An application needs 10GB of RAID 1 for log files, 20GB of RAID 5 for data files, and 20GB of RAID 5 for the operating system. All disks will be 10GB in capacity. Which of the following is the minimum number of disks needed for this application?
An application needs 10GB of RAID 1 for log files, 20GB of RAID 5 for data files, and 20GB of RAID 5 for the operating system. All disks will be 10GB in capacity. Which of the following is the minimum number of disks needed for this application?
To determine the minimum number of disks needed, we can break down the requirements: RAID 1 for log files requires mirroring, so for 10GB of usable space, 2 disks of 10GB each are needed. RAID 5 for data files necessitates at least three disks because one disk’s worth of capacity is used for parity, meaning for 20GB of usable space, 3 disks are required. Similarly, for the operating system, another 3 disks are required for RAID 5 to get 20GB of usable space. Summing these up gives us 2 disks for RAID 1 and 6 disks for RAID 5 (3 for data files and 3 for the operating system), totaling 8 disks. Thus, the correct answer is 8 disks.
To calculate the minimum number of disks needed for the given RAID configurations, let's break down each requirement: RAID 1 for log files: This requires mirroring, so each disk contributes half of its capacity. Since we need a total of 10GB usable space, we'll need 20GB raw capacity (10GB mirrored). Thus, we need 20GB / 10GB = 2 disks. RAID 5 for data files (20GB): RAID 5 requires at least three disks, and one disk worth of capacity is used for parity. So, for 20GB of usable space, we'll need 3 disks. RAID 5 for the operating system (20GB): Same as the data files, we'll need 3 disks for 20GB of usable space. Adding these up: RAID 1 (log files): 2 disks RAID 5 (data files): 3 disks RAID 5 (operating system): 3 disks Total disks needed: 2 + 3 + 3 = 8 disks. So, the correct answer is: C. 8
C is not correct, it does not specify that the 2x 20gb of Raid5 need to be on separate arrays so it is 2 disk for the raid1 = 10gb and 5 disk of raid5 = 40gb.
C. 8 Disks (per AzadOB's explanation)