1. 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)

2. 2 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Outline TCP –Connection management –Reliable data transfer –Flow control –TCP transmission policy –Congestion control Reminder: Quiz 4 –MW session: 11/27 Monday 2-4pm –TTh session: 11/28 Tuesday 4-6pm

3. 3 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Connection Management TCP sender, receiver establish “connection” before exchanging data segments Initialize TCP variables: –Sequence numbers –Buffers, flow control info (e.g. RcvWindow ) Client: connection initiator Socket clientSocket = new Socket("hostname","port number"); Server: contacted by client Socket connectionSocket = welcomeSocket.accept();

4. 4 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

5. 5 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Connection Management Three way handshake: SYN segment is considered as 1 byte SYN/ACK segment is also considered as 1 byte client SYN (seq=x) server SYN/ACK (seq=y, ACK=x+1) ACK (seq=x+1, ACK=y+1) connect accept

6. 6 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Connection Management Closing a connection: client closes socket: clientSocket.close(); Step 1: client end system sends TCP FIN control segment to server Step 2: server receives FIN, replies with ACK. Closes connection, sends FIN. client FIN server ACK FIN close closed timed wait

7. 7 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Connection Management Step 3: client receives FIN, replies with ACK. –Enters “timed wait” - will respond with ACK to received FINs Step 4: server, receives ACK. Connection closed. Note: with small modification, can handle simultaneous FINs client FIN server ACK FIN closing closed timed wait closed

8. 8 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Reliable Data Transfer TCP creates rdt service on top of IP’s unreliable service Pipelined segments Cumulative acks TCP uses single retransmission timer Retransmissions are triggered by: –timeout events –duplicate acks Initially consider simplified TCP sender: –ignore duplicate acks –ignore flow control, congestion control

9. 9 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Sender Events: Data rcvd from app: Create segment with sequence number seq # is byte-stream number of first data byte in segment start timer if not already running (think of timer as for oldest unacked segment) expiration interval: TimeOutInterval Timeout: retransmit segment that caused timeout restart timer Ack rcvd: If acknowledges previously unacked segments –update what is known to be acked –start timer if there are outstanding segment

10. 10 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP: Retransmission Scenarios Host A Seq=100, 20 bytes data ACK=100 time premature timeout Host B Seq=92, 8 bytes data ACK=120 Seq=92, 8 bytes data Seq=92 timeout ACK=120 Host A Seq=92, 8 bytes data ACK=100 loss timeout lost ACK scenario Host B X Seq=92, 8 bytes data ACK=100 time Seq=92 timeout SendBase = 100 SendBase = 120 SendBase = 120 Sendbase = 100

11. 11 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Retransmission Scenarios Host A Seq=92, 8 bytes data ACK=100 loss timeout Cumulative ACK scenario Host B X Seq=100, 20 bytes data ACK=120 time SendBase = 120

12. 12 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP ACK Generation Event at Receiver Arrival of in-order segment with expected seq #. All data up to expected seq # already ACKed Arrival of in-order segment with expected seq #. One other segment has ACK pending Arrival of out-of-order segment higher-than-expect seq. #. Gap detected Arrival of segment that partially or completely fills gap TCP Receiver action Delayed ACK. Wait up to 500ms for next segment. If no next segment, send ACK Immediately send single cumulative ACK, ACKing both in-order segments Immediately send duplicate ACK, indicating seq. # of next expected byte Immediate send ACK, provided that segment starts at lower end of gap

13. 13 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Flow Control Receive side of TCP connection has a receive buffer: Speed-matching service: matching the send rate to the receiving app’s drain rate App process may be slow at reading from buffer Flow control: sender won’t overflow receiver’s buffer by transmitting too much, too fast

14. 14 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Flow Control (Suppose TCP receiver discards out-of-order segments) Spare room in buffer = RcvWindow = RcvBuffer-[LastByteRcvd - LastByteRead] Rcvr advertises spare room by including value of RcvWindow in segments Sender limits unACKed data to RcvWindow –guarantees receive buffer doesn’t overflow

15. 15 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

16. 16 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Transmission Policy Window management not directly tied to ACKs

17. 17 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

18. 18 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

19. 19 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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.

20. 20 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Silly Window Syndrome When receiver application accepts data from TCP 1 byte at a time

21. 21 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

22. 22 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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)

23. 23 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Approaches towards Congestion Control End-end congestion control: no explicit feedback from network congestion inferred from end-system observed loss, delay approach taken by TCP Network-assisted congestion control: routers provide feedback to end systems –single bit indicating congestion (SNA, DECbit, TCP/IP ECN, ATM) –explicit rate sender should send at Two broad approaches towards congestion control

24. 24 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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

25. 25 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Congestion Control Sender limits transmission: LastByteSent-LastByteAcked  cwnd Roughly, cwnd is dynamic, function of perceived network congestion How does sender perceive congestion? loss event = timeout or 3 duplicate acks TCP sender reduces rate ( cwnd ) after loss event three mechanisms: –AIMD –slow start –conservative after timeout events rate = cwnd RTT Bytes/sec

26. 26 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Slow Start When connection begins, cwnd = 1 MSS –Example: MSS = 500 bytes & RTT = 200 msec –Initial rate = 20 kbps Available bandwidth may be >> MSS/RTT –Desirable to quickly ramp up to respectable rate When connection begins, increase rate exponentially fast until first loss event

27. 27 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Slow Start 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 Host A one segment RTT Host B time two segments four segments

28. 28 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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 How to increase cwnd linearly: cwnd (new) = cwnd + mss*mss/cwnd

29. 29 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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  3 dup ACKs indicates network capable of delivering some segments  timeout indicates a “more alarming” congestion scenario Philosophy:

30. 30 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Summary: TCP Congestion Control When cwnd is below Threshold, sender in slow- start 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

31. 31 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Sender Congestion Control StateEventTCP Sender ActionCommentary Slow Start (SS) ACK receipt for previously unacked data CongWin = CongWin + MSS, If (CongWin > Threshold) set state to “Congestion Avoidance” Resulting in a doubling of CongWin every RTT Congestion Avoidance (CA) ACK receipt for previously unacked data CongWin = CongWin+ MSS * (MSS/CongWin) Additive increase, resulting in increase of CongWin by 1 MSS every RTT

32. 32 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Sender Congestion Control StateEventTCP Sender ActionCommentary SS or CALoss event detected by triple duplicate ACK Threshold = CongWin/2, CongWin = Threshold, Set state to “Congestion Avoidance” Fast recovery, implementing multiplicative decrease. CongWin will not drop below 1 MSS. SS or CATimeoutThreshold = CongWin/2, CongWin = 1 MSS, Set state to “Slow Start” Enter slow start SS or CADuplicate ACK Increment duplicate ACK count for segment being acked CongWin and Threshold not changed

33. 33 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Congestion Control

34. 34 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao 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.

35. 35 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Exercise Segment# ActionVariables SendReceiveCommentcwndssthresh Timeout18KB- Retransmit10249216 11:1025(1024)10249216 2ACK 1025Slow start20489216 31025:2049(1024)20489216 42049:3073(1024)20489216 5ACK 2049Slow start30729216 6ACK 3073Slow start40969216 73073:4097(1024)40969216 84097:5121(1024)40969216 95121:6145(1024)40969216 106145:7169(1024)40969216 11ACK 4097Slow start51209216 12ACK 5121Slow start61449216 13ACK 6145Slow start71689216 14ACK 7169Slow start81929216

36. 36 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Exercise Problem #3 157169:8193(1024)81929216 168193:9217(1024)81929216 179217:10241(1024)81929216 1810241:11265(1024)81929216 1911265:13313(1024)81929216 2012289:14337(1024)81929216 2114337:15361(1024)81929216 2215361:16385(1024)81929216 23ACK 8193Slow start9216 24ACK 9217Slow start102409216 25ACK 10241 Cong. Avoid. New cwnd = cwnd + mss*mss/cwnd 103429216 26ACK 11265Cong. Avoid.104439216 27ACK 13313Cong. Avoid.105439216 28ACK 14337Cong. Avoid.106429216 29ACK 15361Cong. Avoid.107409216 30ACK 16385Cong. Avoid.108379216

37. 37 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Causes/Costs of Congestion Two senders, two receivers One router, infinite buffers No retransmission large delays when congested maximum achievable throughput unlimited shared output link buffers Host A in : original data Host B out

38. 38 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Causes/Costs of Congestion One router, finite buffers Sender retransmission of lost packet finite shared output link buffers Host A in : original data Host B out ' in : original data, plus retransmitted data

39. 39 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Causes/Costs of Congestion Always: (goodput) “Perfect” retransmission only when loss: Retransmission of delayed (not lost) packet makes larger (than perfect case) for same in out = in out > in out “Costs” of congestion: more work (retrans) for given “goodput” unneeded retransmissions: link carries multiple copies of pkt R/2 in out R/2 in out R/2 in out R/4 R/3

40. 40 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Causes/Costs of Congestion four senders multihop paths timeout/retransmit in Q: what happens as and increase ? in finite shared output link buffers Host A in : original data Host B out ' in : original data, plus retransmitted data

41. 41 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao Causes/costs of congestion: scenario 3 Another “cost” of congestion: when packet dropped, any “upstream transmission capacity used for that packet was wasted! HostAHostA HostBHostB o u t

42. 42 Spring Semester 2006EEC-484/584: Computer NetworksWenbing Zhao TCP Sending Rate Control To control the sending rate, each sender maintains two windows –Flow control window: window that receiver granted –Congestion window Number of bytes sender can transmit = minimum of two window sizes