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# Scenario: An organization is deploying a new business application in their environment. The new application requires 1TB of storage space for business and application data. During peak workload, the...

Scenario:

An organization is deploying a new business application in their environment. The new application requires 1TB of storage space for business and application data. During peak workload, the application is expected to generate 4900 IOPS (I/O per second) with a typical I/O size of 4KB.

The available disk drive option is 15,000 rpm drive with 100 GB capacity.

Other specifications of the drives are:

• Average seek time = 5 millisecond
• Data transfer rate = 40MB/sec

Directions

Calculate the required number of disk drives that can meet both capacity and performance requirements of an application. See pages 43-44 in textbook for the calculations.

Step 1. - Calculate time required to perform one I/O, which depends on disk service time.

• Disk service time = Av. seek time + rotational latency + data transfer time
• Rotational latency is of the time taken for full rotation. (.5/(rpm/60 sec))
• Data Transfer Time = (Typical I/O)/Data Transfer Rate

Step 2. - Calculate maximum number of IOPS a disk can perform.

(IPOS=1/Disk Service Time) x Utilization

Step 3. - Calculate number of disks required to meet the application requirement.

Review the below video to understand IOPS:

1. Which components constitute the disk service time?
2. Which component contributes the largest percentage of the disk service time in a random I/O operation?

(Application IOPS / Max # IOPS (Step2))

• Show all of your work using all of the steps to arrive at the required number of disks.

Answered Same Day Mar 20, 2020

## Solution

Amit answered on Mar 22 2020
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Table of content
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Step 1
As it is given that
Average seek time = 5 msec.
The rotational latency is calculated as half time in use of complete rotation [Narayanan et al, 2009]. As, it is given that the
Rotation speed = 15000
sec
So, time required for single revaluation = 1/ (15000/60) seconds.
Thus, Â½ revolution = 0.5 / (15000/60)
= 2 Micro second
Av. Seek time (given) = 5 micro second.
The given transfer rate of data= 40 MB/second
So, 4 KB input/output transfer will require time = 4 KB/ 40 MB
= 0.1 micro...
SOLUTION.PDF