1. The M/G/1 Queue
and others

2. Telephone Line Problem
Callers queue for outgoing lines
This system will allow only eight outgoing connections

3. Telephone Line Problem
The outgoing telephone lines could be physical, as in the original version, or virtual, as in ATM, and the decision of the CAC function would decide the number of connections available
This is modeled as a M/M/N/N queue (N servers, no waiting room)

4. Telephone Line Problem
The arrival process is simple Poisson
If there are n < N customers in the queue, then they are all being serviced
The service rate for a single line is m (average duration of phone call = 1/m)
So the service rate for the whole queue is nm for 0 < n ≤ N

5. Telephone Line Problem
As before, the average flow between states must be zero
lnpn = mn+1pn+1 for 0 < n ≤ N
This gives us lpn = (n+1)mpn+1 or
where r = l/m

6. Telephone Line Problem
This gives us
Since we also require Spn = 1, we also get

7. Telephone Line Problem
The designer of the system is interested in the blocking probability
This is simply the probability that n = N, or pN

8. Erlang Formula
The formula on the last slide is called the Erlang-B distribution after the pioneering Danish engineer, AK Erlang
He has given his name to the units we use for traffic intensity, which corresponds to r

9. Telephone Line Problem
The average number of calls in progress is

10. Telephone Line Problem
We would generally expect to be designing systems where PB is low, so E(n) ≈ r
The average number of calls put through per unit time, is

11. Design Problem
You have 1,000 incoming telephone lines, and over an eight hour period, each subscriber makes four phone calls, on the average. The phone calls average five minutes in length. How many outgoing lines must you provide to give a blocking probability of (a) 10% (b) 1%? How many calls would be in progress on the average?

12. Design Problem l = 1000x4/(8x60) = 8.33 calls per minute
m = 1/5 = 0.2 calls per minute
r = l/m = 8.33/0.2 = 41.67
Blocking probability

13. Design Problem
After a spreadsheet calculation, we find that we need at least 43 servers to give less than a 10% chance of blocking, and at least 55 servers to give less than a 1% chance of blocking
The corresponding calls in progress is 37.7 for 43 servers and 41.3 for 55 servers

14. M/G/1 Queue
A memoryless, Poisson process always gives an exponential distribution for inter-arrival or inter-service times
However there are cases where the service process times are not exponentially distributed
For example …

15. M/G/1 Queue
For example, a packet switching system may handle only two different types of packet, one with 100 bytes, and one with 2,000 bytes
The big packets will take longer to serve (transmit)
This gives a dumbell distribution

16. M/G/1 Queue
F(t)
Service time, t
The service times of the small packets would cluster around a low value, and there would be a cluster at a longer time for the larger packets

17. M/G/1 Queue
Another example would be a server (computer) which has to perform different operations on packets, depending on what type of packet it is
Possible operations are encrypting, processing for an on-line game, and simple transmission

18. M/G/1 Queue
And another example would be within an ATM switch, where all packets (or “cells”) are the same size, and thus take the same time to transmit
In these cases, the service time distribution is said to be “general”, and we describe the queue as M/G/1

19. M/G/1 Queue Packet j is completed at time, tj
Service time, customer j
time
tj-1 tj
tj+1
nj-1 nj
nj+1
Packet j is completed at time, tj
The number of packets in the queue at time tj+ is nj

20. M/G/1 Queue
During servicing of customer j, a number of packets, nj, arrives
Then the number of packets in the queue after servicing of packet j is
nj = nj nj for nj-1 > 0
nj =nj for nj-1 = 0

21. M/G/1 Queue These equations can be written using a unit step function
nj = nj-1 – u(nj-1) + nj where u(x) = 1 for x ≥ = 0 for x ≤ 0
The expected values of nj and nj-1 are the same

22. M/G/1 Queue So we must have E(n) = E(u(n))
The average value of arrivals, E(n) must be less than unity for a stable queue, and in fact, must be equal to the utilisation, r, by definition,

23. M/G/1 Queue Then we have by definition Therefore we have
p0 = 1 – r as before (M/M/1)

24. M/G/1 Queue
For the next proof, we need some intermediate results or definitions
Firstly, the definition of the variance of n is

25. M/G/1 Queue Secondly, E(u2(n)) = E(u(n)) = E(n) = r
Thirdly, we assume that nj and nj-1 are independent, so that E(njnj-1) = E(nj)E(nj-1) = rE(nj-1)
We begin the proof with the equation relating nj and nj-1
nj = nj-1 – u(nj-1) + nj

26. M/G/1 Queue
We square both sides of the equation nj2 = nj-12 + u2(nj-1) + nj2 – 2nj-1u(nj-1) – 2u(nj-1)nj + 2njnj-1
We now take the expected value of both sides
r + E(n2) – 2E(n) – 2r2 + 2rE(n) = 0
We now substitute for E(n2)

27. M/G/1 Queue 2E(n) (1 – r) = r - 2r2 + sn2 + r2 = r(1 – r) + sn2
This finally gives us

28. Expected service time
The average service time is E(t), so the average service rate is 1/E(t)
The utilisation, r, is the ratio of average arrival rate to average service rate
So r = lE(t)
That is E(t) = r/l

29. Queue Size To find E(n), we still need to find sn2
We will need a result from the distribution of service times, s2
s2 = E[(t – r/l)2]
= E(t2 – 2tr/l + r2/l2)
= E(t2) – 2r/l x r/l + r2/l2
So E(t2) = s2 + r2/l2

30. Queue Size (find sn2)

31. Queue Size (find sn2)
We reverse the order of summation and integration

32. Queue Size (find sn2)

33. Queue Size
We can now substitute back into the formula for E(n), defining m = l/r
We get
This formula gives expected queue size in terms of server statistics

34. Queue Size
This formula collapses to r/(1-r) for the exponential service distribution
If we know exactly what the service time will be (eg for ATM cells at a switch), then s2 = 0 and

35. Queue Size
This is called a deterministic service time, and the queue is then M/D/1
This gives the lowest possible average queue
The higher the value of s2, the longer the average queue will be
The term 1/(1-r) dominates for high r

36. Time Delay
Average time delay can be found using Little’s formula, so that
E(T) = E(n)/l

37. Problem
Three queuing systems, with Poisson arrivals, are alike, except that the servicing distributions are (a) Poisson (b) deterministic and (c) composed of 50% with a service time of 0.1, and 50% with a service time of 1.0. Compare the average queue levels for all cases when r = 0.5

38. Problem
For case (c) the average service time is 1.1/2 = So m =
s2 = 0.5(0.1 – 0.55) (1.0 – 0.55)2 =
From the formula
E(n)=(0.5/0.5)(1–0.25(1-1.82x0.2025)) = 1.083

39. Problem For case (a) E(n) = 0.5/0.5 = 1
For case (b) E(n) = 0.5/0.5x(1-0.5/2) = 0.75
The distribution of case (c) gives a high variance, and this results in a larger average queue

40. Tutorial Problems