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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.
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
The bit time has nothing to do with the time it takes for a bit to travel on the network medium but has to do with the internals of the NIC. To calculate the bit time at which a NIC ejects bits, use the following: bit time = 1 / NIC speed To calculate the bit time for a 10 Mbit/s NIC, use the formula as follows:
The knee frequency is related to the required bandwidth of a channel, and can be related to the 3 db bandwidth of a system by the equation: [8] / Where Tr is the 10% to 90% rise time, and K is a constant of proportionality related to the pulse shape, equal to 0.35 for an exponential rise, and 0.338 for a Gaussian rise.
For example, if a file is transferred, the goodput that the user experiences corresponds to the file size in bits divided by the file transfer time. The goodput is always lower than the throughput (the gross bit rate that is transferred physically), which generally is lower than network access connection speed (the channel capacity or bandwidth).
Bandwidth commonly measured in bits/second is the maximum rate that information can be transferred; Throughput is the actual rate that information is transferred; Latency the delay between the sender and the receiver decoding it, this is mainly a function of the signals travel time, and processing time at any nodes the information traverses
TCP window scale option is needed for efficient transfer of data when the bandwidth-delay product (BDP) is greater than 64 KB [1].For instance, if a T1 transmission line of 1.5 Mbit/s was used over a satellite link with a 513 millisecond round-trip time (RTT), the bandwidth-delay product is ,, =, bits or about 96,187 bytes.
In order to calculate the data transmission rate, one must multiply the transfer rate by the information channel width. For example, a data bus eight-bytes wide (64 bits) by definition transfers eight bytes in each transfer operation; at a transfer rate of 1 GT/s, the data rate would be 8 × 10 9 B /s, i.e. 8 GB/s, or approximately 7.45 GiB /s.