selective repeat sliding window protocol

Question

Consider a selective repeat sliding window protocol that uses frame size of $2$ kb. The capacity of the link is $2$ Mbps and whose one-way latency is of $100$ msec. To achieve a link utilization of $50$%. What is the minimum number of bits required to represent the sequence number field is__________.

Comments

@aditya you know the ans or not...if know then post it...plzz

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i know only the ans not the solution. the ans is 8

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Given explanation of the question

Answer 1

Time for ACK to reach back for a frame  $=TT_{frame} + PD + PD + TT_{ack} \\= (2 \times 10^6 \times \frac{1}{2 \times 10^6} + 100 + 100 + 0) ms \\=201 ms$. 

Amount of data that could be transferred in $201 ms = 201 \times 10^{-3} \times 2Mbps = 402kb  = 201 $packets. 

So, for at least 50% utilization the number of frames to be sent $= \left \lfloor \frac{201}{2} \right \rfloor = 101$ which requires 7 bits to represent.

Now, the window size in selective repeat protocol is half the sequence number space (given in any text). So, we need one more bit for sequence number field making it at least 8. 

Comments

why we considered Selective Repeat here, why not GBN ?

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It is given in question..

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Oops, missed..

But if nothing given, we consider GBN or SR ?

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Since minimum asked we would take the best case- go back N.

Answer 2

in one rtt it can sen total 400*10³   bits
frame size given 2*10³
so no of frames in window 400/2=200 
200 frames are present in window so we need 8 bits to represent it.

Comments

@kunal in ques they told to achieve link utilization 50% means 1Mbps So window_size = 1Mbps*200ms=200kb. no. of frames in windows 200kb/2kb=100 frame min. no. of bit required to represent Seq.No. is log(100)base2=7bit. plzz correct me if wrong..

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can we solve this q with the help of effeciency formula?

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lu %= (tt/tt+2pt)*100
it is given 50 %  in ques.acc. to which u make half of 2mbps..?? i dont know  that u  r correct or wrong.?? but wt ever i learn till date i used to find how many total bits we can send in one RTT , that much bits by frame size gives no of frames..!! further process u know :)