gatebook

Question

Assume all frames have size F bits, and the propagation delay on the link is P sec, and the bandwidth is B bps. Assume also that the receiver piggybacks acknowledgements on frames going to the sender.

What should the window size n (in frames) be at the sender in order to effectively use the bandwidth?

(A) 2+2PB/F

(B) 1+2PB/F

(C) 2PB/F

(D) 1+PB/F

Comments

if we use the concept of maximum window size=1+2*a where a=pt/tt then answer will be different ???where is mistake ??

PT=P sec(given)

TT=2*F/B( for piggy back)

by using above formula window size=1+PB/F which is option (D)

Answer 1

For effective utilization, Bandwidth will be utilized fully 

so, window size = How much capacity to send / Frame Size

=> Bandwidth * RTT / F

=> B * ( 2 PT + 2 TT ) / F

=> 2 + 2PB / F (Option A)

Comments

why 2TT???

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Assume also that the receiver piggybacks acknowledgements on frames going to the sender

Sender sends (TT) and here it is piggybacking so receiver is also sending a frame with ACK encapsulated (Another TT), so total = 2TT