I'm configuring several network interface and I noticed that some software clients declare jumbo frame at 9000 other at 9216. I'm setting everything at 9216 on all the interfaces.
Everything seems working fine.
However, what kind of networking issue I should expect as issue if MTU was misconfigurated on some appliance?
MTU != max frame size
The MTU is the maximum payload a frame can carry (ie. the maximum IP packet size) while the maximum frame size is the maximum size of the total frame.
If you add the L2 overhead to the MTU you get the maximum frame size - 18 bytes for untagged frames, 22 bytes for 802.1q tagged frames.
Some vendors use dubious terms like "L2 MTU" or don't count the FCS field. Only when you can't clear this up, you should set all NICs to the exact same size and the switches to 22 bytes larger.
Related
MTU (Maximum transmission unit) is the maximum frame size that can be transported.
When we talk about MTU, it's generally a cap at the hardware level and is for the lower level layers - DataLink and Physical layer.
Now, considering the OSI layer, it does not matter how efficient are the upper layers or what kind of magic-sauce they are applying, data-link layer will always construct frames of size < 1500 bytes (or whatever is the MTU) and anything in the "internet" will always be transmitted at that frame size.
Does the internet's transmission rate really capped at 1500 bytes. Now-a-days, we see speeds in 10-100 Mbps and even Gbps. I wonder for such speeds, does the frames still get transmitted at 1500 bytes, which would mean lots and lots and lots of fragmentation and re-assembly at the receiver. At this scale, how does the upper layer achieve efficiency ?!
[EDIT]
Based on below comments, I re-frame my question:
If data-layer transmits at 1500 byte frames, I want to know how is upper layer at the receiver able to handle such huge incoming data-frames.
For ex: If internet speed in 100 Mbps, upper layers will have to process 104857600 bytes/second or 104857600/1500 = 69905 frames/second. Network layer also need to re-assemble these frames. How network layer is able to handle at such scale.
If data-layer transmits at 1500 byte frames, I want to know how is
upper layer at the receiver able to handle such huge incoming
data-frames.
1500 octets is a reasonable MTU (Maximum Transmission Unit), which is the size of the data-link protocol payload. Remember that not all frames are that size, it is just the maximum size of the frame payload. There are many, many things with much smaller payloads. For example, VoIP has very small payloads, often smaller than the overhead of the various protocols.
Frames and packets get lost or dropped all the time, often on purpose (see RED, Random Early Detection). The larger the data unit, the more data that is lost when a frame or packet is lost, and with reliable protocols, such as TCP, the more data must be resent.
Also, having a reasonable limit on frame or packet size keeps one host from monopolizing the network. Hosts must take turns.
For ex: If internet speed in 100 Mbps, upper layers will have to
process 104857600 bytes/second or 104857600/1500 = 69905
frames/second. Network layer also need to re-assemble these frames.
How network layer is able to handle at such scale.
Your statement has several problems.
First, 100 Mbps is 12,500,000 bytes per second. To calculate the number of frames per second, you must take into account the data-link overhead. For ethernet, you have 7 octet Preabmle, a 1 octet SoF, a 14 octet frame header, the payload (46 to 1500 octets), a four octet CRC, then a 12 octet Inter-Packet Gap. The ethernet overhead is 38 octets, not counting the payload. To now how many frames per second, you would need to know the payload size of each frame, but you seem to wrongly assume every frame payload is the maximum 1500 octets, and that is not true. You get just over 8,000 frames per second for the maximum frame size.
Next, the network layer does not reassemble frame payloads. The payload of the frame is one network-layer packet. The payload of the network packet is the transport-layer data unit (TCP segment, UDP datagram, etc.). The payload of the transport protocol is application data (remember that the OSI model is just a model, and OSes do not implement separate session and presentation layers; only the application layer). The payload of the transport protocol is presented to the application process, and it may be application data or an application-layer protocol, e.g. HTTP.
The bandwidth, 100 Mbps in your example, is how fast a host can serialize the bits onto the wire. That is a function of the NIC hardware and the physical/data-link protocol it uses.
which would mean lots and lots and lots of fragmentation and
re-assembly at the receiver.
Packet fragmentation is basically obsolete. It is still part of IPv4, but fragmentation in the path has been eliminated in IPv6, and smart businesses, do not allow IPv4 packet fragments due to fragmentation attacks. IPv4 packets may be fragmented if the DF bit is not set in the packet header, and the MTU in the path shrinks smaller than the original MTU. For example, a tunnel will have a smaller MTU because of the tunnel overhead. If the DF bit is set, then a packet too large for the MTU on the next link, the packet is dropped. Packet fragmentation is very resource intensive on a router, and there is a set of steps that must be performed to fragment a packet.
You may be confusing IPv4 packet fragmentation and reassembly with TCP segmentation, which is something completely different.
When we sent packets from one router to another router on the network layer and the packet size is greater than the MTU (maximum transferable unit) of the router, we have to fragment the packet. My questions is: suppose we need to add padding bits in last fragment, then where do we add padding bits (in the LSB or MSB) and how does the destination router differentiate between packet bits or padding bits?
I want you to consider the following things before:
Limit on the maximum size of IP data-gram is imposed by data link protocol.
IP is the highest layer protocol that is implemented both at routers and hosts.
Reassembly of original data-grams is only done at destination host. This takes off the extra work that need to be done by the routers present in the network core.
I will use the information from the following image to help you get to the answer with an example.
Here initial length of the packet is 2400 bytes which needs to to fragmented according to MTU limit of 1000 bytes.
There are only 13 bits available for the fragment offset and the offset is given as a multiple of eight bytes. This is why the data fields in first and second fragment has size of 976 bytes (It is the highest number divisible by 8, which is smaller than 1000 - 20 bytes). This makes first and second fragment of total size of 996 bytes. The last fragment contains the remaining of 428 bytes of payload (with 448 bytes of total).
Offset can be calculated as 0; 976/8 = 122 and 1952/8 = 244.
When these fragments reach the destination host, reassembly needs to be done. Host uses identification, flag and fragmentation offset for this task. In order to make sure which fragments belong to which data-gram, host uses source, destination addresses and identification to uniquely identify them. Offset values and more fragment bits are used to determine whether all fragments have arrived or not.
Answer to your question
The need to divide payload into multiples of 8 is only required for non-last fragment. Reason of using offset dividing by 8 helps the host to identify the starting address of the next fragment. The host don't need the address of the next fragment if it encounters the last fragment. Thus, no need to worry about payload being multiple of 8 in case of last fragment. Host checks the more fragment flag to identify the last fragment.
A bit of additional information: It is not the responsibility of the network layer to guarantee the delivery of the data-gram. If it encounters that one or more fragment(s) have not arrived then, it simply discards the whole data-gram. Transport layer, which is working above network layer, will take care of this thing, if it is using TCP, by asking the source to re-transmit the data.
Reference: Computer Networking-A Top Down Approach, James F. Kurose, Keith W. Ross (Fifth Edition)
You don't need to add any padding bits. All bits will be push on down the route until the full frame has been sent.
what is the difference between a link MTU max transmission unit and a link bandwidth? MTU refers to the max size of a packet that a link can send. link bandwidth refers to the max number of bits that a link can send. arent they the same thing
MTU = maximum size of 1 packet. For example, ethernet is 1500 bytes.
Bandwidth = bits you can send trough a link, for example, 1 gigabit per second.
So to send 1 megabyte of data over a line, you first need to cut it into small packets that fit the protocol. So you will create about 700 packets of 1500 bytes to fit one megabyte into. Those will go over your line at the bandwidth specified.
But there is overhead: in TCP/IP every packet needs its IP and TCP headers. So the smaller the MTU, the more packets you will need, so the header overhead will become more significant, meaning you can actually use less of the total bandwidth
How do I calculated the minimum and maximum size of an Ethernet frame with split up.We all know maximum size for Ehternet frame is 1518bytes but would be the with split up?
The minimum and maximum frame size is mainly the constraints imposed by the traditional ethernet network at 10Mbps. Since at 10M ethernet shares the same collision domain, a client needs to detect whether a collision had occurred before sending. So, the minimum size is 64bytes at 10Mbps to ensure a collision can be detected within a time limit over certain distance over certain kind of cable specified in the standard.
Nowadays many machines have their own cable/link directly connecting to a switch/router and hence reducing their collision domain (no more collision-detect needed) down to just broadcast domain in the case for a VLAN. Some vendors can support jumbo frames up to 64k in some cases.
What is the size of an empty UDP datagram? And that of an empty TCP packet?
I can only find info about the MTU, but I want to know what is the "base" size of these, in order to estimate bandwidth consumption for protocols on top of them.
TCP:
Size of Ethernet frame - 24 Bytes
Size of IPv4 Header (without any options) - 20 bytes
Size of TCP Header (without any options) - 20 Bytes
Total size of an Ethernet Frame carrying an IP Packet with an empty TCP Segment - 24 + 20 + 20 = 64 bytes
UDP:
Size of Ethernet frame - 24 Bytes
Size of IPv4 Header (without any options) - 20 bytes
Size of UDP header - 8 bytes
Total size of an Ethernet Frame carrying an IP Packet with an empty UDP Datagram - 24 + 20 + 8 = 52 bytes
Himanshus answer is perfectly correct.
What might be misleading when looking at the structure of an Ethernet frame [see further reading], is that without payload the minimum size of an Ethernet frame would be 18 bytes: Dst Mac(6) + Src Mac(6) + Length (2) + Fcs(4), adding minimum size of IPv4 (20) and TCP (20) gives us a total of 58 bytes.
What has not been mentioned yet is that the minimum payload of an ethernet frame is 46 byte, so the 20+20 byte from the IPv4 an TCP are not enough payload! This means that 6 bytes have to be padded, thats where the total of 64 bytes is coming from.
18(min. Ethernet "header" fields) + 6(padding) + 20(IPv4) + 20(TCP) = 64 bytes
Hope this clears things up a little.
Further Reading:
Ethernet_frame
Ethernet
See User Datagram Protocol. The UDP Header is 8 Bytes (64 bits) long.
The mimimum size of the bare TCP header is 5 words (32bit word), while the maximum size of a TCP header is 15 words.
Best wishes,
Fabian
If you intend to calculate the bandwidth consumption and relate them to the maximum rate of your network (like 1Gb/s or 10Gb/s), it is necessary, as pointed out by Useless, to add the Ethernet framing overhead at layer 1 to the numbers calculated by Felix and others, namely
7 bytes preamble
1 byte start-of-frame delimiter
12 bytes interpacket gap
i.e. a total of 20 more bytes consumed per packet.
If you're looking for a software perspective (after all, Stack Overflow is for software questions) then the frame does not include the FCS, padding, and framing symbol overhead, and the answer is 54:
14 bytes L2 header
20 bytes L3 header
20 bytes L4 header
This occurs in the case of a TCP ack packet, as ack packets have no L4 options.
As for FCS, padding, framing symbol, tunneling, etc. that hardware and intermediate routers generally hide from host software... Software really only cares about the additional overheads because of their effect on throughput. As the other answers note, FCS adds 4 bytes to the frame, making it frame 58 bytes. Therefore 6 bytes of padding are required to reach the minimum frame size of 64 bytes. The Ethernet controller adds an additional 20 bytes of framing symbols, meaning that a packet takes a minimum of 84 byte times (or 672 bit times) on the wire. So, for example, a 1Gbps link can send one packet every 672ns, corresponding to a max packet rate of roughly 1.5MHz. Also, intermediate routers can add various tags and tunnel headers that further increase the minimum TCP packet size at points inside the network (especially in public network backbones).
However, given that software is probably sharing bandwidth with other software, it is reasonable to assume that this question is not asking about total wire bandwidth, but rather how many bytes does the host software need to generate. The host relies on the Ethernet controller (for example this 10Mbps one or this 100Gbps one) to add FCS, padding, and framing symbols, and relies on routers to add tags and tunnels (A virtualization-offload controller, such as in the second link, has an integrated tunnel engine. Older controllers rely on a separate router box). Therefore the minimum TCP packet generated by the host software is 54 bytes.