M/m

• average waiting time of a customer in queue
• average delay per customer
• average number of customer in the system
• by Littles theorem

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M/M/s Case Example I
Find p0
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M/M/s Case Example I (cont.)

• 0.429 (_at_ 43 of time, system is empty)
• as compared to m 1 P0 0.20

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M/M/s Case Example I (cont.)

• Find W
• Wq Lq / l 0.152 / (1/10) 1.52 (min)
• W Wq 1 / m 1.52 1 / (1/8) 9.52 min)
• What proportion of time is both repairman busy?
  (long run)
• P(N ³ 2) 1 - P0 - P1 1 - 0.429
  - 0.343 0.228 (Good or Bad?)

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M/M/8

• M/M/8 The infinite server case
• The detailed balance equations
• Then

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M/M/m/m

• M/M/m/m The m server loss system
• when m servers are busy, next arrival will be
  lost
• circuit switched network model

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M/M/m/m

• The blocking probability (Erlang-B formula)

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Moment Generating Function
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Discrete Random Variables
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Continuous Random Variables
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