tcp data corruption

Question

i have a problem with W5100 (192.168.10.100). i uploaded wireshark capture file (w5100extract.pcapng). capture file’s content is “TCP Dup ACK → TCP Fast Retransmission → TCP ZeroWindow → TCP Out-Of-Order → Data Length is wrong”… what’s wrong with this?

Eugeny · Answer

I see no attachments. Please try again.

oxff · Answer

i can’t attach file, so copy & paste packets 1 2018-01-19 19:33:38.700737 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=1 Ack=1 Win=2048 Len=1460 2 2018-01-19 19:33:38.700737 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=1461 Ack=1 Win=2048 Len=588 3 2018-01-19 19:33:38.701783 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=2049 Ack=1 Win=2048 Len=1460 4 2018-01-19 19:33:38.701784 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=3509 Ack=1 Win=2048 Len=588 5 2018-01-19 19:33:38.704211 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=4097 Ack=1 Win=2048 Len=1460 6 2018-01-19 19:33:38.704212 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=5557 Ack=1 Win=2048 Len=588 7 2018-01-19 19:33:38.705582 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=6145 Ack=1 Win=2048 Len=1460 8 2018-01-19 19:33:38.705583 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=7605 Ack=1 Win=2048 Len=588 9 2018-01-19 19:33:38.706630 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=8193 Ack=1 Win=2048 Len=1460 10 2018-01-19 19:33:38.706631 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=9653 Ack=1 Win=2048 Len=588 11 2018-01-19 19:33:38.707732 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=10241 Ack=1 Win=2048 Len=1460 12 2018-01-19 19:33:38.707733 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=11701 Ack=1 Win=2048 Len=588 13 2018-01-19 19:33:38.708325 192.168.10.200 192.168.10.100 TCP 60 47691 → 2111 [ACK] Seq=1 Ack=10241 Win=13140 Len=0 14 2018-01-19 19:33:38.708326 192.168.10.100 192.168.10.200 TCP 1514 2111 → 47691 [PSH, ACK] Seq=12289 Ack=1 Win=2048 Len=1460 15 2018-01-19 19:33:38.708327 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=13749 Ack=1 Win=2048 Len=588 16 2018-01-19 19:33:38.709015 192.168.10.200 192.168.10.100 TCP 60 [TCP Dup ACK 13#1] 47691 →…

oxff · Answer

oxff: TCP Dup ACK -> TCP Fast Retransmission -> TCP ZeroWindow -> TCP Out-Of-Order is this situation normal? <TCP Dup ACK → TCP Fast Retransmission → TCP ZeroWindow → TCP Out-Of-Order> why it happened [zerowindow, out of order]? After this packet, my file transfer is strange. File length received from W5100 is different compared to original file size.

becky · Answer

Hello, I saw the your packet file. In the communication, if the reception rate can not keep up with the transmission rate, there may be Dup ACK packet. However, the ZeroWindow packet is a packet that appears because the client’s window is not released. Make sure the client is receiving properly.

oxff · Answer

thank you for your reply, i’ll check.

Eugeny · Answer

oxff: After this packet, my file transfer is strange. File length received from W5100 is different compared to original file size. What is the TX buffer size for the respective W5100 socket you use? These entries 19 2018-01-19 19:33:38.709729 192.168.10.100 192.168.10.200 TCP 642 2111 → 47691 [PSH, ACK] Seq=15797 Ack=1 Win=2048 Len=588 20 2018-01-19 19:33:38.709732 192.168.10.200 192.168.10.100 TCP 60 [TCP Dup ACK 13#3] 47691 → 2111 [ACK] Seq=1 Ack=10241 Win=13140 Len=0 21 2018-01-19 19:33:38.710831 192.168.10.100 192.168.10.200 TCP 1514 [TCP Fast Retransmission] 2111 → 47691 [PSH, ACK] Seq=10241 Ack=1 Win=2048 Len=1460 make me think that you allocated at least 8 KB for TX buffer. If you would have default 2KB allocated for TX buffer, then W5100 would not be able to re-transmit packet with the position 10241 as its current pointer is at 15797 which is 5KB ahead.