Sliding Window Protocol related

Question

Consider a Sliding Window Protocol for a 10MBps point- to -point link with propagation delay of 2sec. If frame size is 4 kB. The minimum number of bits required for the sequence number is what?

Comments

i am getting no bits is 11.what is the answer?

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The Right Answer is 14.

Answer 1

Sequence(min)=1+2(T prop/ T trans)

Here T prop=2 sec=2000msec

         T trans=4*!0^3/(10*10^6)=4/10000=.0004sec=.4ms

therefore, Sequence(min)=1+2*5000=10001

Bits min=log 10001=upper limit(13.28)=14

Comments

Thanks for your effort. But here minimum number of bits are asked then what is should be.

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Thank you so much

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:)