Question on computer networks


#1

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


#2

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:


#3

We know that
Eeta=N⁄(1+2a)
Propagation time(tp)=30008ms
=24000ms
Transmission time(ta)=(64
8*〖10〗^6)⁄(1.536*〖10〗^6 )
=333.33ms
Lets take the efficiency=100%
1= N⁄((1+2*((24000)⁄(333.33))))
N=145