GO to the Root of Concept: Video Explanation with timestamp 15 Questions on Stop Wait Protocol| GATE PYQs| MIT, Berkeley |Efficiency & Throughput| With NOTES
In stop and wait, a frame is sent and next frame will be sent only after ACK is received. $\text{Efficiency} =\dfrac{\text{Amount of data sent}} {\text{Amount of data that could be sent}}$ $=\dfrac{\text{Amount of data sent}}{ RTT \times 10^6}$
$= \dfrac{\text{Amount of data sent}}{ \left( \text { Prop. delay for data}+\text{Prop.delay for ACK} + \text{Transmission time for data} +\text{Transmission time for ACK} \right) \times 10^6} $
$= \dfrac {1000 \times 8}{ \left( p + p + 1000 \times \dfrac{8}{10^6} + 0 \right) \times 10^6}$ $= \dfrac{8}{2p+8\;\text{ms}} \text{ (where p is the prop. delay in milli seconds)}$ $= \dfrac{4}{p+4} = 0.25 \text{ (given in question)}$ So, $p + 4 = 16, p = 12 \;\text{ms}$.
@Arjun sir why you have multiplied RTT with 10^6 in the first formula.
For do this question in less time..
efficiency(n) = 1/(1+2a)
=> 1/4 = 1/(1+2a)
=> 1+2a = 4 => 2a = 3
=> a=3/2
=> Tp/Tt =3/2
=> Tp = (3/2)*Tt
सबसे बढ़िया explaination है
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