1 Exponential distribution: main limitation So far, we have considered Birth and death process and its direct application To single server queues With the 2 following assumptions  Inter-arrival time is exponential  Service time is exponential Problem with exponential distribution It has the memoryless property (not very good) Since, the longer you get served, => the more likely your service will complete

2 Exponential assumption: example Let us consider Queue in one of the output ports of a router Is it ok to model this queue as an M/M/1 queue?  A lot of people do, but is it correct? The implication of such a model Transmission delay of a packet is exponential?  If you are doing file transfer all packets are 1500 bytes long  The latest statistics from SANs (Storage Area Networks) 1500 bytes P bytes P 2 64 bytes P 3

3 Inter-arrival time: is it exponential? Back in the 80s When the link speeds were very small It was OK to consider packet arrival to be Poisson But as speeds picked up with the advent of SONET Because of the tremendous speed, the utilization of links No longer regular (spaced out with exponential inter-arrival) Instead, traffic is bursty (packets are grouped together)  You go thru an ON period with back to back packets  Followed by an off period  It is better to use Interrupted Poisson distribution

4 Erlang model: mixture of exponentials Erlang Realized the exponential distribution  was not an adequate modeling technique  But it is important to keep it as it leads to easy mathematics Came up with another idea Mixture of exponentials represented by the letter E  Combine several exponentials to make up a service μ Break it up into r exponentials in tandem rμrμrμrμ … rμrμ

5 Erlang model: main concept The arriving customer Takes a sequence of exponential services Instead of a single service time Breaking service time into r service times Is called E r Erlang distribution with r stages However, the r servers are equivalent to one server  Only one customer is accommodated in service at a time  Example: E 2 Why is it important? Arbitrary pdf can be represented by the Erlang model 1/2μ

6 Coxian model: main idea Idea Instead of forcing the customer to get r exponential distributions in an E r model The customer will have the choice to get 1, 2, …, r services Example C 2 : when customer completes the first phase He will move on to 2 nd phase with probability a  Or he will depart with probability b (where a+b=1) a b

7 Laplace transform X is a continuous random variable f X (x) is the probability density function of X Laplace transform of X Since,

8 Main property of Laplace transform Main property Proof: first and second moment

9 Moment generating functions Laplace transform is a special case of Moment generating function ø(t) It is called as such because of all of the moments of X  Can be obtained by successively differentiating it Laplace transform is obtained t = -s

10 Laplace transform of an exponential distribution Let X be a continuous random variable That follows the exponential distribution The Laplace transform of X is given by

11 Convolution of 2 discrete random variables Let X 1 and X 2 be discrete random variables Whose probability distributions are known  P[X 1 = x 1 ] and P[X 2 = x 2 ] Y be the sum of X 1 and X 2 What is the probability distribution of Y = X 1 + X 2 ?  P[Y = y]

12 Convolution of 2 continuous random variables Let X 1 and X 2 be 2 continuous random variables f X1 (x 1 ) and f X2 (x 2 ) are known => f X1 * (s) and f X2 * (s) are known Let Y = X 1 + X 2 What is the pdf of Y?

13 Application to Erlang model X 1 and X 2 are exponentially distributed and independent with mean 1/2μ each 1/2μ X1X1 X2X2 Y=X 1 +X 2

14 Erlang service time: first and second moment The mean of Y? The variance of Y? Exercise at home Obtain the mean and variance of E 2 from its Laplace transform

15 Squared coefficient of variation The squared coefficient of variation Gives you an insight to the dynamics of a r.v. X Tells you how bursty your source is C 2 get larger as the traffic becomes more bursty  For voice traffic for example, C 2 =18  Poisson arrivals C 2 =1 (not bursty)

16 Poisson arrivals vs. bursty arrivals Why do we care If arrival is bursty or Poisson Bursty traffic will place havoc into your buffer Example: router design % loss C2C2

17 Erlang, Hyper-exponential, and Coxian distributions Mixture of exponentials Combines a different # of exponential distributions Erlang Hyper-exponential Coxian μ μμμE4E4 Service mechanism H3H3 μ1μ1 μ2μ2 μ3μ3 P1P1 P2P2 P3P3 μ μμμ C4C4

18 Erlang distribution: analysis Mean service time E[Y] = E[X 1 ] + E[X 2 ] =1/2μ + 1/2μ = 1/μ Variance Var[Y] = Var[X 1 ] + Var[X 2 ] = 1/4μ 2 v + 1/4μ 2 = 1/2μ 2 1/2μ E2E2

19 Squared coefficient of variation: analysis X is a constant X = d => E[X] = d, Var[X] = 0 => C 2 =0 X is an exponential r.v. E[X]=1/μ; Var[X] = 1/μ 2 => C 2 = 1 X has an Erlang r distribution E[X] = 1/μ, Var[X] = 1/rμ 2 => C 2 = 1/r f X * (s) = [rμ/(s+rμ)] r C2C2 0 constant 1 exponential Hypo-exponential Erlang

20 Probability density function of Erlang r Let Y have an Erlang r distribution r = 1 Y is an exponential random variable r is very large The larger the r => the closer the C 2 to 0 E r tends to infintiy => Y behaves like a constant E 5 is a good enough approximation

21 Generalized Erlang E r Classical Erlang r E[Y] = r/μ Var[Y] = r/μ 2 Generalized Erlang r Phases don’t have same μ rμrμrμrμ … rμrμ Y μ1μ1 μ2μ2 … μrμr Y

22 Generalized Erlang E r : analysis If the Laplace transform of a r.v. Y Has this particular structure Y can be exactly represented by An Erlang E r Where the service rates of the r phase  Are minus the root of the polynomials

23 Hyper-exponential distribution P 1 + P 2 + P 3 +…+ P k =1 Pdf of X? μ1μ1 μ2μ2 P1P1 P2P2 PkPk μkμk.... X

24 Hyper-exponential distribution:1 st and 2 nd moments Example: H 2

25 Hyper-exponential: squared coefficient of variation C 2 = Var[X]/E[X] 2 C 2 is greater than 1 Example: H 2, C 2 > 1 ?

26 Coxian model: main idea Idea Instead of forcing the customer to get r exponential distributions in an E r model The customer will have the choice to get 1, 2, …, r services Example C 2 : when customer completes the first phase He will move on to 2 nd phase with probability a  Or he will depart with probability b (where a+b=1) a b

27 Coxian model μ1μ1 μ2μ2 μ3μ3 μ4μ4 a1a1 b1b1 b2b2 b3b3 a2a2 a3a3 μ1μ1 μ1μ1 b1b1 a 1 b 2 μ2μ2 μ1μ1 μ2μ2 μ3μ3 a 1 a 2 b 3 μ1μ1 μ2μ2 μ3μ3 μ4μ4 a 1 a 2 a 3

28 Coxian distribution: Laplace transform Laplace transform of C k Is a fraction of 2 polynomials The denominator of order k and the other of order < k Implication A Laplace transform that has this structure Can be represented by a Coxian distribution  Where the order k = # phases,  Roots of denominator = service rate at each phase

29 Coxian model: conclusion Most Laplace transforms Are rational polynomials => Any distribution can be represented Exactly or approximately By a Coxian distribution

30 Coxian model: dimensionality problem A Coxian model can grow too big And may have as such a large # of phases To cope with such a limitation Any Laplace transform can be approximated by a c To obtain the unknowns (a, μ 1, μ 2 )  Calculate the first 3 moments based on Laplace transform  And match these against those of the C 2 a b=1-a μ1μ1 μ2μ2

31 C 2 : first three moments