Refer to the Exhibit.
A Storage Policy for a Virtual SAN is set to the default policy, as shown in the Exhibit.
Which change would reduce the storage consumption by one third?
A.
Number of failures to tolerate = 1
B.
Number of disk stripes per object = 2
C.
Number of failures to tolerate = 3
D.
Number of disk stripes per object = 1
A, хз почему
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a: Number of failures to tolerate
Defines the number of host, disk, or network failures a virtual machine object can tolerate. For n failures tolerated, n+1 copies of the virtual machine object are created and 2n+1 hosts with storage are required.
Default value is 1. Maximum value is 3.
https://pubs.vmware.com/vsphere-55/index.jsp?topic=%2Fcom.vmware.vsphere.storage.doc%2FGUID-C8E919D0-9D80-4AE1-826B-D180632775F3.html
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How does the “number of failures to tolerate” reduce the storage consumption by one third? but if you changed the “number of disk stripes per object” from 3 to 2 that would reduce the drive usage by one, since there are 3 to begin with it would be reduced by one third. since “number of disk stripes per object” is how many drives are used to stripe the replica of the object. I think the answer is B
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MI,
Because the failures tolerate value is higher than the default (1). You need more copies of the virtual machine, so more disk space is used.
2(failures) + 1 copy = 3
1(failures) + 1 copy = 2
So the storage required is reduced by one third. This is how I understand it.
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