In cloud systems, one tries to scale-out his server-based system by deploying im

Question

In cloud systems, one tries to scale-out his server-based system by deploying images on virtual machines. In some cases the virtual machines are identical in terms of resources. This happens with probability a. So with probability (1-a) the virtual machines are not identical. With identical virtual machines, obviously the performance levels of the deployed images are the same. However, even without identical virtual machines, one could still have the same performance level. Without knowing the virtual machine resources, one cannot exactly determine if the virtual machines are the same. Suppose that the performance are the same with probability b, and not the same with probability (1-b). a) Suppose you are an engineer at Amazing. After several runs within your closer of servers, you have the performance data of virtual machines deployed for a client. In n = 18898 pairs of measurement, n1 = 5.902 pairs all have a high level of performance, whereas n2 = 5.440 pairs all have a low level of performance. The remaining 7.556 pairs have one low and one high level of performance. Estimate a, the probability that a pair of server software instances run on identical virtual machines. b) Estimate b, the probability that the performance of any server instance is high.

Explanation / Answer

b) In the main question, it is written clearly that, b = Prob(the performance of virtual machines are same ). But, in the 2nd question, b is interpreted as the probability of performance of any server instance is high.

Now, if we take b as the probability that performance are same then the estimate of b = (pairs all have high level of performance + pairs all have low level of performance) / n = (n1+n2)/n =(5902+5440)/18898 = 11342/18898 = 60.02% (ans)

a) Let us define two events. SIV = pair of server software instances run on identical virtual machines.

and P(Performance same| SIV) =1 and P(performance different|SIV) = 0

P(SIV|Performance same) = 1/2 and P(performance same)=0.6

So, P(SIV) = P(SIV|Performance same) * P(performance same)/P(Performance same| SIV)

=0.5*0.6 = 0.3 (ans)

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