MTTR} \right )}\)Where t = maintenance TIME (min), MTTR = Mean time to REPAIR (min)Calculation:GIVEN:MTTR = 30 min, t1 = 20 min, t2 = 40 min\(M(t_1)=1-e^{-\left ( \frac{20}{30} \right )}\)M(t1) = 0.4865\(M(t_2)=1-e^{-\left ( \frac{40}{30} \right )}\)M(t2) = 0.7364Net increase in the maintainability = M(t2) - M(t1)= 0.7364 - 0.4865= 0.2499 ≈ 0.25

"> MTTR} \right )}\)Where t = maintenance TIME (min), MTTR = Mean time to REPAIR (min)Calculation:GIVEN:MTTR = 30 min, t1 = 20 min, t2 = 40 min\(M(t_1)=1-e^{-\left ( \frac{20}{30} \right )}\)M(t1) = 0.4865\(M(t_2)=1-e^{-\left ( \frac{40}{30} \right )}\)M(t2) = 0.7364Net increase in the maintainability = M(t2) - M(t1)= 0.7364 - 0.4865= 0.2499 ≈ 0.25

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The mean time to repair (MTTR) for a repairable system is 30 minutes. When maintenance time changes from 20 minutes to 40 minutes, the net increase in maintainability is closest to

General Knowledge General Awareness in General Knowledge 7 months ago

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Concept:Maintainability,\(M(t)=1-e^{-\left ( \frac{t}{MTTR} \right )}\)Where t = maintenance TIME (min), MTTR = Mean time to REPAIR (min)Calculation:GIVEN:MTTR = 30 min, t1 = 20 min, t2 = 40 min\(M(t_1)=1-e^{-\left ( \frac{20}{30} \right )}\)M(t1) = 0.4865\(M(t_2)=1-e^{-\left ( \frac{40}{30} \right )}\)M(t2) = 0.7364Net increase in the maintainability = M(t2) - M(t1)= 0.7364 - 0.4865= 0.2499 ≈ 0.25

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Posted on 08 Dec 2024, this text provides information on General Knowledge related to General Awareness in General Knowledge. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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