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May 11th, 2011 08:00

"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...

180 Posts

May 20th, 2011 09:00