9 Pipelined Protocols and RTT Slides adapted from

9: Pipelined Protocols and RTT Slides adapted from: J. F Kurose and K. W. Ross, 1996 -2010 Transport Layer 3 -1

Stop and Wait v Rdt 3. 0 also called Stop and Wait § Sender sends one packet, then waits for receiver response v What is wrong with stop and wait? § Slow!! Must wait full round trip time between each send v Obvious Fix? § Instead send lots, then stop and wait § Call this a pipelined protocol because many packets in the pipeline at the same time 3: Transport Layer 3 a-2

Pipelined protocols Pipelining: sender allows multiple, “in-flight” yet-tobe-acknowledged packets § range of sequence numbers must be increased to be able to distinguish them all § Additional buffering at sender and/or receiver § Once allow multiple “in-flight” consider that channel may reorder the packets 3: Transport Layer 3 a-3

How bad is Stop and Wait? v v Depends on network conditions example: 1 Gbps link, 15 ms end-to-end prop. delay, 1 KB packet: Ttransmit = Utilization = U = m m r 1 kb = 1 microsec 10**9 b/sec 1 microsec Utilization of the = = 0. 003% channel 30. 001 msec 1 KB pkt every 30 msec -> 33 k. B/sec throughput over 1 Gbps link network protocol limits use of physical resources! In general, smaller packets, longer RTT and higher maximum bandwidth, all make the situation worse 3: Transport Layer 3 a-4

rdt 3. 0: stop-and-wait operation sender receiver first packet bit transmitted, t = 0 last packet bit transmitted, t = L / R RTT first packet bit arrives last packet bit arrives, send ACK arrives, send next packet, t = RTT + L / R Transport Layer 3 -5

Pipelining: increased utilization sender receiver first packet bit transmitted, t = 0 last bit transmitted, t = L / R RTT ACK arrives, send next packet, t = RTT + L / R first packet bit arrives last packet bit arrives, send ACK last bit of 2 nd packet arrives, send ACK last bit of 3 rd packet arrives, send ACK Increase utilization by a factor of 3 with window size of 3 Transport Layer 3 -6

Filling the pipeline How much in-flight data is needed to “fill the pipeline”? v Similar to question of how much water needed to fill a pipe (area of crosssection * length of pipe) v v For networks, it is bandwidth*delay 3: Transport Layer 3 a-7

Pipelined Protocols Go-back-N: big picture: v sender can have up to N unacked packets in pipeline v rcvr only sends cumulative acks § doesn’t ack packet if there’s a gap v sender has timer for oldest unacked packet § if timer expires, retransmit all unack’ed packets Selective Repeat: big pic v sender can have up to N unack’ed packets in pipeline v rcvr sends individual ack for each packet v sender maintains timer for each unacked packet § when timer expires, retransmit only unack’ed packet Transport Layer 3 -8

Go-Back-N Sender: v v v k-bit seq # in pkt header “window” of up to N, consecutive unack’ed pkts allowed ACK(n): ACKs all pkts up to, including seq # n - “cumulative ACK” § may receive duplicate ACKs (see receiver) timer for window of packets timeout(n): retransmit pkt n and all higher seq # pkts in window Transport Layer 3 -9

GBN: sender extended FSM rdt_send(data) L base=1 nextseqnum=1 if (nextseqnum < base+N) { sndpkt[nextseqnum] = make_pkt(nextseqnum, data, chksum) udt_send(sndpkt[nextseqnum]) if (base == nextseqnum) //one timer per window start_timer nextseqnum++ } else refuse_data(data) //or just block the caller Wait rdt_rcv(rcvpkt) && corrupt(rcvpkt) /*don’t do anything if receiver feedback corrupt */ timeout start_timer //resend all packets in window udt_send(sndpkt[base]) udt_send(sndpkt[base+1]) … udt_send(sndpkt[nextseqnum-1]) rdt_rcv(rcvpkt) && notcorrupt(rcvpkt) base = getacknum(rcvpkt)+1 //cummulative ack If (base == nextseqnum) stop_timer //have acked all outstanding else //send any data this allows us room in window to send Transport Layer 3 -10 start_timer //start timer for remaining base to nextseqnum

GBN: receiver extended FSM default udt_send(sndpkt) L Wait expectedseqnum=1 sndpkt = make_pkt(expectedseqnum, ACK, chksum) rdt_rcv(rcvpkt) && notcurrupt(rcvpkt) && hasseqnum(rcvpkt, expectedseqnum) extract(rcvpkt, data) deliver_data(data) sndpkt = make_pkt(expectedseqnum, ACK, chksum) udt_send(sndpkt) expectedseqnum++ ACK-only: always send ACK for correctly-received pkt with highest in-order seq # § may generate duplicate ACKs § need only remember expectedseqnum v out-of-order pkt: § discard (don’t buffer) ->receiver buffering is not required but can help if want to! § Re-ACK pkt with highest in-order seq # Transport Layer 3 -11

GBN in action Loss of one packet (pkt 2) causes retransmissi on of 4 packets (2 -5) Transport Layer 3 -12

Selective Repeat GBN forces sender to retransmit all packets in window even if some have been correctly received v To avoid that we need a finer granularity of acknowledgement v § individual acknowledgements vs cumulative acknowledgements 3: Transport Layer 3 a-13

Selective Repeat v receiver individually acknowledges all correctly received pkts § buffers pkts, as needed, for eventual in-order delivery to upper layer v sender only resends pkts for which ACK not received § sender timer for each un. ACKed pkt v sender window § N consecutive seq #’s § again limits seq #s of sent, un. ACK’ed pkts Transport Layer 3 -14

Selective repeat: sender, receiver windows Transport Layer 3 -15

Selective repeat sender data from above : v if next available seq # in window, send pkt timeout(n): v receiver pkt n in [rcvbase, rcvbase+N-1] v v v resend pkt n, restart timer ACK(n) in [sendbase, sendbase+N]: v v mark pkt n as received if n smallest un. ACKed pkt, advance window base to next un. ACKed seq # send ACK(n) out-of-order: buffer in-order: deliver (also deliver buffered, in-order pkts), advance window to next not-yet-received pkt n in v [rcvbase-N, rcvbase-1] ACK(n) otherwise: v ignore Transport Layer 3 -16

Selective repeat in action Loss of one pkt causes retransmission of just that pkt Transport Layer 3 -17

Selective Repeat vs GBN Selective Repeat requires individual acknowledgements rather than chance for cumulative acknowledgements v GBN results in unnecessary retransmission of data correctly received v Sender can choose to buffer out of order and avoid unnecessary retransmission (but not required) – in selective repeat receiver can acknowledge these out of order packets v 3: Transport Layer 3 a-18

Pipelined protocols Sequence Number Dilemma Example: v v v seq #’s: 0, 1, 2, 3 window size=3 receiver sees no difference in two scenarios! incorrectly passes duplicate data as new in (a) Similar to the problem we saw in stop and wait until we added seq 0 and 1 3: Transport Layer 3 a-19

Sequence Number Space v v Q: what relationship between seq # size and window size? A: sequence number space >= 2 * window size § True for Stop and Wait (2 >= 2*1) § need old and new version of every sequence # v v Still one problem, packets could conceivably delayed for arbitrarily long in the network so could get an old packet N even after the sequence number space has wrapped around Solution? Not really. In practice, assume a maximum time a packet could live in the network 3: Transport Layer 3 a-20

TCP? v TCP is most like GBN § But many TCP implementations will buffer correctly received but of order segments and senders use duplicate acknowledgments to infer which segment dropped. . This is sort of like Selective Repeat v v TCP uses cumulative acknowledgements but counts bytes not packets and receiver ACKS what it wants not last thing it received Window size is not fixed like N in GBN § TCP allows receiver to set a maximum (dynamically) § Effective window size also changed over time in response to signs of congestion in the network v We will discuss TCP specifics next… 3: Transport Layer 3 a-21

Roadmap v Discussed general principles of reliable message delivery over unreliable channel § Lots of it is common sense (like with our flaky fax machine) § But there is a significant degree of subtlety in getting it right! v We are going to move on to talking specifically about TCP § Flow control? Congestion control? v We have most of the tools we need now: receiver feedback, checksums, sequence numbers, cumulative acknowledgments and retransmission timers 3: Transport Layer 3 a-22

Question v Which of the following is true § For a window size of 1, you need at least 2 possible sequence numbers. § For a window size of 1, you need at least 2 bits to store the sequence number. § The number of possible sequence numbers must be >= 2* the window size. § The window size must be >= 2* number of possible sequence numbers. Transport Layer 3 -23