Stop and wait

Question

A channel has a capacity of 256 Mbps, maximum packet size is 1024 bytes and RTT is 200 μsec. So what is efficiency of sender (Assume that channel uses stop & wait protocol?

Your Answer:

13.793

Correct Answer: 2    Status: incorrect

Answer 1

efficiency=Transmission time/(Transmission Time + 2* Propagation Time)

Tt=Length of Pkt / Bandwidth

Tt=1024 *8 (256 * 10^ 6)=32 microsecond

2*Tp= Round  Time Trip=200 microsecond

 

Efficiency= 32/(32+200)=0.1379 or 13.79%

Comments

Packet size is in byte so no need to multiply with 8. Then ans will be 2 (approx)

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Here in the question it is given as  "capacity of channel" is 256 Mbps. But in the given solution it is used as Bandwidth. I am confused. Please suggest.

Answer 2

$T_{t}=\frac{Packet \:size}{Bandwidth}=\frac{1024*8}{256*10^{6}}=32\mu sec$

$RTT=200\mu sec........\left \{ Given \right \}$

So efficiency of sender

$\eta =\frac{1}{1+a}\approx 0.1379\:or\: 13.79........\left \{ where \:a=RTT/T_{t}=200/32=7.25 \right \}$

Comments

then why is the answer is incorrect

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How we will know that, here RTT =2*Tp +Tt   or   RTT=2*Tp? @Bikram sir please guide

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here RTT =2*Tp +Tt 

Generally for large distance like satellite, we ignore Tt value .

But in this question we need to consider it .

Efficiency of sender = 32/(32+200)=0.1379 or 13.79%

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@Bikram

Round trip time(RTT) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgement of that signal to be received.

Ref: https://www.geeksforgeeks.org/program-calculate-round-trip-time-rtt/