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

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive