# How many disks are required to meet both capacity and performance requirements?

An application requires 3.5 TB of capacity. The application generates 2300 IOPS to disks during peak workloads. The vendor indicates that a 146 GB, 15K rpm drive is capable of performing a maximum of 150 IOPS.

How many disks are required to meet both capacity and performance requirements?

A.
15

B.
22

C.
24

D.
60

## 3 Comments on “How many disks are required to meet both capacity and performance requirements?”

1. Nik Mistry says:

Formulas required to calculate:
Disks required to meet an application’s capacity need (Dc):
Dc = (Total capacity required) / (Capacity of a single disk)

Disks required to meet application’s performance need (Dp):
Dp = (IOPS generated by an app at peak workload) / (IOPS serviced by single disk)

IOPS serviced by a disk(s) depends upon disk service time (Ts):
Ts = Seek time + (0.5/(Disk rpm/60)) + (Data block size / Data transfer rate)

Ts is the time taken for an I/O to complete, therefor IOPS serviced by a disk(s) is equal to (1 /Ts)

Disk required for an app = max(Dc, Dp)

Dc = 3.5TB/146 GB = 3500/146 = 23.97 = 24 Drives
Dp = 2300 IOPS / 150 IOPS = 15.33 = 15 Drives

Disk required = max(Dc,Dp) = max(24,15) = 24

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2. Alex says:

I was confused with capacity: 3,5TB = 3.5*1024 = 3584GB
So
Dc = 3.5TB/146 GB = 3584GB/146GB = 24.54 = 25 HDD’s

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3. Arslan 00966571643196 says:

meaning of question is to provide both capacity wise and performance wise drives. capacity wise it is 24 and performance wise 15. both are available in multiple choice then how we can choose correct answer.

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