Hey Everyone, Please post your solutions to the Jan. 31 In-Class Exercise to this thread. Best, Chris

-- Jan. 31 In-Class Exercise

3 Hours = 1/8 = 1/2920 of a year If both disks have to fail in that time then it becomes 1/(2920)^2 = 8,526,400 1/10 x 1/10 x 1/8,526,400 = 1 in 852,640,000 years 852,640,000 / .334 = 2,552,814,371 years

-- Jan. 31 In-Class Exercise

1/(29200^2) = 852640000 3/10, so MTTF is 3.33 10/3 x 85264000 = 2,842,133,333.3 years -Casey Reyes

-- Jan. 31 In-Class Exercise

Odds that one disk out of three fails within a given year will be 3/10. So MTTF for 3 disks will be 3.3 years. 1/29,200 * 1/29,200 = 1 in 852,640,000 years for both disk failures. So the mean time to data loss is 3.3 * 852,640,000 = 2,557,920,000 years.

-- Jan. 31 In-Class Exercise

Assuming that 10% of disks fail in a given year, and each replacement takes 3 hours, we have 1/10 * 1/2920 * 1/29200 failures. The odds that one of those drives fails in a year is 3/10. So the MTTF is 3.33 * 29200 * 29200 years, or 852,640,000 years.

-- Jan. 31 In-Class Exercise

Assuming replacement takes 3 hours = 1/8 of a day = 1/2920 of a year. The odds of both redundant disks failing is 1/10 * 1/2920 * 1/2920 = 1/85264000. The odds that 1 of the 3 disks fails is 3/10. The MTTF is 3.33 years. Mean time to data loss should then be 3.33 * 85264000 = 283,929,120 years.

-- Jan. 31 In-Class Exercise

Odds of first disk fails = @BT@1/2920@BT@. Odds of second disk fails = @BT@(1/2920)^2 = (1/8526400)@BT@ The odds are @BT@(1/10) x (1/8526400)@BT@ or one in 85,264,000 failures. MTTF is 3.33 years for three disks @BT@3.33 * 85264000 = 283929120 years@BT@

-- Jan. 31 In-Class Exercise

3/10 = chance for disk failure with 3 disks @BT@1/(29,200^2) = 1/(852,640,000)@BT@ @BT@1/(852,640,000) times 1/3.33 = 1/(2,839,291,200)@BT@ mean time for data loss

-- Jan. 31 In-Class Exercise

So let say we have one data disk and two redundant disks. If replacement takes (1/2920)^2 of a year, then the odds are 1/10 * (1/2920)^2 or one in 85264000 failures. The odds that one of two disks fails in a given year is 3/10. SoMTTF for 3 disks is 10/3 years. So the mean to data loss is (10/3) * 85264000 = 284,213,333.33 years

-- Jan. 31 In-Class Exercise

If replacement takes 3 hours = @BT@(1/2920)@BT@ of a year, then the odds that two redundant disks fail during replacement time are @BT@(1/10)(1/2920)(1/2920) = (1/85264000)@BT@. The odds that two disks fail in a given year is @BT@(3/10)@BT@. MTTF for 3 disks is @BT@(10/3)@BT@ years. Mean time to data loss is @BT@(10/3)(85264000)@BT@ = 28,421,333 years