Thinking about channel utilization from the point of view of data transmission rate

thinking about channel utilization from the point of view of data transmission rate

@ (computer network)

(2014.36) host A and Host B between the use of GBN protocol: A Send window size is 1000, the data frame length is 1000 bytes, the channel bandwidth is 100Mbps. Each time a data frame is received, a short data frame is used to confirm (transmission delay is ignored). If the one-way transmission delay between A and B is 50ms, then a maximum average data transfer rate can be reached is:80Mbps.

Analysis: What I came up with was to calculate the maximum of channel utilization, then multiply the channel bandwidth by the maximum value. But this relationship does not seem reliable, because channel utilization is defined as the ratio of the time it takes to send data to the sending cycle.
Therefore, it is more intuitive to use direct computing.

A send cycle is: The sender starts with sending the first data, until it receives the first confirmation frame.
Therefore, a transmission period of GBN protocol can be sent 1000 consecutive data frames, a total of 1MB.

When sending a frame: 1000b÷108bps=8⋅10−5=0.08ms 1000b\div 10^8bps = 8 \cdot 10^{-5} = 0.08ms
Send Cycle t = 1000b÷108bps+rtt=100.08ms 1000b\div 10^8bps + RTT = 100.08ms
It is possible to send 1000 frames in a single send cycle for a total of 80ms. So not every tick in a sending cycle is made, and is limited to the window size, so it does not reach 100Mbps.
So the maximum transfer rate is about:
1000⋅1000b÷0.1s=80mbps. 1000\cdot 1000b\div 0.1s = 80Mbps.

The need to think about is that since the continuous transmission, why not reach the channel bandwidth it.

In this way, to reach the bandwidth, it takes a second to send 100Mb, approximately equal to one second to send 12500 frames.
A send cycle is about 0.1 seconds, so that is, a send cycle to send 1250 frames. is already greater than the number of windows, so do not 100Mbps.

Calculate Channel Utilization:
α= send data time sent within a send period =1000⋅0.08100.08=80% \alpha = \frac{Send data time within a send period} {send period} = \frac{1000\cdot 0.08}{100.08} = 80\%

So the maximum data transfer rate is about 80Mbps.

A maximum of all windows are sent within a send cycle, and this restriction is understood correctly.