Vol 8, No 11 (2017)

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Available online throug

ISSN 2229 – 5046

RELIABILITY OF TIME - DEPENDENT STRESS - STRENGTH SYSTEM

WHICH FOLLOW DIFFERENT DISTRIBUTIONS WHEN THE NUMBER

OF CYCLES FOLLOWS BINOMIAL DISTRIBUTION

P. ASHOK*

1

_{, N. SWATHI}

2

_{, M. TIRUMALA DEVI}

3

_{, AND T. SUMATHI UMA MAHESWARI}

₄

1,2,3,4

_{Department of Mathematics,}

Kakatiya University, Warangal, Telangana State, India.

(Received On: 06-09-17; Revised & Accepted On: 03-11-17)

ABSTRACT

R

eliability of time dependent stress strength system is carried out by considering stress/strength variables as deterministic or random fixed or random independent. The number of cycles in any period of time t is assumed to be random. Reliability is obtained for the random models when the numbers of cycles follow binomial distribution where stress and strength follow Pareto distribution and exponential distribution.Reliability computation are made for the above said model

Key words: Binomial distribution, Pareto distribution, Exponential distribution, Deterministic, Random- fixed,

Random- independent and Reliability.

INTRODUCTION

Time dependent stress strength system is defined by Kapur, K.C. and Lamberson L.R [3].In stress-strength models component fails if the stress exceeds strength. The uncertainty about the stress and strength variables is classified into three categories. (1) Deterministic (2) Random fixed (3) Random independent. The components are assumed to be identical and the number of cycles for any time period t is assumed to be random. Expressions for system reliability have been attained when numbers of cycles follow binomial distribution and stress and strength both follow Pareto distribution andexponential distribution for random- fixed stress and random- independent strength, random – independent stress and random- fixed strength

NOTATIONS

f(x): The probability density functions of random variable X. g(y): The probability density function of random variable Y. R (t): Reliability at t time with n number of cycles.

i

R

: Reliability after i cycles. p: probability of success. q: probability of failure.

RELIABILITY EVALUATION

Number of cycles occurring in a given time interval followsbinomial distribution, then

𝜋𝑖(𝑡) =𝑃(𝑋=𝑖) =�𝑛𝑖�𝑝𝑖𝑞𝑛−𝑖, 𝑖= 0,1,2, … . .𝑛.

Case-1: Random- fixed stress and Random- independent strength

Let 𝑓(𝑥) be the probability density function of random fixed stress 𝑋 and 𝑔(𝑦) be the probability density function of random independent strength𝑌.

𝑅𝑖=∫ 𝑓₀∞ (𝑥)�∫ 𝑔_𝑥∞ (𝑦)𝑑𝑦�𝑖𝑑𝑥,𝑖= 0,1,2, … . .𝑛

Corresponding Author: P. Ashok*

1

_,

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𝑅(𝑡) =∑𝑛𝑖=0𝜋𝑖(𝑡)𝑅𝑖

=∑ _{�𝑛𝑖�𝑝}𝑖_𝑞𝑛−𝑖_{∫ 𝑓}₍_𝑥₎_{�∫ 𝑔}∞ ₍_𝑦₎_𝑑𝑦

𝑥 �

𝑖

𝑑𝑥

∞ 0 𝑛

𝑖=𝑜

=𝑞𝑛_{∫ 𝑓}₍_𝑥₎_∑

�𝑛𝑖��𝑝𝑞∫ 𝑔(𝑦)𝑑𝑦 ∞

𝑥 �

𝑖

𝑑𝑥

𝑛 𝑖=0 ∞

0

=𝑞𝑛_{∫ 𝑓}∞ ₍_𝑥₎ 0 �1 +

𝑝

𝑞∫ 𝑔(𝑦)𝑑𝑦 ∞

𝑥 �

𝑛

𝑑𝑥

𝑅(𝑡) =𝑞𝑛_{∫ 𝑓}∞ ₍_𝑥₎ 0 �1 +

𝑝 𝑞𝐺(𝑥)�

𝑛

𝑑𝑥 ,𝑤ℎ𝑒𝑟𝑒 𝐺(𝑥) =∫ 𝑔_𝑥∞ (𝑦)𝑑𝑦

𝑅(𝑡) =∫ 𝑓₀∞ (𝑥)�𝑞+𝑝𝐺(𝑥)�𝑛𝑑𝑥, 𝑤ℎ𝑒𝑟𝑒 𝐺(𝑥) =∫ 𝑔_𝑥∞ (𝑦)𝑑𝑦 (1)

Case-2: Random- independent stress and Random- fixed strength

Let g(y) be the probability density function of random fixed strength 𝑌 and 𝑓(𝑥) be the probability density function of random independent stress𝑋.

𝑅(𝑡) =∑ _{��𝑛𝑖�𝑝}𝑖_𝑞𝑛−𝑖_{∫ 𝑔}₍_𝑦₎_{�∫ 𝑓}𝑦 ₍_𝑥₎_𝑑𝑥

0 �

𝑖

𝑑𝑦

∞

0 �

𝑛

𝑖=𝑜

=𝑞𝑛_{∫ �𝑔}₍_𝑦₎_∑

�𝑛𝑖��𝑝𝑞∫ 𝑓(𝑥)𝑑𝑥 𝑦

0 �

𝑖

𝑑𝑦

𝑛

𝑖=0 �

∞

0

=𝑞𝑛_{∫ 𝑔}∞ ₍_𝑦₎ 0 �1 +

𝑝

𝑞∫ 𝑓(𝑥)𝑑𝑥 𝑦

0 �

𝑛

𝑑𝑦

𝑅(𝑡) =𝑞𝑛_{∫ 𝑔}∞ ₍_𝑦₎ 0 �1 +

𝑝 𝑞𝐹(𝑦)�

𝑛

𝑑𝑥, 𝑤ℎ𝑒𝑟𝑒 𝐹(𝑦) =∫ 𝑓₀𝑦 (𝑥)𝑑𝑥

𝑅(𝑡) =∫ 𝑔₀∞ (𝑦)�𝑞+𝑝𝐹(𝑦)�𝑛𝑑𝑥 ,𝑤ℎ𝑒𝑟𝑒 𝐹(𝑦) =∫ 𝑓₀𝑦 (𝑥)𝑑𝑥 (2)

If Stress follows Pareto Distribution and Strength follows Exponential Distribution

Case-1: Random - fixed stress and Random - independent strength

From equation(1)

𝑅(𝑡) =∫ 𝑓_𝑜∞ (𝑥)�𝑞+𝑝𝐺(𝑥)�𝑛𝑑𝑥 ,𝑤ℎ𝑒𝑟𝑒 𝐺(𝑥) =∫ 𝑔_𝑥∞ (𝑦)𝑑𝑦

𝐺(𝑥) =∫ 𝜇𝑒∞ −𝜇𝑦_𝑑𝑦₌_𝑒−𝜇𝑥

𝑥

𝑅(𝑡) =∫₀∞_𝑥𝜆𝑘𝜆+1𝜆(𝑞+𝑃𝑒−𝜇𝑥)𝑛𝑑𝑥

=𝜆𝑘𝜆_{∫ 𝑥}∞ −(𝜆+1)

0 [𝑛𝑐0𝑞𝑛+𝑛𝑐1𝑞𝑛−1(𝑝𝑒−𝜇𝑥) +𝑛𝑐2𝑞𝑛−2(𝑝2𝑒−2𝜇𝑥) +⋯… … . +𝑛𝑐𝑛𝑞0(𝑝𝑛𝑒−𝑛𝜇𝑥)].

𝑅(𝑡) =𝜆𝑘𝜆_[_{∫ 𝑛}

𝑐0𝑞𝑛𝑥−(𝜆+1) ∞

0 𝑑𝑥+∫ 𝑛𝑐1𝑞𝑛−1𝑝𝑒−𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥+∫ 𝑛𝑐2𝑞𝑛−2𝑝2𝑒−2𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

+. . … +∫ 𝑛∞ _𝑐_𝑛𝑞0𝑝𝑛𝑒−𝑛𝜇𝑥𝑥−(𝜆+1)

0 𝑑𝑥] (1)

𝐼0=∫ 𝑛𝑐0𝑞𝑛𝑥−(𝜆+1) ∞

0 𝑑𝑥=𝑛𝑐0𝑞𝑛∫ 𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐0𝑞𝑛�

𝑥−𝜆−1+1

−𝜆−1+1�₀ ∞

=𝑛𝑐0𝑞𝑛� 𝑥−𝜆

−𝜆�0 ∞

=0

𝐼0= ∫ 𝑛𝑐0𝑞𝑛𝑥−(𝜆+1) ∞

0 𝑑𝑥= 0 … … (𝑖)

𝑙𝑒𝑡 𝐼1=∫ 𝑛𝑐1𝑝𝑞𝑛−1𝑒−𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐1𝑝𝑞𝑛−1∫ 𝑒−𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐1𝑝𝑞𝑛−1��𝑒−𝜇𝑥 𝑥 −𝜆−1+1 −𝜆−1+1�₀

∞

− ∫ 𝑒−𝜇𝑥₍_−𝜇₎ 𝑥 −𝜆

−𝜆

𝑑𝑥

∞

0 �

=𝑛𝑐1𝑝𝑞𝑛−1�0− 𝜇

𝜆∫ 𝑒−𝜇𝑥𝑥−𝜆 ∞

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=𝑛𝑐1𝑝𝑞𝑛−1 𝜆 �−

𝜇

𝜇−𝜆+1Γ(−𝜆+ 1)�

=𝑛𝑐1𝑝𝑞𝑛−1

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�

𝐼1=∫ 𝑛𝑐1𝑝𝑞𝑛−1𝑒−𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥=

𝑛_𝑐1𝑝𝑞𝑛−1

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�… … . . (𝑖𝑖)

𝑙𝑒𝑡 𝐼2=∫ 𝑛𝑐2𝑝2𝑞𝑛−2𝑒−2𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐2𝑝2𝑞𝑛−2∫ 𝑒−2𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐2𝑝2𝑞𝑛−2��𝑒−2𝜇𝑥 𝑥 −𝜆−1+1 −𝜆−1+1�₀

∞

− ∫ 𝑒−2𝜇𝑥₍₋₂_𝜇₎ 𝑥 −𝜆

−𝜆

𝑑𝑥

∞

0 �

=𝑛𝑐2𝑝2𝑞𝑛−2�0− 2𝜇

𝜆 ∫ 𝑒−2𝜇𝑥𝑥−𝜆 ∞

0 𝑑𝑥�

=𝑛𝑐2𝑝2𝑞𝑛−2 𝜆 �−

2𝜇

2𝜇−𝜆+1𝛤(−𝜆+ 1)�

=𝑛𝑐2𝑝2𝑞𝑛−2

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�

𝐼2=∫ 𝑛𝑐2𝑝2𝑞𝑛−2𝑒−2𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥=

𝑛_𝑐2𝑝2𝑞𝑛−2

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�… … . (𝑖𝑖𝑖)

𝑙𝑒𝑡 𝐼3=∫ 𝑛𝑐3𝑝3𝑞𝑛−3𝑒−3𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐3𝑝3𝑞𝑛−3∫ 𝑒−3𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐3𝑝3𝑞𝑛−3��𝑒−3𝜇𝑥 𝑥 −𝜆−1+1 −𝜆−1+1�₀

∞

− ∫ 𝑒−3𝜇𝑥₍₋₃_𝜇₎ 𝑥 −𝜆

−𝜆

𝑑𝑥

∞

0 �

=𝑛𝑐3𝑝3𝑞𝑛−3�0− 3𝜇

𝜆 ∫ 𝑒−3𝜇𝑥𝑥−𝜆 ∞

0 𝑑𝑥�

=𝑛𝑐3𝑝3𝑞𝑛−3 𝜆 �−

3𝜇

3𝜇−𝜆+1𝛤(−𝜆+ 1)�

=𝑛𝑐3𝑝3𝑞𝑛−3

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�

𝐼3=∫ 𝑛𝑐3𝑝3𝑞𝑛−3𝑒−3𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥=

𝑛_𝑐3𝑝3𝑞𝑛−3

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�… … . (𝑖𝑣)

Continuing this process up to n times more, we get

𝑙𝑒𝑡 𝐼𝑛=∫ 𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛𝑒−𝑛𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛∫ 𝑒−𝑛𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥

=𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛��𝑒−𝑛𝜇𝑥 𝑥 −𝜆−1+1 −𝜆−1+1�₀

∞

− ∫ 𝑒−𝑛𝜇𝑥₍_−𝑛𝜇₎ 𝑥 −𝜆

−𝜆

𝑑𝑥

∞

0 �

=𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛�0− 𝑛𝜇

𝜆 ∫ 𝑒−𝑛𝜇𝑥𝑥−𝜆 ∞

0 𝑑𝑥�

=𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛 𝜆 �−

𝑛𝜇

𝑛𝜇−𝜆+1𝛤(−𝜆+ 1)�

=𝑛𝑐3𝑝𝑛𝑞𝑛−𝑛

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�

𝐼𝑛=∫ 𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛𝑒−𝑛𝜇𝑥𝑥−(𝜆+1) ∞

0 𝑑𝑥=

𝑛_𝑐𝑛𝑝𝑛𝑞𝑛−𝑛

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�… … . (𝑣)

Substituting all these values in equation (1)

𝑅(𝑡) =𝜆𝑘𝜆_�_{0 +}𝑛𝑐1𝑝𝑞𝑛−1

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�+

𝑛_𝑐2𝑝2_𝑞𝑛−2

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�+

𝑛_𝑐3𝑝3_𝑞𝑛−3

𝜆 �−𝜇𝜆𝛤(−𝜆+ 1)�+

………+𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛𝜆−𝜇𝜆𝛤(−𝜆+1)

𝑅(𝑡) =𝜆𝑘_𝜆𝜆�−𝜇𝜆_𝛤₍_−𝜆_{+ 1)}_��𝑛

𝑐1𝑝𝑞𝑛−1+𝑛𝑐2𝑝2𝑞𝑛−2+𝑛𝑐2𝑝3𝑞𝑛−3+⋯… +𝑛𝑐𝑛𝑝𝑛𝑞𝑛−𝑛�

𝑅(𝑡) =𝑘𝜆_�−𝜇𝜆_𝛤₍_−𝜆_{+ 1)}_{� ∑} _𝑛 𝑐𝑘𝑝𝑘 𝑛

𝑘=1 𝑞𝑛−𝑘

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Case-2: Random- independent stress and Random- fixed strength From equation (2)

𝑅(𝑡) =∫ 𝑔₀∞ (𝑦)�𝑞+𝑝𝐹(𝑦)�𝑛𝑑𝑦 ,𝑤ℎ𝑒𝑟𝑒 𝐹(𝑦) =∫ 𝑓₀𝑦 (𝑥)𝑑𝑥

𝐹(𝑦) =∫ 𝜆𝑒𝑦 −𝜆𝑥_𝑑𝑥₌_�₁_{− 𝑒}−𝜆𝑦_�

0

𝑅(𝑡) =∫ _𝑦𝜇𝑘𝜇+1𝜇 ∞

0 �𝑞+𝑝�1− 𝑒−𝜆𝑦�� 𝑛

𝑑𝑦=∫ _𝑦𝜇𝑘𝜇+1𝜇 ∞

0 �1− 𝑝𝑒−𝜆𝑦� 𝑛

𝑑𝑦

𝑅(𝑡) =∫ _𝑦𝜇𝑘𝜇+1𝜇 ∞

0 �1− 𝑝𝑒−𝜆𝑦� 𝑛

𝑑𝑦

=𝜇𝑘𝜇_{∫ 𝑦}∞ −(𝜇+1)

0 [𝑛𝑐0− 𝑛𝑐1𝑝𝑒−𝜆𝑦+𝑛𝑐2𝑝2𝑒−2𝜆𝑦−𝑛𝑐3𝑝3𝑒−3𝜆𝑦+⋯… … . +(−1)𝑛𝑛𝑐𝑛𝑝𝑛𝑒−𝑛𝜆𝑦.

𝑅(𝑡) =𝜇𝑘𝜇_[_{∫ 𝑛}

𝑐0𝑦−(𝜇+1) ∞

0 𝑑𝑦+∫ 𝑛𝑐1𝑝𝑒−𝜆𝑦𝑥−(𝜇+1) ∞

0 𝑑𝑦+∫ 𝑛𝑐2𝑝2𝑒−2𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦+⋯… … … +

(−1)𝑛_{∫ 𝑛}

𝑐𝑛𝑝𝑛𝑒−𝑛𝜆𝑦𝑥−(𝜆+1) ∞

0 𝑑𝑦] … … … (1)

𝐼0=∫ 𝑛𝑐0𝑦−(𝜇+1) ∞

0 𝑑𝑦=𝑛𝑐0∫ 𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐0�

𝑦−𝜇−1+1

−𝜇−1+1�₀ ∞

=𝑛𝑐0� 𝑦−𝜆

−𝜇�₀ ∞

= 0

𝐼0=∫ 𝑛𝑐0𝑦−(𝜇+1) ∞

0 𝑑𝑦= 0 … … . . (𝑖)

𝑙𝑒𝑡 𝐼1=∫ 𝑛𝑐1𝑝𝑒−𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐1𝑝 ∫ 𝑒−𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐1𝑝 ��𝑒−𝜆𝑦 𝑦 −𝜇−1+1 −𝜇−1+1�₀

∞

− ∫ 𝑒−𝜆𝑦₍_−𝜆₎ 𝑦 −𝜇

−𝜇

𝑑𝑦

∞

0 �

=𝑛𝑐1𝑝 �0− 𝜆

𝜇∫ 𝑒−𝜆𝑦𝑦−𝜇 ∞

0 𝑑𝑦�

=𝑛𝑐1𝑝 𝜇 �−

𝜆

𝜆−𝜇+1𝛤(−𝜇+ 1)�

=𝑛𝑐1𝑝

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)]

𝐼1=∫ 𝑛𝑐1𝑝𝑒−𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦=

𝑛_𝑐1𝑝

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] … … . . (𝑖𝑖)

𝑙𝑒𝑡 𝐼2=∫ 𝑛𝑐2𝑝2𝑒−2𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐2𝑝2∫ 𝑒−2𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐2𝑝2��𝑒−2𝜆𝑦 𝑦 −𝜇−1+1 −𝜇−1+1�₀

∞

− ∫ 𝑒−2𝜆𝑦₍₋₂_𝜆₎ 𝑦 −𝜇

−𝜇

𝑑𝑦

∞

0 �

=𝑛𝑐2𝑝2�0− 2𝜆

𝜇∫ 𝑒−𝜆𝑦𝑦−𝜇 ∞

0 𝑑𝑦�

=𝑛𝑐2𝑝2 𝜇 �−

2𝜆

2𝜆−𝜇+1𝛤(−𝜇+ 1)�

=𝑛𝑐2_𝜇𝑝2[−𝜆𝜇_𝛤₍_−𝜇_{+ 1)]}

𝐼2=∫ 𝑛𝑐2𝑝2𝑒−2𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦=

𝑛_𝑐2𝑝2

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] … … . . (𝑖𝑖𝑖)

𝑙𝑒𝑡 𝐼3=∫ 𝑛𝑐3𝑝3𝑒−3𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐3𝑝3∫ 𝑒−3𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐3𝑝3��𝑒−3𝜆𝑦 𝑦 −𝜇−1+1 −𝜇−1+1�₀

∞

− ∫ 𝑒−3𝜆𝑦₍₋₃_𝜆₎ 𝑦 −𝜇

−𝜇

𝑑𝑦

∞

0 �

=𝑛𝑐3𝑝3�0− 3𝜆

𝜇∫ 𝑒−3𝜆𝑦𝑦−𝜇 ∞

0 𝑑𝑦�

=𝑛𝑐3𝑝3 𝜇 �−

3𝜆

3𝜆−𝜇+1𝛤(−𝜇+ 1)�

=𝑛𝑐3𝑝3

(5)

𝐼3=∫ 𝑛𝑐3𝑝3𝑒−3𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦=

𝑛_𝑐3𝑝3

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] … … . . (𝑖𝑣)

Continuing this process up to n times more, we get

𝑙𝑒𝑡 𝐼𝑛=∫ 𝑛𝑐𝑛𝑝𝑛𝑒−𝑛𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐𝑛𝑝𝑛∫ 𝑒−𝑛𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦

=𝑛𝑐𝑛𝑝𝑛��𝑒−𝑛𝜆𝑦 𝑦 −𝜇−1+1 −𝜇−1+1�₀

∞

− ∫ 𝑒−𝑛𝜆𝑦₍_−𝑛𝜆₎ 𝑦 −𝜇

−𝜇

𝑑𝑦

∞

0 �

=𝑛𝑐𝑛𝑝𝑛�0− 𝑛𝜆

𝜇 ∫ 𝑒−𝑛𝜆𝑦𝑦−𝜇 ∞

0 𝑑𝑦�

=𝑛𝑐𝑛𝑝𝑛 𝜇 �−

𝑛𝜆

𝑛𝜆−𝜇+1𝛤(−𝜇+ 1)�

=𝑛𝑐𝑛𝑛

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)]

𝐼2=∫ 𝑛𝑐𝑛𝑝𝑛𝑒−𝑛𝜆𝑦𝑦−(𝜇+1) ∞

0 𝑑𝑦=

𝑛_𝑐𝑛𝑝𝑛

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] … … . . (𝑣)

Substituting all these values in equation (1)

=𝜇𝑘𝜇_�0− 𝑛_𝑐1𝑝

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] + 𝑛_𝑐2𝑝2

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)] + 𝑛_𝑐3𝑝3

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)]

+⋯… … .−𝑛𝑐𝑛𝑝𝑛

𝜇 [−𝜆𝜇𝛤(−𝜇+ 1)]

�

𝑅(𝑡) =𝜇𝑘_𝜇𝜇[−𝜆𝜇_𝛤₍_−𝜇_{+ 1)]}_�𝑛

𝑐1𝑝 − 𝑛𝑐2𝑝2+𝑛𝑐3𝑝3− ⋯… . . +(−1)𝑛−1𝑝𝑛�

𝑅(𝑡) =𝑘𝜇_[_−𝜆𝜇_𝛤₍_−𝜇_{+ 1)]}_∑𝑛 ₍₋₁₎𝑘−1 𝑘=1 𝑛𝑐𝑘𝑝𝑘

CONCLUSION

General reliability formula for n cycles is derived when number of cycles follows binomial distribution for the following two cases: (1) random- fixed Stress and random- independent Strength. (2) random- independent Stress and random- fixed Strength.

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Source of support: Nil, Conflict of interest: None Declared.