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in Computer Networks by (9 points) | 43 views
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is it 25%?
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No. Even I think it’s 25 but they gave 34 as the answer.
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we should always use this formula for efficiency = Transmission time/(transmission+RTT). you will get 25%.

I know now a days applied started ignoring Transmission time in denominator which is wrong. so whatever you know is correct

1 Answer

+1 vote

In Stop-Wait we send 1 Frame per RTT.

Now Capacity of channel is

C=RTT*Bandwidth

C=(3ms*2Mbps)=6000 bits(frame size we can send )

so we can send 6000bits but we are sending only 2000 bits

So utilization is (2000/6000)*100 %=33.33 now take ceil

so Ans is 34%

OR

for SW protocol

utilization =(1/RTT)*100%

                = 33.33%(ceil 34%)

as corrected by @

 
ago by (643 points)
edited ago by
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In Stop-Wait we send 1 Frame per RTT. i think we are sending 1 frame per (RTT+Transmission time). so write answer should be 25% not 33.33%
+1

See in genereal there are 3 cases as Past gate standards

  1. If RTT is not  given So we Apply (Tt+2Tp).
  2. if Ques. Mention that RTT is Twice of Propogatioin delayi.e.,(2*Tp) then no need to add Tt here.
  3. if Ques Gives Only RTT (Like in this ques) then you do not need to further manipulate it.

 

refer

https://gateoverflow.in/8363/gate2015-1-53

https://gateoverflow.in/8056/gate2015-2-8

+1
so By default RTT is summation of all the time if nothing is mentioned.
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yes!
+1
Your calculations are wrong. Capacity will be 6000 bits and we will be using only 2000 bits out of those 6000 bits that is why we will be having 1/3*100 as efficiency. So basically we are not adding 2000 bits in the denominator because it has overlapped with the 6000 bits capacity? If yes, can you tell me any case where we need to add the useful_data (frame size) to the denominator or does it never occur.
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thanks for correcting .Edited the ans
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