Chapter 3
Delay models in Data Networks
Section 3.2
Little`s Theorem
3.2 Little`s Theorem



N : average number of customers in
system
: mean arrival rate
T:mean time a customer spends in
system

T
N
Little`s Theorem

Proof



N(t) = number of customers in system at
time t
(t) = number of customers who arrived in
interval [0,t]
Ti = time spent in system by the i-th
customer
Little`s Theorem
Little`s Theorem
 (t )
 (t )
T

 (t )
1
1
N
(

)
d


T


i

t 0
t i 1
t
take
t
lim  N  T
t 
i 1
i
 (t )
3.2.3 Application of Little`s
Theorem

Ex3.1



 : arrival rate in a transmission line
NQ : average number of packets waiting in
queue
W : average waiting time spent by a packet
in queue
NQ = W
Application of Little`s
Theorem


If X = average Tx time
  = X


 : Average number of packets under Tx
I.e. fraction of time that s busy utilization factor
Application of Little`s
Theorem

Ex3.2


N : average number packets in network
T : average delay per packet
N 
n
 T
i 1
i
also
 Ti : average delay of packets arriving at
node i
N i  iTi
3.3 M/M/1 Queuing System

M/M/1



First M : arrival , Poisson
Second M : service , Exponential
1 : server number
M/M/1 Queuing System

Arrival Poisson process



A(t) : number of arrivals from 0 to time t
Number of arrivals that occur in disjoint
intervals are independent
Number of arrivals in any interval of length
 is Poisson distributed with parameter  ,
( )e
PA(t   )  A(t )  n 
n!
 
, n  0,1,
M/M/1 Queuing System

Properties of Poisson process
1. Inter arrival times are independent and
exponentially distributed with parameter

tn : time of the n-th arrival
n  tn 1  tn
 Pn  S   1  e S , S  0
pdf : P(n )  e
  n
1
1

2
 mean , var iance
M/M/1 Queuing System
2. For every t0, 0
PA(t   )  A(t )  0  1    o( )
PA(t   )  A(t )  1    o( )
PA(t   )  A(t )  2  o( )
where lim
 0
Note : e 
o( )

0
(  ) 2
 1   

2!
M/M/1 Queuing System
3. A = A1+A2++AK is also Poisson with
rate  = 1+ 2++ K
A1
merge
………..
A2
Poisson
AK
M/M/1 Queuing System
4.

Poisson split
P
1-P
Also Poisson with
Poisson with
P
(1-P)
M/M/1 Queuing System

Service time : Exponential distribution
with parameter 

Sn : service time of n-th customer
PSn  S   1  e
 S
pdf : P( Sn )  e
,S  0
 Sn
M/M/1 Queuing System

Properties of Exponential : memoryless
Pn  r  t | n  t  P(n  r ), for r, t  0
PSn  r  t | Sn  t  P( Sn  r ), for r, t  0
Markov chain formulation

Let's focus at the times,0,,2,…,k,…

Nk = number of customers in system at
time k = N(k)



Where N(t) is continuous-time Markov Chain
Nk is discrete-time
Let Pij : transition probabilities =
P{Nk+1=j|Nk=i}
Markov chain formulation
P00  1    o( )
Pii  1      o( ), i  1
Pi ,i 1    o( ), i  0
Pi ,i 1    o( ), i  0
Pi , j  o( ), i and j  i, i  1, i  1
Markov chain formulation
steady  state probabilit ies
Pn  lim PN k  n  lim PN (t )  n

Note


k 
t 
During any time interval, the total number
of transitions from state n to n+1 must
differ from the total number of transitions
from n+1 to n by at most 1
I.e. frequency of transitions from n+1 to n
= frequency of transitions from n to n+1
Markov chain formulation
Markov chain formulation
Take ->0  Pn=Pn+1
Pn+1=Pn, n=0, 1, …等比數列
where = /  utilization
 Pn+1=  n+1P0, n=0,1,…
Since <1, and

P
n 0
n
 1  P0  1  
 Pn   n (1   ) n  0,1,...(3.23)
Markov chain formulation


n 0
n 0
N  lim E N (t )   nPn   n n (1  )
t 
1


  (1   )


(3.24)
2
(1   )
1    
N

1
T


(3.25)
  (1   )   
1
1


W 
 
, N Q  W 
    
1 
2
M/G/1 System
Let Ci ： customer I
Wi = waiting time of Ci
Xi = service time of Ci
Ni = # of customers found waiting in
queue when Ci arrives
Ri = residual service time of the
customer in service when Ci
arrives
M/G/1 System
Ri
Xi- Ni
Ci arrives
Ni
Wi  Ri 
Xi-1
Ci start service
i 1
X
j i  N i
j
In steady-state,
1
lim W  R  N Q 
 R  W
i 
u
E (Wi )  E ( Ri )  E ( N i ) E ( X j )
R
W 
1 
( 3.48)
M/G/1 System
To calculate R, by graphical approach:
Residual service time r()
M(t)=# of service completion in [0, t]
X1
X2
X1 X2
Ci starts service
XM(t)
Time 
XM(t)
t
M/G/1 System
Time avg of r() in [0, t]
1
1
  r ( )d 
t 0
t
t
M (t )
X
2
i
1 M ( t ) i 1

2
t
M (t )
M (t )

i 1
1 2
Xi
2
M/G/1 System
Tak e
t  
1
R  lim
t  t

t
0
r ( )d
M (t )

1
M (t )

lim
lim i 1
t 
2 t 
t
M (t )
1

E ( X 2 )
2
P-K
2
R
E ( X ) Formula
W 

1 
2(1   ) (3.53)
X i2
Ex3.15
Consider a go back n ARQ:
1
sender
receiver


2
3 … n-1 n
1
2
3
Prop. delay
1
time
Timeout (n-1) frames
time
Assume that error in the forward channel is p,
return channel is error-free
Packet arrives as a Poisson process with rate
 packets/frame
Ex3.15
Service time X : from when a packet
transmitted until it is
successfully received
1 , if 1st tx is successful (1-p)
X={ 1+n, if 1st tx is un- successful; 2nd is
successful  p(1-p)
1+kn, if 1st k is un- successful;(k+1)th
successful  Pk(1-p)
Ex3.15

 E ( X )   (1  kn) p (1  p )
k
k 0

k
k
 (1  p )  p  n  kp 
k 0
 k 0




E[ X ]   (1  kn) p (1  p )
2
2
k 0
k
Ex3.15
np
 E( X )  1 
1 p
2np n ( p  p )
E[ X ]  1 

2
1 p
(1  p )
2
2
2
E ( X )
W
, T  E( X )  W
2(1  E ( X ))
2
3.5.1 M/G/1 Queue with
vacations
1.
2.
3.
When the server has served all customers, it
goes on vacation
If the system is still idle after a vacation
interval, go on another vacation interval
If a customer arrives during a vacation,
customer waits until the end of vacation.
Chapter 1 section 1.3.1 page 34in Network
or Transport Layer
M/G/1 Queue with vacations
M/G/1 Queue with vacations

Assume vacation intervals v1, v2… are
iid and are independent of customers
arrival & service times.
→A customer must wait for the completion
of the current service or vacation interval,
and then the service of all customers
waiting before it.
M/G/1 Queue with vacations
R
W
1 

Where R is the mean residual time for
completion of service or vacation when
the customer arrives.
M/G/1 Queue with vacations
Let L(t) = # of vacations completed by t
M(t) = # of services completed by t
1 t
1 M (t ) 1
1 L(t ) 1 2
2
r ( )d 
X i   Vi


t 0
t i 1 2
t i 1 2
L(t )
1 M (t ) 1
1 L( t )
2

X


i
t i 1 2
2 t
Vi
2
i 1
L( t )
 E (V 2 )
R
E ( X )
2
2
2
1
E (V )
 (1   )
2
E (V )
Because Fraction of time occupied with vacation = 1-
t (1   )
L( t ) 1  
lim
 E (V ),

t 
L( t )
t
E (V )
R
E ( X ) E (V )
W 


1   2(1   ) 2 E (V )
2
2
Ex3.16 : FDM, SFDM, TDM


m streams of traffic with rate
/m(Poisson)
FDM system – Divide available
bandwidth into m subchannels.
Transmission time for a packet on each
of these subchannels is m.
FDM
  E ( X ) 

m
m  
E ( X )  VarX   E ( X )  0  m  m
2
2

WFDM
2
m
m


(3.56)
2(1   ) 2(1   )
2
Slotted FDM System

Packet trans starts only at time m,
2m,…When the queue is idle, server
takes a vacation of m. (if idle again
after vacation, take another)
WSFDM
m
m

 WFDM  (3.57)
2(1   )
2
TDM System


Look at SFDM queue, ->same queue
WTDM=WSFDM
Summary
TTDM
m
m

 1  TFDM  (  1), better if m  2
2(1   )
2
m
TSFDM 
m
2(1   )
m
TFDM 
m
2(1   )
Service time
Reservations & Polling
Satellite
Collision -> solution:polling or reservation
S1 D1 D1 S2 D2 …
Cycle
S1 D1 D1 S2 D2
Reservation & Polling

1.
2.
3.
M Poisson traffic streams with rate /m
Gated System – only those packets which
arrive prior to the user’s preceding
reservation period are transmitted.
Exhaustive system – all packets are
transmitted including those that arrive
during this data period
Partially gated – all packets that arrive up
to the beginning of the data interval.
Single-User
Gated system:
m=1
Di arrives
S
D
D
S
Wi
D D … D
Di starts
time
Di ends tx
Ri
Vl(I)
l(i)-th reservation interval
Ni : # of packets arrive in front of Di
Single-User


A reservation(vacation)starts when the
system has served all packets which arrive
prior to the preceding reservation interval.
A vacation(M/G/1 queue with vacation) starts
when the system has served all packets
which have arrived.(corresponds to
exhaustive system)
Single-User
 E ( Ri ) 與以前一樣
take i  
E ( X )
2
2
E (V )
R
 (1   )
2
2 E (V )
N  E ( N i )  W
E ( N i )  E ( X )  WE ( X )  W
Single-User
Single-user
gated
E ( X ) E (V ) E (V )
W


(3.61)
2(1   ) 2 E (V ) 1  
2
2
E ( X ) E (V )
W

exhaustive
2(1   ) 2 E (V )
2
2
Multi-User
Packet i starts
Packet i arrives
S
D
D
Wi
D … S D
D … S
Ri
D
D
time
Pakcet i ends
Ni
Sum=Yi
Ni is redefined as # of packets which
must be transmitted before packet i
Multi-user
E (Wi )  E ( Ri )  E ( N i ) E ( X )  E (Yi ) (3.63)
Where Yi : includes all reservation
intervals packet I must want for.
Multi-User

If number the users 0, 1, 2,…,m-1, the
l-th reservation interval is used to make
reservation for user l mod m
m 1
R
E ( X )
2
2

(1   ) E (Vl )
l 0
m 1
2 E (Vl )
l 0
2
Multi-user
W  R  W  Y
R Y
W
(3.65), Y  ?
1 
Multi-user
For an exhaustive system
Let lj=E ( Yj | packet i arrives in user l’s
reservation or data intervals and
belongs to user (l+j) mod m)
0
,
j

0




V ( l 1) mod m  V ( l 2 ) mod m  ...  V ( l  j ) mod m , j  0

Packet i belongs to each user with same prob. = 1/m
Multi-user

m 1

lj
j 1
1

m
m 1

j 1
lj
V ( l 1) mod m

1 V ( l 1) mod m  V ( l  2 ) mod m


m 
V ( l 1) mod m  V ( l  2 ) mod m  ...  V ( l  m 1) mod m

m j

V ( l  j ) mod m
m
j 1
m 1
( 3.66)
Multi-user
All users have equal average data length
in steady state.
P(packet i arrives during user l’s data

interval)

m
P(packet i arrives during user l’s
V
reservation interval)  (1   ) m1 l
V
k 0
k
Multi-user
m 1
E
j 1
(Yi|pkt i arrives in user l’s data or reser. int.)
X P(pkt i arrives in user l’s reser. Or data int. )
m j

(1   )V l
 
V( l  j ) mod m (
 m 1
)
m
m
l  0 j 1
V k
m 1 m 1
k 0
m 1
( m   )V


2
1
where V 
m
(1   ) Vl 2
l 0
2mV
m 1
V
l 0
l
(3.68)
Multi-user
If Vl’s have same dist.
X2
( m   )V  V2
W


2(1   ) 2(1   ) 2V
m 1
where  
2
V
2 V )
(
V
l

2
l
l 0
m
Exhaustive system
(3.69)
Multi-user
- The partially gated system is the same
as the exhaustive system except that if
a packet arrives in its own user’s data
interval (with prob. /m), it is delayed
an extra cycle of reservation periods(mV)
Y is increased by

m
 mV  (1 

m
)  0  V
Multi-user
-
The fully gated system is the same as
partially gated system except if a pkt
arrives during a user’s own reservation
interval (prob. (1-)/m)
-
It is delayed by an additional mV
-
1 
Y is increased by (
)mV
m
Priority Queuing

N classes of customers class i arrives a
Poisson process with rate I
2
1
service time X i  , X i
ui
Each class joins a separate queue
1
Server
2
Priority Queuing
Single server will server customers from the
highest priority queue first

1.
2.
Non-preemptive
- a lower priority customer, once started, is
allowed to finish, when a high priority customer
arrives.
Preemptive resume
- Service for a low priority customer is
interrupted when a high priority customer
arrives and is resumed from the point of
interruption when all higher priority customers
have been served
Non-preemptive
Let NQk=avg. # in queue for priority k
Wk= avg. queueing time for priority k
k = k/k = system utilization for
priority k
R = mean residual service time.
Non-preemptive
W1  R  N
1
Q
1
1
 R  1W1
R
 W1 
1  1
W2  R  N
1
Q
1
1
N
2
Q
1
2

1
1
1W2
Where 1W2 is the avg. # of higher priority customers
that arrives while you are waiting
Non-preemptive
R
 W2 
(1  1 )(1  1  2 )
Similarly,
R
Wk 
(1  1  2  ...   k 1 )(1  1   2  ...   k )
Non-preemptive
R=the residual time
n
1
2
 (  i ) X
2 i 1
Where
X
2
=2nd moment of the service
time avg. over all priority
1 2
n 2
 X 1  ... 
Xn


Non-preemptive
n
代入
1
2
 R  (  i ) X (3.81)
2 i 1
n
Wk 
 X
i 1
i
2
i
2(1  1   2  ...   k 1 )(1  1   2  ...   k )
1
and Tk  Wk 
(3.83)
k
(3.82)
Preemptive
Note that Tk will not be affected by customers
from class k+1 to n
1
Tk  
k
Unfinished work
of Class 1 to k
(A)
Rk
( A) 
1  1  ...   k

Work due to class 1
to k-1 who arrives when
this customer is waiting
(B)
R
(Re call W 
)
1 
k
where Rk 
2

X
 i i
i 1
2
(3.84)
Preemptive
k
( B)  
i 1
1
i
k
iTk   iTk
i 1
k 1
Rk
 Tk 

  iTk
k 1  1  ...   k i 1
1
1
(1  1  ...   k 1 )  Rk
k
 Tk  
(3.87)
(1  1  ...   k 1 )(1  1  ...   k )
1
for k  1, T1  
1
(1  1 )  R1
(1  1 )
(3.86)