6610-l5- - Transmission Control Protocol (TCP) 1 Puzzle...

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1 Transmission Control Protocol (TCP)
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Puzzle Doors numbered 1-100 All doors initially open Toggle switch outside every door If switch is pressed, door will close if it is currently open, and open if it is currently closed For i=1 to 100, you press switches of doors that are multiples of i Which doors are closed at the end of the process?
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3 Transmission Control Protocol (TCP) End-to-end transport protocol Responsible for reliability, congestion control, flow control, and sequenced delivery Applications that use TCP: http (web), telnet, ftp (file transfer), smtp (email), chat Applications that don’t: multimedia (typically) – use UDP instead
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4 Ports, End-points, & Connections Thus, an end-point is represented by (IP address,Port) Ports can be re-used between transport protocols A connection is (SRC IP address, SRC port, DST IP address, DST port) Same end-point can be used in multiple connections IP Layer TCP UDP http   ftp   smtptelnet IP address Protocol ID A1 A2 A3 Port
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5 TCP Connection Establishment Connection Maintenance Reliability Congestion control Flow control Sequencing Connection Termination
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6 Active and Passive Open How do applications initiate a connection? One end (server) registers with the TCP layer instructing it to “accept” connections at a certain port The other end (client) initiates a “connect” request which is “accept”-ed by the server
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7 Connection Establishment & Termination 3-way handshake used for connection establishment Randomly chosen sequence number is conveyed to the other end Similar FIN, FIN+ACK exchange used for connection termination SYN SYN+ACK ACK DATA Server does passive open Accept connection request Send acceptance Start connection Active open Send connection request
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8 Reliability (Loss Recovery) Sequence Numbers TCP uses cumulative Acknowledgments (ACKs) Next expected in-sequence packet sequence number Pros and cons? Piggybacking data 1 2 3 4 3 1 2 3 3 4 Timeout calculation Rtt avg  = k*Rtt avg  + (1-k)*Rtt sample RTO = Rttavg + 4*Rtt deviation
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9 Fundamental Mechanism Simple stop and go protocol Timeout based reliability (loss recovery) data retx ack data data Sliding Window Protocol : 1 2 3 4 5 6 7 8 9 10 11 12 ….  Multiple unacknowledged packets!
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10 Congestion Control Slow Start Start with W=1 For every ACK, W=W+1 Congestion Avoidance (linear increase) For every ACK, W = W+1/W Congestion Control (multiplicative decrease) ssthresh = W/2 W = 1 Alternative: Fall to W/2 and start congestion avoidance directly
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11 Why LIMD? (fairness) W=1 100 10 diff = 90 1 1 diff = 0 Problem? – inefficient W=W/2 100 10 diff = 90 50 5 diff = 45 51 6 diff = 45 52 7 diff = 45 ..
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This note was uploaded on 12/14/2009 for the course ECE 6610 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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6610-l5- - Transmission Control Protocol (TCP) 1 Puzzle...

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