MadeEasy Test Series: Computer Networks - Flow Control Methods

Question

consider a link of length 1000 km with 10^9 bps rate connecting a sender and receiver. assume a fixed packet length of 1250 bytes and sender always has packet to send. packet are never lost or corrupted in the connection. what is the necessary windowsize to achieve 100 % utilization for a sliding window protocol? assume signal propagation is 5 microsec per km

(A) 100 (B) 1000 (C) 110 (D) 1100

Answer 1

1km = 5 microsec so 1000km = 5*1000 microsec

RTT = L/B + 2 * PropagationTime 

here L is packet size

and B is bandwidth

RTT = (1250 * 8 bits)/10⁹ bits + 2*5000 microsec

      = 10micro sec + 10000micro sec

      = 10010 microsec

No of packets = (Bandwidth * delay)/packet size

                     =(10⁹ * 10010 *10^{-6) / (1250 * 8)}

                            ^{= 1001}

Comments

me too getting the same anwer. dont know  why this option isn't available.

Any ways thanks for answering the question.

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ans is B becuase

Ws<=1+2a

1000<=1001

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can u plz derive the formula u used for meeeee plzzzzzzzz