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When talking about circuit bit rates, people will interchangeably use the terms throughput, bandwidth and speed, and refer to a circuit as being a '64 k' circuit, or a '2 meg' circuit — meaning 64 kbit/s or 2 Mbit/s (see also the List of connection bandwidths). However, a '64 k' circuit will not transmit a '64 k' file in one second.
Multiplying the set of processes would give you Rolling throughput yield (RTY). RTY is equal to FPYofA * FPYofB * FPYofC * FPYofD = 0.8500 * 0.8889 * 0.8125 * 0.8267 = 0.5075 Notice that the number of units going into each next process does not change from the original example, as that number of good units did, indeed, enter the next process.
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 channel efficiency, also known as bandwidth utilization efficiency, is the percentage of the net bit rate (in bit/s) of a digital communication channel that goes to the actually achieved throughput. For example, if the throughput is 70 Mbit/s in a 100 Mbit/s Ethernet connection, the channel efficiency is 70%.
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
A rate is said to be achievable if there exists a sequence of codes (,) such that the average probability of error: (^) tends to zero as . The feedback capacity is denoted by C feedback {\displaystyle C_{\text{feedback}}} , and is defined as the supremum over all achievable rates.
rates the fairness of a set of values where there are users, is the throughput for the th connection, and ^ is the sample coefficient of variation. The result ranges from 1 n {\displaystyle {\tfrac {1}{n}}} (worst case) to 1 (best case), and it is maximum when all users receive the same allocation.
To give a practical example, two nodes communicating over a geostationary satellite link with a round-trip delay time (or round-trip time, RTT) of 0.5 seconds and a bandwidth of 10 Gbit/s can have up to 0.5×10 Gbits, i.e., 5 Gbit of unacknowledged data in flight. Despite having much lower latencies than satellite links, even terrestrial fiber ...
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