Question

A system is subject to two types of failure. When a failure occurs, it is type 1 with 40% probability The time from when the system is repaired until a new failure occurs exponentially distributed with mean of 30 days. If the failure is type 1, two different components are repaired by two repairmen in exponentially and independently distributed periods each with a mean of 2 days. If the failure is type 2, one component is repaircd by one repairman in two stages, cach distributed exponentially and independently with a mean of 3 days. Each type 1 failure requires some parts to be changed with new ones which costs $2000 in total, and cach type 2 failure requires new parts which costs $3000 in total. Assume that labor cost is $300 per day per repairman. Compute the long run average repair cost per month

Answer 1

Let denotes the X Tepair cerst per mth3e tays) and y denetes o f faures in ainterval af time t 30(days) we are required to find E(x) T Since the time trm when the system is repaired urtil a mew failure eeCars is eponential with mean of 30 day s ExPlrate o) e. T 上xp(mean=30) i-e. T~ y Prissn(x30!) . E()E(x1 faiure is af tyre 1) P(faiure is f the) +E ( failre is of tyre ) P(failure is eftype is of ty pe ) - E (xlfailare +E (xfaiure is of tabe 2) G0 obtained as fo llows Now, E(x failure ts of type 1) No. 0f day s requinred for repairing is x ExP (2) Leち, .(1) id Where Xi C) x + X,2 X and 2 +2 4 E) 300 x 4x2-24 to tat ezbe cted nd labo ur cost fFor 4 dadS 15 24o E(xfailure is of tare 1) 1 807 1-e is of ty pe 2) 300 x(3+3)- (x1failure Also E(x) -240 x + 1800 x From