Chapter 5. Elementary Performance Analysis

Transcription

1 Chapter 5 Elementary Performance Analysis 1

2 5.0 2

3 5.1 Ref: Mischa Schwartz Telecommunication Networks Addison-Wesley publishing company

4 4

5 p t T m T P(k)= 5

6 6

7 5.2 : arrived rate : service rate 7

8 8

9 9

10 surface 1 surface 2 P P n n P P n1 n1 P n1 n 1 10

11 11

12 12

13 The normalized throughput (input rate/output rate) (the probability that the system is nonempty) = ρ The throughput (customer/see) = λ (absolute) (nonsatulated ρ= λ) (satulated ρ= μ) 13

14 5.3 net arrival rate= = r throughput sec) ( customer / see) (1 PB ) 1 1 P ) 1 1 (1 ( 1 The throughput r/μ= 0 N 1 N (1 ) N ) 14

15 15

16 5.4 16

17 5.5 i i m n0 p n 1 17

18 18 (1 ) 1 ( 1) 1 = Q Q Q Q Q N W P P P m m m 1 1 Q Q P T W m

19 19

20 normalized delay normalized load 20

21 Example 1 Statistical Multiplexing Compared with TDM and FDM Assume m statistically iid Poisson packet streams each with an arrival rate of packets/sec. m The packet lengths for all streams are independent and exponentially distributed. The average transmission time is 1. If the streams are merged into a single Poisson stream, with rate, the average delay per packet is 1 T If, the transmission capacity is divided into m equal portions, as in TDM and FDM, each portion behaves like an M/M/1 queue with arrival rate m and average service rate m. Therefore, the average delay per packet is m T 21

22 Example 2 Using One vs. Using Multiple Channels Statistical MUX(1) A communication link serving m independent Poisson traffic streams with overall rate. Packet transmission times on each channel are exponentially distributed with mean 1. The system can be modeled by the same Markov chain as the M/M/m queue. The average delay per packet is given by 1 PQ T m An M/M/1 system with the same arrival rate and service rate m (statistical multiplexing with one channel having m times larger capacity), the average delay per packet is Pˆ Tˆ 1 Q m m PQ and pˆ Q denote the queueing probability 22

23 When << 1 (light load) P 0, pˆ Q 0, and Q At light load, statistical MUX with m channels produces a delay almost m times larger than the delay of statistical MUX with the m channels combined in one. When, P 1, pˆ Q 1, 1 << 1 ( m ), and 1 Q T Tˆ At heavy load, the ratio of the two delays is close to 1. m T Tˆ 1 23

24 5.6 f 1 1 ( x) 1 x 2 x 2 doesn' t exist 24

25 25

26 26

27 27

28 W R W W R W 28

29 A. 29

30 second moment of service time and load IF X 2 W 30

31 2 31

32 5.7 32

33 an ack 33

34 sec 34

35 bits η i η η 35

36 t out 36

37 37

38 t f t f 38

39 39

40 40

41 41

42 m 42

43 hytes 43

44 5.8 (special case) 44

45 45

46 46

47 5.9 Roll-call Polling Stations are interrogated sequentially, one by one, by the central system, which asks if they have any messages to transmit. time W i t i : walk time : frame transmission time 47

48 The scan or cycle time t c is given by The average scan time t c w i t, are the ave. walk time and the ave. time to transmit pkt at station. i L is the total walk time of the complete poling system. i 48

49 For station i, let i : the ave. pkt arrival rate : the ave. packet length : the number of overhead bits C : the channel capacity in bps : the ave. frame length in time m i m i ( ) / C The average number of packets waiting to be transmitted when station polled is, the time required to transmit is t i c m With the traffic intensity, the average scan time is given i i i t i i tc mi t i c t c i is With channel. N i1 i N i1 m i i representing the total traffic intensity on the common 49

50 For small the average access delay should be / 2. t c Assume that each station has the same w and the same., same frame-length statistics, The average access delay is 2 is the second moment of the frame length, Nm m. The access delay is the average time a packet must wait at a station from the time it first arrives until the time transmission begins. Access delay is thus the average wait time in an M/G/1 queue. Ref: Mischa Schwarty: Telecommunication Networks, Protocols Modeling and Analysis, Addison-Wesley Publishing Company, 1988, PP

51 5.10 Hub Polling Control is passed sequentially from one station to another. Let the polling message be a fixed value, t p sec in length. The time required per station to synchronize to a polling message is t s sec. The total propagation delay for the entire N-station system is sec. 51

52 Hub Polling (Cont ) The total walk time for hub-call polling is L Nt p Nt. s Let the stations all be equally spaced, and the round-trip propagation delay between the controller and station N be sec. The overall propagation delay is just (1 N) 2 The analysis of the hub-polling strategy is identical to that of roll-call polling. The only difference is that the walk time L is reduced through the use of hub polling. For hub polling,. L hub Nt s 52

53 5.11 Throughput Analysis of Nonpersistant CSMA Throughput Analysis of Slotted Nonpersistant CSMA 5.13 Stability of Slotted ALOHA 53

54 5.11 Performance Analysis of Nonpersistent CSMA Definitions Total load=g Vulnerable period=a Packet transmission time=1 Busy period=b Idle period=i Successful utilization period=u The arrival time of the last packet collides with the first one =Y a=propagation delay packet transmission time One cycle = B+I The successful transmission period= 1+a The collision period = 1+Y+a 54

55 55

56 The throughput = s = Eu E B E I Note that if no station transmits duration the first a time unit of the busy period, the transmission is successful. E u ag e The idle period is just the time interval between the end of a busy period and the next arrival. Because the packet arrivals follow a Poisson distribution P k k Ge k! G 56

57 The average duration of an idle period E[I] = 1/G B= 1+a+Y Where y=0 for successful transmissions E[B]=1+a+E[Y] The probability density foundation of Y is the probability that no packet arrival occurs in an interval of length a-y. G( ay) f ( y) Ge for 0 y a = 0 0 a a E Y yf y dy yge G( ay) a Ga Gy Ge ye dy 0 1 ag a 1 e G 1 E B a e G ag Ge s ag G(1 2 a) e ag dy 57

58 58

59 5.12 Throughput Analysis of Slotted Nonpersistant CSMA Slot time=τ All stations are synchronized and required to start transmission only at the beginning of a slot. All packets are assumed to be of a length tp, which is an integral number of time slots. When a packet arrives during a time slot, the station senses the state of the channel at the beginning of the next slot then either transmits if the channel is idle or defers to a later time slot if traffic is present. 59

60 α β γ δ Example: Packets station 1,2,n arrive during the first slot If the next slot is empty, three stations transmit. If packets arrive in the busy period (α,β,γ,δ ) the stations sense the channel to be busy. These stations defer their packets for transmission at a later time. 60

61 The busy period=tp+τ The normalized busy period-(tp+τ )/tp=(1+a) where a= τ /tp Define the average time during a cycle that a transmission is successful E[u] = P s =P[ one packet arrives is slot a some arrival occurs] P[one packet arrives in slot a and some arrival occurs] = P[some arrival occurs] P[one packet arrives in slot a ] = P[some arrival occurs] 61

62 Using Poisson arrival statistics, we have P[one packet arrives in slot a]=age and The busy period is always 1+a E[B]=1+a An idle period always consists of an integral number of time slots I 0 P[some arrival occurs]=1-e age Eu 1-e -ag -ag -ag -ag 62

63 If a packet arrives during the last slot of a busy period, then the next slot immediately starts a new busy period, so that I=0 When there are no arrivals during the last slot of a busy period then the next I-1 slots will be empty until there is an arrival in the final I th slot. This then marks the beginning of a new busy period. Consider the case I=0 The probability p of this occurring is merely the probability of some packet arriving in the interval a PI 0 p 1 e ag Next we look at the case I=1 P[I=1] = the joint probability that no arrival occurs in the last slot of the busy period and that some arrival occurs in the next time slot = (1-p)p 63

64 i P I i (1 p) p i E I a i(1 p) p s i0 a(1 p) p ae = 1 e ag ag Eu E I E B ag age ag 1 e ag ae 1a ag 1 e ag age ag (1 a)(1 e ) ae ag age ag 1 ae a ag 64

65 5.13 Stability of Slotted ALOHA Assumptions a. There are N identical stations, each of which can hold at most one packet. b. If a new packet arrives at a station with a full buffer, the newly arriving packet is lost. (backlogged) c. The station generates new packets according to a Poisson distribution with an average rate p packets/slot. d. A previously collided packet is retransmitted with probabilityαin every succeeding time slot until a successful transmission occurs. 65

66 e. Let k be the number of stations which are backlogged (Each of this stations decides to resend its backlogged packet with probabilityαor to skip the present slot with probability 1-α f. Let r represent the number of retransmission attempts by the k stations during the same slot. g. Let n be the number of new packets generated by the N-k unblocked stations. The probability P k of a successful transmission is k P P r P n P r P n k 1 N k k N k 1 k P N k P P S out, k 66

67 Where P[n=0] is the probability that no new packets are generated, P[n=1] is the probability that a new packet is generated. S out,p is the rate at which packets leave the system. The rate at which packets enter the system is s=(n-k)p For a large N and small p out, k In equilibrium k 1 s 1 1 k s S k e se s 2 e 1 s s 1 s 1 N k p 2! s= S out,p 67

68 68

69 69

70 70

71 71

72 72

73 73