Question on computer networks


A 3000 km long trunk operates at 1.536 Mbps and it is used to transmit 64 Bytes frames. If it uses sliding window protocol then what is the number required sequence numbers. Assume propagation speed of 8 microsec/ km?
A. 63 B. 110 C. 123 D. 145


Can you provide the frame size of sender and receiver in sliding protocol. Or if it is not necessary then please send me the solution. :slight_smile:


We know that
Propagation time(tp)=30008ms
Transmission time(ta)=(64
8*〖10〗^6)⁄(1.536*〖10〗^6 )
Lets take the efficiency=100%
1= N⁄((1+2*((24000)⁄(333.33))))