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Solutions to the Fourth COSC 6360 Quiz for Fall 2012

Solutions to the Fourth COSC 6360 Quiz for Fall 2012. Jehan-François Pâris jfparis@uh.edu. RAID Part A. Consider a RAID level 6 array with ten disks. Which fraction of the array space can be occupied by data? (5 points) . Answer. Consider a RAID level 6 array with ten disks.

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Solutions to the Fourth COSC 6360 Quiz for Fall 2012

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  1. Solutions to theFourth COSC 6360 Quizfor Fall 2012 Jehan-François Pâris jfparis@uh.edu

  2. RAID Part A • Consider a RAID level 6 array with ten disks. • Which fraction of the array space can be occupied by data? (5 points)

  3. Answer • Consider a RAID level 6 array with ten disks. • Which fraction of the array space can be occupied by data? (5 points) • Each stripe contains 2 parity blocks and 10 – 2 = 8 data blocks • 80 percent

  4. RAID Part B • Consider a RAID level 6 array with ten disks. • How would you update a block dand its corresponding parity blocks p and q?

  5. Answer • How would you update a block dand its corresponding parity blocks p and q? • Read old block d, old parity block p and old parity blockq • Computenew p = old p old d  new d new q = old q old d  new d • Write new d, new p and new q

  6. LFS • What is the cost of a write in a log-structured file system, when its segment cleaner has to clean four segments to produce two clean segments? (10 points)

  7. Answer • What is the cost of a write in a log-structured file system, when its segment cleaner has to clean four segments to produce two clean segments? (10 points) • We use the formula Cost = 1/(1 – u) and note that u = 50 percent • Cost is 4 disk accesses per new block being written

  8. Answer (long way) • When we clean up 4 segments, we must • Read the contents of the 4 segments • Write somewhere else the 2 segments containing live data • The process gives us two clean segments that we can use later for new writes • The total cost of these new writes is • 4 + 2 + 2 = 8 • 4 disk accesses per block being written

  9. Journaling file systems • What are the main advantages and disadvantages of using the data mode in journaling file systems?(210 points)

  10. Answer • What are the main advantages and disadvantages of using the data mode in journaling file systems?(210 points) • Advantage: Safest solution • Disadvantage: Slowest

  11. Soft updates • Which dependency information do soft updates maintain? (10 points)

  12. Answer • Which dependency information do soft updates maintain? (10 points) • They maintain dependency information about cached pieces of metadata at the directory entry level • This i-node must be updated before/after this directory entry

  13. NFS • How can non-volatile RAM (NVRAM) improve the response time of an NFS server? (10 points) • Is there a cheaper way to achieve the same result? (10 points) (Hint: just mention it)

  14. Answer • How can non-volatile RAM (NVRAM) improve the response time of an NFS server? (10 points) • NVRAM allows servers to respond to write requests without waiting for the completion of their own write requests • Is there a cheaper way to achieve the same result? (10 points) (Hint: just mention it)

  15. Answer • How can non-volatile RAM (NVRAM) improve the response time of an NFS server? (10 points) • NVRAM allows servers to respond to write requests without waiting for the completion of their own write requests • Is there a cheaper way to achieve the same result? (10 points) (Hint: just mention it) • Safe asynchronous writes

  16. NFS • Why must NFS client requests be both self-contained and idempotent? (210 points)

  17. Answer • Why must NFS client requests be both self-contained and idempotent? (210 points) • They must be self-contained becauseNFS is stateless and keeps no record of previous requests • They must be idempotent to allow clients to resend any request for which they did not get a reply

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