+2:1 is N+M/b, where N == data stripe units, M == FEC stripe units, and b == # of unique drives we'll use on a node (within any given stripe).  A "stripe unit" is 128 KiB in size, being 16 x 8 KiB blocks.  On a 5 node cluster at +2:1, N = 8 (given a sufficiently large file), M = 2 and b = 2, per stripe.   A 1 MiB file, then, on that 5 node cluster, at +2:1, has 8 fully formed data stripe units and 2 fully formed FEC stripe units, in a single stripe.  The amount of FEC is thus 2 * 128 KiB, or 256 KiB, which is M/(N+M) or 20% towards protection overhead.  So you'll need at least 256 KiB of additional space to store this 1 MiB file, not including space for inodes, metadata, etc, which isn't accounted for in the above theory...
Hi, I am working on this calculation. Not sure I am right. 1024 KiB File, on 5 nodes cluster, N+2:1 protection, how much space do we need for this file.  My calculation: 128KiB strip x 8 to get 1024 KiB and will take 1024KiB. For protection, 1024 x 20% since 4+1. So the total is 1024 + (1024x 20%). Am I right?
Space Calculation on Isilon