5V0-22.21 · Question #15
An organization wants to configure a new storage policy based on the following requirements: Failures to tolerate = FTT 1/RAID-5 (Erasure Coding) Number of disk stripes per object = 8 IOPS limit for…
The correct answer is A. 2. Figure 2. Object using RAID-5, with a stripe width of 8, where there are 2 components per host. With vSAN 7 U1, a stripe width setting for erasure codes must be done in multiples of 4 or 6 respectively for the effective stripe width to be increased. This new method of…
Question
An organization wants to configure a new storage policy based on the following requirements:
Failures to tolerate = FTT 1/RAID-5 (Erasure Coding) Number of disk stripes per object = 8 IOPS limit for object = 0 Object Space Reservation = Thin provisioning Flash read cache reservation = 0% Disable object checksum = No Force provisioning = No The administrator creates the policy using storage policy based management and assigns it to a 100GB virtual machine on a 4-node vSAN cluster to test the results of the new storage policy. How many components will be created per host for the storage objects of the virtual machine on the vSAN datastore?
Options
- A2
- B1
- C8
- D32
How the community answered
(16 responses)- A81% (13)
- B6% (1)
- C13% (2)
Explanation
Figure 2. Object using RAID-5, with a stripe width of 8, where there are 2 components per host. With vSAN 7 U1, a stripe width setting for erasure codes must be done in multiples of 4 or 6 respectively for the effective stripe width to be increased. This new method of calculation is much more practical, as a RAID-1 mirror was far more likely to need a higher stripe width value than RAID-5/6. Objects using RAID-5/6 erasure codes would most likely not benefit from stripe widths beyond 2 or 3, if at all. The table below shows the stripe width settings for vSAN 7 U1 as it relates to the data placement scheme used. https://blogs.vmware.com/virtualblocks/2021/01/21/stripe-width-improvements-in-vsan-7-u1/
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