Probability and Random Processes

1
Probability and Random Processes

• STA 3533
• Module 6
• Networks of Queues
• Continue

2
Topics

• Case Study Solution
• Introduction to Queuing
• Open queuing networks
• Closed queuing networks

3
Introduction to Queuing

• Problem 1
• A data communication line delivers a block of
  information every 10 microseconds. A decoder
  checks each block for errors and corrects the
  errors if necessary. It takes 1 ms to determine
  whether a block has any errors. If the block has
  error, it takes 5 ms to correct it, and if it has
  more than one error it takes 20 ms to correct the
  error. Blocks wait in a queue when the decoder
  falls behind. Suppose that the decoder is
  initially empty and that the numbers of errors in
  the first 10 blocks are 0, 1, 3, 1, 0, 4, 0, 1,
  0, 0.
• a. Plot the number of blocks in the decoder as a
  function of time.
• b. Find the mean number of blocks in the decoder.
• c. What percentage of the time is the decoder
  empty?

4
Introduction to Queuing

• Solution 1
• 1. Interarrivals are constant with interarrival
  times 10 msec
• 2. Service time
• if 0 error 1 msec
• 1 error 15 msec
• gt1 error 120 msec

5
Introduction to Queuing

• a. The plot
• b.

N(t)
10
30
20
40
60
50
70
80
90
100
110
TObservation 100 msec
6
Introduction to Queuing

• c.
• Server is working during 65 msec S service
  times
• proportion idle time 1 65/100
• 0.35

7
Introduction to Queuing

• Problem 2
• Three queues are arranged in the loop as shown
  below. Assume that the mean service time in the
  queue i is mi 1/ m.
• a. Suppose the queue has a single customer
  circulating in the loop. Find the mean time ET
  it takes the customer to cycle around the loop.
  Deduce from ET the mean arrival rate l at each
  of the queues. Verify that Littles formula holds
  for these two quantities.
• b. If there are N customers circulating in the
  loop, how are the mean arrival rate and mean
  cycle time related?

m3
m1
m2
8
Introduction to Queuing

• Solution
• a. One customer ? no waiting
• ?
• Littles Formula ?

9
Introduction to Queuing

• b. Let eT mean cycle time per customer, then
• Total in system N leT by Littles
  formula

10
Open Queuing Networks

• Consider a network of K queues in which customers
  arrive from outside the network to queue k
  according to independent Poisson processes of
  rate ak. We assume that the service time of a
  customer in queue k is exponentially distributed
  with rate mk and independent of all other service
  times and arrival processes. We also suppose that
  queue k has ck servers.
• After completion of service in queue k, a
  customer proceeds to queue i with probability Pki
  and exists the network with probability

11
Open Queuing Networks

• The total arrival rate lk into queue k is the sum
  of the external arrival rate and the internal
  arrival rates
• It can be shown that above equation has a unique
  solution if no customer remains in the network
  indefinitely.
• Such networks are known as Open Queuing Networks

12
Open Queuing Networks

• The vector of the number of customers in all
  queues,
• 
  is a Markow process.
• Now Jackson s theorem gives steady state pmf for
  N(t).
• Jacksons Theorem if lk lt ckmk, then for any
  possible state n (n1, n2,,nK ),
• Where P Nk nk is the steady state pmf of an
  M/M/ck system with arrival rate lk and service
  rate mk

13
Open Queuing Networks

• Jacksons Theorem states that the numbers of
  customers in the queues at time t are independent
  random variables.
• In addition, it states that the steady state
  probabilities of the individual queues are those
  of an M/M/ck system. This is an amazing result
  because in general the input process to a queue
  is not Poisson, as was demonstrated in the sample
  queue with feedback discussed in the beginning of
  this section.

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Open Queuing Networks

• Example Messages arrive at a concentrator
  according to a Poisson process of rate a. The
  time required to transmit a message and receive
  an acknowledgment is exponentially distributed
  with mean 1/m. Suppose that a message need to be
  retransmitted with probability p. Find the steady
  state pmf for the number of messages in the
  concentrator.
• Solution The overall system can be represented
  by the simple queue with feedback shown below.

.9
l
a
.1
m
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Open Queuing Networks

• The net arrival rate into the queue is l a
  lp, that is,
• Thus, the pmf for the number of messages in the
  concentrator is

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Open Queuing Networks

• Example New programs arrive at a CPU according
  to a Poisson process of rate a as shown in below
  fig. A program spends as exponentially
  distributed execution time of mean 1/m1 in the
  CPU. At the end of this service time, the program
  execution is complete with probability p or it
  requires retrieving additional information from
  secondary storage with probability 1-p. Suppose
  that the retrieval of information from secondary
  storage requires an exponentially distributed
  amount of time with mean 1/m2. Find the mean time
  that each program spends in the system.

1-p
I/O
CPU
a
m2
p
m1
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Open Queuing Networks

• Solution the net arrival rates into the two
  queues are
• Thus
• Each queue behaves like an M/M/1 system so,
• Where r1 l1/m1 and r2 l2/m2 . Littles
  formula then gives the mean for total time spent
  in the system

18
Close Queuing Networks

• In some problems, a fixed number of customers,
  say I, circulate endlessly in a closed queuing
  network.
• For Example, some computer system models assume
  that at any time a fixed number of programs use
  the CPU and input/ output (I/O) resources of a
  computer as show in below figure.

p
I/O
CPU
1-p
m2
m1
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Close Queuing Networks

• We now consider queuing networks that are
  identical to the previously discussed Open
  Network except that the external arrivals rates
  are zero and the networks always contain a fixed
  number of customers I. We show that the steady
  state pmf for such system is product from but
  that the states of the queues are no longer
  independent.
• The net arrival rate into the queue k is now
  given by
• 
  (1)
• Note that these equations have the same from as
  the set of equations that define the stationary
  pmf for a discrete-time Markow chain with
  transition probabilities Pjk.

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Close Queuing Networks

• The only difference is that the sum of the lk s
  need not be one. Thus the solution vector to eq.
  (1) must be proportional to the stationary pmf
  pj corresponding to Pjk
• 
  (2)
• And where l(I) is a constant that depends on I,
  the number of customers in the queuing network.
  If we sum both sides of eq. (2) over k, we see
  that l(I) is the sum of the arrival rates in all
  the queues un the network, and pk lk / l(I) is
  the fraction of the total arrivals to queue k.

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Close Queuing Networks

• Theorem Let lk l(I) pk be a solution to
  eq.(1), and let n (n1, n2, ,nk) by any state
  of the network for which n1, n2, ,nk ? 0 and
• Then
• where P Nk nk is the steady state pmf of an
  M/M/ck system with arrival rate lk and service
  rate mk, and where S(I) is the normalization
  constant given by

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Close Queuing Networks

• Equation 4 states that P N(t) n has a product
  form. However P N(t) n is no longer equal to
  the product of the marginal pmf s because of the
  normalization constant S(I).
• This constant arises because the fact that there
  are always I customers in the network implies
  that the allowable states n must satisfy equation
  (3).
• The theorem can be proved by taking the approach
  used to prove Jackson s theorem.

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Close Queuing Networks

• Example Suppose the computer system example that
  we have seen in open queuing network is operated
  so that there are always I programs in the
  system. The resulting network of queue is shown
  below. Note that the feedback loop around the CPU
  signifies the completion of the one job and its
  instantaneous replacement by another one. Find
  the steady state pmf of the system. Find the rate
  at which programs are completed.

p
I/O
CPU
1-p
m2
m1
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Close Queuing Networks

• Solution The stationary probabilities associated
  with equation (1) are found by solving
• The stationary probabilities are then
• And the arrival rates are
• Solution Continues

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Close Queuing Networks

• The stationary pmf for the network is then
• Solution Continues

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Close Queuing Networks

• The rate at which programs are completed is pl1 .
  We find l1 from the relation between server
  utilization and probability of an empty system