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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.
Some other computer architectures use different modules with a different bus width. In a single-channel configuration, only one module at a time can transfer information to the CPU. In multi-channel configurations, multiple modules can transfer information to the CPU at the same time, in parallel.
The latter will increase the data transfer rate for a given RPM speed. Improvement of data transfer rate performance is correlated to the areal density only by increasing a track's linear surface bit density (sectors per track). Simply increasing the number of tracks on a disk can affect seek times but not gross transfer rates.
For example, in the case of file transfer, the goodput corresponds to the achieved file transfer rate. The file transfer rate in bit/s can be calculated as the file size (in bytes) divided by the file transfer time (in seconds) and multiplied by eight. As an example, the goodput or data transfer rate of a V.92 voiceband modem is affected by the ...
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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.
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).
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