Given :

Arrival rate (a) = 80 IOPS i.e. $R_a = \frac{1}{a}$

service time = 6 ms = $\frac{1}{80} \times 10^3$

= 12.5 ms

To calculate :

1) Utilization of I/O controller $\therefore U = \frac{Rs}{Ra}$

$\frac{6}{12.5} = 0.48$

= 48 %

2) Total Response time (R) = $\frac{Rs}{1-U}$

= $\frac{6}{1-0.48} = 11.53$

= $\approx 12 ms$

3) Average Queue size : $\frac{U^2}{1-U}$

= $\frac{(0.48)^2}{(1-0.48)} = 0.44$

4) Total time spent by a request in a Queue

= U * R

= 0.48 * 12

= 5.76 ms