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📄 rfc2525.txt

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   Implications      Entering slow start upon timeout forms one of the cornerstones of      Internet congestion stability, as outlined in [Jacobson88].  If      TCPs fail to do so, the network becomes at risk of suffering      "congestion collapse" [RFC896].   Relevant RFCs      RFC 1122 requires use of slow start after loss.  RFC 2001 gives      the specifics of how to implement slow start.  RFC 896 describes      congestion collapse.      The retransmission timeout discussed here should not be confused      with the separate "fast recovery" retransmission mechanism      discussed in RFC 2001.   Trace file demonstrating it      Made using tcpdump recording at the sending TCP (A).  No losses      reported by the packet filter.Paxson, et. al.              Informational                      [Page 6]RFC 2525              TCP Implementation Problems             March 1999   10:40:59.090612 B > A: . ack 357125 win 33580 (DF) [tos 0x8]   10:40:59.222025 A > B: . 357125:358585(1460) ack 1 win 32768   10:40:59.868871 A > B: . 357125:358585(1460) ack 1 win 32768   10:41:00.016641 B > A: . ack 364425 win 33580 (DF) [tos 0x8]   10:41:00.036709 A > B: . 364425:365885(1460) ack 1 win 32768   10:41:00.045231 A > B: . 365885:367345(1460) ack 1 win 32768   10:41:00.053785 A > B: . 367345:368805(1460) ack 1 win 32768   10:41:00.062426 A > B: . 368805:370265(1460) ack 1 win 32768   10:41:00.071074 A > B: . 370265:371725(1460) ack 1 win 32768   10:41:00.079794 A > B: . 371725:373185(1460) ack 1 win 32768   10:41:00.089304 A > B: . 373185:374645(1460) ack 1 win 32768   10:41:00.097738 A > B: . 374645:376105(1460) ack 1 win 32768   10:41:00.106409 A > B: . 376105:377565(1460) ack 1 win 32768   10:41:00.115024 A > B: . 377565:379025(1460) ack 1 win 32768   10:41:00.123576 A > B: . 379025:380485(1460) ack 1 win 32768   10:41:00.132016 A > B: . 380485:381945(1460) ack 1 win 32768   10:41:00.141635 A > B: . 381945:383405(1460) ack 1 win 32768   10:41:00.150094 A > B: . 383405:384865(1460) ack 1 win 32768   10:41:00.158552 A > B: . 384865:386325(1460) ack 1 win 32768   10:41:00.167053 A > B: . 386325:387785(1460) ack 1 win 32768   10:41:00.175518 A > B: . 387785:389245(1460) ack 1 win 32768   10:41:00.210835 A > B: . 389245:390705(1460) ack 1 win 32768   10:41:00.226108 A > B: . 390705:392165(1460) ack 1 win 32768   10:41:00.241524 B > A: . ack 389245 win 8760 (DF) [tos 0x8]      The first packet indicates the ack point is 357125.  130 msec      after receiving the ACK, A transmits the packet after the ACK      point, 357125:358585.  640 msec after this transmission, it      retransmits 357125:358585, in an apparent retransmission timeout.      At this point, A's cwnd should be one MSS, or 1460 bytes, as A      enters slow start.  The trace is consistent with this possibility.      B replies with an ACK of 364425, indicating that A has filled a      sequence hole.  At this point, A's cwnd should be 1460*2 = 2920      bytes, since in slow start receiving an ACK advances cwnd by MSS.      However, A then launches 19 consecutive packets, which is      inconsistent with slow start.      A second trace confirmed that the problem is repeatable.   Trace file demonstrating correct behavior      Made using tcpdump recording at the sending TCP (C).  No losses      reported by the packet filter.   12:35:48.442538 C > D: P 465409:465921(512) ack 1 win 4608   12:35:48.544483 D > C: . ack 461825 win 4096   12:35:48.703496 D > C: . ack 461825 win 4096   12:35:49.044613 C > D: . 461825:462337(512) ack 1 win 4608Paxson, et. al.              Informational                      [Page 7]RFC 2525              TCP Implementation Problems             March 1999   12:35:49.192282 D > C: . ack 465921 win 2048   12:35:49.192538 D > C: . ack 465921 win 4096   12:35:49.193392 C > D: P 465921:466433(512) ack 1 win 4608   12:35:49.194726 C > D: P 466433:466945(512) ack 1 win 4608   12:35:49.350665 D > C: . ack 466945 win 4096   12:35:49.351694 C > D: . 466945:467457(512) ack 1 win 4608   12:35:49.352168 C > D: . 467457:467969(512) ack 1 win 4608   12:35:49.352643 C > D: . 467969:468481(512) ack 1 win 4608   12:35:49.506000 D > C: . ack 467969 win 3584      After C transmits the first packet shown to D, it takes no action      in response to D's ACKs for 461825, because the first packet      already reached the advertised window limit of 4096 bytes above      461825.  600 msec after transmitting the first packet, C      retransmits 461825:462337, presumably due to a timeout.  Its      congestion window is now MSS (512 bytes).      D acks 465921, indicating that C's retransmission filled a      sequence hole.  This ACK advances C's cwnd from 512 to 1024.  Very      shortly after, D acks 465921 again in order to update the offered      window from 2048 to 4096.  This ACK does not advance cwnd since it      is not for new data.  Very shortly after, C responds to the newly      enlarged window by transmitting two packets.  D acks both,      advancing cwnd from 1024 to 1536.  C in turn transmits three      packets.   References      This problem is documented in [Paxson97].   How to detect      Packet loss is common enough in the Internet that generally it is      not difficult to find an Internet path that will force      retransmission due to packet loss.      If the effective window prior to loss is large enough, however,      then the TCP may retransmit using the "fast recovery" mechanism      described in RFC 2001.  In a packet trace, the signature of fast      recovery is that the packet retransmission occurs in response to      the receipt of three duplicate ACKs, and subsequent duplicate ACKs      may lead to the transmission of new data, above both the ack point      and the highest sequence transmitted so far.  An absence of three      duplicate ACKs prior to retransmission suffices to distinguish      between timeout and fast recovery retransmissions.  In the face of      only observing fast recovery retransmissions, generally it is not      difficult to repeat the data transfer until observing a timeout      retransmission.Paxson, et. al.              Informational                      [Page 8]RFC 2525              TCP Implementation Problems             March 1999      Once armed with a trace exhibiting a timeout retransmission,      determining whether the TCP follows slow start is done by      computing the correct progression of cwnd and comparing it to the      amount of data transmitted by the TCP subsequent to the timeout      retransmission.   How to fix      If the root problem is that the implementation lacks a notion of a      congestion window, then unfortunately this requires significant      work to fix.  However, doing so is critical, for reasons outlined      above.2.3.   Name of Problem      Uninitialized CWND   Classification      Congestion control   Description      As described above for "No initial slow start", when a TCP      connection begins cwnd is initialized to one segment (or perhaps a      few segments, if experimenting with [RFC2414]).  One particular      form of "No initial slow start", worth separate mention as the bug      is fairly widely deployed, is "Uninitialized CWND".  That is,      while the TCP implements the proper slow start mechanism, it fails      to initialize cwnd properly, so slow start in fact fails to occur.      One way the bug can occur is if, during the connection      establishment handshake, the SYN ACK packet arrives without an MSS      option.  The faulty implementation uses receipt of the MSS option      to initialize cwnd to one segment; if the option fails to arrive,      then cwnd is instead initialized to a very large value.   Significance      In congested environments, detrimental to the performance of other      connections, and likely to the connection itself.  The burst can      be so large (see below) that it has deleterious effects even in      uncongested environments.   Implications      A TCP exhibiting this behavior is stressing the network with a      large burst of packets, which can cause loss in the network.   Relevant RFCs      RFC 1122 requires use of slow start.  RFC 2001 gives the specifics      of slow start.Paxson, et. al.              Informational                      [Page 9]RFC 2525              TCP Implementation Problems             March 1999   Trace file demonstrating it      This trace was made using tcpdump running on host A.  Host A is      the sender and host B is the receiver.  The advertised window and      timestamp options have been omitted for clarity, except for the      first segment sent by host A.  Note that A sends an MSS option in      its initial SYN but B does not include one in its reply.   16:56:02.226937 A > B: S 237585307:237585307(0) win 8192         <mss 536,nop,wscale 0,nop,nop,timestamp[|tcp]>   16:56:02.557135 B > A: S 1617216000:1617216000(0)         ack 237585308 win 16384   16:56:02.557788 A > B: . ack 1 win 8192   16:56:02.566014 A > B: . 1:537(536) ack 1   16:56:02.566557 A > B: . 537:1073(536) ack 1   16:56:02.567120 A > B: . 1073:1609(536) ack 1   16:56:02.567662 A > B: P 1609:2049(440) ack 1   16:56:02.568349 A > B: . 2049:2585(536) ack 1   16:56:02.568909 A > B: . 2585:3121(536) ack 1      [54 additional burst segments deleted for brevity]   16:56:02.936638 A > B: . 32065:32601(536) ack 1   16:56:03.018685 B > A: . ack 1      After the three-way handshake, host A bursts 61 segments into the      network, before duplicate ACKs on the first segment cause a      retransmission to occur.  Since host A did not wait for the ACK on      the first segment before sending additional segments, it is      exhibiting "Uninitialized CWND"   Trace file demonstrating correct behavior      See the example for "No initial slow start".   References      This problem is documented in [Paxson97].   How to detect      This problem can be detected by examining a packet trace recorded      at either the sender or the receiver.  However, the bug can be      difficult to induce because it requires finding a remote TCP peer      that does not send an MSS option in its SYN ACK.   How to fix      This problem can be fixed by ensuring that cwnd is initialized      upon receipt of a SYN ACK, even if the SYN ACK does not contain an      MSS option.Paxson, et. al.              Informational                     [Page 10]RFC 2525              TCP Implementation Problems             March 19992.4.   Name of Problem      Inconsistent retransmission   Classification      Reliability   Description      If, for a given sequence number, a sending TCP retransmits      different data than previously sent for that sequence number, then      a strong possibility arises that the receiving TCP will      reconstruct a different byte stream than that sent by the sending      application, depending on which instance of the sequence number it      accepts.      Such a sending TCP exhibits "Inconsistent retransmission".   Significance      Critical for all environments.   Implications      Reliable delivery of data is a fundamental property of TCP.   Relevant RFCs      RFC 793, section 1.5, discusses the central role of reliability in      TCP operation.   Trace file demonstrating it      Made using tcpdump recording at the receiving TCP (B).  No losses      reported by the packet filter.   12:35:53.145503 A > B: FP 90048435:90048461(26)                             ack 393464682 win 4096                                        4500 0042 9644 0000                    3006 e4c2 86b1 0401 83f3 010a b2a4 0015                    055e 07b3 1773 cb6a 5019 1000 68a9 0000   data starts here>504f 5254 2031 3334 2c31 3737*2c34 2c31                    2c31 3738 2c31 3635 0d0a   12:35:53.146479 B > A: R 393464682:393464682(0) win 8192   12:35:53.851714 A > B: FP 90048429:90048463(34)                          ack 393464682 win 4096                                        4500 004a 965b 0000                    3006 e4a3 86b1 0401 83f3 010a b2a4 0015                    055e 07ad 1773 cb6a 5019 1000 8bd3 0000   data starts here>5041 5356 0d0a 504f 5254 2031 3334 2c31                    3737*2c31 3035 2c31 3431 2c34 2c31 3539                    0d0aPaxson, et. al.              Informational                     [Page 11]RFC 2525              TCP Implementation Problems             March 1999      The sequence numbers shown in this trace are absolute and not      adjusted to reflect the ISN.  The 4-digit hex values show a dump      of the packet's IP and TCP headers, as well as payload.  A first      sends to B data for 90048435:90048461.  The corresponding data      begins with hex words 504f, 5254, etc.      B responds with a RST.  Since the recording location was local to      B, it is unknown whether A received the RST.      A then sends 90048429:90048463, which includes six sequence

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