My application connects to the server using TCP and system is working fine. Using Wireshark I noticed that some TCP packets of 60 bytes are sent with no data. Is this normal?
as part of TCP transmissions and handshake are there some packets sent with no data?
Thanks
Kumar
There are the ACK packets that carry no data and only acknowledge received data. When using Wireshark it should display these "flags" on the empty packets.
To be more accurate you should show a screenshot of the wireshark capture, so we know what the size you mean is.
Meanwhile I dissected an ACK packet of IPv4 TCP traffic here and these are my results:
Protocol Size Description
Ethernet protocol 14 byte 2 MAC adresses (6 byte), Protocoll type (2 byte)
Internet protocol 20 byte The default header size
TC protocol 32 byte The default header size (20 byte) + options (12 byte)
_____________________________________________________________________________________
Total 66 byte
without EP 52 byte (Probably the size the OP is talking about)
without EP, Opts 40 byte (The size talked about in the comments)
The options in my case were 2 NOPs (each 1 byte) and two timestamps (each 5 byte??). The order of the protocols is the same as in Wireshark.
Wireshark splits the information down to each field, so you can see what takes up how much space.
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
I've encountered the following statement somewhere in the net: "Although
theoretically only 8 bytes of L4 info may be guaranteed in a fragment,
assume complete L4 info is available...". I don't understand how possible
that only 8 bytes of a transport are guaranteed in a fragment, as the IP
fragment can't be less than 46 bytes (minimum payload size for Ethernet
frame), and this includes 20 bytes of IP header and 20bytes of TCP header
(not considering variable-lngth options), UDP will be less.
Thus for the first IP fragment we always can expect IP header at tcp header,
while other fragments will carry only IP header + payload.
I believe I'm missing something, but I still can't understand why only 8
bytes can be guaranteed in a fragment? I would appreaciate if someone helps
to clarify this issue. Thanks !
Mark
Imagine a router receives a TCP packet containing one more byte of data than will fit in the MTU of the destination network. It must split on an 8-byte boundary because that's an IP fragmentation rule.
It won't split into more than two fragments because that would be silly. So it must include at least one byte of data in the first fragment..
Thus the minimum number of data bytes you can place in the first fragment of an IP datagram is 8.
I am looking at an application which transmits precisely 4380 bytes over HTTP, including headers, before the connection stalls. 4380 happens to be exactly 3 times the maximum segment size of a 1500-byte Ethernet frame (1500 bytes minus 40 bytes of TCP/IP headers times 3).
The connection appears to remain open; it's just that no data is being received.
Why would it stall after precisely 3 ethernet frames? Where should I start looking for a reason?
How do you get the maximum number of bytes that can be passed to a sendto(..) call for a socket opened as a UDP port?
Use getsockopt(). This site has a good breakdown of the usage and options you can retrieve.
In Windows, you can do:
int optlen = sizeof(int);
int optval;
getsockopt(socket, SOL_SOCKET, SO_MAX_MSG_SIZE, (int *)&optval, &optlen);
For Linux, according to the UDP man page, the kernel will use MTU discovery (it will check what the maximum UDP packet size is between here and the destination, and pick that), or if MTU discovery is off, it'll set the maximum size to the interface MTU and fragment anything larger. If you're sending over Ethernet, the typical MTU is 1500 bytes.
On Mac OS X there are different values for sending (SO_SNDBUF) and receiving (SO_RCVBUF).
This is the size of the send buffer (man getsockopt):
getsockopt(sock, SOL_SOCKET, SO_SNDBUF, (int *)&optval, &optlen);
Trying to send a bigger message (on Leopard 9216 octets on UDP sent via the local loopback) will result in "Message too long / EMSGSIZE".
As UDP is not connection oriented there's no way to indicate that two packets belong together. As a result you're limited by the maximum size of a single IP packet (65535). The data you can send is somewhat less that that, because the IP packet size also includes the IP header (usually 20 bytes) and the UDP header (8 bytes).
Note that this IP packet can be fragmented to fit in smaller packets (eg. ~1500 bytes for ethernet).
I'm not aware of any OS restricting this further.
Bonus
SO_MAX_MSG_SIZE of UDP packet
IPv4: 65,507 bytes
IPv6: 65,527 bytes