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.
Related
I have been learning wireshark lately. While inspecting TCP segments, I saw a strange situation, at least for me. There was a mismatching SEQ,ACK numbers. Then I realized that difference between two ACK's same as 1 and half package size. However, as far as I know, ACKs are only growing by the full packet size. So what happened here?
SEQ ACK
1 1
2897 8689
5793 13033 <--
8689 14481
11585
14481
Well, there's not much to go by here, but I'm going to guess that you are capturing packets on the machine sending the large packets, i.e., the packets with 2896 bytes of TCP payload. As such, you are seeing the packets before they are IP-fragmented and actually transmitted out on the wire.
But you can't just send 2896 bytes of data onto the wire; Ethernet links typically impose a 1500 byte MTU (Maximum Transmission Unit), and when you account for IP and TCP header overhead, you usually end up with 1460 bytes of available payload or MSS (Maximum Segment Size). In your case it looks like you're only getting 1448 bytes of available MSS, most likely due to the addition of one or more IP and/or TCP header options.
In any case, the 2896 bytes of payload are going to be fragmented over 2 IP fragments, each containing 1448 bytes of TCP payload. I'm pretty sure that what you're seeing is an ACK from the receiver after having received 1 full segment plus 1 IP fragment from the next segment.
The previous ACK number was 8689, and 8689 + 2896 = 11585. Now add 1/2 of the data segment (2896 / 2 = 1448) and you get 11585 + 1448 = 13033. That's the ACK number you're seeing. Now add the other 1/2 and you get 13033 + 1448 = 14481, which is the ACK number of the next packet.
I hope that makes sense?
For an in depth look at the drawbacks of local packet captures, I direct you to a well-written blog by Jasper Bongertz titled, "The drawbacks of local packet captures".
Since I do not have reputation to comment the answer of Christopher Maynard, I comment it here.
As Christopher stated correctly the MTU (Maximum Transmission Unit) is standard Ethernet networks is 1500Bytes. However, crafting 1460Bytes (1500Bytes - 20Bytes IP Header - 20Bytes TCP Header, ignoring possible TCP options) TCP segments imposes a high load on the network stack since every segment must be handled separately.
As an example, if I want to transmit 1Mb (equals to 1024*1024Bytes = 1048576Bytes) of data, it would result in 1Mb/1460Bytes = 719 segments. In contrast to 1Mb/65525Bytes (theoretical max TCP segment size) = 16 segments. Since the overhead in the Kernel is mostly independent of the segment/packet size, small segments require much more processing.
To counteract this fact, TCP Segmentation Offloading (TSO) was developed. TSO allows the Kernel to craft TCP segments with maximum size and the NIC (Network Interface Card) uses hardware acceleration to split these segments into the separate TCP segments of 1460Bytes. Therefore, no IP fragmentation is required.
I have a system that sends "many" (hundreds) of UDP datagrams in bursts, every once in awhile (say, 10 times a minute). According to nload, this averages about 222kBit/s. The content of these datagrams is JSON. I've considered altering the system so that it waits some time (500ms?) and combines many of the JSON objects into one datagram, before sending. But I'm not sure it's worth the effort (bandwidth, protocol, frequency of sending considered.) Would the new approach provide any real benefits over the current one?
The short answer is that it's up to you to decide that.
The long version is that it depends on your use case. Since we don't know what you're building, it's hard to say what's more important - latency? Throughput? Reliability? Something else? Let's analyze some pros and cons. Here's what I came up with:
Pros to sending larger packets:
Fewer messages means fewer system calls and less I/O against the network. That means fewer blocked/waiting threads and less time spent on interrupts.
Fewer, larger packets means less overhead for each individual packet (stuff like IP/UDP headers that's send with each packet). Therefore a higher data rate is (theoretically) achievable, although keep in mind that all of these headers (L2+IP+UDP) typically add up to no more than 60-70 bytes per packet since the UDP header is only 8 bytes long.
Since UDP doesn't guarantee ordering, larger packets with more time between them will reduce any existing reordering.
Cons to sending larger packets:
Re-writing existing code, and making it (slightly) more complicated.
UDP is unreliable, so a loss of a single (large) packet would be more significant compared to the loss of a small packet.
Latency - some data will have to wait 500ms to be sent. That means that a delay is added between the sender and the receiver.
Fragmentation - if one of the packets you create crosses the MTU boundary (typically 1450-1500 bytes including the IP+UDP header, which is normally 28 bytes long), the IP layer would need to fragment the packet into several smaller ones. IP fragmentation is considered bad for a multitude of reasons.
Processing of larger packets might take longer
Two words commonly used in networking world - Packets and frames.
Can anyone please give the detail difference between these two words?
Hope it might sounds silly but does it mean as below
A packet is the PDU(Protocol Data Unit) at layer 3 (network layer - ip packet) of the networking OSI model.
A frame is the PDU of layer 2 (data link) of the OSI model.
Packets and Frames are the names given to Protocol data units (PDUs) at different network layers
Segments/Datagrams are units of data in the Transport Layer.
In the case of the internet, the term Segment typically refers to TCP, while Datagram typically refers to UDP. However Datagram can also be used in a more general sense and refer to other layers (link):
Datagram
A self-contained, independent entity of data carrying sufficient information to be routed from the source to the destination computer without reliance on earlier exchanges between this source and destination computer andthe transporting network.
Packets are units of data in the Network Layer (IP in case of the Internet)
Frames are units of data in the Link Layer (e.g. Wifi,
Bluetooth, Ethernet, etc).
A packet is a general term for a formatted unit of data carried by a network. It is not necessarily connected to a specific OSI model layer.
For example, in the Ethernet protocol on the physical layer (layer 1), the unit of data is called an "Ethernet packet", which has an Ethernet frame (layer 2) as its payload. But the unit of data of the Network layer (layer 3) is also called a "packet".
A frame is also a unit of data transmission. In computer networking the term is only used in the context of the Data link layer (layer 2).
Another semantical difference between packet and frame is that a frame envelops your payload with a header and a trailer, just like a painting in a frame, while a packet usually only has a header.
But in the end they mean roughly the same thing and the distinction is used to avoid confusion and repetition when talking about the different layers.
Actually, there are five words commonly used when we talk about layers of reference models (or protocol stacks): data, segment, packet, frame and bit. And the term PDU (Protocol Data Unit) is a generic term used to refer to the packets in different layers of the OSI model. Thus PDU gives an abstract idea of the data packets. The PDU has a different meaning in different layers still we can use it as a common term.
When we come to your question, we can call all of them by using the general term PDU, but if you want to call them specifically at a given layer:
Data: PDU of Application, Presentation and Session Layers
Segment: PDU of Transport Layer
Packet: PDU of network Layer
Frame: PDU of data-link Layer
Bit: PDU of physical Layer
Here is a diagram, since a picture is worth a thousand words:
Consider TCP over ATM. ATM uses 48 byte frames, but clearly TCP packets can be bigger than that. A frame is the chunk of data sent as a unit over the data link (Ethernet, ATM). A packet is the chunk of data sent as a unit over the layer above it (IP). If the data link is made specifically for IP, as Ethernet and WiFi are, these will be the same size and packets will correspond to frames.
Packet
A packet is the unit of data that is routed between an origin and a destination on the Internet or any other packet-switched network. When any file (e-mail message, HTML file, Graphics Interchange Format file, Uniform Resource Locator request, and so forth) is sent from one place to another on the Internet, the Transmission Control Protocol (TCP) layer of TCP/IP divides the file into "chunks" of an efficient size for routing. Each of these packets is separately numbered and includes the Internet address of the destination. The individual packets for a given file may travel different routes through the Internet. When they have all arrived, they are reassembled into the original file (by the TCP layer at the receiving end).
Frame
1) In telecommunications, a frame is data that is transmitted between network points as a unit complete with addressing and necessary protocol control information. A frame is usually transmitted serial bit by bit and contains a header field and a trailer field that "frame" the data. (Some control frames contain no data.)
2) In time-division multiplexing (TDM), a frame is a complete cycle of events within the time division period.
3) In film and video recording and playback, a frame is a single image in a sequence of images that are recorded and played back.
4) In computer video display technology, a frame is the image that is sent to the display image rendering devices. It is continuously updated or refreshed from a frame buffer, a highly accessible part of video RAM.
5) In artificial intelligence (AI) applications, a frame is a set of data with information about a particular object, process, or image. An example is the iris-print visual recognition system used to identify users of certain bank automated teller machines. This system compares the frame of data for a potential user with the frames in its database of authorized users.
I am working on a project in which i have to send data in bits.
Is it possible to send 1 bit from one computer to another through internet.Most of the people
said to me that minimum internet packet size is 64bytes.If i send 1bit from one computer to another then packet bandwidth is 64bytes.
A TCP or UDP packet consists of a header and data. Maybe you could have one bit of data in the data section, but you would need the header as well. Without the header it would be impossible to send the packet. The header contains all the information required for sending the packet where it is supposed to go and making sure it arrives safely.
I will take "internet packet" to mean Ethernet frame based on the value you give.
An Ethernet frame has a minimum total size of 64 bytes including both header and payload, this is to ensure that the time to transmit a single frame is greater than the round trip time between nodes.
This requirement is a feature of any network that uses CSMA/CD (specifically the CD (collision detection) part), it allows a sensing node to detect a collision whilst still transmitting a frame.
Whilst Ethernet can be used to send a frame smaller than 64 bytes "padding" will be added to ensure the frame is at least 64 bytes.
suppose i have a 4MBits network and i want to calculate the data throughput, this is considering the max transfer rate minus overhead from ethernet/IP/TCP headers.
Reading on the web i found out that the MSS ( maximum segment size) of a TCP segment is 576 - 20 - 20, these last two being TCP and IP headers overhead, resulting in a 93% of data, meaning i will be only using 93% of my 4MBits link to transfer data. Now where's the link ayer overhead? Shouldn't it be added as well? If im not wrong an ethernet header is around 46 bytes so the final sum would be 576 - 20 - 20 - 46 = 490, resulting in an 85% data throughput, but am i doing something wrong?
Just work bottom up. Regular ethernet frames (no jumbo frames, no vlan tagging) are 1542 bytes in total and can have a payload of 1500 bytes. An Ipv4 header without options is 20 bytes and a TCP header without options also 20 bytes. So you end up with 1460 bytes possible payload of a 1542 byte link-layer frame. So your efficiency is 1460/1542=0.9468223086900129, resulting in a maximum throughput of 3.7872892347600517Mbps.
Notice however this will usually be lower. This is the theoretical maximum rate for a continuous stream you can get on a full duplex link, after the TCP session is established and when you're the only user of that link. Also note that as soon as you're sending at a slightly higher rate for some time your link will get congested, you will see drops and your actual TCP throughput might drop significantly because of slow-start.
If the link is wireless (802.11) the calculation becomes a lot more complex because of RTS/CTS mechanisms, but it's about /2 for only one active user and that's without incorporating loss, which is unrealistic.
In general, the protocol can impact network throughput and much more than simply the packet overhead. You mention that you want to measure throughput on an Ethernet/IP/TCP network but the impact of packet overhead of those protocols is NOT the only thing to consider. TCP is a connection-oriented protocol and uses ACK's to signal if a packet has been received or not. user1777914 missed the mark about ACK's but was on to something - they do not take up any more SPACE but they can DELAY the transmission of packets. As latency increases the overall network throughput can decrease based on how often the application or hosting OS expects a response.
W. Richard Stevens has written an AMAZING book on TCP/IP. Here is an except that explains theoretical TCP performance, what impacts it and how it is calculated.
There too is the Nagle algorithm helps with latency but if disabled can slow down throughput.