A Result for a Counter Problem

Transcription

1 A Result for a Counter Problem M. Hlynka and P.H. Brill Department of Mathematics and Statistics University of Windsor Windsor, Ontario, Canada N9B 3P4 hlynka@uwindsor.ca January 2, 2008 Keywords: counter model, stochastic, M/D/1/1 queue Date: January 2, Do not quote without the authors permission. Abstract We present present a solution for the probability of a Type I counter being free at time t if the locked time is constant and the arrivals follows a Poisson process, and the system is free at time zero. This is the same as finding the probability of zero customers at time t in an M/D/1/1 queueing system, if the system starts empty. 1

2 1 Introduction Several authors in stochastic processes discuss electronic counter problems (such as Geiger counters). Examples include [2] and [1]. In [1], electronic counters are classified as one of two types. Signals, in the form of pulses, arrive at the electronic counter according to some process. A type I counter is either free or locked (dead) at any time t. If it is free then, it can accept an arriving signal. Once a signal is accepted, the counter is locked and no further signals can be accepted until the counter becomes free again. Any signals arriving, while the counter is locked, are lost to the system. For this paper, we assume that the locked time is a fixed positive constant D (for dead time). A type II counter, does not count signals that arrive during the locked time, but the locked time is extended by each arrival, even those during locked time. If the locked time contributed by each arrival is a constant D, then the counter remains locked for D time units after an arrival and that locked time may be extended if there is an arrival during a locked time.

3 Assume that a counter is free at time zero. Assume pulses arrive at a counter according to a Poisson process at rate λ. Let p(t) be probability that the counter is free at time t. For a type II counter, [1] indicates that p(t) = e λt e λd if 0 <t<d if t>d. This is reasonable. If t D, then the counter is unlocked if there are no arrivals in the period (0,D]. If t>d, then the counter is free at time t iff there are no arrivals in the previous D time units. Thus p(t) =e Dλ in that case. In section 2, we consider the Type II counter model. In Section 3, we give some numerical results. 2 Type I counter Solution We next consider the more complex Type I counter case. Property 2.1. For a type I counter, assume that signals arrive according to a Poisson process at rate λ. The counter is either locked or free at any time t. If the signal arrives when the counter is free, then the counter becomes locked for a time D. Any arriving signals

4 during the locked time do not count and do not affect the locked time. After the counter becomes free again, the next arriving signal will lock the counter for time D and so on. Let p(t) be the probability that the counter is free at time t. Thenp(t) = e λt p 1 (t) if 0 <t D (λ(t D)) 0 e λ(t D) p 1 (D)+ (λ(t D))1 e λ(t D) p 2 (t) if D<t D i 1 j=0 (λ(t id))j e λ(t id) p i j ((i j)d)+ (λ(t id))i e λ(t id) j! i! p i+1 (t) if id < t (i +1)D, i =1, 2,... Proof. Assume that the counter is free at time 0. CASE 1: 0 <t D. The counter will be locked at time t iff there is an arrival before t. So the probability that it is free at time t is the probability of no arrivals before t, namely p 1 (t) =e λt. CASE 2: D<t 2D. In this case, there are two ways that the counter could be free at time t. One way: There could be no arrivals over the entire interval (0,t). This could be viewed as 0 arrivals in (0,t D) and no arrivals in

5 the interval (t D, t) oflengthd. The numbers of arrivals in these nonoverlapping intervals are independent so the probability is (λ(t D)) 0 e λ(t D) p 1 (D) Second way: There could be at least one arrival in (0,t D), after which the counter is locked for an interval of length D. Let the first such arrival be at time s. After clearing, in the remaining time interval (D + s, t) oflengtht (D + s), there can be no arrivals (because an arrival would lock the counter at time t). But the intervals (0,s) and (D + s, t) have total length t D. So we are really getting the probability of exactly one arrival in an interval of length t D. This has probability (λ(t D))1 e λ(t D). Thus our total probability for case 2 is p 2 (t) = (λ(t D))0 e λ(t D) p 1 (D)+ (λ(t D))1 e λ(t D). Note that p 2 (D) =p 1 (D) so we can extend both functions to include the endpoints. CASE 3: 2D <t<3d In this case, there could be 0 or 1 or 2 arrivals while the counter is free. One way: There could be 0 arrivals in (0,t 2D) followed by the situation in Case 2. The probability is

6 (λ(t 2D)) 0 e λ(t 2D) p 2 (2D). Second way: There could be at least one arrival in (0,t 2D), the first at time s, then the counter is locked for the interval (s, s + D), followed by no arrivals in (s + D, t D), followed by no arrivals in the interval (t D, t). But the intervals (0,s)and(s + D, t D) have total length t 2D. This situation is equivalent to exactly one arrival in an interval of length t 2D followed by locking time D, and no arrivals in the final interval of length D. The probability of this second way is (λ(t 2D)) 1 e λ(t 2D) p 1 (D). Third way. There could be at least one arrival in (0,t 2D), say the first is at s, then the counter is locked until time s + D, then at least another arrival in (s + D, t D), say the first is at r, then counter is locked until r + D, followed by no arrival in the final interval (r + D, t). But the three small intervals (0,s), (s + D, r), (r + D, t) have total length t 2D so this is equivalent to getting exactly 2 arrivals in an interval of length t 2D. The probability of the third way is (λ(t 2D)) 2 e λ(t 2D). 2!

7 Summing the probabilities of the three ways gives p 3 (t) = (λ(t 2D))0 e λ(t 2D) p 2 (2D)+ (λ(t 2D))1 e λ(t 2D) p 1 (D)+ (λ(t 2D)) 2 e λ(t 2D). 2! The pattern is clear at this point. 3 Numerical Results We give a partial check of the results using the statstical language R by considering a specific case. Let D =2,λ = 1. We compute p 1 (D), p 2 (2D), p 3 (2D), p 4 (4D), p 5 (5D), p 6 (6D), p 7 (7D). We know that the limiting result should be are the results. 1/λ D +1/λ =1/3. Here >D=2;L=1 >test=(1/l)/(d+1/l); test >p1=exp(-l*d);p >p2=exp(-l*d)*p1+l*d*exp(-l*d):p >p3=exp(-l*d)*p2+l*d*exp(-l*d)*p1+((l*d)ˆ2)*exp(-l*d)/2;p

8 >p4=exp(-l*d)*p3+l*d*exp(-l*d)*p2+p1*((l*d)ˆ2)*exp(-l*d)/2 +((L*D)ˆ3)*exp(-L*D)/6 >p >p5=exp(-l*d)*p4+l*d*exp(-l*d)*p3+p2*((l*d)ˆ2)*exp(-l*d)/2 +p1*((l*d)ˆ3)*exp(-l*d)/6+((l*d)ˆ4)*exp(-l*d)/24 >p >p6=exp(-l*d)*p5+l*d*exp(-l*d)*p4+p3*((l*d)ˆ2)*exp(-l*d)/2 +p2*((l*d)ˆ3)*exp(-l*d)/6+p1*((l*d)ˆ4)*exp(-l*d)/24 +((L*D)ˆ5)*exp(-L*D)/120 >p >p7= exp(-l*d)*p6+l*d*exp(-l*d)*p5+p4*((l*d)ˆ2)*exp(-l*d)/2+ p3*((l*d)ˆ3)*exp(-l*d)/6+p2*((l*d)ˆ4)*exp(-l*d)/24+ p1*((l*d)ˆ5)*exp(-l*d)/120+((l*d)ˆ6)*exp(-l*d)/720 >p

9 References [1] S. Karlin and H.M. Taylor. A First Course in Stochastic Processes. Second Edition Academic Press. [2] E. Parzen. Stochastic Processes Holden Day.