CS 43: Computer Networks. 18: Transmission Control Protocol October 12-29, 2018

Transcription

1 CS 43: Computer Networks 18: Transmission Control Protocol October 12-29, 2018

2 Reading Quiz Lecture 18 - Slide 2

3 Midterm topics Abstraction, Layering, End-to-end design HTTP, DNS, , BT, etc. (app protocols) Basics of distributed systems The transport layer, UDP, TCP (minus flow and congestion control) Concurrency, application structure, and blocking The functions we highlighted in labs Anything else we talked about in class Lecture 18 - Slide 3

4 Midterm format Closed book, closed notes, no need of a calculator 1 page each with: definitions (1 point) short answers (1-3 points) 2-3 multi-part questions (1-5 points) Answer length: short answer: approx. 2 sentences. long answer: approx. 4-5 sentences. Lecture 18 - Slide 4

5 Last class The Two-Generals Problem communicate via an unreliable link the two armies cannot reliably coordinate their attack Automatic Repeat Request (ARQ) Protocols Stop-and-Wait Protocol: send data, wait for response Go-Back-N Selective Repeat Lecture 16 - Slide 5

6 Today: TCP Practical Reliability Questions What does connection establishment look like? How do we choose sequence numbers? How do the sender and receiver keep track of outstanding pipelined segments? How should we choose timeout values? How many segments should be pipelined? Lecture 18 - Slide 6

7 Five-layer Internet Model Application: the application (e.g., the Web, ) Transport: end-to-end connections, reliability Network: routing Link (data-link): framing, error detection Physical: 1 s and 0 s/bits across a medium (copper, the air, fiber) Lecture 18 - Slide 7

8 Transmission Control Protocol Break message into packets (TCP segments) Should be delivered reliably & in-order GET HTTP/1.1 Host: 3 google.c 2 p://www 1 GET htt Lecture 3 - Slide 8

9 Transmission Control Protocol Slide 9

10 TCP Overview Point-to-point, full duplex One pair of hosts Messages in both directions Reliable, in-order byte stream No discrete message Connection-oriented Handshaking (exchange of control messages) before data transmitted Pipelined Many segments in flight Flow control Don t send too fast for the receiver Congestion control Don t send too fast for the network Lecture 18 - Slide 10

11 URG: urgent data (generally not used) ACK: ACK # valid PSH: push data now (generally not used) RST, SYN, FIN: connection estab (setup, teardown commands) Internet checksum (as in UDP) head len TCP Segments 32 bits source port # dest port # not used sequence number acknowledgement number U A P R S F checksum receive window application data (variable length) Urg data pointer options (variable length) counting by bytes of data (not segments!) # bytes rcvr willing to accept Lecture 18 - Slide 11

12 TCP Segments 32 bits 20 Bytes (UDP was 8) head len source port # dest port # not used sequence number acknowledgement number U A P R S F checksum receive window Urg data pointer options (variable length) application data (variable length) Lecture 18 - Slide 12

13 Practical Reliability Questions What does connection establishment look like? How should we choose timeout values? How do the sender and receiver keep track of outstanding pipelined segments? How do we choose sequence numbers? How many segments should be pipelined? Lecture 18 - Slide 13

14 A connection 1. Requires stored state at two hosts. 2. Requires stored state within the network. 3. Establishes a path between two hosts. A. 1 B. 1 & 3 C. 1, 2 & 3 D. 2 E. 2 & 3 Lecture 18 - Slide 14

15 Connections In TCP, hosts must establish a connection prior to communicating. Opportunity to exchange initial protocol state. Which sequence numbers to use. What the maximum segment size should be. Initial window sizes, etc. (several parameters) Lecture 18 - Slide 15

16 Connection Establishment (three-way handshake) SYN_SENT Active participant (client) SYN, SequenceNum = x Passive participant (server) LISTEN ESTABLISHED SYN + ACK, SequenceNum = y, Acknowledgment = x + 1 SYN_RCVD ACK, Acknowledgment = y + 1 +data ESTABLISHED Lecture 18 - Slide 16

17 TCP Segments 32 bits ACK: ACK # valid RST, SYN, FIN: connection estab (setup, teardown commands) head len source port # dest port # not used sequence number acknowledgement number U A P R S F checksum receive window application data (variable length) Urg data pointer options (variable length) Lecture 18 - Slide 17

18 Connection Establishment (three-way handshake) connect() SYN_SENT (Client) SYN, SequenceNum = x (Server) LISTEN bind(), listen() accept() ESTABLISHED connect() returns eventually, send() SYN + ACK, SequenceNum = y, Acknowledgment = x + 1 ACK, Acknowledgment = y + 1 (piggybacked data optional) SYN_RCVD ESTABLISHED accept() returns Both sides agree on connection. Lecture 18 - Slide 18

19 Piggybacking So far, we ve assumed distinct sender and receiver roles Client Request ACK ACK Response Server Client Request Server Response + ACK Request Request + ACK In reality, usually both sides of a connection send some data request/response is a common pattern ACK Response Without Piggybacking Response + ACK With Piggybacking Lecture 18 - Slide 19

20 Connection Teardown Orderly release by sender and receiver when done Delivers all pending data and hangs up Cleans up state in sender and receiver Each side may terminate independently Lecture 18 - Slide 20

21 TCP Connection Teardown close() close() Lecture 18 - Slide 21

22 Why does one side need to wait before transitioning to CLOSED state? A. Random protocol artifact there is no reason for it to wait. B. There is a reason for it to wait the reason is Lecture 18 - Slide 22

23 The TIME_WAIT State We wait 2*MSL (maximum segment lifetime) before completing the close. The MSL is arbitrary (usually 60 sec) ACK might have been lost and so FIN will be resent Could interfere with a subsequent connection This is why we used SO_REUSEADDR socket option in lab 2 Says to skip this waiting step and immediately abort the connection Lecture 18 - Slide 23

24 Practical Reliability Questions What does connection establishment look like? How do we choose sequence numbers? How do the sender and receiver keep track of outstanding pipelined segments? How should we choose timeout values? How many segments should be pipelined? Lecture 18 - Slide 24

25 How should we choose the initial sequence number? A. Start from zero B. Start from one What can go wrong with sequence numbers? -How they re chosen? -In the course of using them? C. Start from a random number D. Start from some other value (such as?) Lecture 18 - Slide 25

26 Sequencing Initial sequence numbers (ISN) chosen at random Does not start at 0 or 1 (anymore). Helps to prevent against forgery attacks. TCP sequences bytes rather than segments Example: if we re sending 1500-byte segments Randomly choose ISN (suppose we picked 1150) First segment (sized 1500) would use number 1150 Next would use 2650 Lecture 18 - Slide 26

27 Sequence Prediction Attack (1996) Attacker Target Server Trusted Client Lecture 18 - Slide 27

28 Sequence Prediction Attack (1996) Attacker (From: Forged IP of Trusted Client) SYN (From: Forged IP of Trusted Client) ACK (Guess the ISN of server) Target Server Evil commands SYN ACK Trusted Client Lecture 18 - Slide 28

29 Practical Reliability Questions What does connection establishment look like? How do we choose sequence numbers? How do the sender and receiver keep track of outstanding pipelined segments? How should we choose timeout values? How many segments should be pipelined? Lecture 18 - Slide 29

30 Windowing (Sliding Window) At the sender: What s been ACKed What s still outstanding What to send next At the receiver: Go-back-N Highest sequence number received so far. (Selective repeat) Which sequence numbers received so far. Buffered data. Lecture 18 - Slide 30

31 Go-back-N At the sender: At the receiver: Keep track of largest sequence number seen. If it receives ANYTHING, sends back ACK for largest sequence number seen so far. (Cumulative ACK) Lecture 18 - Slide 31

32 Cumulative Acknowledgements An ACK for sequence number N implies that all data prior to N has been received. Sender Receiver Data 0 Data 1500 Data 3000 ACK 1500 ACK 3000 ACK 4500 Data 4500 Data 6000 Data 7500 Lecture 18 - Slide 32

33 Cumulative Acknowledgements An ACK for sequence number N implies that all data prior to N has been received. Sender Receiver Sender Receiver Data 0 Data 0 Data 1500 Data 3000 ACK 1500 ACK 3000 ACK 4500 Data 1500 Data 3000? Data 4500 Data 6000 Data 7500 Lecture 18 - Slide 33

34 What should we do with an out-of-order segment at the receiver? A. Drop it. B. Save it and ACK it. C. Save it, don t ACK it. D. Something else (explain). Lecture 18 - Slide 34

35 Selective Repeat Lecture 18 - Slide 35

36 If you were building a transport protocol, which would you use? A. Go-back-N B. Selective repeat C. Something else (explain) Lecture 18 - Slide 36

37 Practical Reliability Questions What does connection establishment look like? How do we choose sequence numbers? How do the sender and receiver keep track of outstanding pipelined segments? How should we choose timeout values? How many segments should be pipelined? Lecture 18 - Slide 37

38 Timeouts How long should we wait before timing out and retransmitting a segment? Too short: needless retransmissions Too long: slow reaction to losses Should be (a little bit) longer than the RTT Lecture 18 - Slide 38

39 Estimating RTT Problem: RTT changes over time Routers buffer packets in queues Queue lengths vary Receiver may have varying load Sender takes measurements Use statistics to decide future timeouts for sends Estimate RTT and variance Apply smoothing to account for changes Lecture 18 - Slide 39

40 Estimating RTT For each segment that did not require a retransmit (ACK heard without a timeout) Consider the time between segment sent and ACK received to be a sample of the current RTT Use that, along with previous history, to update the current RTT estimate Exponentially Weighted Moving Average (EWMA) Lecture 18 - Slide 40

41 EWMA EstimatedRTT = (1 a) * EstimatedRTT + a * SampleRTT a is usually 1/8. In other words, our current estimate is a blend of 7/8 of the previous estimate plus 1/8 of the new sample. DevRTT = (1 B) * DevRTT + B * SampleRTT EstimatedRTT B is usually 1/4 Lecture 18 - Slide 41

42 Example Suppose EstimateRTT = 64, Dev = 8 Latest sample: 120 New estimate = 7/8 * /8 * 120 = = 71 New dev = 3/4 * 8 + 1/4 * = = 18 Another sample: 400 New estimate = 7/8 * /8 * 400 = = 112 New dev = 3/4 * /4 * = = 85 Lecture 18 - Slide 42

43 Book Example (Smoothing) RTT: gaia.cs.umass.edu to fantasia.eurecom.fr RTT (milliseconds) time (seconnds) SampleRTT Estimated RTT Lecture 18 - Slide 43

44 TCP Timeout Value TimeoutInterval = EstimatedRTT + 4*DevRTT estimated RTT safety margin Lecture 18 - Slide 44