Transport Layer ECE544: Communication Networks-II, Spring 2013 Tam Vu WINLAB, Dept. of Computer Science Rutgers University Includes teaching materials from, L. Peterson, Sumathi Gopal and Sumit Rangwala, D. Raychaudhuri, Mike Freedman

IP Protocol Stack: Key Abstractions Application Best-effort local packet delivery Best-effort global packet delivery Reliable streams Applications Messages Transport Network Link Problem: Network Layer (IP) provides only best-effort communication services

Applications requirements vs. IP layer limitations Guarantee message delivery Network may drop messages. Deliver messages in the same order they are sent Messages may be reordered in networks and incurs a long delay. Delivers at most one copy of each message Messages may duplicate in networks. Support arbitrarily large message Network may limit message size. Support synchronization between sender and receiver Allows the receiver to apply flow control to the sender Support multiple application processes on each host Network only support communication between hosts Many more

IP Protocol Stack: Key Abstractions Application Best-effort local packet delivery Best-effort global packet delivery Reliable streams Applications Messages Transport Network Link Transport layer: Provide applications with good abstractions Without support or feedback from the network Is the lowest layer in the network stack that is an end-to-end protocol

Transport Protocols Logical communication between processes Sender divides a message into segments Receiver reassembles segments into message Transport services (De)multiplexing packets Detecting corrupted data Optionally: reliable delivery, flow control, …

Two Basic Transport Features Demultiplexing: port numbers Error detection: checksums Server host 128.2.194.242 Client host Service request for 128.2.194.242:80 (i.e., the Web server) Web server (port 80) Client OS Echo server (port 7) IP payload detect corruption 6

Most Popular Transport Protocols User Datagram Protocol (UDP) Support multiple applications processes on each host Option to check messages for correctness with CRC check Transmission Control Protocol (TCP) Ensures reliable delivery of packets between source and destination processes Ensures in-order delivery of packets to destination process Other services Real Time Protocol (RTP) Serves real-time multimedia applications Moves decision making to the applications Runs over UDP

User Datagram Protocol (UDP) Service: Support for multiple processes on each host to communicate Issue: IP only provides communication between hosts (IP addresses) Solution Add port number and associate a process with a port number 4-Tuple Unique Connection Identifier: [SrcPort, SrcIPAddr, DestPort, DestIPAddr ] Lightweight communication between processes Send and receive messages Avoid overhead of ordered, reliable delivery No connection setup delay, in-kernel connection state Used by popular apps Query/response for DNS Real-time data in VoIP 16 31 SrcPort DesPort Length Checksum Payload

User Datagram Protocol (UDP): Error Detection Service: Ensure message correctness Issue: Packet corruption in transit Solution Use Checksum. Includes UDP header, payload, pseudo header Pseudo header Protocol number, source IP address, destination IP address, and UDP length 16 31 SrcPort DesPort Length Checksum Payload

Transmitting a stream of bytes ? Connection oriented Explicit set-up and tear- down of TCP connection Flow control Prevent overflow of the receiver’s buffer space Congestion control Adapt to network congestion for the greater good Stream-of-bytes service Sends and receives a stream of bytes Reliable, in-order delivery Corruption: checksums Detect loss/reordering: sequence numbers Reliable delivery: acknowledgments and retransmissions

Transmission Control Protocol (TCP) First proposed by Vinton Cerf and Robert Kahn, 1974 TCP/IP enabled computers of all sizes, from different vendors, different OSs, to communicate with each other. Used by 80% of all traffic on the Internet Reliable, in-order delivery, connection-oriented, bye-stream service

Starting and Ending a Connection: TCP Handshakes

Establishing a TCP Connection SYN Each host tells its Initial Sequence Number (ISN) to the other host. SYN ACK ACK Data Data Three-way handshake to establish connection Host A sends a SYN (open) to the host B Host B returns a SYN acknowledgment (SYN ACK) Host A sends an ACK to acknowledge the SYN ACK

TCP Header Source port Destination port Sequence number Flags: SYN FIN RST PSH URG ACK Acknowledgment HdrLen Flags Advertised window Checksum Urgent pointer Options (variable) Data

Step 1: A’s Initial SYN Packet A’s port B’s port A’s Initial Sequence Number Flags: SYN FIN RST PSH URG ACK Acknowledgment 20 Flags Advertised window Checksum Urgent pointer Options (variable) A tells B it wants to open a connection…

Step 2: B’s SYN-ACK Packet B’s port A’s port B’s Initial Sequence Number Flags: SYN FIN RST PSH URG ACK A’s ISN plus 1 20 Flags Advertised window Checksum Urgent pointer Options (variable) B tells A it accepts, and is ready to hear the next byte… … upon receiving this packet, A can start sending data 17

Step 3: A’s ACK of the SYN-ACK A’s port B’s port Sequence number Flags: SYN FIN RST PSH URG ACK B’s ISN plus 1 20 Flags Advertised window Checksum Urgent pointer Options (variable) A tells B it is okay to start sending … upon receiving this packet, B can start sending data 18

SYN Loss and Web Downloads Upon sending SYN, sender sets a timer If SYN lost, timer expires before SYN-ACK received Sender retransmits SYN How should the TCP sender set the timer? No idea how far away the receiver is Some TCPs use default of 3 or 6 seconds Implications for web download User gets impatient and hits reload … Users aborts connection, initiates new socket Essentially, forces a fast send of a new SYN! 19

Tearing Down the Connection B ACK SYN SYN ACK Data ACK FIN ACK FIN ACK A time Closing (each end of) the connection Finish (FIN) to close and receive remaining bytes And other host sends a FIN ACK to acknowledge Reset (RST) to close and not receive remaining bytes 20

Sending/Receiving the FIN Packet Sending a FIN: close() Process is done sending data via socket Process invokes “close()” Once TCP has sent all the outstanding bytes… … then TCP sends a FIN Receiving a FIN: EOF Process is reading data from socket Eventually, read call returns an EOF 21

Data transmission

TCP: Byte-stream Service: Byte-stream Application reads or writes a stream of bytes to the transport Issue: IP is packet-oriented Solution: TCP maintains a local buffer Chop the stream into packets and transmit (sender) Coalesce data from packets to form a stream (receiver)

TCP “Stream of Bytes” Service Host A Byte 0 Byte 1 Byte 2 Byte 3 Byte 80 Host B Byte 0 Byte 1 Byte 2 Byte 3 Byte 80 24

…Emulated Using TCP “Segments” Host A Byte 0 Byte 1 Byte 2 Byte 3 Byte 80 Segment sent when: Segment full (Max Segment Size), Not full, but times out, or “Pushed” by application TCP Data TCP Data Host B Byte 0 Byte 1 Byte 2 Byte 3 Byte 80 25

TCP Segment IP packet TCP packet TCP segment IP Hdr IP Data TCP Hdr TCP Data (segment) IP packet No bigger than Maximum Transmission Unit (MTU) E.g., up to 1500 bytes on an Ethernet link TCP packet IP packet with a TCP header and data inside TCP header is typically 20 bytes long TCP segment No more than Maximum Segment Size (MSS) bytes E.g., up to 1460 consecutive bytes from the stream 26

Sequence Number Sequence number = 1st byte Host A Host B ISN (initial sequence number) Byte 81 Sequence number = 1st byte TCP Data TCP Data Host B 27

Reliable Delivery on a Lossy Channel With Bit Errors

Challenges of Reliable Data Transfer Over a perfectly reliable channel: Done Over a channel with bit errors Receiver detects errors and requests retransmission Over a lossy channel with bit errors Some data missing, others corrupted Receiver cannot easily detect loss Over a channel that may reorder packets Receiver cannot easily distinguish loss vs. out-of-order 30

An Analogy Alice and Bob are talking What if Alice couldn’t understand Bob? Bob asks Alice to repeat what she said What if Bob hasn’t heard Alice for a while? Is Alice just being quiet? Has she lost reception? How long should Bob just keep on talking? Maybe Alice should periodically say “uh huh” … or Bob should ask “Can you hear me now?” 31

Take-Aways from the Example Acknowledgments from receiver Positive: “okay” or “uh huh” or “ACK” Negative: “please repeat that” or “NACK” Retransmission by the sender After not receiving an “ACK” After receiving a “NACK” Timeout by the sender (“stop and wait”) Don’t wait forever without some acknowledgment 32

TCP Support for Reliable Delivery Detect bit errors: checksum Used to detect corrupted data at the receiver …leading the receiver to drop the packet Detect missing data: sequence number Used to detect a gap in the stream of bytes ... and for putting the data back in order Recover from lost data: retransmission Sender retransmits lost or corrupted data Two main ways to detect lost packets 33

TCP Acknowledgments Host A Sequence number = 1st byte ISN (initial sequence number) Sequence number = 1st byte TCP Data ACK sequence number = next expected byte TCP Data Host B 34

Automatic Repeat reQuest (ARQ) ACK and timeouts Receiver sends ACK when it receives packet Sender waits for ACK and times out Simplest ARQ protocol Stop and wait Send a packet, stop and wait until ACK arrives Time Packet ACK Timeout Sender Receiver

Quick TCP Math Initial Seq No = 501. Sender sends 4500 bytes successfully acknowledged. Next sequence number to send is: (A) 4501 (B) 5000 (C) 5001 (D) 5002 Next 1000 byte TCP segment received. Receiver acknowledges with ACK number: (A) 5001 (B) 6000 (C) 6001

Flow Control: TCP Sliding Window

Sliding Window: Motivation Stop-and-wait is inefficient Only one TCP segment is “in flight” at a time Consider: 1.5 Mbps link with 50 ms round-trip-time (RTT) Assume segment size of 1 KB (8 Kbits) 8 Kbits/segment at 50 msec/segment  160 Kbps That’s 11% of the capacity of 1.5 Mbps link 39

Sliding Window Allow a larger amount of data “in flight” Allow sender to get ahead of the receiver … though not too far ahead Sending process Receiving process TCP TCP Last byte written Last byte read Next byte expected Last byte ACKed Last byte received Last byte sent 40

Receiver Buffering Receive window size Amount that can be sent without acknowledgment Receiver must be able to store this amount of data Receiver tells the sender the window Tells the sender the amount of free space left Window Size Data ACK’d Outstanding Un-ack’d data Data OK to send Data not OK to send yet 41

TCP: Flow Control Flow Control “Prevent sender from overrunning the capacity (buffer) of the receiver” Solution: Use adaptive receiver window size Goal is to keep (C) – (A) < MaxRcvBuffer Every packet carries ACK and AdvertisedWindow Sending Appl Receiving Appl LastByteAcked (J) (K) LastByteSent (I) LastByteWritten (B) NextByteExpected (C) LastByteRcvd LastByteRead (A) TCP LastByteSent (K) – LastByteAcked (J) <= AdvertisedWindow EffWin = AdvertisedWin - (LastByteSent-LastByteAcked) AdvertisedWindow = MaxRcvBuffer- ((NextByteExp-1)-LastByteRead) LastByteWritten – LastByteAcked <= MaxSendBuffer

Optimizing Retransmissions

Reasons for Retransmission Packet Packet Packet Timeout Timeout Timeout ACK ACK Packet Packet Packet ACK Timeout Timeout Timeout ACK ACK ACK lost DUPLICATE PACKET Early timeout DUPLICATE PACKETS Packet lost

How Long Should Sender Wait? Sender sets a timeout to wait for an ACK Too short: wasted retransmissions Too long: excessive delays when packet lost TCP sets timeout as a function of the RTT Expect ACK to arrive after an “round-trip time” … plus a fudge factor to account for queuing But, how does the sender know the RTT? Running average of delay to receive an ACK 45

TCP Timeout Issue: RTT in a wide area network varies substantially Solution: Adaptive Timeout Original Algorithm: EstimatedRTT = a x EstimatedRTT + (1-a) x SampleRTT Timeout = β x EstimatedRTT (β = 2) Problem Does not distinguish whether the ACK is for original transmission or retransmission Constant β is not good. Assumes constant variance

TCP Timeout Karn/Partridge Algorithm Whenever TCP retransmits a segment, it stops taking samples of the RTT Only measure SampleRTT for segments that have been sent only once Each time TCP retransmits, set the next timeout to be twice the last timeout Relieves congestion Jacobson/Karels Algorithm: Adaptive variance (uses mean variance) Difference = SampleRTT - EstimatedRTT EstimatedRTT = EstimatedRTT + (d x Difference) → (same as in original) Deviation = Deviation + d(|Difference|- Deviation) Timeout = m x EstimatedRTT + f x Deviation (default: set m = 1 and f= 4 )

TCP Deadlock TCP Deadlock To solve it: receiver advertises a window size of 0, the sender stops sending data the window size update from the receiver is lost To solve it: the sender starts the persist timer when AdvertisedWindow = 0 When the persist timer expires, the sender sends a small packet

Triggering Transmission When to transmit a segment: small segments subject to large overhead Reach max segment size (MSS): the size of the largest segment TCP can send without causing the local IP to fragment MSS = local MTU – IP & TCP header The sending process explicitly ask the TCP to transmit, “push”

Even with flow control packets might not reach the destination Congestion Source 1 Even with flow control packets might not reach the destination Dest 1 Source 2 Dest 2 Source 3 When the network cannot support the sender’s rate Queues at the network elements overflow

Congestion Control vs. Flow Control Mechanism to prevent sender from overrunning the capacity of the network When network is the bottleneck Flow Control Mechanism to prevent sender from overrunning the capacity of the receiver When receiver is the bottleneck

Congestion Control: Design Approach Maintain another window at the sender called CongestionWindow (cwnd) CongestionWindow is the max number of packets allowed in the network Number of unACKed packets at the sender. Key: How to calculate congestion window (cwnd) Various approaches possible TCP estimates it based on observed packet losses Assumes packet loss as indication of congestion Since we don’t know whether the network or the receiver is the bottleneck MaxWindow = MIN(CongestionWindow, AdvertisedWindow) EffectiveWin = MaxWindow – (LastByteSent – LastByteAcked)

Congestion Avoidance: (AIMD) If no congestion in the network (increase conservatively) Increase the congestion window additively every RTT If congestion in the network (decrease aggressively) Decrease the congestion window multiplicatively, immediately How is congestion detected? Estimated (more later) Every ACK reception cwnd = cwnd + MSS*(MSS/cwnd) cwnd in bytes Every RTT w = w + 1 w = cwnd in segments Every ACK reception w = w + 1/w w = cwnd in segments cwnd = cwnd/2 cwnd in bytes

Congestion Avoidance: (AIMD) CongestionWindow Size Startup time Time TCP’s saw tooth pattern Issues with additive increase takes too long to ramp up a connection from the beginning The entire advertised window may be reopened when a lost packet retransmitted and a single cumulative ACK is received by the sender

TCP “Slow Start”: To start quickly! Maintain another variable slow start threshold (ssthresh) Last known stable rate If (cwnd > ssthresh) State = congestion avoidance Else State = slow start In Slow start Increase the congestion window exponentially every RTT Key: How is ssthresh calculated? Every ACK reception cwnd = cwnd + MSS cwnd in bytes Every ACK reception w = w + 1 w = cwnd in segments

TCP: Congestion Detection and Retransmit Loss of packet indicates congestion Timer Timeouts (No ACK) Set according to Jacobson/Karels algorithm On timer timeout ssthresh = max(2*MSS, effwin/2); cwnd = MSS Notice this will cause TCP to go into slow start Issue: takes a long time to detect a packet loss Affects throughput Any other quicker way of detecting a packet loss?

Fast Retransmit Observation: A series of duplicate ACKs might mean a packet loss Solution Every time receiver receives a packet (out-of-order), sends a duplicate ACK Sender retransmit the missing packet after it receives some number of duplicate ACKs (e.g. 3 duplicate ACKs) Fast Retransmit does not replace timeouts Issue: Reduces latency (early retransmit) but still incurs loss in throughput (slow start after packet loss ) PKT 1 PKT 2 ACK 1 PKT 3 ACK 2 PKT 4 ACK 2 PKT 5 PKT 6 ACK 2 ACK 2 PKT 3 Retran ACK 6

Fast Recovery Transmit a packet for every ACK received till the retransmitted packet is ACK’d ssthresh= (2*MSS, cwdn/2); cwnd = sshthred + 3 On every ACK will the ACK of retransmitted packet cwnd = cwnd + 1 On reception of ACK of retransmitted packet Start congestion avoidance instead of slow start cwnd = ssthresh

Homework 5.13 (3rd ed and 4th ed) 5.16 5.28 5.34 5.39 Due 4/5