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The speed of light imposes a minimum propagation time on all electromagnetic signals. It is not possible to reduce the latency below = / where s is the distance and c m is the speed of light in the medium (roughly 200,000 km/s for most fiber or electrical media, depending on their velocity factor).
By convention, bus and network data rates are denoted either in bits per second (bit/s) or bytes per second (B/s). In general, parallel interfaces are quoted in B/s and serial in bit/s. The more commonly used is shown below in bold type.
In SI, this slope or derivative is expressed in the units of meters per second per second (/, usually termed "meters per second-squared"). Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the ...
Snap, [6] or jounce, [2] is the fourth derivative of the position vector with respect to time, or the rate of change of the jerk with respect to time. [4] Equivalently, it is the second derivative of acceleration or the third derivative of velocity, and is defined by any of the following equivalent expressions: = ȷ = = =.
Throughput is usually measured in bits per second (bit/s, sometimes abbreviated bps), and sometimes in packets per second (p/s or pps) or data packets per time slot. The system throughput or aggregate throughput is the sum of the data rates that are delivered over all channels in a network. [1] Throughput represents digital bandwidth consumption.
People are often concerned about measuring the maximum data throughput in bits per second of a communications link or network access. 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.
In this graph, the widest path from Maldon to Feering has bandwidth 29, and passes through Clacton, Tiptree, Harwich, and Blaxhall. In graph algorithms, the widest path problem is the problem of finding a path between two designated vertices in a weighted graph, maximizing the weight of the minimum-weight edge in the path.
Given these overheads, the maximum goodput is 1460/1526 × 100 Mbit/s which is 95.67 megabits per second or 11.959 megabytes per second. Note that this example doesn't consider additional Ethernet overhead, such as the interframe gap (a minimum of 96 bit times), or collisions (which have a variable impact, depending on the network load).