# 15.1 Some random selection problems  (Page 2/6)

 Page 2 / 6

## Message routing

A junction point in a network has two incoming lines and two outgoing lines. The number of incoming messages N 1 on line one in one hour is Poisson (50); on line 2 the number is ${N}_{2}\sim$ Poisson (45). On incoming line 1 the messages have probability ${p}_{1a}=0.33$ of leaving on outgoing line a and $1-{p}_{1a}$ of leaving on line b. The messages coming in on line 2 have probability ${p}_{2a}=0.47$ of leaving on line a. Under the usual independence assumptions, what is the distribution of outgoing messages on line a?What are the probabilities of at least 30, 35, 40 outgoing messages on line a?

SOLUTION

By the Poisson decomposition, ${N}_{a}\sim$ Poisson $\left(50·0.33+45·0.47=37.65\right)$ .

ma = 50*0.33 + 45*0.47 ma = 37.6500Pa = cpoisson(ma,30:5:40) Pa = 0.9119 0.6890 0.3722

VERIFICATION of the Poisson decomposition

1. ${N}_{k}=\sum _{i=1}^{N}{I}_{{E}_{ki}}$ .
This is composite demand with ${Y}_{k}={I}_{{E}_{ki}}$ , so that ${g}_{{Y}_{k}}\left(s\right)={q}_{k}+s{p}_{k}=1+{p}_{k}\left(s-1\right)$ . Therefore,
${g}_{{N}_{k}}\left(s\right)={g}_{N}\left[{g}_{{Y}_{k}}\left(s\right)\right]={e}^{\mu \left(1+{p}_{k}\left(s-1\right)-1\right)}={e}^{\mu {p}_{k}\left(s-1\right)}$
which is the generating function for ${N}_{k}\sim$ Poisson $\left(\mu {p}_{k}\right)$ .
2. For any ${n}_{1},\phantom{\rule{0.166667em}{0ex}}{n}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{n}_{m}$ , let $n={n}_{1}+{n}_{2}+\cdots +{n}_{m}$ , and consider
$A=\left\{{N}_{1}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2}={n}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{N}_{m}={n}_{m}\right\}=\left\{N=n\right\}\cap \left\{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\right\}$
Since N is independent of the class of ${I}_{{E}_{ki}}$ , the class
$\left\{\left\{N=n\right\},\phantom{\rule{0.166667em}{0ex}}\left\{{N}_{1n}={n}_{1},\phantom{\rule{0.166667em}{0ex}}{N}_{2n}={n}_{2},\cdots ,\phantom{\rule{0.277778em}{0ex}}{N}_{mn}={n}_{m}\right\}\right\}$
is independent. By the product rule and the multinomial distribution
$P\left(A\right)={e}^{-\mu }\frac{{\mu }^{n}}{n!}·n!\prod _{k=1}^{m}\frac{{p}_{k}^{{n}_{k}}}{\left({n}_{k}\right)!}=\prod _{k=1}^{m}{e}^{-\mu {p}_{k}}\frac{{p}_{k}^{{n}_{k}}}{{n}_{k}!}=\prod _{k=1}^{m}P\left({N}_{k}={n}_{k}\right)$
The second product uses the fact that
${e}^{\mu }={e}^{\mu \left({p}_{1}+{p}_{2}+\cdots +{p}_{m}\right)}=\prod _{k=1}^{m}{e}^{\mu {p}_{k}}$
Thus, the product rule holds for the class $\left\{{N}_{k}:1\le k\le m\right\}$ , so that it is independent.

## Extreme values

Consider an iid class $\left\{{Y}_{i}:1\le i\right\}$ of nonnegative random variables. For any positive integer n we let

${V}_{n}=min\left\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{n}=max\left\{{Y}_{1},\phantom{\rule{0.166667em}{0ex}}{Y}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{Y}_{n}\right\}$

Then

$P\left({V}_{n}>t\right)={P}^{n}\left(Y>t\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({W}_{n}\le t\right)={P}^{n}\left(Y\le t\right)$

Now consider a random number N of the Y i . The minimum and maximum random variables are

${V}_{N}=\sum _{n=0}^{\infty }{I}_{\left\{N=n\right\}}{V}_{n}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{W}_{N}=\sum _{n=0}^{\infty }{I}_{\left\{N=n\right\}}{W}_{n}$

$\square$

Computational formulas

If we set ${V}_{0}={W}_{0}=0$ , then

1. ${F}_{V}\left(t\right)=P\left(V\le t\right)=1+P\left(N=0\right)-{g}_{N}\left[P\left(Y>t\right)\right]$
2. ${F}_{W}\left(t\right)={g}_{N}\left[P\left(Y\le t\right)\right]$

These results are easily established as follows. $\left\{{V}_{N}>t\right\}=\underset{n=0}{\overset{\infty }{\bigvee }}\left\{N=n\right\}\left\{{V}_{n}>t\right\}$ . By additivity and independence of $\left\{N,\phantom{\rule{0.166667em}{0ex}}{V}_{n}\right\}$ for each n

$P\left({V}_{N}>t\right)=\sum _{n=0}^{\infty }P\left(N=n\right)P\left({V}_{n}>t\right)=\sum _{n=1}^{\infty }P\left(N=n\right){P}^{n}\left(Y>t\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{since}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({V}_{0}>t\right)=0$

If we add into the last sum the term $P\left(N=0\right){P}^{0}\left(Y>t\right)=P\left(N=0\right)$ then subtract it, we have

$P\left({V}_{N}>t\right)=\sum _{n=0}^{\infty }P\left(N=n\right){P}^{n}\left(Y>t\right)-P\left(N=0\right)={g}_{N}\left[P\left(Y>t\right)\right]-P\left(N=0\right)$

A similar argument holds for proposition (b). In this case, we do not have the extra term for $\left\{N=0\right\}$ , since $P\left({W}_{0}\le t\right)=1$ .

Special case . In some cases, $N=0$ does not correspond to an admissible outcome (see [link] , below, on lowest bidder and [link] ). In that case

${F}_{V}\left(t\right)=\sum _{n=1}^{\infty }P\left({V}_{n}\le t\right)P\left(N=n\right)=\sum _{n=1}^{\infty }\left[1-{P}^{n}\left(Y>t\right)\right]P\left(N=n\right)=\sum _{n=1}^{\infty }P\left(N=n\right)-\sum _{n=1}^{\infty }{P}^{n}\left(Y>t\right)P\left(N=n\right)$

Add $P\left(N=0\right)={P}^{0}\left(Y>t\right)P\left(N=0\right)$ to each of the sums to get

${F}_{V}\left(t\right)=1-\sum _{n=0}^{\infty }{P}^{n}\left(Y>t\right)P\left(N=n\right)=1-{g}_{N}\left[P\left(Y>t\right)\right]$

$\square$

## Maximum service time

The number N of jobs coming into a service center in a week is a random quantity having a Poisson (20) distribution. Suppose the service times (in hours) for individualunits are iid, with common distribution exponential (1/3). What is the probability the maximum service time for the units is no greater than 6, 9, 12, 15, 18 hours?SOLUTION

## Solution

$P\left({W}_{N}\le t\right)={g}_{N}\left[P\left(Y\le t\right)\right]={e}^{20\left[{F}_{Y}\left(t\right)-1\right]}=exp\left(-20{e}^{-t/3}\right)$
t = 6:3:18; PW = exp(-20*exp(-t/3));disp([t;PW]')6.0000 0.0668 9.0000 0.369412.0000 0.6933 15.0000 0.873918.0000 0.9516

where we get a research paper on Nano chemistry....?
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
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
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive By By Sandhills MLT By OpenStax By OpenStax By Madison Christian By Cath Yu By Richley Crapo By Stephen Voron By Madison Christian By OpenStax