University:
Massachusetts Institute of Technology
Course:
22.38 | Probability and Its Applications to Reliability, Quality Control, and Risk Assessment
Academic year:

2021
Views:

65

Pages:

3
Author:

Florisav5zl1

4.5) For parallel systems: X = Πᵢxᵢ → R = 1 - (1-p)⁵ = 1 - (0.65)⁵ = 0.951 4.6) Pump: 992, .968 a) System 1: A₁ = [(0.992)(0.968)]² = 0.9221 System 2: A₂ = [(0.992)(0.968)]³ + 3(1 - (0.992)(0.968)) = 0.995 b) S1: (15+60)2 + 410000(1-0.922) = 4980 S2: (15+60)3 + 100000(1-0.996) = 275 → Clearly the 2-out-of-3 system is better in this case where the cost of failure is so high. 4.7) Psystem = p³ + 3p²(1-p) = 3p²-2p³ a) R(s) = 3e^(-2λt) - 2e^(-3λt) f(t) = -R'(s) = 6λ(e^(-2λt) - e^(-3λt)) b) MTTF: ∫₀^∞ tf(t)dt = (5/6λ) mode: df/dt = 0 => -2λe^(-2λt) + 3λe^(-3λt) = 0 => t = ln(2/3) / λ f(tm) = 3e^(-2λtm) - 2e^(-3λtm) Substituting x = e^(-λtm) f(tm) = 3x² - 2x³ x = 1/2 is the only root that makes sense tm = ln(2)/λ 4.12 cont) c) P(vessel fails to operated as intended) = P(failure of v) * P(burner works) + P(burner fails) 3-out-of-4 system: R(t) = 4(0.97)³(0.03) + (0.97)⁴ → λ = 2.02 * 10^(-4) + 446 P(burners work) = (0.97)⁴ + 4(0.97)³(0.03) + 6(0.97)²(0.03)² + 4(0.97)(0.03)³ = 0.888 P(burners fail) = 1 - 0.888 = 0.112 P(vessel fail) = 50% = 1 - e^(-λpt) → λp = 2.02 * 10^(-4) P(Fail) = 0.252(0.387) + 0.112 = 0.1335 => P(Success starts run 2 months) = 1 - P(Fail) = 1 - 0.1335 = 0.8665 f) P(40-50%) = P(≥2 burners fail) + P(≥2 boilers fail) = [0.97(0.03)]² + 0.97² = 0.9991