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A link has transmission speed of $10^6$ bits/sec. It uses data packets of size $1000$ $\text{bytes}$ each. Assume that the acknowledgment has negligible transmission delay and that its propagation delay is the same as the data propagation delay. Also, assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly $25$$\text{%}$. The value of the one way propagation delay (in milliseconds) is_____.
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What is the significance and meaning of one way propogation delay?
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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, $p + 4 = 16, p = 12 \;\text{ms}$.

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sir, efficiency=useful time /total time and throughput =amount of data sent/amount of data that could be sent.
but you have used throughput's formula for efficiency.why?
correct me if i am wrong.
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@Arjun sir why you have multiplied RTT with 10^6 in the first formula.

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Cz it’s a round trip !!
Sender to receivr( data transmission ) and from receiver to sender( acknowledgement transmission)
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16 votes
16 votes
Transmission time Tt=1000*8/10^6=8/10^3 sec=8 ms

efficiency =25%=1/4=Tt/(Tt+2*Tp)

8/(8+2*Tp)=1/4

Tp=12

so propagation Delay is 12 ms
3 votes
3 votes
as per the question Bandwidth$= 10^{6}bps$

data size$=1000 Bytes$

by these two we can calculate transmission time by using formula $\frac{Data size}{BW}$

as data size in bytes convert to bits hence $T_{t}= 8ms$

Efficiency$=25\% i.e$ $0.25$

efficiency$= \frac{T_{t}}{T_{t}+2T_{p}}$         (I am not taking td ack time as it is negligible)

$0.25= \frac{T_{t}}{T_{t}+2T_{p}}$

$0.25(T_{t}+2T_{p}) = T_{t}$

$0.25T_{t} + 0.5T_{t} = T_{t}$

Hence $0.5T_{p} = 0.75T_{t}$

$T_{p} = 0.75T_{p}/0.5$

$T_{p} = 1.5 T_{t}$

$1.5\times8 = 12ms$

$T_{p} = 12ms$
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Given Bandwidth (B) = 10^6 bits/sec , L = 1000 * 8 bits ,  η = 25% = 1/4

Tt = L/ B = 1000* 8 / 10^6 = 8ms  (Transmisson delay of acknowledgement is negligible not  Packet )

 η=1/1+2*a

1/4=1/1+2*a => a = 3/2 => Tp = 3/2*Tt => Tp = 3/2 * 8 = 12 ms
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