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I am working on homework assignment and I found a similar question and the answer that I can refer to.

On this website I am being presented the question and the solution. Question 1A:

http://www.ee.uwa.edu.au/~roberto/teach/ind426/handouts/tutorials/solutions/tute3.pdf

A channel has a data rate of 4kbps and a propagation delay of 20ms. Find the frame size in Stop-and-Wait to reach at least 50% of efficiency.

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Where does the 80 come from? What is the value of L from here?

Please help!

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1 Answer

up vote 1 down vote accepted

$$a = \frac{T_\text{prop}}{T_\text{frame}}=\dfrac{20 \times 10^{-3}}{\tfrac{L}{4 \times 10^3}} = \dfrac{20 \times 10^{-3} \; \times \; 4 \times 10^3}{L} = \dfrac{80}{L}$$

so $$U \ge 0.5 \iff a = \frac{1}{2U} - \frac{1}{2} \le 0.5 \iff L=\frac{80}{a} \ge 160.$$

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Ah! This explain a lot! Thank you. –  HelpNeeder Mar 28 '12 at 0:35
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