Communication Networks I
Lecture No. 11, October 8, 1998
Lecture summary by Jia Wei
Main points: This lecture is dealing with M/G/1 queue, a single-server queueing system
where customers arrive according to a Poisson process with rate but the customer
service times have a general distribution. The objective is to derive and understand the
Pollaczek-Khinchin (P-K) formula.
1. Queue discipline: In a queueing system with a single-server, suppose that customers
arrive according to a Poisson process with rate  , and the customers are served in the
order they arrive. We denote Xi as the service time of the ith arrival and assume that
the random variables (X1, X2, etc) are identically distributed, mutually independent,
and independent of the interarrival times.
2. Pollaczek-Kinchin (P-K) formula
Let
_
X = E[X] = Average service time
_
X2 = E[X2] = Second moment of service time
W : Average customer waiting time
_
X, the utilization factor.
We have P-K formula:
X
 W
)   1(2
2
In the M/M/1 case, service times are exponentially distributed , when x >0
f ( X )  e  X
E X  
1

2
VarX  


2
W 



2(1  ) 1 


2
1

2
 
E X 2  VarX   ( E X ) 2 
2
2
In the M/D/1 case, service times are deterministic,
f (X )   (X 
E X  
W'
1

Xi 
)
 
1
E X2 


2

2(1  )

1

1
2
1
 W
2

3. Proof of P-K formula
Denote
Wi = Waiting time for customer i
Ri = Residual service time seen by i
Ni = Number of customers waiting in queue when customer i arrives.
When i arrives, waiting customers are i-1, i-2, … i-Ni. Customer i-(Ni+1) is in
service
Then
i 1
X
Wi  Ri 
j i  N i
j
 X i 1  X i 2    X i  Ni  Ri 

Ni is independent of X i 1 , X i 2 , 

E Wi   E Ri   E X i 1  X i 2    X i  Ni

E E X i 1  X i 2    X i  N i | N i  n

 E Ni X

 E N i X
 NQ X
W  R  NQ X
According to Little’s theorem, N Q  W
W  R   XW


W
R
R

1 X 1 
We can calculate R by a graphical argument.
Residual service time r(t)
Denote
r (t )  residual service time at time t
M (t )  number of service completions by time t
X3
X1
X4
X2
X1
R(t ) 
X2
X3
t 0
t i 1 2
t
r  d   X i 
1 t
1 M ( t ) 1 2 M (t )
X4
t
M (t )
 2X i2
1
M (t )
i 1
When t -> 
1
X2
2
R   X  (1   X )  0   X 
2
2X
P[system is empty] = 1   X , service time is 0
P[systeim is busy] =  X , service time is
X2
2X
In P-K equation, all long term average quantities should be viewed as limits when
time or customer index increases to infinity. We assume the limits of W, R, NQ exist,
and this is true of almost all systems of interest to us provided that    .
P-K formula is valid for any order of servicing customers as long as the order is
determined independently of the required service time.
4. P-K formula for M/G/1 queue with vacations
The analysis of this system is the same as that of the P-K formula except that
vacations must be included in the graph of residual service times r(t).
Denote
V1,V2, …: the durations of the successive vacations taken by the server. They are
independent and identically distributed random variables. They are also independent
of the customer interarrival times and service times.
R’ : residual service time or possibly residual vacation time
We have
W  R '  N Q X  R '  W X
R'
W
1 X
Residual service time r(t)
R’ can be calculated by graph argument
X3
V1
X1
X2
X1
Denote
V2
X2
V1
X4
X5
V2
X3
X4
t
M(t) = number of completed services
L(t) = number of completed vacations
For any t where a service or vacation is just completed, we have
When t->  ,
M (t )

t
V1  V2    VL ( t )
t
 fraction of time on vacation
= 1-fraction of time busy = 1  
L(t ) V1  V2  VL( t )
L(t )
V
t
L(t )
t
L( t )
1 

t
V
R' 
W 
X 2
2

1  1 2
V
V 2
X2
V2

2(1   ) 2V