Why is TCP receive window considered to be a multiple of MSS Maximum Segment Size?
Wiki states that in order to fully utilize the packet lengths and avoid IP fragmentation , an integral multiple of the Maximum Segment Size (MSS) is generally recommended for the receive window and the value is therefore often only given as a factor/multiple.
Here it is stated that segments exceeding the MSS size is discarded.
For alignment... ergo it fits in a nic buffer on [ROM] # a place in its memory which is then what you read from via a port or DMA
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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.
I understand from this discussion that boost::asio::async_write writes data to the kernel buffers only. It does not mean that the peer has received the data. But if I am sending big packets of size let's say 200000 bytes each, and then I pull the network cable to kill the connection abruptly. Will it still keep reporting on and on saying 200000 bytes written into kernel buffers for each async_write? My testing says that it doesn't. It gives up with a large buffer like 200000 bytes and does not report all bytes sent. But if its a small buffer like 30-40 bytes, it keeps reporting okay?
Question:
The primary point of raising this question is: Is there an underlying buffer size which gets filled up at one point for async_write to say that now its not able to write anymore because the earlier scheduled data has not gone out? If yes then what is the size of this underlying buffer? Can I query it from the boost::asio::ip::tcp::socket?
You can query/change the underlying system socket buffer size with send_buffer_size socket option.
The operating system though can dynamically adjust the socket buffer size and limit its maximum size:
tcp_wmem (since Linux 2.4)
This is a vector of 3 integers: [min, default, max]. These
parameters are used by TCP to regulate send buffer sizes. TCP
dynamically adjusts the size of the send buffer from the
default values listed below, in the range of these values,
depending on memory available.
I have a peripheral over USB that is sending data samples at a rate of 183 MBit/s. I would like to send this data over ethernet, which is limited to < 100 Mbit/s. Is it possible to send this data without overflow (i.e missing data) by increasing the TCP socket buffer?
It also depends on the receiver window size. Even if there is 100mbits, sender will push data depending on the window size available on the receiver. TCP window size without scaling enabled can go only upto 64kb. In your case, this size is not sufficient as it needs at least (100-183Mbits)10MB buffer. In Windows 7 & newer Linux OS, TCP by default enables window scaling which can extend the size upto 1GB. After enabling the TCP window scaleing option, you can increase the socket buffer to a bigger size say 50MB which should provide the required buffering.
The short answer is, it depends.
Increasing buffers (at transmitter) can help if the data is bursty. If the average rate is <100MBit (actually less, you need to allow for network contention and overhead), then buffering can help. You can do this by increasing the size of the buffers internally to the TCP stack, or by buffering internally to your application.
If the data isn't bursty, or the average is still too high, you might need to compress the data before transmission. Dependant on the nature of the data, you may be able to achieve significant compression.
I submit text file to server capturing by wireShark. my computer's window size is steady to 17408. but server's window size is increasing 6912, 9856, 12800 ...
I want to know why server's window size is increasing. and first TCP segment data is 502 bytes. and the other TCP segment is 1460 bytes.
why window size is increasing? why first TCP segment data is different the other?
As far as I know, the growth of TCP window size is related to the so-called slow start algorithm, which is described in detail here TCP Slow start
Also in the program Wireshark is a definite option to recompile the TCP-packets in accordance with the useful content L7, so that the size of the package there may be, in principle, any
For example, if a set of TCP-packets, which are encapsulated huge HTTP-request, instead of breaking in fragments, it can be shown in one huge fake TCP-packet
This option is called TCP reassemble
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.