Consider the use of 10K-bit size frames on a 10 Mbps satellite channel with 270 ms delay.

Question

What is the link utilization for STOP-AND-WAIT ARQ technique assuming P = 10-3 ?

PLEASE ANSWER WITH WORKING

Answer 1

We know link utilisation of a network =  1 / (1 + 2a) where

                                                 a   =  propogation time / transmission time

In calculating transmission time as we know in stop n wait ARQ , only 1 frame can be sent at a time hence the data size is taken of 1 frame only.

So transmission time   =   Data size / Data rate

                                  =   10 * 10³  bits / 10 * 10⁶  bps

                                  =    1 ms

Propogation delay given  =  270 ms

So  the value of a as mentioned earlier  =  270 / 1   =  270

So link utilisation         =     1 / ( 1 + 2 * 270 )

                                  =     1 / (541)

                                  =     0.19 %