Exam DEA-1TT4 topic 1 question 19 discussion

IOPS needed is 4500 Each drive is 90 IOPS So to reach the 4500 IOPS, we need 50 drive

Answer is C

Ans : 50 To calculate the minimum number of disk drives needed for a RAID 0 configuration to meet the capacity and performance requirements, let's start by determining the number of drives required for capacity: The application requires 2 TB of storage space, and each available disk drive has a capacity of 250 GB. So, to achieve 2 TB (2000 GB), you'd need 2000 GB / 250 GB per drive = 8 disk drives. Next, let's consider the IOPS requirement during peak workloads: The application is expected to generate 4500 IOPS with a typical I/O size of 4 KB. Each drive can perform a maximum of 90 IOPS with a 70% utilization. To calculate the number of drives needed for IOPS, divide the required IOPS by the maximum IOPS per drive: 4500 IOPS / 90 IOPS per drive = 50 drives Therefore, for a RAID 0 configuration to meet both the capacity and performance requirements (considering the higher number between capacity and performance), you would need a minimum of 50 disk drives.

For req performance (70% utilization - 90 IOPS): 4500/90=50

Shoul be 50. IOPS are already defined at 70%. "Consider an example in which the capacity requirement for an application is 1.46 TB. The number of IOPS generated by the application at peak workload is estimated at 9,000. IOPS. The vendor specifies that a 146-GB, 15,000-rpm drive is capable of doing a maximum 180 IOPS. In this example, the number of disks required to meet the capacity requirements will be 1.46 TB/146 GB = 10 disks. To meet the application IOPS requirements, the number of disks required is 9,000/180 = 50. <<However, if the application is response-time sensitive, the number of IOPS a disk drive can perform should be calculated based on 70-percent disk utilization. Considering this, the number of IOPS a disk can perform at 70 percent utilization is 180 × 0.7 = 126 IOPS. Therefore, the number of disks required to meet the application IOPS requirement will be 9,000/126 = 72.>> Information Storage and Management . Wiley. Kindle Editio.

from course ISM v4: Step 2: Calculate the maximum number of IOPS the drive can perform − Maximum number of IOPS = 1 / 6.25 ms = 160 IOPS − Maximum number of IOPS at 70% utilization = 160 x 0.7= 112 IOPS • Step 3: Calculate the number of drives for the application − Drives required to meet performance requirement = 5200 / 112 = 47 − Drives required to meet capacity requirement = 1.5 TB / 250 GB = 6 • Number of drives required = Maximum (Capacity, Performance) − Maximum (6, 47) = 47 disk drives

Should be D , You want it to perform even with 100% utulization so add +25% slack