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Question 46

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?

    Correct Answer: A

    When a storage object is configured with RAID-5 (Erasure Coding) on a 4-node vSAN cluster, the data is distributed across 3 nodes for data and 1 node for parity. Given that the number of disk stripes per object is set to 8, and considering the redundancy provided by RAID-5, this means each object is split into 8 stripes and distributed across the hosts. Since there are 4 nodes, this setup results in 2 components being created per host (8 stripes / 4 nodes = 2 components per host).

Discussion
jb844Option: A

I Think..."A" not "B" because: per oswaldo from link- 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.

oswaldo13Option: A

the correct option is 2. https://blogs.vmware.com/virtualblocks/2021/01/21/stripe-width-improvements-in-vsan-7-u1/

BroTigerOption: C

it's C: A 200GB object using a storage policy with RAID-5 and a stripe width of 8 would result in an object distributed across 4 hosts totaling 8 components.

BroTiger

sorry, my bad... it's A :)

GodMan114Option: A

I think A makes the most sense. 2 objects per host. Stripe width 8 across 4 hosts (3 storage, 1 parity).

ranauroOption: C

in a vsan 7 U1 we have this document https://blogs.vmware.com/virtualblocks/2021/01/21/stripe-width-improvements-in-vsan-7-u1/ A 200GB object using a storage policy with RAID-5 and a stripe width of 8 would result in an object distributed across 4 hosts totaling 8 components. so in this case, options C makes more sense

ranauro

sorry, my bad... the question wants to know about components per HOST, so 8/4 =2 components..

faissalmanOption: A

2 Components per each host, which means 8 components in total. the correct answer as per the question is: A.

jb844Option: B

per oswaldo from link- 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.