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+19 votes
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_____.
asked in Computer Networks by Veteran (96.2k points)
edited by | 3.5k views

5 Answers

+31 votes
Best answer

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} $

$= \dfrac {1000 \times 8}{ \left( p + p + 1000 \times \dfrac{8}{10^6} + 0 \right) \times 10^6}$

$= \dfrac{8}{2p+8ms} \text{ (where p is the prop. delay in milli seconds)}$

$= \dfrac{4}{p+4} = 0.25 \text{ (given in question)}$

So, $p + 4 = 16, p = 12 ms$.

answered by Veteran (408k points)
edited by
I am getting ans upto 4 but after than what you did I can't understand plz explain.

How  p + 4=16 I didnt get this
Is it clear now? p in seconds is converted to p in milliseconds by multiplying by 1000.
Thank you sir

I got it
Transmission delay = $\frac{1000 \times 8 \; bits \;}{10^6 \text {bits per sec }} = 8 \;ms\;$
 Efficiency of stop and wait =  $\frac{1}{1+2(\frac{ \text {propagation delay }}{\text{ transmission delay }})} \\ \implies \frac{1}{1+2(\frac{\text {propagation delay }}{8 \;ms\;})} = \frac{1}{4} \\ p=12 \;ms\;$
can tell me is it necessary to add propogation delay of acknowlegment as you have not considered ??
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.

@Arjun sir why you have multiplied RTT with 10^6 in the first formula.

+10 votes
Transmission time Tt=1000*8/10^6=8/10^3 sec=8 ms

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



so propagation Delay is 12 ms
answered by Active (4k points)
+2 votes
Answer: 12
answered by Boss (33.8k points)
+1 vote
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 $0.5T_{p} = 0.75T_{t}$

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

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

$1.5\times8 = 12ms$

$T_{p} = 12ms$
answered by (437 points)
edited by
0 votes
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 )


1/4=1/1+2*a => a = 3/2 => Tp = 3/2*Tt => Tp = 3/2 * 8 = 12 ms
answered by (221 points)

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