📄 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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