How Do Stochastic and Renewal Processes Determine System Performance and Costs?

[llovyna]

I really need your help for a solution to these exercises. I will be so grateful.1/ passengers arrive at a train station according to a poisson process of rate lambda per minute and trains depart station according to a renewal process with inter-departure times uniformly distributed between a and b minutes. Find the long run fraction of time when the station is empty.2/ An item with exponential life time distribution of rate lambda is installed in a system. It is inspected periodically, and is replaced immediately by a new item of same life time distribution if found defective at an inspection. The inspection is performed every h units of time, so they occur at times t=h,2h,3h..., and the time to perform an inspection may be ignored. Suppose each inspection costs a, a failed item in system incurs a continuous cost at rate of b per unit time, and there is no replacement cost. Find the long run cost per unit time.

Thank you so much for your time.

 

[Valkarie]

1/ The fraction of time when the station is empty is equal to the probability that the station is empty at any given minute. This can be calculated using the Poisson process formula: P(X=0) = e^(-λ)where λ is the arrival rate (in this case, the value of lambda per minute).2/ The long run cost per unit time is equal to the cost of performing an inspection (a) plus the cost of a failed item in the system (b) multiplied by the probability that the item is defective at any given inspection. This can be calculated using the exponential distribution formula: P(X>t) = e^(-λt) where λ is the rate of failure (in this case, the value of lambda) and t is the inspection interval (in this case, the value of h). Therefore, the long run cost per unit time is given by: Cost = a + b * e^(-λh)