Assuming:
TCP Window Size of client: 2000 bytes
MSS (maximum segment size) of client: 1500 bytes
Now server want to send 3000 bytes to client B1..B3000. Server sends:
Packet-1: Contains B1..B1500 bytes, SeqNo: 1, Length: 1500
Packet-2: Contains B501..B2000 bytes, SeqNo: 501, Length: 1500
So in effect:
Data B1....B501....B1500....B2000....B3000
Packet-1 B1....B501....B1500
Packet-2 B501....B1500....B2000
Is it valid and allowed by TCP protocol?
Related
I've noticed in wireshark that I'm able to send 4096 bytes of data to a HTTP webserver (from uploading a file) however the server only seems to be acknowledging data 1460 bytes at a time. Why is this the case?
The size of TCP segments is restricted to the MSS (Maximum Segment Size), which is basically the MTU (Maximum Transmission Unit) less the bytes comprising the IP and TCP overhead. On a typical Ethernet link, the MTU is 1500 bytes and basic IP and TCP headers comprise 20 bytes each, so the MSS is 1460 (1500 - 20 - 20).
If you're seeing packets indicated with a length field of 4096 bytes, then it almost certainly means that you're capturing on the transmitting host and Wireshark is being handed the large packet before it's segmented into 1460 byte chunks. If you were to capture at the receiving side, you would see the individual 1460 byte segments arriving and not a single, large 4096 byte packet.
For further reading, I would encourage you to read Jasper Bongertz's blog titled, "The drawbacks of local packet captures".
TCP by default uses path MTU discovery:
When system send packet to the network it set don't fragment flag (DF) in IP header
When IP router or you local machine see DF packet that should be fragmented to match MTU of the next hop link it sends feedback (RTCP fragmentation need) that contains new MTU
When system receives fragmentation needed ICMP it adjusts MSS and send data again.
This procedure is performed to reduce overall load on the network and increase probability of each packet delivery.
This is why you see 1460 packets.
Regarding to you question: the server only seems to be acknowledging data 1460 bytes at a time. Why is this the case?
TCP keep track window that defines "how many bytes of data you can send without acknowledge". Its purpose is to provide flow control mechanisms (sender can't send too much data that can't be processed) and congestion control mechanisms (sender can't send too much data to overload network). Window is defined by receiver side and may be increased during connection when TCP will estimate real channel bandwidth. So you may see one ACK that acknowledges several packets.
I am building an app where my phone frequently send data to my server. Since I would be using my mobile data, I was wondering how much data it cost to set up (and tear down?) a TCP connection to my server.
TCP Three-way handshake
Device 1 sends its TCP sequence number and maximum segment size to Device 2.
Device 2 responds by sending its sequence number and maximum segment size to Device 1.
Device 1 acknowledges receipt of the sequence number and segment size information.
Each packet is composed of an IP header and data (payload). In this case, the data section contains TCP. The TCP header contains various fields including the source and destination ports, sequence and acknowledgment numbers, window size, TCP flags, urgent pointer, and reserved bits.
Like the IP header, the TCP header may also contain options. (Note that TCP options and IP options are two different things.) Because the TCP options change the length of the TCP header, the length is set in the header.
IPv4 header is five 4-byte chunks, or 20 bytes total.
TCP typically usually uses 24 bytes of the header for handshake (first two packets) and about 20 for normal packet transmission.
Maximum Segment Size (MSS): 4 bytes
Window Scale (WSCALE): 3 bytes
Timestamp (TS): 10 bytes
No Operation (NOP): 1 byte
Selective Acknowledgment Permitted (SackOK): 2 bytes
Selective Acknowledgment Data: 10 bytes (plus 8 bytes for each additional pair of sequence numbers)
Terminating a Connection
Even though establishing a connection using 3-way handshake requires only 3 packets to be transmitted, tearing down one requires 4!
In the first frame the client sends a FIN that is accompanied by an ACK. The FIN parameter is set, it will inform the server that it has no more data to send.
The response (2nd frame) would be simply the server acknowledging the FIN sent from the client.
Even though TCP has established connections between the two computers, the connections are still independent of one another. Therefore, the server will also transmit a FIN to the client.
You guessed it right ... the client would ACK the FIN of the server in the last forth packet.
The offset of each of the frames is typically 20 bytes.
To sum it up.
Establishing a connection: ~ 128-136 bytes
Tearing down a connection: ~ 160 bytes
If you plan to use TLS / SSL handshake, this is estimated to be between 4.5k-6.5k.
Note: Please also take a look at TCP/IP Header Compression
Sources:
Inside the TCP Handshake
Explanation of the Three-Way Handshake via TCP/IP
Studying Normal Traffic, Part Three: TCP Headers | Symantec Connect
When send a UDP datagram larger than the MTU size, only the last fragment of the UDP datagram is putted out to the destination. The rest of fragments are silently discarded. Sendto() return the sendlength (no error).
Envirenment:
OS: vxWorks6.8
CPU: ARM9
MTU: 1500
The vxWorks board is direct connected via ethernet cable to the Win7 PC (with wireshark).
The DF flag isn't set.
Example 1:
sendto with 1400 Bytes
wireshark shows: udp with 1400 Databytes4
Example 2:
sendto with 1800 Bytes
wireshark shows: Fragmented IP protocol (proto=udp 0x11, off=1496, id0a00) -> Data (312 bytes)
Example 3:
sendto with 4000 Bytes
wireshark shows: Fragmented IP protocol (proto=udp 0x11, off=1496, id0a00) -> Data (1016 bytes)
Example 4:
sendto with 7800 Bytes
wireshark shows: Fragmented IP protocol (proto=udp 0x11, off=1496, id0a00) -> Data (328 bytes)
I use the same test on a second vxworks board with CPU PPC and this work fine and the ip-fragmentation work properly.
Why the first fregments of the udp packet are always discarded?
Many Thanks
I have a lot of pcap files
in each pcap file there are some tcp packets
I want to the following steps:
1 filter the pcap file to get expected tcp packets(if with tshark, the filter is like tcp.srcport==80 and tcp.dstport==20000). The tcp data in each tcp packet are of tens of bytes,
1) the first 2 bytes are sequence number
2) the second 2 bytes indicate a length: **n**
3) the following **n** bytes are random (if **n** is 2, then there are 2 random bytes)
4) immediately after the **n** bytes is an IP address
2 from each tcp packet, I want to get the source IP and the IP address string as mentioned in 4)
Above all, I just want to get an IP pair from each tcp packet, the source IP is easy to be obtained from tshark, but the IP address in the tcp data needs some effort
are there any good source code snippets or good ways that can be used for my target?
thanks!
Can someone please explain why maximum legal size of ICMP echo packet is calculated as follows:
65535 - 20 - 8 = 65507
Thanks.
65535 bytes is the maximum allowed size of a IPv4 network packet, while 20 and 8 are the sizes of the IP and ICMP headers, which leaves up to 65507 bytes for the ICMP data.