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Basic queueing system components Arrival patterns What is common to these arrival patterns? What is different? How can we describe/specify an arrival.

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Presentation on theme: "Basic queueing system components Arrival patterns What is common to these arrival patterns? What is different? How can we describe/specify an arrival."— Presentation transcript:

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2 Basic queueing system components

3 Arrival patterns What is common to these arrival patterns? What is different? How can we describe/specify an arrival pattern?

4 Describing an arrival pattern 040302010 N(t) = number of arrivals upto time t {N(t), 0<t<40} is called a process N(t) is a random variable for every timepoint t To specify an arrival pattern, we must specify the probability distribution of N(t) for every time t

5 Describe an arrival process E.g. N(t) = N(  t,  t ) What are reasonable  t,  t ? Is this a reasonable process? Let N(0) = 0, h a small interval of time What is a reasonable distribution for N(h)? 040302010

6 The Poisson process Small time interval h. Prob(event in interval h) = h The probability of an event in h depends only on the length of h and not on its position, t. e.g, it does not matter how long since an event last occurred. The process has no memory. (sometimes called the Markovian property). We choose to make h and  small enough so that at most one event will occur in h.  is called the event rate or arrival rate

7 No events P(event in interval h) = h P{0 events in interval [t, t+h] } = 1 - h (here  h  Define p 0 (t) = P{0 events in [0, t] } Then p 0 (t+h) = P{ 0 events in [0, t+h]} Since probability of an event in h is independent of any other: p 0 (t+h) = p 0 (t).(1 - h 

8 Poisson distribution P(N=k) = m k  e -m / k! Poisson process (rate = ) P(N(t)=k) = ( t) k e - t / k!

9 How long before the next arrival? T= time of next arrival f T (t) = e  t  exponential distribution) the average time between events = 1/  P(T>t) = P(N(t)=0) = e - t

10 Exponential distribution E(x)=

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12 Merging two Poisson streams

13 Prob{event in stream 1 is in h } =   h Prob{event in stream 2 is in h    h  Prob{some event is in h} = Prob{stream 1 event} + Prob{stream 2 event} =  h    h = (     h Merged stream acts like a Poisson process with rate    

14 Partitioning two streams Whenever an event (arrival) occurs we divert it into A or B stream according to fixed probabilities, p A and p B. Prob{A event in h} = prob{event in h} × Prob{choosing A stream} = h pp  The A stream acts like a Poisson stream with a rate of events = p A

15 Poisson arrivals  Random arrivals Given: arrival has occurred in [0,t] Let X = time to arrival Then X ~ U[0,t] Interpretation: arrival has occurred at random in [0,t] t0 X

16 How long do you have to wait for service? Red job already in service. Blue job comes into the queue (length 1). How long will blue have to wait (how much is R) ? Assume job lengths are Exponentially distributed with average length  Then E(R) =  (renewal paradox) R T

17 Residual time distribution P(R>r) =P(T>r+t |T>t) =P(T>r+t)/P(T>t) = e - (r+t) /e - t= = e - r = P (T > r) R T t additional time to complete a service is independent of how long it has already been in service time up to next arrival is independent of when the previous arrival occurred

18 How long between arrivals ? Since no memory the zero point can be an event. Thus time between events is also exponential,  f(t) = e  t the average time between events = 1/ 


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