In contemporary safety engineering, quantifying reliability system within stress/strength environments has changed into a critical analytical need, especially when modeling independent statistical states. To maximize model dependability in harsh operations, students and engineers frequently deploy standby redundancy. This format systematically presents backup components to support the prime functional. Although classic reliability models routinely consider that any replacing unit functions in identical stresses and ambient environments as its predecessor, actual field operations constantly invalidate this supposition.^1,2

In the broader context of reliability concept, (stress) and (strength) denote concepts rather than narrow mechanical parameters. (Stress) encompasses any destructive agency be it shifting power loads, mechanical strain, or environmental volatility capable of causing a component to fail. (Strength), conversely, reflects the system’s intrinsic threshold of resistance, determined by the mean (strength) necessary to start a failure. So, interference theory dictates that failure occurs at the thorough intersection where operational (stress) pauses over this inherent (strength) barrier.^3–5

The working architecture of a standard standby system configuration groups a main component with its auxiliary backups into a united set. Scheme functionality remains uninterrupted as long as all set contains at least one active unit. Imagine an engineering basis containing (m) same components, designed such that only a single unit works under load at any known moment. This strategy keeps one component active while preserving the remaining (m-1) units as spares. Upon the failure of the active unit, the switching mechanism immediately triggers a standby replacement. Mathematical ideals consider this detection and transfer process is instant and error__free, and that idle units involvement (zero wear) though waiting. This exact defines a standby redundancy.^6,7 Then these backup components carry no load prior to beginning, the assets are split into active and standby, all governed by a distinct failure distribution. A (hot standby) organization applies if the component’s idling failure rate matches its active operational rate. Conversely, if the failure rate of the inactive unit drops to absolute (zero), the setup is classified as (gold standby) system which forms the core basic premise of this article.⁸ Reliability system variety extensive use of the exponential-distribution, Since of the memory less property of this distribution, it is well-suited to system the constant hazard ratio portion of the bathtub curve used in reliability system. A popular system is the exponential-distribution which is useful when the ‘no ageing’ phenomenon is evident. As against this, several units exhibit positive ageing phenomena.

Primary details in this discipline, notably by Sriwastav and Kakaty (1981),⁹ supposed that (stress) and (strength) distributions across totally components were perfectly identical. But, actual industrial applications are influenced by localized factors, meaning distinct physical variables often govern individual elements. This reality justifies shifting to non identical distributions. Over the times, the academic community has explored diverse statistical pathways for standby configurations. Kakati (1983) started this by relating a mixture of exponential distributions to calculate (stress-strength) reliability. Future, Rekha and Chechu Raju (1997)¹⁰ established closed solutions for stress-reliability in (m) cascade structures, coupling exponential (stress) with standby strengths derivative from Rayleigh and exponential distributions. More developments by Gogioi et al. (2010)¹¹ studied an (m) standby stress/strength model where cumulative (stress) shocks followed a Poisson process, calculating reliability crossways Exponential, Gamma, Normal, and Weaibull distributions. This was future added by the work of Gogioi and Baorah (2012)¹² and Sanidhya and U.mamahdseswari (2013)¹³ utilizing Exponential and Lindley distributions. Kala J. Kantu (2018)¹⁴ in this work a stress-strength system is expressed for a multi-component model involving of r identical components also, the estimation of system reliability of Lindley-Exponential Stress-Strength system with multi-component model with more than two stresses is studied.

By these literature gaps and the clear necessity to map realistic environments where components face separate external factors, this work rejects the traditional identical distribution constraint. Instead, we give the stress/strength performance of all system components as independent but non-identically distributed random variables. the core objectives of this research are:

Consider that Y_i (i = 1,2,…,n) is the strength, which is an independent random variable (r.v.) arranged in the order of activation. In addition, Ψ _i (i = 1,2,…,n) is the stress that is independent r.v. so, R ₄ is given by^15,16

Where R (i) = P (Y₁ < Ψ ₁, Y₂ < Ψ ₂,…, Y _i-1 < Ψ _i-1, Y_i ≥Ψ _i) is reliability.

Consider f_i(y), g _i( Ψ ) as the p.d.f. (i = 1,2,…,n); therefore, we have

Where Ti(Ψ) is the cumulative distribution function and T¯i(Ψ)=1−Ti(Ψ) . In this study, we examine four different distributions.

Let the strength follow a combination of tow exponential distributions in all the four cases with p.d.f.;

Therefore;

F-1(y)=p1exp(−ϑ1y)+(1−p1)exp(−ϑ2y)

2.1.1 (Stress) follows one parameter exponential combinations

The stress follows one parameter exponential combinations with p.d.f.;

gi(Ψ)=μie−μiΨμi>0,Ψ≥0∀i=1,…,n

So,

R(1)=P(Y1≥Ψ)=∫0∞T¯1(Ψ)g1(Ψ)dΨ=∫0∞[P1e−ϑ1Ψ+(1−P1)e−ϑ2Ψ]μie−μiΨdΨ=P1μ1∫0∞e−(ϑ1+μ1)ΨdΨ+(1−P1)μ1∫0∞e−(ϑ2+μ1)ΨdΨ=P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1

Now,

R(2)=P(Y1<Ψ1,Y2≥Ψ2)=[∫0∞T1(Ψ)g1(Ψ)dΨ][∫0∞T¯2(Ψ)g2(Ψ)dΨ]=[1−(P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1)]∫0∞[P3e−ϑ3Ψ+(1−P3)e−ϑ4Ψ]μ2e−μ2ΨdΨ=[1−(P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1)][P3μ2ϑ3+μ2+(1−P3)μ2ϑ4+μ2]

R(3)=P(Y1<Ψ1,Y2<Ψ2,Y3≥Ψ3)=[∫0∞T1(Ψ)g1(Ψ)dΨ][∫0∞T2(Ψ)g2(Ψ)dΨ][∫0∞T¯3(Ψ)g3(Ψ)dΨ]=[1−(P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1)][1−(P3μ2ϑ3+μ2+(1−P3)μ2ϑ4+μ2)][∫0∞[P5e−ϑ5Ψ+(1−P5)e−ϑ6Ψ]μ3e−μ3ΨdΨ][1−(P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1)][1−(P3μ2ϑ3+μ2+(1−P3)μ2ϑ4+μ2)][P5μ3ϑ5+μ3+(1−P5)μ3ϑ6+μ3]

In the same way;

R(4)=P(Y1<Ψ1,Y2<Ψ2,Y3<Ψ3,Y4≥Ψ4)=[∫0∞T1(Ψ)g1(Ψ)dΨ][∫0∞T2(Ψ)g2(Ψ)dΨ][∫0∞T3(Ψ)g3(Ψ)dΨ][∫0∞T¯4(Ψ)g4(Ψ)dΨ]=[1−(P1μ1ϑ1+μ1+(1−P1)μ1ϑ2+μ1)][1−(P3μ2ϑ3+μ2+(1−P3)μ2ϑ4+μ2)][1−(P5μ3ϑ5+μ3+(1−P5)μ3ϑ6+μ3)][P7μ4ϑ7+μ4+(1−P7)μ4ϑ8+μ4]

By using Equation (1), we get;

R4=R(1)+R(2)+R(3)+R(4)

2.1.2 (Stress) follows two parameter exponential combinations

The stress follows two parameter exponential combinations with p.d.f.;

gi(Ψ)=1£ie−(Ψ−αi)/£iαi,£i,Ψ>0∀i=1,…,n

Now,

R(1)=P(Y1≥Ψ1)=∫0∞[P1e−ϑ1Ψ+(1−P1)e−ϑ2Ψ]1£1e−(Ψ−α1)/£1dΨ=P1£1eα1£1∫α∞e−(ϑ1+1£1)ΨdΨ+(1−P1)£1eα1£1∫α∞e−(ϑ2+1£1)ΨdΨ=eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)]

R(2)=P(Y1<Ψ1,Y2≥Ψ2)=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])][P3£2eα2£2∫α∞e−(ϑ3+1£2)ΨdΨ+(1−P3)£2eα2£2∫α∞e−(ϑ4+1£2)ΨdΨ]=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])]eα2£2£2[P3e−(ϑ3+1£2)α2(ϑ3+1£2)+(1−P3)e−(ϑ4+1£2)α2(ϑ4+1£2)]

So,

R(3)=P(Y1<Ψ1,Y2<Ψ2,Y3≥Ψ3)=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])][1−(eα2£2£2[P3e−(ϑ3+1£2)α2(ϑ3+1£2)+(1−P3)e−(ϑ4+1£2)α2(ϑ4+1£2)])][P5£3eα3£3∫α3∞e−(ϑ5+1£3)ΨdΨ+(1−P5)£3eα3£3∫α3∞e−(ϑ6+1£3)ΨdΨ]=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])][1−(eα2£2£2[P3e−(ϑ3+1£2)α2(ϑ3+1£2)+(1−P3)e−(ϑ4+1£2)α2(ϑ4+1£2)])](eα3£3£3[P5e−(ϑ5+1£3)α2(ϑ3+1£3)+(1−P5)e−(ϑ6+1£3)α3(ϑ4+1£3)])

In the same way;

R(4)=P(Y1<Ψ1,Y2<Ψ2,Y3<Ψ3Y4≥Ψ4)=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])][1−(eα2£2£2[P3e−(ϑ3+1£2)α2(ϑ3+1£2)+(1−P3)e−(ϑ4+1£2)α2(ϑ4+1£2)])][1−(eα3£3£3[P5e−(ϑ5+1£3)α3(ϑ3+1£3)+(1−P5)e−(ϑ6+1£3)α3(ϑ4+1£3)])].[P5£4eα4£4∫0∞e−(ϑ5+1£4)ΨdΨ+(1−P5)£4eα4£4∫0∞e−(ϑ6+1£4)ΨdΨ=[1−(eα1£1£1[P1e−(ϑ1+1£1)α1(ϑ1+1£1)+(1−P1)e−(ϑ2+1£1)α1(ϑ2+1£1)])][1−(eα2£2£2[P3e−(ϑ3+1£2)α2(ϑ3+1£2)+(1−P3)e−(ϑ4+1£2)α2(ϑ4+1£2)])].[1−(eα3£3£3[P5e−(ϑ5+1£3)α2(ϑ3+1£3)+(1−P5)e−(ϑ6+1£3)α3(ϑ4+1£3)]).(eα4£4£4[P7e−(ϑ7+1£4)α4(ϑ7+1£4)+(1−P7)e−(ϑ8+1£4)α4(ϑ8+1£4)])]

By using Equation (1), we get;

R4=R(1)+R(2)+R(3)+R(4)

2.1.3 (Stress) follows one parameter Lindley distribution

The stress follows a one-parameter Lindley distribution with p.d.f.⁶

gi(Ψ)=ωi2(1+ωi)(1+Ψ)e−ωiΨωi,Ψ>0∀i=1,…,n

R(1)=∫0∞[P1e−ϑ1Ψ+(1−P1)e−ϑ2Ψ]ω12(1+ω1)(1+Ψ)e−ω1ΨdΨ=P1ω12(1+ω1)∫0∞(1+Ψ)e−(ϑ1+ω1)ΨdΨ+(1−P1)ω12(1+ω1)∫0∞(1+Ψ)e−(ϑ2+ω1)ΨdΨ=P1ω12(1+ω1)[e−(ϑ1+ω1)ΨdΨ+∫0∞Ψe−(ϑ1+ω1)Ψ]+(1−P1)ω12(1+ω1)[e−(ϑ2+ω1)ΨdΨ+∫0∞Ψe−(ϑ2+ω1)Ψ]=ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2]

R(2)=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])]P3ω22(1+ω2)∫0∞(1+Ψ)e−(ϑ3+ω2)ΨdΨ+(1−P3)ω12(1+ω2)∫0∞(1+Ψ)e−(ϑ4+ω2)ΨdΨ=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])](ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])

So,

R(3)=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])][1−(ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])].[P5ω32(1+ω3)∫0∞(1+Ψ)e−(ϑ5+ω3)ΨdΨ+(1−P5)ω32(1+ω3)∫0∞(1+Ψ)e−(ϑ6+ω3)ΨdΨ]=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])][1−(ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])][(ω32(1+ω3)[P5ϑ5+ω3+1(ϑ5+ω3)2+(1−P5)ϑ6+ω3+1(ϑ6+ω3)2])]

In the same way;

R(4)=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])].[1−(ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])].[1−(ω32(1+ω3)[P5ϑ5+ω3+1(ϑ5+ω3)2+(1−P5)ϑ6+ω3+1(ϑ6+ω3)2])][P7ω42(1+ω4)∫0∞(1+Ψ)e−(ϑ7+ω4)ΨdΨ+(1−P7)ω42(1+ω4)∫0∞(1+Ψ)e−(ϑ8+ω4)ΨdΨ]=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])][1−(ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])][1−(ω32(1+ω3)[P5ϑ5+ω3+1(ϑ5+ω3)2+(1−P5)ϑ6+ω3+1(ϑ6+ω3)2])][ω42(1+ω4)[P7ϑ7+ω4+1(ϑ7+ω4)2+(1−P7)]ϑ8+ω4+1(ϑ8+ω4)2]

So,

R(4)=[1−(ω12(1+ω1)[P1ϑ1+ω1+1(ϑ1+ω1)2+(1−P1)ϑ2+ω1+1(ϑ2+ω1)2])][1−(ω22(1+ω2)[P3ϑ3+ω2+1(ϑ3+ω2)2+(1−P3)ϑ4+ω2+1(ϑ4+ω2)2])][1−(ω32(1+ω3)[P5ϑ5+ω3+1(ϑ5+ω3)2+(1−P5)ϑ6+ω3+1(ϑ6+ω3)2])](ω42(1+ω4)[P7ϑ7+ω4+1(ϑ7+ω4)2+(1−P7)ϑ8+ω4+1(ϑ8+ω4)2])

By using Equation (1), we have;

R4=R(1)+R(2)+R(3)+R(4)

2.1.4 (Stress) follows two parameter generalized exponential combinations

The stress follows a two-parameter generalized exponential combinations with p.d.f.;

gi(Ψ)=αiτie−τiΨ(1−e−τiΨ)αi−1αi,τi,Ψ>0∀i=1,…,n

R(1)=α1τ1P1∫0∞e−θ1Ψe−τ1Ψ(1−e−τ1Ψ)α1−1dΨ+α1τ1(1−P1)∫0∞e−θ2Ψe−τ1Ψ(1−e−τ1Ψ)α1−1dΨ=α1P1∫01Ψα1−1(1−Ψ)ϑ1τ1dΨ+α1(1−P1)∫01Ψα1−1(1−Ψ)ϑ2τ1dΨ=α1P1Beta(α1,1+ϑ1τ1)+α1Beta(α1,1+ϑ2τ1)=α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)]

R(2)=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])]α2P3∫01Ψα2−1(1−Ψ)ϑ3τ1dΨ+α2(1−P3)∫01Ψα2−1(1−Ψ)ϑ4τ2dΨ=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])]α2Г(α2)[P3Г(1+ϑ3τ2)Г(α2+1+ϑ3τ2)+(1−P3)Г(1+ϑ4τ2)Г(α2+1+ϑ4τ2)]

So,

R(3)=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])][1−(α2Г(α2)[P3Г(1+ϑ3τ2)Г(α2+1+ϑ3τ2)+(1−P3)Г(1+ϑ4τ2)Г(α2+1+ϑ4τ2)])]α2P3∫01Ψα2−1(1−Ψ)ϑ3τ1dΨ+α2(1−P3)∫01Ψα2−1(1−Ψ)ϑ4τ2dΨ=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])][1−(α2Г(α2)[P3Г(1+ϑ3τ2)Г(α2+1+ϑ3τ2)+(1−P3)Г(1+ϑ4τ2)Г(α2+1+ϑ4τ2)])]α3Г(α3)[P5Г(1+ϑ5τ2)Г(α3+1+ϑ5τ3)+(1−P5)Г(1+ϑ6τ3)Г(α3+1+ϑ6τ3)]

In the same way:

R(4)=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])]=[1−(α1Г(α1)[P1Г(1+ϑ1τ1)Г(α1+1+ϑ1τ1)+(1−P1)Г(1+ϑ2τ1)Г(α1+1+ϑ2τ1)])][1−(α2Г(α2)[P3Г(1+ϑ3τ2)Г(α2+1+ϑ3τ2)+(1−P3)Г(1+ϑ4τ2)Г(α2+1+ϑ4τ2)])][1−(α3Г(α3)[P5Г(1+ϑ5τ2)Г(α3+1+ϑ5τ3)+(1−P5)Г(1+ϑ6τ3)Г(α3+1+ϑ6τ3)])]α4Г(α4)[P7Г(1+ϑ7τ4)Г(α4+1+ϑ7τ4)+(1−P7)Г(1+ϑ8τ4)Г(α4+1+ϑ8τ4)]

Therefore, we have;

R4=R(1)+R(2)+R(3)+R(4)

In this section, we will estimate R ₄ of the different distributions mentioned above using the method of maximum likelihood estimation. For all four distributions, we consider P and (1- P) as two sub-populations with mixing. Also we assume f₁(y) and f₂(y) be P.d.f. with parameters ϑ1 and ϑ2 .

Generated on the new experimental- simulation data and analytical results, this work establishes critical insights round the (R4) reliability system performance of the standby ( R4) in four distinct (stress) distributions, while keeping the (strength) distribution constant: It is considered a combination of two exponentials distributions.

Grant information

The author(s) declared that no grants were involved in supporting this work.

Copyright

© 2026 Ahmed I et al. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Version 1

Version 2

VERSION 2 PUBLISHED 22 Dec 2025

Comment