# Sliding Window Protocol

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A sender is sending data to the receiver over a half-duplex link.Transmission time for data packets sent from sender to receiver is 80 μs. The propagation delay over the link is 280 μs. What frame size is required for the sliding window protocol with a window size of 8 packets to achieve throughput of 40 x 106 bits per second.

A) 1800 B

B) 225 B

C) 250 bits

D) 2000 bits

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i am getting 400 Bytes .. whats wrong ??

here efficiency is 100 % ...

1+2a is 8 and we have 8 window size ..so efficiency is 100 %

Now effective bandwidth=Efficiency*Bandwidth ....

We can write bandwidth as L/Transmission delay ..by putting all values it comes up to 400 Bytes
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n/(1+2a) -> this formula is for full dupex. Here in question connections are half duplex.
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Are the options correct?
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I don't know as per test series correct answer is 225 B.
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The correct answer is 225 B.
Here the link is given as half-duplex therefore efficiency=N/(1+a)=8/4.5=(16/9).
Hence,(16/9) * Bandwidth=40*10^6
Hence,Bandwidth=22.5*10^6 bps
Hence M=B*Transmission time=22.5*10^6*80*10^-6=1800 bits=225 B.
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How you got this formula N/(1+a) ,what's the logic????
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I think that in half-duplex link we could send data only during propogation time therefore total time=Tt+Tp and useful time =Tt.Hence link utilisation=N*Tt/(Tt+Tp).But in full duplex link we can transmit the data during both the propogation times when the sender is sending data as well as when receiver is sending acknowledgment.Hence total time=Tt+2*Tp.But still we are able to send only during Tt time hence,useful time =Tt.Hence link utilisation=N*Tt/(Tt+2*Tp).

Please correct me if I am wrong.
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Yes it is true in half-duplex link we could send data only during propogation time. But is total time Tt+tp only??? I mean we need to consider whole trip time.
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I guess we should consider the total time as time only when data could be sent so that we can find out the efficiency as useful time/total time.In full duplex we are able to send the data during whole trip time.
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If we need to transmit 8 packets then it takes 8*Tt+Tp Time. Only after that acnowledgement needs to be sent by reciever.Correct me if i am wrong??

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That would be the case for cumulative acknowledgement but for individual acknowledgment we would receive an acknowledgement after every packet ie Tt+2*Tp time and after that we would be able to send the next packet.
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So half duplex is efficient than full duplex, is it possible ??
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Maybe in this case because even though the link might be full duplex and we can send data for Tt+2*Tp we are able to send only for Tt time due to wait for acknowledgement.But if the link is half duplex we are able to send for Tt time out of possible Tt+Tp time since when the acknowledgement is being sent we can't send the data.Hence in case of full duplex efficiency may be less.
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Okk thanks
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@Rutvik one more issue. If you consider Efficiency as N/(1+a) then this value is greater than 1,is it possible??
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Here N/(1+a) is more precisely link utilization.We are getting link utilization as 8/4.5.The efficiency is 1/(1+a) which is 1/4.5 and link utilization is N/(1+a).
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What is the source of

## link utilisation=N*Tt/(Tt+2*Tp) for Full Duplex

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this is such a low quality question. How can efficiency be more than 100%?
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@sushmita What is formula to calculate efficiency , for half duplex link ?? And what's the logic behind that formula.
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many have told 1+a is used here bcz channel is half - duplex but Even in STOP AND WAIT protocol which works on half duplex we use 1+2a there , why not here .Please clear my doubt

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