By I. Good

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Du = 1· m! /! 1=0 Hence, for r = 1, 2, ... , 1 {ra+a-1 prob(N,Ce) = r) = ~ m~a 6-m=~a 6 m ra-1 a 1 (pt /~ -pi (pt)l e- pi 1 ra+a-1 =- i m} ~ (ra+a-1)---~~ /! (pfi e- pt ra-1 +~ ~ I! (1-ra+a) l=ra-a ra+a-1 { = (r+l) pt a +- ra+a-2 ra-1 1~ -~~=~ 1 -(r-1) 1 =~a r~a-2 }(ptie-pl I! l=ra-a-1 . (11) Further . -:-:--a 1=0 /! = {~ -~~}(pt):~-pl. 1=0 (12) 1=0 For small values of r and a, these formulae are convenient for numerical calculation, working from tables of cumulative sums of the Poisson distribution.

Ra +a- I. Hence, since the number of stages completed in time t follows a Poisson distribution of mean pt, ra+a-1 prob (Njo> = r) = "" ~ m=ra (p )m -pt t e • m! 5 (ii). >, the number of renewals in the corresponding equilibrium renewal process. 2. 3) into an equation for the probability generating function of N,. Let 00 G(t,O = l: ''prob(N, r=O = r) (1) 00 = 1+ l: ,r-l(,-l)K,(t). (2) r=l Now if the Laplace transform with respect to t of k,(t) is k~(s), then that of K,(t) is k~(s)fs. Hence, on applying a Laplace transformation to (2), we have that G*(s,Q = ~+~ ~ ''- 1 (,-l)k~(s).

The following examples are of renewal processes that are not, in general, Poisson processes. Example. , a component being immediately replaced on failure. Example. Consider a labour force of m rr:en. Let a' component' be a man, 'failure' occurring when the man leaves the job, a replacement being made immediately. Suppose that at timet= 0, all the men are new to the job. Make the further, rather dubious, assumption that all failure-times are independent and with the same distribution. Then we have m independent ordinary renewal processes in operation simultaneously.

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