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

# Question on computer networks

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.

We know that

Eeta=N⁄(1+2a)

Propagation time(tp)=3000*8ms
=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