Little\'s law, queuing systems A large machine consists of many identical compon

Question

Little's law, queuing systems A large machine consists of many identical components. Suppose that the number of components that will fail forms a Poisson process with rate 1 per hour, i.c., the time until a component will fail is exponential. Suppose that the repair time for a failed component is also exponential with average 40 minutes. i. Suppose that the whole machine is nonoperational if more than 2 components failed. What is the proportion of the time that the machine is nonoperational? ii. Suppose now that we would like to make the average number of failed components to be less than 1.5. What should be the average repair time in order to satisfy this requirement?

Explanation / Answer

= Average arrival rate = 1 per hr.
= Average service rate = 60/40 = 1.5 per hr.

(i)

We need to find, Prob(more than 2 components in system)
Pn>2 = ( / )2+1 = (1/1.5)^3 = 0.296

(ii)

Average number of components in system = Ls = 1.5

So, / ( - ) = 1.5
or, 1 / ( - 1) = 1.5 (given = 1)
or, = 1.67 per hr.

Refer: Queuing formulae for M/M/1 system

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