mmcinfinity

8/2/2019 mmcinfinity

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M/M/C/ Model

This queuing system deals with queries whichare being served by the parallel server in whicheach server has an independent and identically

distributed exponential serving line distributionwith the arrival process assumed to followpoisson distribution. This model is ageneralization of M/M/1model in the sense that

here we consider C service channels instead of1 service channel. Here input is poisson andservice is exponential so this is a birth & deathprocess. Here the mean arrival rate is given by

= +a. For mean service rate if there are more

{


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DIFFERENCE EQUATION

For1


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For n=op0(t+t)= p (t)[1- t+0(t)]

+p (t)[1- t)( t)+ O(t)-(3)

0

Combining the terms involving(t) in O(t)

1

From(1) For 1


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For n=0 From(3):-p (t+ t)= (1- t) p (t) + t

p (t)+ 0 (t)

0 1 (

6)

Here (4), (5), (6) represents the set of difference equation forthe M/M/C/infinitymodel.

0

Difference differential equations:-

Dividing equations by (t) and taking limits t 0 we get:-lim p (t+ t)-p (t)/ t = lim (+n) tp (t)/ t

+lim ( t)p (t)/ t

+lim (n+1)p (t) t/t

n nt 0 t 0 n

t 0 n-1

t 0 n+

1


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d/dt p (t)= -(+n )p (t) + p (t) + p (t)

(7)

Similarly from eq.(5) and eq.(6),we get

d/dt p (t) = -(+c)pn(t) + pn-1(t) + cpn+1(t)(8)

And

d/dt p0(t)= -p0(t) + p1(t)(9)

So equations (7),(8),(9) represents difference

n n n-1

n+1

n


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Steady State equations:-

As t ,we reach a steady state which isindependent of time t

Lim t 0 d/dt pn(t) = 0

0= -( +n)pn + pn-1+ (n+1)pn+1 -(10)

0= -( +n)pn + pn-1 + cpn+1 -(11)

Pn+1=( +n)pn/c - pn-1/c


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p1= ( /)p0

Equations (10),(11),(12) are known as Steady Statedifferential equations for this model.

for n=1-( +)p1 + p0 + 2p2 =0

-( +)p0/ + p0 + 2p2 =0

P2= [ ( +) /- /2] p0

= [( +) - /(2)] p0


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For n=2

[ +2/3]p2 - /3 p1

+2/3 * /2 * p0/ - 2/32 p0

2/32 p0 [ -1+2/2]

P3= 2 /3!2 p0 Therefore,pn= 1/n! [ /]n p0

1n


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Pc = 1/c! ( /)c p0

Put n=c and n=c-1 in eq.(13),we get:

[ +c/c] .1/c! . ( /)c p0 /c .1/(c-1)! .( /)c-1 p0

Pc+1 = 1/c! . c/c p0 [ +c/c - 1]

= 1/c!. ( /)c p0 . [ +c-c/c]

= 1/c! ( /)c p0 . ( /c)

= 1/c! ( /)c+1 p0 . 1/cor

1/c.c! ( /)c+1 p0


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Similarly for n=c+2 we get,

Pc+2 = 1/c2(c!) . ( /)c+2 p0

For n=c+3

Pc+3 = 1/c3(c!) . ( /)c+3 p0

Thus,pn = 1/cn-c(c!) . ( /)n p0 ;nc

Pn = 1/n! . ( /)n p0 ;1n


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n=0 pn=1

c-1 n=1 pn + n=c pn =1

c-1

-> n=1 1/n! .( /)n p0 + n=c 1/cn-c(c!) . ( /)n p0 =1

Let /=r and /c= or r=c

c-1 P0 =[n=0 1/n! .rn + n=c 1/cn-c(c!) .rn ]-1

c-1

P0 =[n=0 1/n! .(c )n + n=c 1/cn-c(c!) .(c )n]-1


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c-1P0 = [n=0 (c )n/n! + (c )c/c! .(1/1- )]-1

Hence


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=p 0 rc/c!(1-r/c)

Probability that an arrival has to stand in a queue whenwe have c services

=p0 rc/c!(1-r/c) ,r= /

2.Expected no.of units in the queuelet Nq be a random variable.

We knowLq= E(Nq)

= n=c(n-c) pn

= n=c (n-c) ( /)n/c!cn-c p0


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Pn= {( /)n/n! p01n


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Lq = rc+1/(c-1)! .(c-r)2 .p0 {r=( /), =( /c)}

3.Probability that the service is immediatelyavailable on arrival

Let Tq be the random variable denoting waitingtime of an arriving unit. Then above performancemeasure is equivalent to:=prob [Tq=0)

Prob[there are less than (c-1)units in the system}

c-1

=n=0 pn =1 - n=c pn=1- n=c n cn-c.c! 0


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END


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OPERATIONS RESEARC

PROJECT

By:KartikGupta6135

mmcinfinity

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