Bandwidth: b=64kbits/second the frame length

Дата канвертавання25.04.2016
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  1. The Problem

Consider an error-free 64kbps satellite channel used to send 512-byte data frames in one direction, with very short acknowledgements coming back the other way. What is the maximum throughput for window sizes of 1, 7, 15 and 127? We assume that the earth-satellite propagation time is 270 msec.
  1. The Solution

  1. 2.1 Elementary Quantities

What we know:

  • the bandwidth: b=64kbits/second

  • the frame length: l=512 bytes=4096 bits

  • the propagation time: tprop=270msec

We can also compute the time required to transmit one frame on the channel:

Let us consider a window size of N and see that happens:

t=0t=ttranst=N* ttranst=tpropt= ttrans +tpropt= ttrans +2*tpropAfter this, all the frames are acknowledged one after the other, and the process re-starts. So, one can see that the on the line there are always N or N-1 frames in transit. The sender sends N frames during the time interval t= ttrans +2*tprop. Thus the throughput is:

However, there is a condition to be met in order for things to work the way they have been described: when the sender receives the first acknowledge, it should have send its entire window, that is,


or equivalently:

Hence, the general formula for the throughput is given by:

  1. 2.3. The Numerical Solution

In our case, the condition becomes: 9, that is, the top throughput is reached for N=9.

Hence, the numeric solution for our problem is:

NThroughputEfficiency16798 bits/second10.37%747586 bits/second72.61%1564 kbits/second100%12764 kbits/second100%
  1. Conclusion

We can conclude that when using a sliding window protocol on an error-free channel, the throughput grows linearly with the size of the window until the bandwidth of the transmission line is reached.

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