MadeEasy Workbook: Computer Networks - Sliding Window

Question

Consider a sliding window protocol operating at the data link layer between two stations 1000
kilometers apart, directly connected by an error-free 1.0 Mbps link. The frame sizes used are
500 bits, of which 50 bits are header and 450 bits are data. Ack frames are 100 bits, and frame
processing time is 100 microseconds. Assume that the signal propagation delay is 10 µsec/km.
If a window size of W = 10 is used, then the efficiency of the channel is

Comments

I am getting answer of this 24.15%. Is it correct?

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we have to follow stepwise to solve this problem

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1.Trasmission of packet from source.(A)
2.propogation of packet from source to destination  in error free medium.(B)
3.Processing of data recieved at reciever.(C)
4.Transmission of acknowledgement packet.(D)
5.propogation of packet from destination to source in error free medium.(E)

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$(A)=\frac{\text{frame size}}{\text{data rate}}=\frac{500}{1 \times 10 ^{6}}=500 \mu sec$

$(B)=100 \mu sec$

$(C)=10 \times 1000=10000 \mu sec$

$(D)=\frac{100+50}{1 \times 10^{6}} \mu sec=150 \mu sec$

$(E)=(B)=10000 \mu sec$

Efficiency of channel=$\frac{A \times w}{A+B+C+D+E}=\frac{500 \times 10}{20750}=24.09 \%$

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yes $24$ wil be right answer!

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i have already explained!

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@Anand, in ack transmission time why have you included 50 because whole ack frame size is given?

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I ALSO GOT 24.15%...

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@Anand

I guess the steps B and C are interchanged..

Also, while calculating the time taken for transmission of ACK packet (D), why have you added 50 bits ? I am asking this since the ACK frame size is already given in the question..

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how the processing time is 10*1000 micro sec