# 15.1 Some random selection problems  (Page 6/6)

 Page 6 / 6

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

1. $N\left(t+h\right)-N\left(t\right)=N\left(h\right)$ for all $t,\phantom{\rule{0.277778em}{0ex}}h>0$ , and
2. For ${t}_{1}<{t}_{2}\le {t}_{3}<{t}_{4}\le \cdots \le {t}_{m-1}<{t}_{m}$ , the class $\left\{N\left({t}_{2}\right)-N\left({N}_{1}\right),\phantom{\rule{0.277778em}{0ex}}N\left({t}_{4}\right)-N\left({t}_{3}\right),\phantom{\rule{0.277778em}{0ex}}\cdots ,\phantom{\rule{0.277778em}{0ex}}N\left({t}_{m}\right)-N\left({t}_{m-1}\right)\right\}$ is independent.

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

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 $\left(n,\phantom{\rule{0.166667em}{0ex}}\lambda \right)$ .

## Gamma arrival times

Suppose the arrival times ${S}_{n}\sim$ gamma $\left(n,\phantom{\rule{0.166667em}{0ex}}\lambda \right)$ and g is such that

${\int }_{0}^{\infty }|g|<\infty \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}E\left[\sum _{n=1}^{\infty },|g\left({S}_{n}\right)|\right]<\infty$

Then

$E\left[\sum _{n=1}^{\infty },g,\left({S}_{n}\right)\right]=\lambda {\int }_{0}^{\infty }g$

VERIFICATION

We use the countable sums property (E8b) for expectation and the corresponding property for integrals to assert

$E\left[\sum _{n=1}^{\infty },g,\left({S}_{n}\right)\right]=\sum _{n=1}^{\infty }E\left[g\left({S}_{n}\right)\right]=\sum _{n=1}^{\infty }{\int }_{0}^{\infty }g\left(t\right){f}_{n}\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{f}_{n}\left(t\right)=\frac{\lambda {e}^{-\lambda t}{\left(\lambda t\right)}^{n-1}}{\left(n-1\right)!}$

We may apply (E8b) to assert

$\sum _{n=1}^{\infty }{\int }_{0}^{\infty }g{f}_{n}={\int }_{0}^{\infty }g\sum _{n=1}^{\infty }{f}_{n}$

Since

$\sum _{n=1}^{\infty }{f}_{n}\left(t\right)=\lambda {e}^{-\lambda t}\sum _{n=1}^{\infty }\frac{{\left(\lambda t\right)}^{n-1}}{\left(n-1\right)!}=\lambda {e}^{-\lambda t}{e}^{\lambda t}=\lambda$

the proposition is established.

## Discounted replacement costs

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 $\left(n,\phantom{\rule{0.166667em}{0ex}}\lambda \right)$ and the present value of all future replacements is

$C=\sum _{n=1}^{\infty }c{e}^{-\alpha {S}_{n}}$

The expected replacement cost is

$E\left[C\right]=E\left[\sum _{n=1}^{\infty },g,\left({S}_{n}\right)\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}g\left(t\right)=c{e}^{-\alpha t}$

Hence

$E\left[C\right]=\lambda {\int }_{0}^{\infty }c{e}^{-\alpha t}\phantom{\rule{0.166667em}{0ex}}dt=\frac{\lambda c}{\alpha }$

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

$E\left[C\right]=\frac{1200·4}{0.08}=60,000$

## Random costs

Suppose the cost of the n th replacement in [link] is a random quantity C n , with $\left\{{C}_{n},\phantom{\rule{0.166667em}{0ex}}{S}_{n}\right\}$ independent and $E\left[{C}_{n}\right]=c$ , invariant with n . Then

$E\left[C\right]=E\left[\sum _{n=1}^{\infty },{C}_{n},{e}^{-\alpha {S}_{n}}\right]=\sum _{n=1}^{\infty }E\left[{C}_{n}\right]E\left[{e}^{-\alpha {S}_{n}}\right]=\sum _{n=1}^{\infty }cE\left[{e}^{-\alpha {S}_{n}}\right]=\frac{\lambda c}{\alpha }$

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 .

## Finite horizon

Under the conditions assumed in [link] , above, let N t be the counting random variable for arrivals in the interval $\left(0,\phantom{\rule{0.166667em}{0ex}}t\right]$ .

$\text{If}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{Z}_{t}=\sum _{n=1}^{{N}_{t}}g\left({S}_{n}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{then}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}E\left[{Z}_{t}\right]=\lambda {\int }_{0}^{t}g\left(u\right)\phantom{\rule{0.166667em}{0ex}}du$

VERIFICATION

Since ${N}_{t}\ge n$ iff ${S}_{n}\le t$ , $\sum _{n=1}^{{N}_{t}}g\left({S}_{n}\right)=\sum _{n=0}^{\infty }{I}_{\left(0,\phantom{\rule{0.166667em}{0ex}}t\right]}\left({S}_{n}\right)g\left({S}_{n}\right)$ . In the result of [link] , replace g by ${I}_{\left(0,t\right]}g$ and note that

${\int }_{0}^{\infty }{I}_{\left(0,t\right]}\left(u\right)g\left(u\right)\phantom{\rule{0.166667em}{0ex}}du={\int }_{0}^{t}g\left(u\right)\phantom{\rule{0.166667em}{0ex}}du$

## Replacement costs, finite horizon

Under the conditions of [link] , consider the replacement costs over a two-year period.

SOLUTION

$E\left[C\right]=\lambda c{\int }_{0}^{t}{e}^{-\alpha u}\phantom{\rule{0.166667em}{0ex}}du=\frac{\lambda c}{\alpha }\left(1-{e}^{-\alpha t}\right)$

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 $E\left[C\right]=60000·0.1479=8,871.37$ .

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.

## General interarrival, exponential g

Suppose ${S}_{0}=0$ and ${S}_{n}=\sum _{i=1}^{n}{W}_{i}$ , where $\left\{{W}_{i}:1\le i\right\}$ is iid. Let $\left\{{V}_{n}:1\le n\right\}$ be a class such that each $E\left[{V}_{n}\right]=c$ and each pair $\left\{{V}_{n},\phantom{\rule{0.166667em}{0ex}}{S}_{n}\right\}$ is independent. Then for $\alpha >0$

$E\left[C\right]=E\left[\sum _{n=1}^{\infty },{V}_{n},{e}^{-\alpha {S}_{n}}\right]=c·\frac{{M}_{W}\left(-\alpha \right)}{1-{M}_{W}\left(-\alpha \right)}$

where M W is the moment generating function for W .

DERIVATION

First we note that

$E\left[{V}_{n}{e}^{-\alpha {S}_{n}}\right]=c{M}_{{S}_{n}}\left(-\alpha \right)=c{M}_{W}^{n}\left(-\alpha \right)$

Hence, by properties of expectation and the geometric series

$E\left[C\right]=c\sum _{n=1}^{\infty }{M}_{W}^{n}\left(-\alpha \right)=\frac{{M}_{W}\left(-\alpha \right)}{1-{M}_{W}\left(-\alpha \right)},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{provided}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}|{M}_{W}\left(-\alpha \right)|<1$

Since $\alpha >0$ and $W>0$ , we have $0<{e}^{-\alpha W}<1$ , so that ${M}_{W}\left(-\alpha \right)=E\left[{e}^{-\alpha W}\right]<1$ .

## Uniformly distributed interarrival times

Suppose each ${W}_{i}\sim$ uniform $\left(a,\phantom{\rule{0.166667em}{0ex}}b\right)$ . Then (see Appendix C ),

${M}_{W}\left(-\alpha \right)=\frac{{e}^{-a\alpha }-{e}^{-b\alpha }}{\alpha \left(b-a\right)}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}E\left[C\right]=c·\frac{{e}^{-a\alpha }-{e}^{-b\alpha }}{\alpha \left(b-a\right)-\left[{e}^{-a\alpha }-{e}^{-b\alpha }\right]}$

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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