Asm voting diskgroup created in normal redundancy size difference

The checkresults is correct.
It is the way ASM works. The redundancy level is the critical prameter.

external redundancy => no data mirror
normal redundancy => each block has 1 mirror in a different failgroup
high redundancy => each block has 2 mirrors in different failgroups

It doesn’t matter how many failgroups are existing. You have to copy each block in a 2nd failgroup when normal redundancy is used.
Each failgroup should have the same size. Otherwise it is hard to calculate the useable space in the diskgroup.

The calculation for normal redundancy is:
Calculate the summary of all failgroups and devide it by 2.
=> 3*10GB / 2 = 15GB

That’s it for the space calculation.

The voting diskgroup is often is misunderstand in RAC environments…
We always need an unequal numer of failgroups for voting - the redundancy level defines the minimum number of failgroups for voting. (1 for external, 3 for normal and 5 for high)

Important! That does not affect the space calculation in the diskgroup itself!

Summary:
The check is fine. It is a misunderstand of the concept of data mirroring in ASM.

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