Question

The distance between two stations M and N is L kilometres. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:

• A. $\left[lo{g}_2\frac{2LtR+K}{2K}\right]$
• B. $\left[lo{g}_2\frac{2LtR}{K}\right]$
• C. $\left[lo{g}_2\frac{2LtR+2K}{K}\right]$
• D. $\left[lo{g}_2\frac{2LtR+K}{K}\right]$

Answer 1

The correct option is D $\left[lo{g}_2\frac{2LtR+K}{K}\right]$



Frame size K bit long
Propagation delay t sec/km
Channel capacity = R bits/ sec

$U=\frac{w.\frac{K}{R}sec}{\frac{K}{R}sec+2Lt}$
$1=\frac{\frac{wK}{R}sec}{\frac{K+2LtR}{R}},w=\frac{K+2LtR}{K}$
$2^n=\frac{K+2LtR}{k},n=\left[lo{g}_2\frac{K+2LtR}{K}\right]$