5) Calculate the reliability of the system shown in Fig. 5 for a 1000 hrs. mission....

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5) Calculate the reliability of the system shown in Fig. 5 for a 1000 hrs. mission. The mission starts at T = 1000 hr. in the life of the system. Component Useful failure rate, f/hr. Mean wear-out time, hrs. Standard deviation, hrs. 1 1 x 10-5 800 400 1 3 2 10 x 10-5 1600 600 2 4 3 2 x 10-5 5000 200 4 5 x 10-5 6000 1000 Example The useful life failure rate of a component is 10 f/106 hours and the mean wear our life is 1000 hours with a standard deviation of 100 hours. What is the component reliability for a mission of 750 hours starting from 300 hours from now? = 10-5 f/hr R(t) = R(t) x R (t) -t = e R(t) = e z = (t - M) / o M = 1000 hr -750 x 10-5 = 0.9925 o=200 hr From the Table: z = 0.5 = 7.0 R(1050) = 0.5-0.1915 = 0.3085 R(300) = 0.5 +0.5 = 1.0 R(t) = 0.3085/1.0 = 0.3085 R(t) = 0.9925 x 0.3085 = 0.306 R(t) = F(z) = 0.1915 = 0.5 T = 300 hr R. (T+1) R (T) KO 0300 = R (1050) R (300) 1000 0 t = 750 hr 1050 0.5 Lin hours 9 5) Calculate the reliability of the system shown in Fig. 5 for a 1000 hrs. mission. The mission starts at T = 1000 hr. in the life of the system. Component Useful failure rate, f/hr. Mean wear-out time, hrs. Standard deviation, hrs. 1 1 x 10-5 800 400 1 3 2 10 x 10-5 1600 600 2 4 3 2 x 10-5 5000 200 4 5 x 10-5 6000 1000 Example The useful life failure rate of a component is 10 f/106 hours and the mean wear our life is 1000 hours with a standard deviation of 100 hours. What is the component reliability for a mission of 750 hours starting from 300 hours from now? = 10-5 f/hr R(t) = R(t) x R (t) -t = e R(t) = e z = (t - M) / o M = 1000 hr -750 x 10-5 = 0.9925 o=200 hr From the Table: z = 0.5 = 7.0 R(1050) = 0.5-0.1915 = 0.3085 R(300) = 0.5 +0.5 = 1.0 R(t) = 0.3085/1.0 = 0.3085 R(t) = 0.9925 x 0.3085 = 0.306 R(t) = F(z) = 0.1915 = 0.5 T = 300 hr R. (T+1) R (T) KO 0300 = R (1050) R (300) 1000 0 t = 750 hr 1050 0.5 Lin hours 9