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A customer wants to add Flash drives to their VNX array and use them as FAST Cache. They
want at least 250 GB of FAST Cache capacity on each SP. There are no Flash drives in the array
and they want to follow EMC best practices on spare disks. What should be used?
A.
Five 200 GB Flash drives
B.
Six 100 GB Flash drives
C.
Seven 200 GB Flash drives
D.
Eight 100 GB Flash drives
The answer should be A?
200GB X2 = 400GB per SP, mirror with dual controller = 4 x 200GB SSDs.
Additional one SSD for HS so total should be 5.
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Following your calculation, you forgot a factor 2 :
200GB X2 = 400GB per SP
x2 for the 2 SP
x2 for The mirror
+1 spare
That’s 6+1 disks … which is proposed in C
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It’s 250GB / SP, that means a total of 500GB, and you can only get that with 6 x 200 GB in RAID10 + 1 hot spare
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All of the answers are incorrect. The right way to calculate it for the total number of drives for both SPs and a hot spare requires meeting the capacity needs, doubling the value for mirrors (not adding like kerkael, and doubling the value again for each SP, then adding one disk for a hot spare.
** remember that a 100 GB Flash drive holds around 93 GB, and a 200 GB Flash drive holds about 186 GB
the FAST Cache requirement is 250 GB. FAST cache is done in raid 1 sets and a hot spare disk is required.
drives required == # of drives to meet requirement + mirror of required drives + hot spare drive
for 100GB drives, we need 3 to meet the 250GB requirement.
==> that is 3+3+1 == 7x 100GB drives
for 200GB drives, we need 2 to meet the 250GB requirement.
==> that 2+2+1 == 5x 200GB drives.
since two SPs are needed, multiply the required +mirrors by 2
ie, 2(3+3)+1 = 13 drives @ 100GB
or 2(2+2)+1 = 9 drives @ 200GB
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Hi All,
How much % is this dummy valid? I want to attempt it in the coming weeks.
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