1 / 40

ECTE461 – Telecommunications Queuing Theory

ECTE461 – Telecommunications Queuing Theory. Mathematical analysis of telecommunications switch performance. Assessment. Queuing Systems. Queue at supermarket – FIFO (first in first out) with single server Queue at NAB – FIFO with multiple servers (tellers)

malo
Download Presentation

ECTE461 – Telecommunications Queuing Theory

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ECTE461 – Telecommunications Queuing Theory Mathematical analysis of telecommunications switch performance

  2. Assessment

  3. Queuing Systems • Queue at supermarket – FIFO (first in first out) with single server • Queue at NAB – FIFO with multiple servers (tellers) • Inventories at warehouse or grocer – FIFO • Packets arriving at switch – FIFO • Phone callers trying to get a line – FIFO with multiple servers

  4. FIFO Queues Single server Multiple servers

  5. Nature of Queue • Usually FIFO. Newcomers get on the end of the queue. When they reach the front of the line, they are served • Queue can be empty, but in general it has some non-negative integer number of members • There may be a maximum number of spaces

  6. New Arrivals • We never know just when a new member will arrive to be added to a queue • We assume that new arrivals can be described in terms of probability • For example, the probability of a new member arriving in the next second is 0.4

  7. Probability • Something which we are certain will not happen has a probability of 0 • Something which will happen for certain has a probability of 1.0 • A probability of 0.4 means that after a long (theoretically infinite) run of tests, the event will have happened in 40% of cases

  8. Probability • If there are n possible outcomes, and the probability of the ith outcome is pi, then • It is possible for n to go to ∞

  9. Probability • If each possible outcome represents a number, ki, then the expected value, on the average is • Again, n may go to ∞

  10. Combined Events • If the probabilities of two independent events are p1 and p2, then the probability of both of them occurring is p1p2 • Consider a total of N pairs of these events. In p1N cases, event 1 has occurred. In a fraction, p2 of these, event 2 has also occurred; ie p1p2N

  11. Conditional Probability • Event 1 = e1, event 2 = e2 • The probability of e1 and e2 occurring, given that we know that e1 has already occurred, is simply p2 • P(e1 and e2) = p1p2 • P(e1 and e2|e1) = p2 = p1p2/p1 = P(e1 and e2)/P(e1)

  12. Arrival of Packets • e1 = packet arrives after period, a • e2 = packet arrives after a + b • e3 = packet arrives after b a b t=0 time, t

  13. Arrival of Packets • If the packet does not arrive until a + b, then it necessarily does not arrive before a. • That is if e2 occurs, then e1 has necessarily occurred, since a + b > a. • So P(e1 and e2) = P(e2) and • P(e1 and e2|e1) = P(e2|e1)

  14. Arrival of Packets • If we know that no packet has arrived at t = a, then the probability that a packet will not arrive until t = a + b is simply p3, the probability that a packet will not arrive before a time of b has elapsed • Ie, P(e2|e1) = p3

  15. Arrival of Packets • The conditional probability formula gives • P(e1 and e2|e1) = P(e1 and e2)/P(e1) • Ie, P(e3) = P(e2)/P(e1) • Ie, p2 = p1p3 • We can re-write this as follows

  16. Exponential Distribution • P(packet does not arrive before a + b) = P(packet does not arrive before a) x P(packet does not arrive before b) • We could write this as • P(a + b) = P(a) P(b) • Then put a = t and b = dt, to give • P(t + dt) = P(t) P(dt)

  17. Exponential Distribution • Now probability that packet arrives in time dt is ldt • So P(dt) = 1 – ldt • And P(t + dt) = P(t) + dt • So we have • P(t) + dt = P(t)(1 – ldt)

  18. Exponential Distribution • This gives us • = -l P(t) • which leads to • P(t) = e-lt + const • Since P(0) = 1, const = 0

  19. Exponential Distribution • P(t) = e-lt • That is, if the probability of a packet arriving in any small time interval, dt is ldt, then the probability of the next packet arriving after time, t, is e-lt • This is called the exponential distribution

  20. Exponential Distribution • This distribution is “memoryless”: that is, it is not changed by what has happened before. At every point in time, the probability of the next packet arriving after t is e-lt • Many situations can be described by this process, to a reasonable degree of accuracy

  21. Exponential Distribution • Real time traffic is often close to exponential distribution • This is not always the case. If the traffic is “bursty”, then earlier events change the probability • This is frequently the case with data traffic

  22. Exponential Distribution • If we have a time, t = a + b, then • P(t) = e-lt = e-l(a+b) = e-lae-lb = P(a)P(b) (as we assumed to begin)

  23. Poisson and Markov • The process which produces the exponential distribution is called a “Poisson” process • A Markov “chain” is a chain of events described by an exponential probability distribution • Both of these terms will be used

  24. Poisson Process • How many packets do we expect to arrive in an interval, T? • What is the average arrival rate of packets in a Poisson process? • To answer these questions, and others, we need to find the probability of k packets arriving in T

  25. Poisson Process • We divide T into m very short intervals, Dt • The probability of one packet arriving in one of these intervals is lDt • The probability of no packets arriving is 1 – lDt • We ignore the possibility of more than one packet arriving (O(Dt2))

  26. Poisson Process • In order to get k packets arriving, we need one packet to arrive in each of k of the m short intervals • For any particular combination of these intervals, the probability is • (lDt)k x (1 – lDt)m-k

  27. Poisson Process • The number of different combinations of k intervals out of a total of m intervals is • This is the coefficient for the binomial expansion

  28. Poisson Process • So the probability is • As m becomes very large, and Dt k becomes very small compared to m • So m!/(m-k)! ≈ mk

  29. Poisson Process • Since mDt = T, the probability can be re-written

  30. Poisson Process • So we can write the probability

  31. Poisson Process • Remember, this was the probability that k packets would arrive in a period, T • It is called the Poisson distribution

  32. Example • To three decimal places, what are the probabilities of 0, 1, 2, …10 packets arriving in the period T = 1/l? • Here we have • The first 11 values are 0.368, 0.368, 0.184, 0.061, 0.015, 0.003, 0.000,…

  33. Exponential and Poisson • If we put k = 0, we have the probability that no packets arrive in the interval, T • We see that the exponential distribution is a special case of the Poisson • P(0) = e-lT

  34. Poisson Process • We can now calculate the average, or expected number of packets arriving during T • We use the formula for the average of a set of possible outcomes (from before)

  35. Poisson Process

  36. Poisson Process • The term for k = 0 is zero. We then replace k-1 with j to get

  37. Poisson Process • That is, the expected, or average number of packets arriving in time, T is just lT • This means that the average rate of arrivals is l packets per second

  38. And Finally • We can use a similar method to show that • And the variance is

  39. Tutorial Questions

More Related