Neat and comprehensive cheat sheet of all the formulas that you need during the examination 1CV40 - Quality and Reliability Engineering. On the last page is still a bit of room for additions.
𝑛 Normal distribution:
Chapter 2 λ(𝑡) = ∑ λ𝑖 (𝑡)
𝑖=1 1 1 (𝑡−𝜇) 2 𝑡−𝜇
− ∙
𝑅(𝑡 ) = 𝑃 (𝑇 ≥ 𝑡) 𝑛 𝑓 (𝑡) = ∙ 𝑒 2 𝜎2 𝑧=
√2𝜋 ∙ 𝜎 𝜎
𝑅(𝑡 ) = ∏ 𝑅𝑖 (𝑡 )
𝐹 (𝑡) = 𝑃 (𝑇 < 𝑡) = 1 − 𝑅(𝑡) 𝑖 =1
𝑀𝑇𝑇𝐹 = 𝜇 𝑠𝑡. 𝑑 = 𝜎
− ∑𝑛 ( )
𝑖=1 λ𝑖 𝑡 ∙𝑡
𝑑𝐹 (𝑡) 𝑑𝑅 (𝑡 ) =𝑒
𝑓 (𝑡) = = − 𝐶𝐷𝐹 = 𝑃(𝑍 ≤ 𝑧) = Φ(𝑧) 𝑅(𝑡) = 1 − Φ
𝑑𝑡 𝑑𝑡 Poisson distribution: The chance that n fa ilures
Lognormal distribution (log(t) has a normal distribution):
𝑡′ in range t=0 to t.
𝐹 (𝑡) = ∫ 𝑓 (𝑡 ′ )𝑑𝑡′ 1 −
1
∙(ln(
𝑡
)) 2
0 𝑒 −𝜆∙𝑡 ∙ (𝜆 ∙ 𝑡)𝑛 𝑓 (𝑡) = ∙𝑒 2∙𝑠2 𝑡𝑚𝑒𝑑
𝑝𝑛 (𝑡) = √2𝜋 ∙ 𝑠 ∙ 𝑡
∞ 𝑛!
𝑅(𝑡 ) = ∫ 𝑓 (𝑡 ′ )𝑑𝑡′ 1 𝑡
𝑡′
𝑝0 (𝑡) = 𝑅(𝑡) = 𝑒 −𝜆∙𝑡 𝐹 (𝑡) = Φ ( ln( ))
𝑠 𝑡𝑚𝑒𝑑
𝑃 (𝑎 ≤ 𝑇 ≤ 𝑏) = 𝑅(𝑎 ) − 𝑅(𝑏) 𝐸[𝑛] = 𝜎 2 = 𝜆 ∙ 𝑡
1 𝑡
= 𝐹 (𝑏) − 𝐹(𝑎) Calculate the probability that there have been
𝑅 (𝑡 ) = 1 − Φ ( ln( ))
𝑠 𝑡𝑚𝑒𝑑
∞ n failures in time t=0 to t:
2 2
𝑀𝑇𝑇𝐹 = ∫ 𝑡 ∙ 𝑓(𝑡 )𝑑𝑡 𝑠 −𝜆∙𝑡 𝑛 𝜎 2 = 𝑡𝑚𝑒𝑑 2 ∙ 𝑒 𝑠 ∙ [𝑒 𝑠 − 1]
0 𝑒 ∙ (𝜆 ∙ 𝑡)
𝑅𝑠 (𝑡) = ∑ 2 2/2
∞ 𝑛=0 𝑛! 𝑡𝑚𝑜𝑑𝑒 = 𝑡𝑚𝑒𝑑/ 𝑒 𝑠 𝑀𝑇𝑇𝐹 = 𝑡𝑚𝑒𝑑/ 𝑒 𝑠
𝑀𝑇𝑇𝐹 = ∫ 𝑅 (𝑡) 𝑑𝑡 Gamma distribution:
0 The Gamma distribution is closely related to
the Poisson, and calculates the probability that
∞ 𝑡𝑟−1 ∙𝑒 −𝑡/𝛼
2 2
the kth failure will occur by time t (Yk = t of kth): 𝑓 (𝑡) = 𝑀𝑇𝑇𝐹 = 𝛼𝛾 𝜎 2 = 𝛾𝛼 2
𝜎 = ∫ 𝑡 ∙ 𝑓(𝑡 ) 𝑑𝑡 − (𝑀𝑇𝑇𝐹 )2 𝑡𝛾 ∙𝛤 (𝛾)
0 𝑘 (𝜆 ∙ 𝑡)𝑖 𝐼(𝑡/𝛼,𝛾) 𝐼 (𝑡/𝛼,𝛾)
𝐹𝑌 (𝑡) = 1 − 𝑒 −𝜆∙𝑡 ∙ ∑ 𝐹 (𝑡) = 𝑅(𝑡) = 1 −
∞ 2 𝑖 =0 𝑖! 𝛤(𝛾) 𝛤(𝛾)
= ∫0 (𝑡 − 𝑀𝑇𝑇𝐹 ) 𝑑𝑡
𝑡𝑚𝑜𝑑𝑒 = 𝛼(𝛾 − 1) 𝑓𝑜𝑟 𝛾 > 1
𝑡′ 𝐸[𝑌𝑘 ] = 𝑘/𝜆 𝜎 2 = 𝑘/𝜆2
− ∫0 𝜆 (𝑡′)𝑑𝑡′
𝑅(𝑡 ) = 𝑒
𝑘−1
𝑡′
𝑌𝑚𝑜𝑑𝑒 =
𝜆
Chapter 5
∫0 𝜆 (𝑡′)𝑑𝑡′
𝐴𝐹𝑅 = 𝑛
𝑡 If the system is guaranteed to work until t0: 𝑅𝑠 (𝑡)𝑠𝑒𝑟𝑖𝑒𝑠 = ∏ 𝑅𝑖 (𝑡)
𝑖=1
𝑅 (𝑇0 + 𝑡) 𝑅(𝑡 ) = 𝑒 −𝜆∙(𝑡−𝑡0)
𝑅(𝑡 | 𝑇0 ) = 𝑛
𝑅(𝑇0 ) 𝑅𝑠 (𝑡)𝑝𝑎𝑟 = 1 − ∏ (1 − 𝑅𝑖 (𝑡))
𝑑𝑅 (𝑡) 𝑖=1
∞ 𝑓 (𝑡) = − = 𝜆 ∙ 𝑒 −𝜆∙(𝑡−𝑡0 )
𝑑𝑡 1 1
𝑀𝑇𝑇𝐹 (𝑡 | 𝑇0 ) = ∫ 𝑅 (𝑡 | 𝑇0 )𝑑𝑡 𝑀𝑇𝑇𝐹𝑠𝑒𝑟𝑖𝑒𝑠 = ∑𝑛 =
1 𝑙𝑛0,5 𝑖=1 λ𝑖 (𝑡) ∑𝑛
𝑖=1 1/ 𝑀𝑇𝑇𝐹𝑖
0
1 ∞ 𝑀𝑇𝑇𝐹 = 𝑡0 + 𝑡𝑚𝑒𝑑 = 𝑡0 +
λ λ
= ∙ ∫𝑇 𝑅(𝑡′) 𝑑𝑡 1 1 1
𝑅 (𝑇0 ) 0
𝑀𝑇𝑇𝐹𝑝𝑎𝑟 = + −
Chapter 4 𝜆1 𝜆2 𝜆1 + 𝜆2
Chapter 3 High and low-level redundancy indicate how a system is
Weibull distribution: designed. The variable m indicates the number of horizontal
Exponential distribution (CFR with λ(t)=λ): components and n indicates the number of vertical
used to calculate the chance that the next 𝛽 𝑡 𝛽−1 components:
failure will happen in range t=0 to t. 𝜆 (𝑡) = ∙( )
𝜃 𝜃
𝑅(𝑡 ) = 𝑒 −𝜆∙𝑡 𝑡′ 𝑡 𝛽
∫0 𝜆 (𝑡′)𝑑𝑡′ −( )
𝑅(𝑡 ) = 𝑒 − = 𝑒 𝜃
𝐹 (𝑡) = 1 − 𝑒 −𝜆∙𝑡
𝛽 𝑡 𝛽−1 −( 𝑡 ) 𝛽
𝑑𝐹 (𝑡) −𝜆∙𝑡 𝑓 (𝑡) = ∙( ) ∙𝑒 𝜃
𝑓 (𝑡) = = 𝜆 ∙𝑒 𝜃 𝜃 The chance that x=k out of n identical components work is:
𝑑𝑡
𝑀𝑇𝑇𝐹 = 𝜃 ∙ Γ(1 + 1/𝛽) 𝑛
1 1 𝑃(𝑥 ) = ( ) 𝑅𝑥 (1 − 𝑅)𝑛−𝑥
𝑀𝑇𝑇𝐹 = 𝜎 = 𝜎2 = 𝑥
λ λ2
2 1 2
1 𝜎 2 = 𝜃2 ∙ [Γ (1 + ) − [Γ (1 + )] ] 𝑛!
𝛽 𝛽 = 𝑅𝑥 (1 − 𝑅)𝑛−𝑥
𝑡𝑚𝑒𝑑 = ∙ ln(0,5) 𝑥! (𝑛 − 𝑥 )!
λ 1/𝛽
−𝜆∙𝑡
𝑡𝑅 = 𝜃 ∙ (− ln(𝑅) ) 𝑛
𝑅(𝑡 | 𝑇0 ) = 𝑅(𝑡) = 𝑒 𝑅𝑠 = ∑ 𝑃(𝑥)
1/𝛽 𝑥=𝑘
𝑡′ 𝑡𝑚𝑜𝑑𝑒 = 𝜃 ∙ (1 − 1/𝛽) 𝛽> 1
𝜆 𝑖 (𝑡′ )𝑑𝑡′
𝑅𝑖 (𝑡 ) = 𝑒 − ∫0 ∞
𝑀𝑇𝑇𝐹 = ∫0 𝑅𝑠 (𝑡) 𝑑𝑡 = ∙ ∑𝑛𝑥=𝑘
1 1
1 𝛽 𝜆 𝑥
λ(𝑡) = 𝛽 ∙ 𝑡𝛽−1 ∙ ∑𝑛𝑖=1 ( )
𝜃𝑖
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