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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
In actuality, a 64 kilobyte file is 64 × 1,024 × 8 bits in size and the 64 k circuit will transmit bits at a rate of 64 × 1,000 bit/s, so the amount of time taken to transmit a 64 kilobyte file over the 64 k circuit will be at least (64 × 1,024 × 8)/(64 × 1,000) seconds, which works out to be 8.192 seconds.
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.
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
Routing metrics are configuration values used by a router to make routing decisions. A metric is typically one of many fields in a routing table. Router metrics help the router choose the best route among multiple feasible routes to a destination. The route will go in the direction of the gateway with the lowest metric.
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:
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.
The time is measured from the end of the frame check sequence of one frame to the start of the preamble for the next. [ 2 ] : 5 During data reception, some interpacket gaps may be smaller due to variable network delays, clock tolerances (all speeds), and the presence of repeaters (10 Mbit/s only).