POISSONPROCESSPOISSONPROCESS

Counting Counting Process:Process:

•State of the State of the system is given system is given by the total by the total number of number of customers in the customers in the system .system .

c ont..c ont..

N(t)=state of system at N(t)=state of system at time t .time t .

(1)(1) N(0)=0N(0)=0

(2)(2) events are independant events are independant if they relate to non-if they relate to non-overlapping intervals .overlapping intervals .4pm 5pm 6pm

(3)(3) stationary process :stationary process :

i.e Pdf depends only i.e Pdf depends only on the interval on the interval length , not starting length , not starting point .point .

(4)(4) probablity that one probablity that one event occurs in an event occurs in an interval of length h is interval of length h is

λh+ 0(h)

f (h) = 0(h)

if Lf (h)h

→ 0

asL h→ 0

5)5) probablity probablity that more than that more than one event one event occurs in an occurs in an interval of interval of length h is 0(h) .length h is 0(h) .

•example :example :

for hfor h5/25/2 hh5/2 5/2 /h = h/h = h3/23/2 --->0 as h--- --->0 as h---> 0> 0

(5)(5) probablity that probablity that more than one event more than one event occurs in an interval of occurs in an interval of length h is 0(h) .length h is 0(h) .

Set up a Differential Set up a Differential Differency Equation :Differency Equation :

LetLet ::

Pn(t) = the probability Pn(t) = the probability that N(t)=nthat N(t)=n

pn (t) =P N(t)=n[ ]Pn(t+h)=Pn(t)P0 (h) + Pn−1P1(h) +Pn−2P2 (h).....

P0 (h)=prob no event P0 (h)=prob no event occurs in an interval of occurs in an interval of length h.length h.

[ ]P h p

P h h h0

0

1

1 0

( )

( ) ( )

= −

= − +λ

Pn(t + h) =Pn(t)[1−λh+ 0(h)] +Pn−1(t)λh+ 0(h)(......)

Pn(t+ h) =Pn(t)−λhPn(t) +λhPn−1(t) +o(h)(......)

limh→ 0

Pn(t +h)−Pn(t)h

=−λPn(t)−λPn−1(t)

+0(h)h

(.....)

P t h P t P t for nn n n/ ( ) ( ) ( )... .. . . .+ =− + ≥−λ λ 1 1

t2 tntn-1t10

[ ]P t h P t h h

P t h P t

hP t

h

h

P t P t

h

0 0

0

0 00

0 0

1 0

0

( ) ( ) ( )

lim( ) ( )

( )( )

( ) ( )/

+ = − +

+ −= − +

⎡

⎣⎢⎤

⎦⎥

= −

→

λ

λ

λ

[ ]P s ensτ λ> = − −1

CDF for Pdf of τ :

δδ

λ

λ

λ

( )

( )

( )

( ) .... ...

/

\

s e

Pdf s

f s

e

or

f s for s

s

s

1

0 0

−

=

=

= <

⎧

⎨⎪

⎩⎪

−

M is used :

Eτλ

1

λ220

Pn(t)=prob. n arrivals occur in an interval of lengtht .

BIRTH & DEATH PROCESS

Let λn be t :

Birth Rate when t state at the system is n

n be t dent

Pn (t +h)=Pn(t) 1−λnh+ 0(h)[ ] 1−nh+ 0(h)[ ]

+Pn−1(t)λn−1h1−nh+ 0(h)[ ]....................

+Pn+1(t) 1−λn−1h+ 0(h)[ ].(n +h) +o(h)(......)

λnh.nh=0(h)

lim( ) ( ) ( )

hPn t h Pn t

hP P P p

h

hn n n n n n n n

→+ −

=− − + + +⎡

⎣⎢

⎤

⎦⎥− − + +

01 1 1 1

0λ λ

P t P P P Pn n n n n n n n n

/( ) = − + + −

+ + − −λ λ

1 1 1 1

at equilibrium (i.e steady at equilibrium (i.e steady state)state)

Pn/(t) =0

QλnPn −n+1Pn+1 =λn−1Pn−1 −nPn

λkPk −k+1Pk+1 =constant

P0 (t +h)=P0(t) 1−λ0h+ 0(h)[ ]+P1(t)1h+ 0(h)

λ−1 =00 =

P0/(t) =−λ0P0(t) + 1P1(t)

At steady state:At steady state:

P t

get

P P

or

P P

P Pk k k k

0

0 0 1 1

0 0 1 1

1 1

0

0

0

0

/ ( ) =

− + =

− =− =+ +

λ

λ λ

P Pn

P P

P P

P

nn

n

to n

s

++

=

=

+ =

+ + + + =

11

10

10

0 1

00

1

0 1

1 2

1

1 1

λ

λ

λ

λ λ

K K

1 24 4 4 4 34 4 4 4[ ...... ......]

P0 =1s

suppose...all...λn =λand..all..n =

Pn+1 =λ

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

n

P0 =λn

n P0

ρ

=

c untiligation factor

ρ

λ

μ

=

traffic inquiry

s p

P s

P

P

P P

n

nn n

= + + + =−

=

−=

= −

= = −

11

1

11

11

1

1

0

0

0

0

ρρ

ρρ

ρ ρ ρ

....

( )

E N[ ] = KPK = Kρ K (1−ρ) =K=0

∞

∑k=0

∞

∑

ρ(1−ρ) KρK−1

K=1

∞

∑

Kx xK K

K

K

n

−

=

∞

=

=−

−⎛⎝⎜

⎞⎠⎟ =

−

∑

1

0

11

11

11

( )

( )

\

/

ρρ

ρ ρ

L E Nn

= + −−

=−

( ) ( )( )

ρ ρρ

ρρ

11

1 1

L =λω

[ ]L E N

Arrival Rate

Arrival Time spent By n customer in the system

average serving time t average waiting in array

=

=

=

=

=

λ

ω

ω

ωμ

..

.. .. .. . . .. .. . . . .

.. .. .. . . .. . . ..

1

L

q

=

= +

λω

ω

ω1

ωλ λ

ρρ λ

λλ

λ

ω ω λ

λ λ

ρ λ

ω ρω

= =−

=−

=−

= − =−

− =−

−−

=

1 1

1

1

1

1

1 1 1

L

q

q

( )