1. Delay models in Data Networks
Chapter 3
Delay models in Data Networks

2. Section 3.2
Little`s Theorem

3. 3.2 Little`s Theorem : average number of customers in system
: mean arrival rate
T:mean time a customer spends in system

4. 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

5. Little`s Theorem

6. Little`s Theorem

7. 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

8. Application of Little`s Theorem
If = average Tx time
  = 
 : Average number of packets under Tx
I.e. fraction of time that s busy utilization factor

9. Application of Little`s Theorem
Ex3.2
N : average number packets in network
T : average delay per packet
also
Ti : average delay of packets arriving at node i

10. 3.3 M/M/1 Queuing System M/M/1 First M : arrival , Poisson
Second M : service , Exponential
1 : server number

11. 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  ,

12. M/M/1 Queuing System Properties of Poisson process
Inter arrival times are independent and exponentially distributed with parameter 
tn : time of the n-th arrival

13. M/M/1 Queuing System
For every t0, 0

14. M/M/1 Queuing System A = A1+A2++AK is also Poisson with
rate  = 1+ 2++ K
Poisson A1
merge A2
……….. AK

15. M/M/1 Queuing System P  1-P Also Poisson with P split Poisson

16. M/M/1 Queuing System
Service time : Exponential distribution with parameter 
Sn : service time of n-th customer

17. M/M/1 Queuing System
Properties of Exponential : memoryless

19. 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}

20. Markov chain formulation

21. Markov chain formulation
Note
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

22. Markov chain formulation

23. 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

24. Markov chain formulation

25. 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

26. M/G/1 System Ri Xi- Ni Xi-1 Ci start service Ni Ci arrives
In steady-state,

27. M/G/1 System To calculate R, by graphical approach:
Residual service time r()
M(t)=# of service completion in [0, t] X1 X2
XM(t)
Time  X2 X1
XM(t)
Ci starts service t

28. M/G/1 System
Time avg of r() in [0, t]

29. M/G/1 System
P-K
Formula
(3.53)

30. Ex3.15 Consider a go back n ARQ:
sender 1 2 3 …
n-1 n 1
time
Timeout (n-1) frames 1 2 3
receiver
time
Prop. delay
Assume that error in the forward channel is p, return channel is error-free
Packet arrives as a Poisson process with rate  packets/frame

31. 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)

32. Ex3.15

33. Ex3.15

34. 3.5.1 M/G/1 Queue with vacations
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 page 34in Network or Transport Layer

35. M/G/1 Queue with vacations

36. 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.

37. M/G/1 Queue with vacations
Where R is the mean residual time for completion of service or vacation when the customer arrives.

38. M/G/1 Queue with vacations
Let L(t) = # of vacations completed by t
M(t) = # of services completed by t

40. Because Fraction of time occupied with vacation = 1-

41. 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.

42. FDM

43. 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)

44. TDM System
Look at SFDM queue, ->same queue
WTDM=WSFDM

45. Summary
Service time

46. Reservations & Polling
Satellite
Collision -> solution:polling or reservation … S1 D1 D1 S2 D2 S1 D1 D1 S2 D2
Cycle

47. Reservation & Polling 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.

48. Single-User Gated system: m=1 Di arrives Di starts Wi time S D D S D D… D
Di ends tx Ri
Vl(I)
l(i)-th reservation interval
Ni : # of packets arrive in front of Di

49. 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)

50. Single-User

52. Single-User
Single-user gated

53. Multi-User
Packet i arrives
Packet i starts Wi
time S D D D … S D D … S D D Ri
Pakcet i ends Ni
Sum=Yi
Ni is redefined as # of packets which must be transmitted before packet i

54. Multi-user
Where Yi : includes all reservation intervals packet I must want for.

55. 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

56. Multi-user

57. Packet i belongs to each user with same prob. = 1/m
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)
Packet i belongs to each user with same prob. = 1/m

58. Multi-user

59. Multi-user All users have equal average data length in steady state.
P(packet i arrives during user l’s data interval)
P(packet i arrives during user l’s reservation interval)

60. Multi-user (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. )

61. Multi-user
If Vl’s have same dist.
Exhaustive system
(3.69)

62. 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

63. 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
Y is increased by

64. Priority Queuing
N classes of customers class i arrives a Poisson process with rate I
service time
Each class joins a separate queue 1
Server 2

65. Priority Queuing
Single server will server customers from the highest priority queue first
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

66. 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.

67. Non-preemptive Where 1W2 is the avg. # of higher priority customers
that arrives while you are waiting

68. Non-preemptive
Similarly,

69. Non-preemptive R=the residual time Where =2nd moment of the service
time avg. over all priority

70. Non-preemptive 代入

71. Preemptive Note that Tk will not be affected by customers
from class k+1 to n
Work due to class 1
to k-1 who arrives when
this customer is waiting
(B)
Unfinished work
of Class 1 to k
(A)

72. Preemptive