EEC484584 Computer Networks Lecture 16 Wenbing Zhao wenbingieee
EEC-484/584 Computer Networks Lecture 16 Wenbing Zhao wenbing@ieee. org (Part of the slides are based on Drs. Kurose & Ross’s slides for their Computer Networking book, and on materials supplied by Dr. Louise Moser at UCSB and Prentice-Hall) Spring Semester 2006 EEC-484/584: Computer Networks
2 Outline • TCP – – – Connection management Reliable data transfer Flow control TCP transmission policy Congestion control • Reminder: Quiz 4 – MW session: 11/27 Monday 2 -4 pm – TTh session: 11/28 Tuesday 4 -6 pm Spring Semester 2006 EEC-484/584: Computer Networks 2
3 TCP Connection Management TCP sender, receiver establish “connection” before exchanging data segments • Initialize TCP variables: – Sequence numbers – Buffers, flow control info (e. g. Rcv. Window) • Client: connection initiator Socket client. Socket = new Socket("hostname", "port number"); • Server: contacted by client Socket connection. Socket = welcome. Socket. accept(); Spring Semester 2006 EEC-484/584: Computer Networks 3
4 TCP Connection Management Three way handshake: Step 1: client host sends TCP SYN segment to server – specifies initial sequence number – no data Step 2: server host receives SYN, replies with SYN/ACK segment – server allocates buffers – specifies server initial sequence number Step 3: client receives SYN/ACK, replies with ACK segment, which may contain data Spring Semester 2006 EEC-484/584: Computer Networks 4
5 TCP Connection Management Three way handshake: client connect • SYN segment is considered as 1 byte • SYN/ACK segment is also considered as 1 byte SYN ( accept seq=x ) 1) CK ( YN/A K=x+ C A y, seq= S ACK Spring Semester 2006 server EEC-484/584: Computer Networks (seq= x+1, A CK=y +1) 5
6 TCP Connection Management client Closing a connection: close client closes socket: server FIN client. Socket. close(); Step 1: client end system sends ACK TCP FIN control segment to server replies with ACK. Closes connection, sends FIN timed wait Step 2: server receives FIN, close ACK closed Spring Semester 2006 EEC-484/584: Computer Networks 6
7 TCP Connection Management client Step 3: client receives FIN, replies with ACK. closing – Enters “timed wait” - will respond with ACK to received FINs server FIN ACK closing FIN Step 4: server, receives ACK. Note: with small modification, can handle simultaneous FINs timed wait Connection closed. ACK closed Spring Semester 2006 EEC-484/584: Computer Networks 7
8 TCP Reliable Data Transfer • TCP creates rdt service on top of IP’s unreliable service • Pipelined segments • Cumulative acks • TCP uses single retransmission timer Spring Semester 2006 • Retransmissions are triggered by: – timeout events – duplicate acks • Initially consider simplified TCP sender: – ignore duplicate acks – ignore flow control, congestion control EEC-484/584: Computer Networks 8
9 TCP Sender Events: Data rcvd from app: Timeout: • Create segment with • retransmit segment that sequence number caused timeout • seq # is byte-stream • restart timer number of first data byte Ack rcvd: in segment • If acknowledges • start timer if not already previously unacked running (think of timer segments as for oldest unacked – update what is known to segment) be acked • expiration interval: – start timer if there are outstanding segment Time. Out. Interval Spring Semester 2006 EEC-484/584: Computer Networks 9
10 TCP: Retransmission Scenarios Host A Host B tes da t X ACK a =100 loss Seq=9 2, 8 by tes da Sendbase = 100 Send. Base = 120 ta 0 K=10 AC Send. Base = 100 time lost ACK scenario Spring Semester 2006 Host B Seq=92 timeout 2, 8 by Send. Base = 120 2, 8 by tes da ta Seq= 100, 2 0 byte s data 0 10 = K 120 = C K A AC Seq=92 timeout Seq=9 Host A time EEC-484/584: Computer Networks 2, 8 by tes da ta 20 K=1 AC premature timeout 10
11 TCP Retransmission Scenarios Host A Host B Seq=9 timeout 2, 8 by t Send. Base = 120 es dat a =100 K C A 00, 20 bytes data Seq=1 X loss 120 = ACK time Cumulative ACK scenario Spring Semester 2006 EEC-484/584: Computer Networks 11
12 TCP ACK Generation Event at Receiver TCP Receiver action Arrival of in-order segment with expected seq #. All data up to expected seq # already ACKed Delayed ACK. Wait up to 500 ms for next segment. If no next segment, send ACK Arrival of in-order segment with expected seq #. One other segment has ACK pending Immediately send single cumulative ACK, ACKing both in-order segments Arrival of out-of-order segment higher-than-expect seq. #. Gap detected Immediately send duplicate ACK, indicating seq. # of next expected byte Arrival of segment that partially or completely fills gap Immediate send ACK, provided that segment starts at lower end of gap Spring Semester 2006 EEC-484/584: Computer Networks 12
13 TCP Flow Control • Receive side of TCP connection has a receive buffer: • App process may be slow at reading from buffer Spring Semester 2006 Flow control: sender won’t overflow receiver’s buffer by transmitting too much, too fast • Speed-matching service: matching the send rate to the receiving app’s drain rate EEC-484/584: Computer Networks 13
14 TCP Flow Control (Suppose TCP receiver discards out-of-order segments) Spare room in buffer = Rcv. Window = Rcv. Buffer-[Last. Byte. Rcvd - Last. Byte. Read] Spring Semester 2006 • Rcvr advertises spare room by including value of Rcv. Window in segments • Sender limits un. ACKed data to Rcv. Window – guarantees receive buffer doesn’t overflow EEC-484/584: Computer Networks 14
15 TCP Transmission Policy • Window management not directly tied to ACKs – The sender can send new segments only if the receiver has room to receive them • What if the receiver’s window drops to 0 ? – Sender may not normally send segments with two exceptions – Exception 1: urgent data may be sent, e. g. , to allow user to kill process running on the remote machine – Exception 2: sender may send a 1 -byte segment to make the receiver re-announce the next byte expected and window size Spring Semester 2006 EEC-484/584: Computer Networks 15
TCP Transmission Policy 16 • Window management not directly tied to ACKs Spring Semester 2006 EEC-484/584: Computer Networks 16
17 TCP Transmission Policy • Nagle’s algorithm – To address the 1 -byte-at-a-time sender problem • Clark’s algorithm – To address the 1 -byte-at-a-time receiver problem Spring Semester 2006 EEC-484/584: Computer Networks 17
18 1 -byte-at-a-time Sender Problem • Sender sends 1 byte (e. g. , typed one character in an editor) • A segment of 1 byte is sent to the remote machine (41 -byte IP packet) • Remote machine acks immediately (40 -byte IP packet) • Editor (in remote machine) program reads the received 1 byte, a windows update segment is sent to user (40 -byte IP packet) • Editor program echoes the 1 byte received to the user terminal (41 -byte IP packet) • In all, 162 bytes of bandwidth used, 4 segments are sent for each character typed Spring Semester 2006 EEC-484/584: Computer Networks 18
19 Nagle’s Algorithm • When sender application passes data to TCP one byte at a time – – Send first byte Buffer the rest until first byte ACKed Then send all buffered bytes in one TCP segment Start buffering again until all ACKed • Implemented widely in TCP, can be disabled/enabled by using socket options • For some application, it is necessary to disable the Nagle’s algorithm, e. g. , X Windows program over Internet, to avoid erratic mouse movement, etc. Spring Semester 2006 EEC-484/584: Computer Networks 19
20 Silly Window Syndrome • When receiver application accepts data from TCP 1 byte at a time Spring Semester 2006 EEC-484/584: Computer Networks 20
21 Clark’s Algorithm • Receiver should not send window update until – It can handle max segment size it advertised when connection established, or, – Its buffer is half empty, whichever is smaller • Sender should wait until – It has accumulated enough space in window to send full segment, or, – One containing at least half of receiver’s buffer size • Nagle’s algorithm and Clark’s algorithm are complementary Spring Semester 2006 EEC-484/584: Computer Networks 21
22 Principles of Congestion Control Congestion: • Informally: “too many sources sending too much data too fast for network to handle” • Different from flow control! • Manifestations: – lost packets (buffer overflow at routers) – long delays (queueing in router buffers) Spring Semester 2006 EEC-484/584: Computer Networks 22
23 Approaches towards Congestion Control Two broad approaches towards congestion control End-end congestion control: Network-assisted congestion control: • no explicit feedback from network • congestion inferred from end-system observed loss, delay • approach taken by TCP • routers provide feedback to end systems – single bit indicating congestion (SNA, DECbit, TCP/IP ECN, ATM) – explicit rate sender should send at Spring Semester 2006 EEC-484/584: Computer Networks 23
24 TCP Congestion Control: Additive Increase, Multiplicative Decrease • Approach: increase transmission rate (window size), probing for usable bandwidth, until loss occurs – Additive increase: increase cwnd by 1 MSS every RTT until loss detected – Multiplicative decrease: cut cwnd in half after loss Saw tooth behavior: probing for bandwidth Spring Semester 2006 EEC-484/584: Computer Networks 24
25 TCP Congestion Control How does sender perceive congestion? Last. Byte. Sent-Last. Byte. Acked • loss event = timeout or cwnd 3 duplicate acks • Roughly, • TCP sender reduces cwnd rate (cwnd) after loss rate = Bytes/sec RTT event • cwnd is dynamic, function of three mechanisms: • Sender limits transmission: perceived network congestion Spring Semester 2006 – AIMD – slow start – conservative after timeout events EEC-484/584: Computer Networks 25
26 TCP Slow Start • When connection begins, cwnd = 1 MSS – Example: MSS = 500 bytes & RTT = 200 msec – Initial rate = 20 kbps • When connection begins, increase rate exponentially fast until first loss event • Available bandwidth may be >> MSS/RTT – Desirable to quickly ramp up to respectable rate Spring Semester 2006 EEC-484/584: Computer Networks 26
27 TCP Slow Start Host A RTT • When connection begins, increase rate exponentially until first loss event: – Double cwnd every RTT – Done by incrementing cwnd for every ACK received • Summary: initial rate is slow but ramps up exponentially fast Spring Semester 2006 EEC-484/584: Computer Networks Host B one segme nt two segme nts four segme nts time 27
28 Congestion Avoidance Q: When should the exponential increase switch to linear? A: When cwnd gets to 1/2 of its value before timeout Implementation: • Variable Threshold • At loss event, Threshold is set to 1/2 of cwnd just before loss event Spring Semester 2006 How to increase cwnd linearly: cwnd (new) = cwnd + mss*mss/cwnd EEC-484/584: Computer Networks 28
29 Congestion Control • After 3 duplicated ACKs: – cwnd is cut in half – window then grows linearly • But after timeout event: – cwnd instead set to 1 MSS – window then grows exponentially – to a threshold, then grows linearly Spring Semester 2006 Philosophy: q 3 dup ACKs indicates network capable of delivering some segments q timeout indicates a “more alarming” congestion scenario EEC-484/584: Computer Networks 29
30 Summary: TCP Congestion Control • When cwnd is below Threshold, sender in slowstart phase, window grows exponentially • When cwnd is above Threshold, sender is in congestion-avoidance phase, window grows linearly • When a triple duplicate ACK occurs, Threshold set to cwnd/2 and cwnd set to Threshold • When timeout occurs, Threshold set to cwnd/2 and cwnd is set to 1 MSS Spring Semester 2006 EEC-484/584: Computer Networks 30
TCP Sender Congestion Control State Event TCP Sender Action 31 Commentary Slow Start ACK receipt Cong. Win = Cong. Win + MSS, (SS) for previously unacked data If (Cong. Win > Threshold) set state to “Congestion Avoidance” Resulting in a doubling of Cong. Win every RTT Congestion Avoidance Additive increase, resulting in increase of Cong. Win by 1 MSS every RTT (CA) ACK receipt Cong. Win = Cong. Win+ for previously MSS * (MSS/Cong. Win) unacked data Spring Semester 2006 EEC-484/584: Computer Networks 31
TCP Sender Congestion Control State Event TCP Sender Action Commentary SS or CA Loss event detected by triple duplicate ACK Threshold = Cong. Win/2, Cong. Win = Threshold, Set state to “Congestion Avoidance” Fast recovery, implementing multiplicative decrease. Cong. Win will not drop below 1 MSS. SS or CA Timeout Threshold = Cong. Win/2, Cong. Win = 1 MSS, Set state to “Slow Start” Enter slow start SS or CA Duplicate ACK Increment duplicate ACK Cong. Win and count for segment being Threshold not acked changed Spring Semester 2006 EEC-484/584: Computer Networks 32 32
33 TCP Congestion Control Spring Semester 2006 EEC-484/584: Computer Networks 33
34 Exercise • Suppose that the TCP congestion window is set to 18 KB and a timeout occurs. How big will the window be if the next four transmission bursts are all successful? Assume that the maximum segment size is 1 KB. Spring Semester 2006 EEC-484/584: Computer Networks 34
35 Exercise Segment# Action Receive Send Comment Timeout Retransmit 1 1: 1025(1024) 2 ACK 1025 Slow start Variables cwnd ssthresh 18 KB 1024 9216 2048 9216 3 1025: 2049(1024) 2048 9216 4 2049: 3073(1024) 2048 9216 5 ACK 2049 Slow start 3072 9216 6 ACK 3073 Slow start 4096 9216 7 3073: 4097(1024) 4096 9216 8 4097: 5121(1024) 4096 9216 9 5121: 6145(1024) 4096 9216 10 6145: 7169(1024) 4096 9216 11 ACK 4097 Slow start 5120 9216 12 ACK 5121 Slow start 6144 9216 13 ACK 6145 Slow start 7168 9216 14 ACK 7169 Slow start 8192 9216 Spring Semester 2006 EEC-484/584: Computer Networks 35
36 Exercise Problem #3 15 7169: 8193(1024) 8192 9216 16 8193: 9217(1024) 8192 9216 17 9217: 10241(1024) 8192 9216 18 10241: 11265(1024) 8192 9216 19 11265: 13313(1024) 8192 9216 20 12289: 14337(1024) 8192 9216 21 14337: 15361(1024) 8192 9216 22 15361: 16385(1024) 8192 9216 23 ACK 8193 Slow start 9216 24 ACK 9217 10240 9216 25 ACK 10241 10342 9216 26 ACK 11265 Slow start Cong. Avoid. New cwnd = cwnd + mss*mss/cwnd Cong. Avoid. 10443 9216 27 ACK 13313 Cong. Avoid. 10543 9216 28 ACK 14337 Cong. Avoid. 10642 9216 29 ACK 15361 Cong. Avoid. 10740 9216 30 ACK 16385 Cong. Avoid. 10837 9216 Spring Semester 2006 EEC-484/584: Computer Networks 36
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