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In computer networking, the maximum transmission unit (MTU) is the size of the largest protocol data unit (PDU) that can be communicated in a single network layer transaction. [1]: 25 The MTU relates to, but is not identical to the maximum frame size that can be transported on the data link layer, e.g., Ethernet frame.
The default TCP Maximum Segment Size is for IPv4 is 536. For IPv6 it is 1220. [1]: §3.7.1 Where a host wishes to set the maximum segment size to a value other than the default, the maximum segment size is specified as a TCP option, initially in the TCP SYN packet during the TCP handshake. The value cannot be changed after the connection is ...
In the seven-layer OSI model of computer networking, packet strictly refers to a protocol data unit at layer 3, the network layer. [2] A data unit at layer 2, the data link layer, is a frame. In layer 4, the transport layer, the data units are segments and datagrams.
A packet sniffer, which intercepts TCP traffic on a network link, can be useful in debugging networks, network stacks, and applications that use TCP by showing the user what packets are passing through a link. Some networking stacks support the SO_DEBUG socket option, which can be enabled on the socket using setsockopt.
An IPv6 network does not perform fragmentation in network elements, but requires end hosts and higher-layer protocols to avoid exceeding the path MTU. [20] The Transmission Control Protocol (TCP) is an example of a protocol that adjusts its segment size to be
This may require breaking large protocol data units or long data streams into smaller chunks called "segments", since the network layer imposes a maximum packet size called the maximum transmission unit (MTU), which depends on the maximum packet size imposed by all data link layers on the network path between the two hosts. The amount of data ...
The packet transmission time in seconds can be obtained from the packet size in bit and the bit rate in bit/s as: Packet transmission time = Packet size / Bit rate. Example: Assuming 100 Mbit/s Ethernet, and the maximum packet size of 1526 bytes, results in Maximum packet transmission time = 1526×8 bit / (100 × 10 6 bit/s) ≈ 122 μs
A typical method of performing a measurement is to transfer a 'large' file from one system to another system and measure the time required to complete the transfer or copy of the file. The throughput is then calculated by dividing the file size by the time to get the throughput in megabits , kilobits , or bits per second.