Question

2. A manufacturing engineer observes the wear-out rate of a milling machine tool fits a Weibull hazard model to the tool wear data. The parameters of the model are Determine the reliability of the tool after 10 h, the MTTF and the mean residual life at 10 h.

Answer 1

Given that beta = 2.25 alpha = 30 For weibull distribution The probability function is f(t) = 1.e^(beta t)alpha and density function is f(t) = d/dt f(t) = {alpha beta^alpha t^alpha - 1 e^-(beta t) alpha t > 0 0 t > 0} Reliability function R(t) = 1 - f(t) = e^(beta t) alpha MTTF = integral^infinity_0 R(t) dt given beta = 2.25 if alpha = 30 values will tends to infinity we use alpha = 0.3 assuming typo cssoD The reliability of the tool after 10h = e^-(beta t)alpha = e^(2.25 times 10)(0.3) = 0.078 \r\n MTTF = integral^infinity_0 R(T) dt integral is invalid for integer range mean residual life at 10h failure rate function = alpha + (a - 1) 0.3(2.25)^0.3 10(0.7) = 0.3 times 1.275 times 0.1995 = 0.07633269