A $3000\ km$ long trunk operates at $1.536\ mbps$ and is used to transmit $64$ bytes frames and uses sliding window protocol. The propagation speed is $6\mu sec/km$

If only $6$ bits are reserved for sequence number field, then the efficiency of the system is

Given,

Sequence number field $=6$ bits

So, total sequence numbers are:

$2^6=64$

In sliding window protocol, maximum usable sender window size is:

$W=2^6-1$

$W=64-1=63$

From previous calculation,

$a=54$

Efficiency of sliding window protocol is:

$\eta=\frac{W}{1+2a}$

$\eta=\frac{63}{1+2(54)}$

$\eta=\frac{63}{109}$

$\eta=0.578$