"Information Storage and Management" Book Pg 50 Qno. 6

Q6: Consider a disk I/O system in which an I/O request arrives at the rate of 80 IOPS. The disk service time is 6ms.

     a. Compute the following:

     -Utilization of the I/O Controller.

     -Total response time.

     -Average queue size.

     -Total time spent by a request in a queue.

     b. Compute the preceeding parameter if the service time is halved..

The solution I have calculated are as follows: Please let me know if I have committed any mistakes...thanks.

a.

     1. Utilization=Rs/Ra; Given: a=80 IOPS=> 0.08 I/O per millisec.

          therefore, Ra=1/(a)=> 1/(0.08)=>12.5

          U=6/12.5 => 0.48 or 48%.

     2. Response Time(R)= Rs/(1-U)

          R= 6/(1-.48) => 6/(0.52) => 11.5 milliseconds.

     3. Average Queue Size= (U*U)/(1-U)

          => (.48*.48)/(1-0.48)

          => (.2304)/(.52)

          => 0.44.

     4. Total Time Spent by a request in a Queue= U*R

          => 0.48*11.5

          => 5.52 milliseconds.

b. If service time is halved, i.e Rs=3millisecons.

     1. utlization=> 3/12.5=> 0.24 or 24%

     2. Response Time(R) => 3/(1-0.24)=> 3/(0.76)=>3.94.

     3. Avg. Queue Size, (0.24*0.24)/(1-0.24)=>(0.0576)/(0.76)=>0.075.

     4. Total Time Spent by a request in a Queue= U*R

          => 0.24*3.94

          => 0.946 milliseconds.

please guide me if I am wrong anywhere...thanx...