Question: Some RAID systems improve reliability with intra-disk redundancy to protect against nonrecoverable read failures. For example, each individual disk on such a system might reserve one 4KB parity block in every 32 KB extent and then store 28KB (7 4KB blocks) of data and 4 KB (1 4KB block) of parity in each extent. In this arrangement, each data block is protected by two parity blocks: one interdisk parity block on a different disk and on intradisk parity block on the same disk. This approach may reduce a disk's effective nonrecoverable read error rate because if one block in an extent is lost, it can be recovered from the remaining sectors and parity on the disk. Of course, if multiple blocks in the same extent are lost, the system must rely on redundancy from other disks.
(a) Assuming that a disk's nonrecoverable read errors are independent and occur at a rate of one lost 512 byte sector per 1015 bits read, what is the effective nonrecoverable read error rate if the operating system stores one parity block per seven data blocks on the disk? Hint: You may find the bc or dc arbitrary-precision calculators useful. These programs are standard in many Unix, Linux, and OSX distributions. See the man pages for instructions.
(b) Why is the above likely to significantly overstate the impact of intradisk redundancy?