Western Digital gives the rate of non-recoverable read errors per bits read of i

Question

Western Digital gives the rate of non-recoverable read errors per bits read of its 8 TB WD80EFZX hard drive as less than 1 in 1014. Suppose such a drive is full. What can you say about the probability of encountering at least one non-recoverable read error while copying the entire drive, based on this information? 1. a. Give the exact answer and explain it. Consider that hard drive manufacturers use decimal, not binary prefixes in their capacity specifications, i.e. 1 TB is one trillion (1012) bytes. b. Give a decimal approximation of the probability in a. using the calculus fact that for small lxl and large integers n. You may give your answer as a percentage.

Explanation / Answer

(a) Total Hard drive = 8TB

1 TB holds = 240 bytes

1 byte = 8 bit

1 TB = 243 bits

Total bits in the hard drive = 8 * 243 = 246 bits

Probability that read errord per bit = 1/1014

so the given distribution is n = 246

p = 10-14

so Pr( at least one non- recoverable error) = 1 - P(no non- recoverable error)

= 1 - 246 C 0 * (10-14) 0 ( 1- 10-14)2 ^ 46

= 1 - ( 1- 10-14)2 ^ 46

= 1 - 0.4950

= 0.505

(b) ( 1 + x/n) n = ex

so expected number of non - recoveraable error in 8 TB hardrive = 10-14 * 246 = 0.7037

so Pr (at least one non - recoverable error) = 1 - Pr( X = 0)

Pr( X =0) when by poisson distiubtion = 0.7037

so Pr(X =0) = e- x/x! = 0.4948

Pr (at least one non - recoverable error) = 1 - Pr( X = 0) = 1 - 0.4948 = 0.5052

so. 50.5% that there is atleast one non - recoverable error.

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