This is just an Excerpt from a larger document, click here to view the entire document. Numerical Examples The concepts discussed are best explained and understood by working out simple numerical examples. Let a computer system be composed of five identical terminals in series. Let the required system reliability, for unit mission time (T = 1) be R(1) = 0.999. We will now calculate each component's reliability, unreliability, and failure rate values. From the data and formulas just given, each terminal reliability Ri(T) can be obtained by inverting the system reliability R(T) equation for unit mission time (T = 1): Component unreliability is: Ui(1) = 1 - Ri(1) = 1 - 0.9998 = 0.0002. Component FR is obtained by solving for λ in the equation for component reliability: Now, assume, that component reliability for mission time T = 1 is given: Ri(1) = 0.999. Now, we are asked to obtain total system reliability, unreliability, and FR, for the (computer) system and mission time T = 10 hours. First, for unit time: Hence, system FR is: If we require system reliability for mission time T = 10 hours, R(10), and the unit time reliability is R(1) = 0.995, we can use either the 10th power or the FR λs: If mission time T is arbitrary, then R(T) is called "Survival Function" (of T). R(T) can then be used to find mission time "T" that accomplishes a pre-specified reliability. Assume that R(T) = 0.98 is required and we need to find out maximum time T: Hence, a Mission Time of T = 4.03 hours (or less) meets the requirement of reliability 0.98 (or more). Let's now assume that a new system, a ship, will be propelled by five identical engines. The system must meet a reliability requirement R(T) = 0.9048 for a mission time T = 10. We need to allocate reliability by engine (component reliability), for the required mission time T. We invert the formula for system reliability R(10), expressed as a function of component reliability. Then, we solve for component reliability Ri(10): We now calculate system FR (λs) and MTTF (μ) for the fiveengine system. These are obtained for mission time T = 10 hours and required system reliability R(10) = 0.9048: FR and MTTF values, equivalently, can be obtained using FR per component, yielding the same results: Finally, assume that the required ship FR λs = 5 × λ = 0.010005 is given. We now need component reliability, Unreliability and FR, by unit mission time (T = 1): R(1) = Exp{-λs} = Exp {-0.010005} = 0.99 = Exp{-5 × λ} = [Exp(-λ)]5 = [Ri(1)]5 Component reliability: Ri (1) = [R(1)]1/5 = [0.99]0.2 = 0.998M Component unreliability: Ui (1) = 1 - Ri (1) = 1 - 0.998 = 0.002 Component FR: λ = [- ln (R(1))]/n × 1 = [-ln(0.99)]/5 = 0.002