CSE3213 Computer Network I

Transcription

1 SE33 omputer Network I Service Model, Error ontrol, Flow ontrol, and Link Sharing (h and 5.7.) ourse page: Slides modiied om lberto Leon-Garcia and Indra Widjaja

2 Peer-to to-peer Protocols and Service Models

3 Peer-to to-peer Protocols n + peer process SDU n peer process n peer process PDU n + peer process SDU n peer process n peer process Peer-to-Peer processes execute layer-n protocol to provide service to layer-(n+) Layer-(n+) peer calls layer-n and passes Service Data Units (SDUs) or transer Layer-n peers exchange Protocol Data Units (PDUs) to eect transer Layer-n delivers SDUs to destination layer-(n+) peer 3

4 Service Models The service model speciies the inormation transer service layer-n provides to layer-(n+) The most important distinction is whether the service is: onnection-oriented onnectionless Service model possible eatures: rbitrary message size or structure Sequencing and Reliability Timing, Pacing, and Flow control Multiplexing Privacy, integrity, and authentication 4

5 onnection-oriented Transer Service onnection Establishment onnection must be established between layer-(n+) peers Layer-n protocol must: Set initial parameters, e.g. sequence numbers; and llocate resources, e.g. buers Message transer phase Exchange o SDUs Disconnect phase Example: TP, PPP n + peer process send n + peer process receive SDU Layer n connection-oriented service SDU 5

6 onnectionless Transer Service No onnection setup, simply send SDU Each message send independently Must provide all address inormation per message Simple & quick Example: UDP, IP n + peer process send n + peer process receive SDU Layer n connectionless service 6

7 Message Size and Structure What message size and structure will a service model accept? Dierent services impose restrictions on size & structure o data it will transer Single bit? Block o bytes? Byte stream? Ex: Transer o voice mail = long message Ex: Transer o voice call = byte stream (a) voice mail= message = entire sequence o speech samples call = sequence o -byte messages (b) 7

8 Segmentation & Blocking To accommodate arbitrary message size, a layer may have to deal with messages that are too long or too short or its protocol Segmentation & Reassembly: a layer breaks long messages into smaller blocks and reassembles these at the destination Blocking & Unblocking: a layer combines small messages into bigger blocks prior to transer long message or more short messages or more blocks block 8

9 Reliability & Sequencing Reliability: re messages or inormation stream delivered error-ee and without loss or duplication? Sequencing: re messages or inormation stream delivered in order? RQ protocols combine error detection, retransmission, and sequence numbering to provide reliability & sequencing Examples: TP and HDL 9

10 Pacing and Flow ontrol Messages can be lost i receiving system does not have suicient buering to store arriving messages I destination layer-(n+) does not retrieve its inormation ast enough, destination layern buers may overlow Pacing & Flow ontrol provide backpressure mechanisms that control transer according to availability o buers at the destination Examples: TP and HDL 0

11 Timing pplications involving voice and video generate units o inormation that are related temporally Destination application must reconstruct temporal relation in voice/video units Network transer introduces delay & jitter Timing Recovery protocols use timestamps & sequence numbering to control the delay & jitter in delivered inormation Examples: RTP & associated protocols in Voice over IP

12 Multiplexing Multiplexing enables multiple layer-(n+) users to share a layer-n service multiplexing tag is required to identiy speciic users at the destination Examples: UDP, IP

13 Privacy, Integrity, & uthentication Privacy: ensuring that inormation transerred cannot be read by others Integrity: ensuring that inormation is not altered during transer uthentication: veriying that sender and/or receiver are who they claim to be Security protocols provide these services and are discussed in hapter Examples: IPSec, SSL 3

14 End-to to-end vs. Hop-by by-hop service eature can be provided by implementing a protocol end-to-end across the network across a single hop in the network Example: Perorm error control at every hop in the network or only between the source and destination? Perorm low control between every hop in the network or only between source & destination? We next consider the tradeos between the two approaches 4

15 Error control in Data Link Layer (a) Packets Data link layer Physical layer Frames Packets Data link layer Physical layer B Data Link operates over wire-like, directly-connected systems Frames can be corrupted or lost, but arrive in order (b) 3 Medium B 3 Data link perorms error-checking & retransmission Ensures error-ee packet transer between two systems Physical layer entity Data link layer entity 3 Network layer entity 5

16 Error ontrol in Transport Layer Transport layer protocol (e.g. TP) sends segments across network and perorms end-to-end error checking & retransmission Underlying network is assumed to be unreliable End system Messages Transport layer Network layer Data link layer Physical layer Network layer Data link layer Physical layer Segments Network layer Data link layer Physical layer Messages Transport layer Network layer Data link layer Physical layer End system B Network 6

17 Segments can experience long delays, can be lost, or arrive out-o-order because packets can ollow dierent paths across network End-to-end error control protocol more diicult 3 End System α End System β Medium B Network 3 Network layer entity 4 Transport layer 7entity

18 End-to to-end pproach Preerred Hop-by-hop Data Data Data Data / N / N / N / N Hop-by-hop cannot ensure EE correctness Faster recovery End-to-end /N Simple inside the network Data Data Data Data More scalable i complexity at the edge 8

19 RQ Protocols and Reliable Data Transer 9

20 utomatic Repeat Request (RQ) Purpose: to ensure a sequence o inormation packets is delivered in order and without errors or duplications despite transmission errors & losses We will look at: Stop-and-Wait RQ Go-Back N RQ Selective Repeat RQ Basic elements o RQ: Error-detecting code with high error coverage s (positive acknowledgments Ns (negative acknowlegments) Timeout mechanism 0

21 Stop-and and-wait RQ Transmit a ame, wait or Packet Timer set ater each ame transmission Transmitter (Process ) Inormation ame ontrol ame Receiver (Process B) Error-ee packet Header Inormation packet R Header R Inormation ame ontrol ame: s

22 Need or Sequence Numbers (a) Frame lost B Frame 0 Time-out Frame Frame Frame Time (b) lost B Frame 0 Time-out Frame Frame Frame Time In cases (a) & (b) the transmitting station acts the same way But in case (b) the receiving station B accepts ame twice Question: How is the receiver to know the second ame is also ame? nswer: dd ame sequence number in header S last is sequence number o most recent transmitted ame

23 (c) Premature Time-out Sequence Numbers B Time-out Frame 0 Frame 0 Frame Frame Time The transmitting station misinterprets duplicate s Incorrectly assumes second acknowledges Frame Question: How is the receiver to know second is or ame 0? nswer: dd ame sequence number in header R next is sequence number o next ame expected by the receiver Implicitly acknowledges receipt o all prior ames 3

24 -Bit Sequence Numbering Suices S last Timer R next Transmitter S last Receiver B R next Global State: (S last, R next ) (0,0) Error-ee ame 0 (0,) arrives at receiver or ame arrives at transmitter Error-ee ame (,0) arrives at receiver (,) or ame 0 arrives at transmitter 4

25 Stop-and and-wait RQ Transmitter Ready state wait request om higher layer or packet transer When request arrives, transmit ame with updated S last and R Go to Wait State Wait state Wait or or timer to expire; block requests om higher layer I timeout expires retransmit ame and reset timer I received: I sequence number is incorrect or i errors detected: ignore I sequence number is correct (R next = S last +): accept ame, go to Ready state Receiver lways in Ready State Wait or arrival o new ame When ame arrives, check or errors I no errors detected and sequence number is correct (S last =R next ), then accept ame, update R next, send ame with R next, deliver packet to higher layer I no errors detected and wrong sequence number discard ame send ame with R next I errors detected discard ame 5

26 First ame bit enters channel Stop-and and-wait Eiciency Last ame bit enters channel hannel idle while transmitter waits or arrives t B t First ame bit arrives at receiver Last ame bit arrives at receiver Receiver processes ame and prepares 0000 bit Mbps takes 0 ms to transmit I wait or = ms, then eiciency = 0/= 9% I wait or = 0 ms, then eiciency =0/30 = 33% 6

27 Stop-and and-wait Model t 0 = total time to transmit ame t proc B t prop ame t time t proc t ack t prop t 0 = = t t prop prop + + t t proc proc + + t n R + t + ack n R a bits/ino ame bits/ ame channel transmission rate 7

28 S&W Eiciency on Error-ee channel Eective transmission rate: R 0 e bits or header & R number o inormation bits delivered to destination n = = total time required to deliver the inormation bits t 0 n o, Transmission eiciency: η 0 = R e R = n n t0 R o = + n n a + n n ( t o prop + t n proc ) R. Eect o ame overhead Eect o ame Eect o Delay-Bandwidth Product 8

29 Example: Impact o Delay-Bandwidth Product n =50 bytes = 0000 bits, n a =n o =5 bytes = 00 bits xdelayxbw Eiciency ms 00 km 0 ms 000 km 00 ms 0000 km sec km Mbps % % 0 5 9% 0 6 % Gbps 0 6 % % % % Stop-and-Wait does not work well or very high speeds or long propagation delays 9

30 S&W Eiciency in hannel with Errors Let P = probability ame arrives w/o errors vg. # o transmissions to irst correct arrival is then / ( P ) I -in-0 get through without error, then avg. 0 tries to success vg. Total Time per ame is then t 0 /( P ) η SW = R e R = n t 0 n o P R = + n n a + n n ( t o prop + t n proc ) R ( P ) Eect o ame loss 30

31 Example: Impact Bit Error Rate n =50 bytes = 0000 bits, n a =n o =5 bytes = 00 bits Find eiciency or random bit errors with p=0, 0-6, 0-5, 0-4 P = ( p) n e n p or large n and small p P Eiciency Mbps & ms 88% 86.6% 79.% 3.% Bit errors impact perormance as n p approach 3

32 Go-Back Back-N Improve Stop-and-Wait by not waiting! eep channel busy by continuing to send ames llow a window o up to W s outstanding ames Use m-bit sequence numbering I or oldest ame arrives beore window is exhausted, we can continue transmitting I window is exhausted, pull back and retransmit all outstanding ames lternative: Use timeout 3

33 Go-Back Back-N N RQ Go-Back-4: 4 ames are outstanding; so go back Time B 3 out o sequence ames R next Frame transmission are pipelined to keep the channel busy Frame with errors and subsequent out-o-sequence ames are ignored Transmitter is orced to go back when window o 4 is exhausted 33

34 Window size long enough to cover round trip time Stop-and-Wait RQ 0 Time-out expires 0 Time B Receiver is looking or R next =0 Go-Back-N RQ 0 Four ames are outstanding; so go back Time B Receiver is looking or R next =0 Out-osequence ames

35 Go-Back Back-N N with Timeout Problem with Go-Back-N as presented: I ame is lost and source does not have ame to send, then window will not be exhausted and recovery will not commence Use a timeout with each ame When timeout expires, resend all outstanding ames 35

36 Go-Back Back-N N Transmitter & Receiver Frames transmitted and ed Timer Timer Timer Transmitter Send Window... S last S recent S last +W s - Buers S last S last +... S recent... S last +W s - oldest un- ed ame most recent transmission max Seq # allowed Receiver Frames received Receive Window R next Receiver will only accept a ame that is error-ee and that has sequence number R next When such ame arrives R next is incremented by one, so the receive window slides orward by one 36

37 Sliding Window Operation Transmitter Send Window... m-bit Sequence Numbering Frames transmitted and ed S last S recent S last +W s - m 0 Transmitter waits or error-ee ame with sequence number S last When such ame arrives, S last is incremented by one, and the send window slides orward by one S last send window i + W s i + i 37

38 Maximum llowable Window Size is W s = m - M = = 4, Go-Back - 4: Transmitter goes back Time B R next Receiver has R next = 0, but it does not know whether its or ame 0 was received, so it does not know whether this is the old ame 0 or a new ame 0 M = = 4, Go-Back-3: 0 Transmitter goes back 3 0 Time B R next Receiver has R next = 3, so it rejects the old ame 0 38

39 Piggybacking in Bidirectional GBN Transmitter Receiver S B recent RB next S recent R next Receiver Transmitter Receive Window B Receive Window R next Send Window... R B next B Send Window... S last S last +W s - Timer Timer Timer Timer Buers S last S last +... S recent... S last +W s - Note: Out-osequence error-ee ames discarded ater R next examined S B last S B last +WB s - Timer Buers S B last Timer S B last +... Timer S B recent... Timer S B last +WB s - 39

40 Required Timeout & Window Size T out T proc T prop T T T prop Timeout value should allow or: Two propagation times + processing time: T prop + T proc ame that begins transmission right beore our ame arrives T Next ame carries the, T W s should be large enough to keep channel busy or T out 40

41 Eiciency o Go-Back Back-N GBN is completely eicient, i W s large enough to keep channel busy, and i channel is error-ee ssume P ame loss probability, then time to deliver a ame is: t i irst ame transmission succeeds ( P ) T + W s t /(-P ) i the irst transmission does not succeed P t η GBN GBN = t ( P n no tgbn = R ) + P { t s Wst + P no n = + ( W ) P } = t ( P + ) P W t s P and Delay-bandwidth product determines W s 4

42 Example: Impact Bit Error Rate on GBN n =50 bytes = 0000 bits, n a =n o =5 bytes = 00 bits ompare S&W with GBN eiciency or random bit errors with p = 0, 0-6, 0-5, 0-4 and R = Mbps & 00 ms Mbps x 00 ms = bits = 0 ames Use W s = Eiciency S&W 8.9% 8.8% 8.0% 3.3% GBN 98% 88.% 45.4% 4.9% Go-Back-N signiicant improvement over Stop-and-Wait or large delay-bandwidth product Go-Back-N becomes ineicient as error rate increases 4

43 Selective Repeat RQ Go-Back-N RQ ineicient because multiple ames are resent when errors or losses occur Selective Repeat retransmits only an individual ame Timeout causes individual corresponding ame to be resent N causes retransmission o oldest un-acked ame Receiver maintains a receive window o sequence numbers that can be accepted Error-ee, but out-o-sequence ames with sequence numbers within the receive window are buered rrival o ame with R next causes window to slide orward by or more 43

44 44 B 0 Time N Selective Repeat RQ Selective Repeat RQ

45 Selective Repeat RQ Transmitter Send Window... Receiver Receive Window Frames transmitted and ed S last S recent S last + W s - Frames received R next R next + W r - Timer Buers S last Buers R next + Timer S last + R next + Timer... S recent R next + W r - max Seq # accepted S last + W s - 45

46 Send & Receive Windows Transmitter Receiver m - 0 m - 0 S last send window i + W s i + Moves k orward when arrives with R next = S last + k k =,, W s - i j + W r R next receive window Moves orward by or more when ame arrives with Seq. # = R next i j 46

47 What size W s and W r allowed? Example: M= =4, W s =3, W r =3 Send Window {0,,} {,} {} {.} 0 0 Frame 0 resent Time Receive Window B 3 {0,,} {,,3} {,3,0} {3,0,} Old ame 0 accepted as a new ame because it alls in the receive window 47

48 W s + W r = m is maximum allowed Example: M= =4, W s =, W r = Send Window {0,} {} {.} 0 Frame 0 resent 0 Time Receive Window B {0,} {,} {,3} Old ame 0 rejected because it alls outside the receive window 48

49 Why W s + W r = m works Transmitter sends ames 0 to Ws-; send window empty ll arrive at receiver ll s lost Transmitter resends ame 0 Receiver window starts at {0,, W r } Window slides orward to {W s,,w s +W r -} Receiver rejects ame 0 because it is outside receive window m - 0 m - 0 S last W s +W r - send window W s - receive window R next W s 49

50 pplications o Selective Repeat RQ TP (Transmission ontrol Protocol): transport layer protocol uses variation o selective repeat to provide reliable stream service Service Speciic onnection Oriented Protocol: error control or signaling messages in TM networks 50

51 Eiciency o Selective Repeat ssume P ame loss probability, then number o transmissions required to deliver a ame is: t / (-P ) η SR = t n R n o /( P ) = ( n n o )( P ) 5

52 Example: Impact Bit Error Rate on Selective Repeat n =50 bytes = 0000 bits, n a =n o =5 bytes = 00 bits ompare S&W, GBN & SR eiciency or random bit errors with p=0, 0-6, 0-5, 0-4 and R= Mbps & 00 ms Eiciency S&W 8.9% 8.8% 8.0% 3.3% GBN 98% 88.% 45.4% 4.9% SR 98% 97% 89% 36% Selective Repeat outperorms GBN and S&W, but eiciency drops as error rate increases 5

53 omparison o RQ Eiciencies ssume n a and n o are negligible relative to n, and L = (t prop +t proc )R/n =(W s -), then Selective-Repeat: no ηsr = ( P )( ) ( P n Go-Back-N: P P ηgbn = = + ( W ) P + LP Stop-and-Wait: ( P ηsw = n ( t a + + n S prop ) + t n proc ) R ) P + L For P 0, SR & GBN same For P, GBN & SW same 53

54 RQ Eiciencies Eiciency RQ Eiciency om parison LOG(p) p Delay-Bandwidth product = 0, 00 Selective Repeat Go Back N 0 Stop and Wait 00 Go Back N 00 Stop and Wait 0 54

55 Flow ontrol 55

56 Flow ontrol buer ill Inormation ame Transmitter Receiver ontrol ame Receiver has limited buering to store arriving ames Several situations cause buer overlow Mismatch between sending rate & rate at which user can retrieve data Surges in ame arrivals Flow control prevents buer overlow by regulating rate at which source is allowed to send inormation 56

57 X ON / X OFF threshold Inormation ame Transmitter Receiver Transmit X OFF Transmit Time on o o on B Time T prop Threshold must activate OFF signal while T prop R bits still remain in buer 57

58 t cycle Window Flow ontrol Return o permits Time B Sliding Window RQ method with W s equal to buer available Transmitter can never send more than W s ames s that slide window orward can be viewed as permits to transmit more an also pace s as shown above Return permits (s) at end o cycle regulates transmission rate Problems using sliding window or both error & low control hoice o window size Interplay between transmission rate & retransmissions TP separates error & low control Time 58

59 Link Sharing Using Statistical Multiplexing 59

60 Statistical Multiplexing Multiplexing concentrates bursty traic onto a shared line Greater eiciency and lower cost Header Data payload B Buer Output line Input lines 60

61 Tradeo Delay or Eiciency (a) Dedicated lines B B (b) Shared lines B B Dedicated lines involve not waiting or other users, but lines are used ineiciently when user traic is bursty Shared lines concentrate packets into shared line; packets buered (delayed) when line is not immediately available 6

62 Multiplexers inherent in Packet Switches N N Packets/ames orwarded to buer prior to transmission om switch Multiplexing occurs in these buers 6

63 Multiplexer Modeling B Input lines Buer Output line rrivals: What is the packet interarrival pattern? Service Time: How long are the packets? Service Discipline: What is order o transmission? Buer Discipline: I buer is ull, which packet is dropped? Perormance Measures: Delay Distribution; Packet Loss Probability; Line Utilization 63

64 Delay = Waiting + Service Times Packet completes transmission Packet begins transmission P P P3 P4 P5 Service time Packet arrives at queue P P P3 P4 Waiting time P5 Packets arrive and wait or service Waiting Time: om arrival instant to beginning o service Service Time: time to transmit packet Delay: total time in system = waiting time + service time 64

65 Fluctuations in Packets in the System (a) Dedicated lines B B (b) Shared line B B (c) N(t) Number o packets in the system 65

66 Packet Lengths & Service Times R bits per second transmission rate L = # bits in a packet X = L/R = time to transmit ( service ) a packet Packet lengths are usually variable Distribution o lengths Dist. o service times ommon models: onstant packet length (all the same) Exponential distribution Internet Measured Distributions airly constant See next chart 66

67 Measure Internet Packet Distribution Dominated by TP traic (85%) ~40% packets are minimum-sized 40 byte packets or TP s ~5% packets are maximum-sized Ethernet 500 ames ~5% packets are 55 & 576 byte packets or TP implementations that do not use path MTU discovery Mean=43 bytes Stand Dev=509 bytes Source: caida.org 67