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Consider a disk I/Osystem in which an I/0 request arrives at a rate 100 IOPS. The service time Rs=8ms.

The service time Rs=8ms. Calculate the following measures of disk performances:
1) Utilisation of I/O controller (U).
2) Total Response Time (R).
3) Average Queue Size.
4) Total time spent by request in the Queue. If the service time is half i.e. Rs=4ms and I/O request arrives at a rate of 100 OPS, them calculate the following measures of disk performance:
1) Utilization of I/O controller (U)
2) Total Response Time (R)
3) Average Queue Size.
4) Total time spent by request in the Queue

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Given :

Arrival Rate (a) = 100 IOPS, So

$R_a = \frac{1}{a} = \frac{1}{100}$

$= 0.01 \times 10^2 = 10 ms$

Rs = 8 ms Given

(a) Utilization of i/o controller is

$U = \frac{Rs}{Ra} = \frac{8}{10} = 0.8$ i.e. 80%

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

$\frac{8}{1-0.8} = 40 ms$

(C) Average Queue Size $= \frac{U^2}{1-U}$

$\frac {(0.8)^2}{1-0.8} = 3.2 ms$

(d) Total time spent by request in Queue = U x R

= 0.8 x 40 = 32 ms

Now, Service time is half i.e. Rs = 4 ms

Therefore,

(a) Utilization U = Rs/Ra = 4/10 i.e. 0.4 ie. 40%

(b) Total Response time (R) = $Rs/_1-u = \frac{4}{1-0.4} = 6.67 ms$

(C) Average Queue Size $= \frac{u^2}{1-u} = \frac{(0.4)^2}{1-0.4} = 0.26 ms$

(d) Total time spent by request = $U \times R = 0.4 \times 6.67 = 2.67 ms$