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Remark . The counting process is a Poisson process in the sense that ${N}_{t}\sim $ Poisson $\left(\lambda t\right)$ for all $t>0$ . More advanced treatments show that the process has independent, stationary increments. That is
In words, the number of arrivals in any time interval depends upon the length of the interval and not its location in time, and the numbers of arrivals in nonoverlappingtime intervals are independent.
Emergency calls arrive at a police switchboard with interarrival times (in hours) exponential (15). Thus, the average interarrival time is 1/15 hour (four minutes).What is the probability the number of calls in an eight hour shift is no more than 100, 120, 140?
p = 1 - cpoisson(8*15,[101 121 141])p = 0.0347 0.5243 0.9669
We develop next a simple computational result for arrival processes for which ${S}_{n}\sim $ gamma $(n,\phantom{\rule{0.166667em}{0ex}}\lambda )$ .
Suppose the arrival times ${S}_{n}\sim $ gamma $(n,\phantom{\rule{0.166667em}{0ex}}\lambda )$ and g is such that
Then
VERIFICATION
We use the countable sums property (E8b) for expectation and the corresponding property for integrals to assert
We may apply (E8b) to assert
Since
the proposition is established.
A critical unit in a production system has life duration exponential $\left(\lambda \right)$ . Upon failure the unit is replaced immediately by a similar unit. Units fail independently. Cost of replacement of a unit is c dollars. If money is discounted at a rate α , then a dollar spent t units of time in the future has a current value ${e}^{-\alpha t}$ . If S _{n} is the time of replacement of the n th unit, then ${S}_{n}\sim $ gamma $(n,\phantom{\rule{0.166667em}{0ex}}\lambda )$ and the present value of all future replacements is
The expected replacement cost is
Hence
Suppose unit replacement cost $c=1200$ , average time (in years) to failure $1/\lambda =1/4$ , and the discount rate per year $\alpha =0.08$ (eight percent). Then
Suppose the cost of the n th replacement in [link] is a random quantity C _{n} , with $\{{C}_{n},\phantom{\rule{0.166667em}{0ex}}{S}_{n}\}$ independent and $E\left[{C}_{n}\right]=c$ , invariant with n . Then
The analysis to this point assumes the process will continue endlessly into the future. Often, it is desirable to plan for a specific, finite period. The result of [link] may be modified easily to account for a finite period, often referred to as a finite horizon .
Under the conditions assumed in [link] , above, let N _{t} be the counting random variable for arrivals in the interval $(0,\phantom{\rule{0.166667em}{0ex}}t]$ .
VERIFICATION
Since ${N}_{t}\ge n$ iff ${S}_{n}\le t$ , $\sum _{n=1}^{{N}_{t}}g\left({S}_{n}\right)=\sum _{n=0}^{\infty}{I}_{(0,\phantom{\rule{0.166667em}{0ex}}t]}\left({S}_{n}\right)g\left({S}_{n}\right)$ . In the result of [link] , replace g by ${I}_{(0,t]}g$ and note that
Under the conditions of [link] , consider the replacement costs over a two-year period.
SOLUTION
Thus, the expected cost for the infinite horizon $\lambda c/\alpha $ is reduced by the factor $1-{e}^{-\alpha t}$ . For $t=2$ and the numbers in [link] , the reduction factor is $1-{e}^{-0.16}=0.1479$ to give $E\left[C\right]=60000\xb70.1479=8,871.37$ .
In the important special case that $g\left(u\right)=c{e}^{-\alpha u}$ , the expression for $E\left[\sum _{n=1}^{\infty},g,\left({S}_{n}\right)\right]$ may be put into a form which does not require the interarrival times to be exponential.
Suppose ${S}_{0}=0$ and $S}_{n}=\sum _{i=1}^{n}{W}_{i$ , where $\{{W}_{i}:1\le i\}$ is iid. Let $\{{V}_{n}:1\le n\}$ be a class such that each $E\left[{V}_{n}\right]=c$ and each pair $\{{V}_{n},\phantom{\rule{0.166667em}{0ex}}{S}_{n}\}$ is independent. Then for $\alpha >0$
where M _{W} is the moment generating function for W .
DERIVATION
First we note that
Hence, by properties of expectation and the geometric series
Since $\alpha >0$ and $W>0$ , we have $0<{e}^{-\alpha W}<1$ , so that ${M}_{W}(-\alpha )=E\left[{e}^{-\alpha W}\right]<1$ .
Suppose each ${W}_{i}\sim $ uniform $(a,\phantom{\rule{0.166667em}{0ex}}b)$ . Then (see Appendix C ),
Let $a=1$ , $b=5$ , $c=100$ and $\alpha =0.08$ . Then,
a = 1;
b = 5;c = 100;
A = 0.08;MW = (exp(-a*A) - exp(-b*A))/(A*(b - a))
MW = 0.7900EC = c*MW/(1 - MW)
EC = 376.1643
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