CS4/MSc Computer Networking. Lectures 4-5 Transport layer protocols TCP/UDP automatic repeat request

Transcription

1 S4/MSc omputer Networking Lectures 4-5 Transport layer protocols TP/UDP automatic repeat request omputer Networking, opyright University o Edinburgh 005

2 Transport services and protocols provide logical communication between app processes running on dierent hosts Multiplexing/demultiplexing transport protocols run in end systems send side: breaks app messages into segments, passes to network layer rcv side: reassembles segments into messages, passes to app layer network layer provides logical communication between hosts application transport network data link physical network data link physical logical end-end transport network data link physical network data link physical network data link physical network data link physical application transport network data link physical

3 Basic services o transport protocols Depends on what is oered by the network layer Error control build a reliable channel between two peers Flow control allow the receiver to regulate the low o data Synchronization/time recovery or media streams (audio/video) they must be played back at the correct rate. Security: Privacy ensure no-one can read the inormation Integrity no-one can change what is transmitted uthentication veriy the identity o sender/receiver Implementation eiciency: network/link utilisation Idle time Header overhead 3

4 End-to-end or hop-by-hop? service eature can be provided by implementing a protocol end-to-end across the network across every hop in the network End-to-end advantages Simplicity, scalability, maintainability, interoperability, cost» Only two nodes are involved» Hop-by-hop only works i the service is provided at every hop an provide a service over channels that do not/cannot provide the same service: e.g. encryption Hop-by-hop advantages Speed cannot provide a ast service over a slow channel Reliability when using noisy links» Transmission o a large ile over a number o links with high bit error rate 4

5 Internet transport layer protocols: UDP Stands or User Datagram Protocol Unreliable, connectionless transport layer protocol Services beyond IP : demultiplexing and (optional) error checking Source Port Destination Port UDP Length UDP hecksum Data Port numbers added in the header» Identiy the particular application in the given host Optional checksum on the whole datagram» 0 checksum value means no checksum wanted» a checksum calculated to be 0 has to be included as all s» IP checksum algorithm uses s complement arithmetic; so all s = 0 5

6 UDP (cont) The datagram padded out to a multiple o 6 bits or checksum calculation» but the padding not transmitted pseudo-header also included in the checksum calculation» and not transmitted with the datagram Source IP ddress Destination IP ddress Protocol = 7 UDP Length Permits a check that the datagram has reached the correct destination orrupted datagrams are discarded» no error message returned to source 6

7 Error control protocols What is needed: Error detection special data encoding, R, etc Timeouts timers, interrupts, reverse channel to receive acknowledgements, positive or negative utomatic Repeat Request (RQ) amily o protocols Stop and wait Go back N Selective repeat ssumption (or now): channels are like wires, PDUs cannot arrive out-o-order Types o PDUs: Inormation transer user data ontrol acknowledgements, etc. 7

8 Stop and wait Basic idea: Send ame Wait or acknowledgement o receipt I ack does not arrive by some reasonable time, re-transmit ame What can go wrong? Frame is lost or has errors cknowledgement is lost/garbled 8

9 Inormation ames must be numbered B Frame 0 Time-out Frame Frame Frame Time B Frame 0 Time-out Frame Frame Frame Time Neither transmitter nor receiver have a global view o the situation The transmitter cannot distinguish between a lost ame or lost ack it just retransmits the ame The receiver does not know i a ame is new or a retransmission Solution: the inormation ames are numbered 9

10 ck ames must be numbered too B Time-out Frame 0 Frame 0 Frame Frame Time Premature timeouts may result in multiple acks or the same inormation ame The transmitter may misinterpret a duplicate ack as an ack or a subsequent ame I that ame is lost, the sender will never know about it Solution: cks are numbered too 0

11 Stop and wait (with sequencing numbers) Transmitter keeps track o the sequence number S last o the ame being sent plus the ame itsel, in case it needs to be retransmitted Receiver keeps track only o the sequence number o the next ame it is expecting to receive, R next Sequence number cannot get too large Limited, ixed space in the header -bit sequence number adequate in this case ombination o S last and R next orms the state o the transmission link S last will be 0 or ; R next will be 0 or thereore our states : (0,0), (0,), (,0), (,) depending on which ame has been transmitted and which s received

12 Stop and wait: state machine 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 One veriied ame transmission corresponds to hal a cycle in the FSM

13 Link utilisation The link is an important resource, its utilisation must be maximised First ame bit enters channel 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 3

14 Stop and wait: ame transmission time t 0 = total time to transmit ame t proc B t prop ame t time t proc t prop t ack 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 4

15 Stop and wait: eiciency 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: 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 5

16 Stop and wait: Impact o transmission errors in eiciency Previous calculations assume error-ee transmission In presence o errors the eective bit rate drops P probability o an error in a ame (data or ack) transmission verage number o retransmissions: /(-P ) Total time to transer ame becomes t = t 0 P = (t prop + t proc Transmission eiciency multiplied (drops) by (-P ) + n R + n R a ) P Link error rate is usually quoted as probability o error or single bit 6

17 Go-back-N The source o Stop-and-Wait ineiciency is that it does not ill the channel with data Improve Stop-and-Wait by not waiting or an ack beore sending the next packet i.e. pipeline the ame transmission Transmitter has a limited number o ames (called a window) that can be outstanding without acknowledgment : W s W s is chosen to allow the channel to be ully utilised Frames are numbered cks are also numbered 7

18 Example: Go-back-4 Go-Back-4: 4 ames are outstanding; so go back Time B 3 out o sequence ames R next When there are ewer than Ws- subsequent packets to send retransmissions are not triggered, since the window is not exhausted Need to associate a timer with every packet 8

19 Go-back-N with timeouts Transmitter Send Window... Receiver Receive Window Frames transmitted and ed Timer Timer S last S recent S last +W s - Timer Buers S last S last +... S recent... S last +W s - oldest un- ed ame most recent transmission max Seq # allowed When #X is received, assume all ames up to X are received, even i previous s did not reach the xmitter Frames received 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 Receiver sends an with seq. number R next or error-ee but out o sequence ames 9

20 Go-back-N: window size I there are m bits available in the header or sequencing, which is the largest possible value or W S? receiver must be able to determine unambiguously which ame has been received taking into account the wrapping around when count reaches 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 onclusion: W S <= m - 0

21 Go-back-N with negative acknowledgments Go-Back-7: Transmitter goes back to ame time B N Out-o-sequence ames N with sequence number R next acknowledges all ames up to R next - and hints the transmitter that an error has been detected in ame R next Go-back-N with N, results in having the transmitter go back less than N ames, so speeds up error recovery

22 Go-back-N with bidirectional inormation low Xmitter and receiver unctions o the protocol implemented at both ends cknowledgement ames can be piggybacked into headers o I-ames Signiicant improvement in use o bandwidth: separate acknowledge ames can oten be avoided I no ames are yet ready to be transmitted to piggyback into, receiver can set an timer» that deines the maximum time it will wait or a suitable I-ame» i it expires, a separate control ame can be sent with the acknowledgment receiver handles out-o sequence packets slightly dierently: corrupted ame is ignored Out o sequence, error-ee ames are discarded, but the sequence number is extracted and processed

23 Go-back-N perormance 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 Propagation delay is not part o t GEN because go-back-n does not wait or the acknowledgement beore sending the next ame When errors occur W S ames are transmitted Best possible eiciency is Just header overhead n n o 3

24 Selective-Repeat Go-back-N perorms badly in noisy channels Retransmits more than the minimum required Selective-repeat improvements: Reception window is made larger so it accepts (error-ee) out-o-order ames Only individual ames are retransmitted More storage is required at the receiver omplexity is higher 4

25 Selective-Repeat: Operation example Time B Ns are not essential; speed-up the retransmission o a speciic ame I N not used (or lost) a timeout will eventually cause retransmission The basic objective remains to advance the values o R next and S last by delivery o the oldest outstanding ame N

26 Selective-Repeat: Window sizes Example: M= =4, Ws=3, Wr=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 6

27 Selective-Repeat: max window sizes Worst case scenario: Try to ool receiver into believing that a retransmission o ame 0 is a new ame, numbered 0 because the sequence # wrapped around ll ames (0 W S -) are received correctly ll s (and Ns) are lost Receive window is then W S W S +W R - Frame W S +W R will not be accepted until the transmitter gets ame 0 acknowledged (by an # in the range to W S ) Thus, max window sizes: W S +W R = m m - 0 m - 0 send window S last W s - W s +W r - R next W s receive window 7

28 Selective repeat perormance t SR = t P η SR = t n R n o /( P ) = ( n n o )( P ) Propagation delay depends only on probability o error in transmission ssuming channel is always ull, i.e. send window > bandwidth delay product Best possible eiciency is Same as go-back-n n n o 8

29 RQ protocols, comparison o eiciency ssume header, ack overheads are negligible # ames in transit: L = (t prop +t proc )R/n =(W s -) η SR Eiciency ( P ) η GBN P + LP Bit error rate (log 0 ) η SW P + L Selective Repeat Go Back N 0 Stop and Wait 0 Go Back N 00 Stop and Wait 00 9

30 Flow control protocols The receiver has limited buer space to store ames I the transmitter sends data at a higher rate, the buer can overlow and some o the transerred data will be dropped mechanism is needed so that the receiver can tell the transmitter to slow down (or stop) or speed-up again Implementation requires: Detection o potential buer overlow reverse channel or low control messages to transmitter Two types o ames: Inormation transer user data Flow-control ames 30

31 X ON \ X OFF protocol threshold Inormation ame Transmitter Receiver Transmit X OFF Transmit Time on o o on B Time T prop Receiver must activate OFF signal while T prop R bits still remain in buer 3

32 Sliding window low control The sliding window protocols can be used or low control; set W s equal to receiver buer size Transmitter can never send more than W s ames s slide window orward = permit transmission o new ames s are called credits in this case Flow control can be combined with error control in a sliding window RQ protocol Window size depends on bandwidth-delay product and size o receiver s buer lternative: extend ame header with an extra ield or control-low credits; decouples credits om acks 3

33 ombining selective repeat with low control Send Window S last + W a R last R last + W R ames transmitted & ed S last S last + W s S recent R next R new Receive Window S last oldest unacknowledged ame S recent highest-numbered transmitted ame S last +W a - highest-numbered ame that can be transmitted S last +W s - highest-numbered ame that can be accepted om the application R last highest-numbered ame not yet read by the application R next next expected ame R new highest numbered ame received correctly R last +W R - highest-numbered ame that can be accommodated in receive buer dvertised window: W a = W R -(R new - R last ) the number o currently available buers, included in the ame header 33

34 End-to-end protocols Problems (only when lower layer provides datagram service) Packets arrive out o order Old packets (even om previous connections) reappear» The correct packet could be wrongly rejected as a duplicate Solution Packets have a maximum lietime areully selected initial sequence number» Implies agreement, implies connection establishment procedure Sequence number space is large, but legal windows are (relatively) small Wait or a reasonable time beore re-establishing a connection End-to-end delays are subject to more variation How to select a reasonable timeout? 34

35 Sequence numbers ssume T is (a multiple o) maximum packet (and ack) lietime Must ensure two packets (or the same pair o hosts and sockets) with same sequence number are never outstanding or a time dierence o T Thus i they do, they must be duplicates o the same packet retransmitted Each computer has a timer (counter) Not synchronised. Synchronisation is very hard/expensive! Sequence numbers created om the timer Sequence space large, so or a single connection, wrap around time is much larger than T The two peers agree on an initial sequence number 35

36 Sequence numbers - Problems rashes cause problem: which was the last sequence number? Expensive solution: wait or T beore doing anything lternatively, impose urther restrictions on sequence numbers The current timer (= potential initial seq number) determines a set o sequence numbers (in existing connections) that should not be used. Sequence number Forbidden region T Forbidden region T Time Rate o actual sequence numbers cannot be steeper than the timer Fast timers needed Eventually will enter the orbidden region on the right 36

37 End-to-end retransmission timeout Timeout depends on round trip time (RTT): time om when segment is sent to when is received Round trip time (RTT) across a network (esp. Internet) is highly variable Routes vary and can change in mid-connection Traic luctuates Timeout too short: excessive number o retransmissions Timeout too long: recovery too slow daptive estimation o RTT used in TP Measure RTT each time received: τ n α = 7/8 typical t RTT (new) = α t RTT (old) + ( α) τ n 37

38 Timeout using RTT variability RTT (milliseconds) time (seconnds) SampleRTT Estimated RTT Estimate variance σ o RTT variation Estimate or timeout: t out = t RTT + k σ RTT I RTT highly variable, timeout increase accordingly I RTT nearly constant, timeout close to RTT estimate 38

39 TP: Overview RFs: 793,, 33, 08, 58 point-to-point: one sender, one receiver reliable, in-order byte steam: no message boundaries pipelined: TP congestion and low control set window size send & receive buers Transmitter pplication byte stream segments pplication byte stream Receiver ull duplex data: bi-directional data low in same connection MSS: maximum segment size connection-oriented: handshaking (exchange o control msgs) init s sender, receiver state beore data exchange low controlled: sender will not overwhelm receiver Send s buer Receive buer 39

40 TP segment structure URG: urgent data (generally not used) : # valid PSH: push data now (generally not used) RST, SYN, FIN: connection estab (setup, teardown commands) Internet checksum (as in UDP) 3 bits source port # dest port # head len sequence number acknowledgement number not used U P checksum R S F Receive window application data (variable length) Urg data pnter Options (variable length) counting by bytes o data (not segments!) # bytes rcvr willing to accept 40

41 onnection Establishment Three-way handshake Host Host B SYN, Seq_no = x SYN, Seq_no = y,, ck_no = x+ Seq_no = x+,, ck_no = y+ sends connection request to B» SYN=; initial sequence number = x B acknowledges connection request» =; SYN=; initial sequence number = y; next data byte expected=x+» s SYN consumes the irst sequence number acknowledges the request om B» =; sequence number = x+; next data byte expected=y+» on receipt at B, connection is established 4

42 onnection Termination There is no protocol that can do this 00% saely! The two army problem: the side that sends the last message cannot know i it has been received at the other side The inal message (ack) is allowed to be lost 4

43 TP onnection Termination Each end o a TP connection terminates independently Host receives the or the last data, it sends a segment with FIN Host B receives the FIN segment, inorms its application that the other entity has terminated its connection and s the FIN Host B can still transmit in the other direction until it has inished Host B eventually also sends a FIN segment Host receives FIN, replies with and goes into a wait state Starts a TIME-WIT timer set to x maximum segment lietime» st MSL accounts or time a segment in one direction can remain in the network» nd MSL accounts or the transit time o the reply The only valid segment that can arrive in this interval is a retransmission o the FIN segment (e.g. i the was lost)» i a FIN retransmission is detected, the is retransmitted and the timer restarted Wait state means old session segments are gone beore termination 43