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Exercises –Availability 17/ 6 / 2014 1. Exercise: failure probability of a disk Compute the probability that a disk with MTTF = 100000 hours fails at least once every 3 years . If instead we have 2 disks, which is the probability that at least one of them fails? Solution: For one disk:   100000 26280 1   etXP ≈ 23% (also valid the approximate solution: 26280/100000  26%, since 0,26 << 1) For two disks:   2 100000 26280 1   etXP ≈ 41% 2. Exercise: computation of Reliability A load balancer(A) is connected in series with a group(B) of three parallel web servers. The load balancer have MTTF of 1000 days and MTTR of 2 days. Each web server have MTTF of 350 days and MTTR of 1 day. Compute the probability of n o failures in a t = 7 days period for both (A) an d (B) as well as for the whole system (A+B). Solution: RA(7) = 1-7/1000 = 0.993 RB(7) = 1- (7/350)3 = 0.999992 (parallel) RA+B(7) = RA(7) × RB(7) = 0.992992 (serial) 3. Exercise: Availability A load balancer(A) is connected in seri es with a group(B) of three parallel web servers. The load balancer have MTTF of 1000 days and MTTR of 2 days. Each web server (WS) have MTTF of 800 days and MTTR of 1 day. Compute the availability of the whole system. Solution: A = MTTF / (MTTF+MTTR) We can compute the availability through availability block formulas: ASerial = A1A2…An AParallel = 1-∏ (1-Ai)n MTTFA = 1000; MTTFA= 2 AA=1000/(1000+2)=0.998 MTTFws = 800; MTTRws= 1; AWS= 800/(800+1)=0.99875 AB = 1-(1-AWS)3 = 0.99999 AA+B = AA AB = 0.998 0.999999 = 0.99799 4. Exercise: availability Consider the following structure where MTTF and MTTR of the components are shown. Compute the availability of each component and of the whole infrastructure. Solution: We can compute the availability through availability block formulas:…