point to point link to the moon

Question

Suppose you are designing the sliding window protocol for a 1 Mbps point to point link to the moon, which has one way latency(delay) of 1.25 seconds. assuming that each frame carries 1 KB of data, What is minimum number of bits you need for sequence number?

Comments

If not mentioned then we will go with Go Back N

Answer 1

window size(w)=(bandwidth*two way propagation delay)/frame size

so here w=(10^6*2*1.25)/(1*1024*8)

=305

by default sliding window means go_back n ARQ

so here 2^n-1=305

n=9 bits

Comments

In satellite communication we sent next packet only when ack of ack is recieved correctly so should be 4*one way propogation delay

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If Go Back N is used then why  is $2^{n-1}$=305?

And why not $2^{n}$=305?

Answer 2

Bandwidth delay product=1.25*2*10^6

Size of sender window=1.25*2*10^6/8*10^3

                                         =312.5(313 aprox)

In sliding window sws=rws

No of seq no=312.5*2=625

To represent 625 10 bits are required

Comments

By default, We always use GBN, so why we are considering here SWS = RWS??

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1st Thing is point to point link is given then effeciany we will take is maximum

and here minimum number of bits are asked so we will take the case of GBN because in this case it is 9 bits and in case of SR it is 10 bits.

so final answer will be 9 bits.

@srestha confirmation needed.

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@ reena_kandari 

Why point to point link means max efficiency?