This is wikipedia's explanation of the length field of the UDP header:
Length
A field that specifies the length in bytes of the UDP header
and UDP data. The minimum length is 8 bytes because that is the length
of the header. The field size sets a theoretical limit of 65,535 bytes
(8 byte header + 65,527 bytes of data) for a UDP datagram. The
practical limit for the data length which is imposed by the underlying
IPv4 protocol is 65,507 bytes (65,535 − 8 byte UDP header − 20 byte IP
header).
The practical limit for the data length should minus 20 byte IP header, why is that?
Take a good look at the explanation of the IP header at this link :
https://www.ietf.org/rfc/rfc791.txt
I quote :
Total Length: 16 bits
Total Length is the length of the datagram, measured in octets, including internet header and data. This field allows the length of a datagram to be up to 65,535 octets. Such long datagrams are impractical for most hosts and networks. All hosts must be prepared to accept datagrams of up to 576 octets (whether they arrive whole or in fragments). It is recommended that hosts only send datagrams larger than 576 octets if they have assurance that the destination is prepared to accept the larger datagrams.
The number 576 is selected to allow a reasonable sized data block to be transmitted in addition to the required header information. For example, this size allows a data block of 512 octets plus 64 header octets to fit in a datagram. The maximal internet header is 60 octets, and a typical internet header is 20 octets, allowing a margin for headers of higher level protocols.
So the maximum total length is 65535 but this includes the IP header itself.
Therefore you have an IP payload that can be 65535 - 20 = 65515.
But the payload of IP in your case is UDP and UDP has a header of its own which is 8 bytes. Hence you get to the theoretical limit of the payload of a UDP packet : 65,535 − 8 byte UDP header − 20 byte IP header
Note the use of theoretical instead of practical. The practical limit of a UDP packet takes into account the probability of fragmentation and thus considers the mtu of the network layer. The link above also has an interesting sentence containing the value 576. 576 - 20 - 8 = 548 which is not quite 534 but getting close. This might explain this practical limit.
Because UDP packets are encapsulated in IP packets which headers are 20 bytes. You can't send UDP packets without a encapsulated IP packet. Usually the actual limit is way less and it depends on the MTU of the routers between the two endpoints transmitting the UDP packet.
Because the IP header has to be (a) sent and (b) counted in the 16-bit length word. See RFC 791 #3.1.
However the real practical limit is generally accepted to be 534 bytes, to avoid fragmentation at the IP layer, which increases the risk of datagram loss.
Related
I have a question about the UDP protocol.
I want to stream data from a particle sensor (Xilinx FPGA Dev Board) to a raspberry pi (or Windows 10 laptop) via UDP byte socket as binary data. I don't care if some packets are lost, because there are many particles coming after anyways... But if a packet gets lost, all of the particles information should get lost.
The connection is a short lan cable over the 1 Gbit/s Ethernet port.
The "minimum" amount of data is 192 Bit (24 Byte) per particle (16 x 12 Bit value) and the maximum amount of particles is 3300 per second.
So I have to transfer max. 192 x 3300 = 79200 Byte/s plus Header etc.
Maximum packet size of UDP 65.507 Byte
As I understand the packet size has to be devidable by 24 Byte in my application.
Which leaves me with a packet size range of 24 to 65.496 Byte.
But if the concentration is lower I don't want to wait minutes until a packet is filled and ready to be send.
What would you guys suggust in regard of repitition rate and size?
E.g. a 1008 Byte packet has to be send about every 13 ms at max. particle concentration.
best regards
I just tested a UDP socket with 1024 Byte Buffer, which works nice with strings or Bitvectors for single test packets.
If the datagram bigger than MTU， 2 layer switch will drop it? Dose 2 layer switch can report a ICMP? If not report ICMP, how can I determine the data size to pass the switch successfully?
If the datagram bigger than MTU， 2 layer switch will drop it?
Yes. A switch does not forward frames larger than the (configured) maximum size and drops them. For standard Ethernet, that's 1500 bytes payload plus 18 bytes L2 overhead. Note that MTU is an L3 term referring to the maximum packet size that an underlying network can transport.
Does 2 layer switch report a ICMP?
No. A layer-2 switch generally sends no ICMP messages nor is there an ICMP message to report oversized frames in L2.
A layer-3 switch used as gateway should return an ICMP Fragmentation required when the destination network's MTU does not admit the IP packet without fragmentation and its DF bit is set or IPv6 is used. For IPv4 without DF, the gateway just fragments the packet.
If not report ICMP, how can I determine the data size to pass the switch successfully?
On an unmanaged switch, see above for the maximum standard size. A few support jumbo frames, check their documentation. On some managed switches you can configure the maximum frame size globally or by VLAN. Methods and syntax vary.
I’ve been searching for a while in order to find out the typical/minimum size of payload analysis while configuring security devices such as IDS/IPS.
I know it is possible to configure within SNORT rules both Offset and Depth parameters. Sort speech: a SNORT rule configured with a 1 byte Offset and 7 bytes depth will analyze incoming packets from 1-7 bytes of payload + Header size.
I know depth parameter value varies from 1-65535 but i’d like to know what is the minimum size of bytes for an accurate traffic analysis.
As an example, if I receive a meterpreter payload which is roughly 700 bytes size, do I need to scan the whole packet in order to raise an alert?.
Thanks in advance.
I have a Linux UDP Server written in C and I am sending a UDP datagram of 16 bytes. All the data is received correctly by the client, but the wireshark log is showing that two extra bytes are being added:
00 16 00 00 c8 44 01 14 01 01 02 01 02 00 00 10 00 00
The two bytes are all zeros, I'm not sure where they are coming from, I have the data of 16 bytes sent in my sendto() function.
These must be added on from padding at the Linux Kernel layer? Is there anyway to stop this from happening?
Screen shot of wireshark, apparently this is Ethernet padding bytes... Why are they here?
the Ethernet frames need to be minimum of 64 octets, including the 4 octets of Frame Check Sequence (FCS) that is not shown in the screenshot. Without those 2 zero padding octets, that frame would be only 62 octets long. Also notice that they're not in the UDP packet, they're outside. The length of the UDP frame on the outside is given as 24, this includes 16 octets of the data that you sent; the remaining 8 octets are 4 16-bit numbers: source port, destination port, length and checksum. The Ethernet padding bytes are outside both the UDP and the IP framing.
The reason for the minimum 64-octet frame length is old - Ethernet was originally designed for bus topology, where media access was regulated with a method called carrier-sense multiple access with collision detection (CSMA/CD). Any device may transmit on the bus at any time when the bus is free, but they must monitor the bus for simultaneous transmissions. If a collision occurs, then each currently sending party backs off for a random delay. The 64-byte minimum was set so that the sending device would notice a collision on a cable run that is hundreds of metres long before it has finished sending the frame.
From Wireshark documentation via Google search for "wireshark minimum ethernet frame":
Ethernet packets with less than the minimum 64 bytes for an Ethernet packet (header + user data + FCS) are padded to 64 bytes, which means that if there's less than 64-(14+4) = 46 bytes of user data, extra padding data is added to the packet.
Beware: the minimum Ethernet packet size is commonly mentioned at 64 bytes, which is including the FCS. This can be confusing as the FCS is often not shown by Wireshark, simply because the underlying mechanisms simply don't supply it. [...]
By the way this is in no way related to the C programming language.
Does UDP packet loss percentage might increase considering packet size? For example if I send 100'000 packets, in first try byte[] size is 30, but second 300. Could packet size play role in it's drop ability or packet loss percentage is not its size dependent?
The packet loss is depending on the size of the packet. This has several reasons.
IP packets can go up to 64k approximately, but they are fragmented up to the MTU of ethernet and if one of those packets gets lost , the whole IP packet is dropped. For larger packets if the traffic is high the probability is higher that the larger packet will be dropped. The MTU is around 1500 bytes.
There is more to it than just that. Internally a protocol stack is implemented using internal buffers that are a lot smaller than the mtu, this can vary from 300 bytes and larger. But the point is that these buffers are also a limited resource. If the network device runs out of buffers, then the packet will be dropped as well.
If you don't know the MTU on the network in question according to the link below a 512-byte UDP payload is considered reasonable to allow a margin for other header information that you may not have anticipated.
What is the largest Safe UDP Packet Size on the Internet
Because you're sending larger packets, yes it could increase the chances that packets are dropped.
Now if you compare sending 100000 packets of 30 bytes or 10000 packets of 300 bytes, even though the user data is the same the total size of the packets is larger due to the headers.