Problem 3.3 In a data center, it is critical to store information redundantly so

Question

Problem 3.3 In a data center, it is critical to store information redundantly so that no data is lost if a single hard drive fails. In this problem, we will examine a few methods1 for adding redundancy across the n hard drives in a single server rack:

• No Redundancy: Each drive in a rack stores different data. Data is lost if one or more out of the n drives fail.

• Repetition: Each drive in a rack stores a copy of the same data. Data is lost only if all n drives in the rack fail.

• Single Loss Recovery: The first n 1 drives in a rack store different data and the last drive stores the XOR of the data from these drives. Data is lost if two or more out of the n drives fail.

• Double Loss Recovery: This is a more complex scheme that can tolerate two drive failures. Data is lost if three or more out of the n drives fail.

Below, we have illustrated examples of the first three strategies for n = 3 drives and the fourth strategy for n = 8 drives. The symbols b1, b2, b3, b4 refer to blocks of bits and the symbol refers to the XOR operation. You may assume that drives fail independently of one another and the probability that a drive is working at the end of a day is p.

(a) For each of the four strategies described above, write down an expression for the probability that the rack loses no data at the end of the day. Your expressions should be in terms of n and p. Simplify as much as you can.

(b) Evaluate the probabilities from part (a) for n = 8 and p = 0.9.

(c) To further increase the reliability, you have decided to maintain 3 identical copies of each rack. Using your expressions from part (a) as a building block, determine, for each of the four strategies, the probability that at least one of these three racks has lost no data at the end of the day.

(d) Evaluate the probabilities from part (c) for n = 8 and p = 0.9.

No Redundancy Repetition Single Loss Recovery b1 b2 b3 b b b1 Double Loss Recovery | b1 II b2 || b3 11 b4 11b1 b211bg(D b411b2 (D b311b1b2ball

Explanation / Answer

a.) Stratergy A: Probability that 0 drives fail at the end of the day:p^n

Stratergy B: Probability that upto n-1 drives fail at the end of the day= 1- Probability that all n drives fail at the end of the day= 1-(1-p)^n

Stratergy C: Probability that less than 2 drives fail at the end of the day= probability that 0 drives fail+ probability that 1 drive fails = p^n+n*(1-p)(p)^(n-1)

Stratergy D: Probability that 0, 1 or 2 drives fail at he end of the day=

p^n+n*(1-p)(p)^(n-1)+(n)(n-1)/2(1-p)^2(p)^(n-2)

(b) 1.) .4305

2.) .9999

3.) 0.8131

4.) .9619

(c) Probability that at least one of the three do not loose that is that 1- pr. that all loose data. Assuming that all the three racks behave independently of each other and as a wholeit is: 1- (1-Pr. calculated in part a)^3

(d) The evaluated probabilities are as follows:

1.) 0.8153

2.)1

3.)0.9935

4.)0.9999

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