<i>R</i> <sub>4</sub>- STANDBY- STRENGTH-STRESS MODEL

Isaam Ahmed; Humam A Abdulrazzaq; M. Jasim Mohammed

doi:10.12688/f1000research.172210.1

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Research Article

R ₄- STANDBY- STRENGTH-STRESS MODEL

[version 1; peer review: awaiting peer review]

Isaam Ahmed ¹, Humam A Abdulrazzaq², M. Jasim Mohammed²

PUBLISHED 22 Dec 2025

Author details Author details

¹ of Mathematics, University of Anbar, Ramadi, Al Anbar Governorate, 009647812524220, Iraq
² Mathematics, University of Anbar, Ramadi, 009647812524220, Iraq

Isaam Ahmed
Roles: Conceptualization, Data Curation, Formal Analysis, Project Administration, Software, Writing – Original Draft Preparation

Humam A Abdulrazzaq
Roles: Formal Analysis

M. Jasim Mohammed
Roles: Formal Analysis

OPEN PEER REVIEW

REVIEWER STATUS AWAITING PEER REVIEW

This article is included in the Fallujah Multidisciplinary Science and Innovation gateway.

Abstract

Background

A standby system is defined as a system consisting of several components, where only one component operates at a time while the others remain in standby mode. The system fails when the applied stress exceeds the strength of the active component, at which point one of the standby components is activated. It is assumed that the inactive components do not undergo any change in their properties while in standby mode, and the system is considered to have failed when all its components have failed.

Methods

In this study, a mathematical formula of the standby system was derived for more than one component. Y is subject to independent random variable Ψ with Distribution of stress, which is considered a combination of two exponentials and the stress. It follows the following distributions: one-parameter exponential, two-parameter exponential, one parameter Lindley, and two-parameter generalized exponential. Hence, the dependent functions of the standby system are found { R ( 1 ) , R ( 2 ) , R ( 3 ) , R ( 4 ) , R 4 } depending on the mathematical formulas derived for each of the four distributions mentioned above.

Results

In this section, we will obtain the stress-strength of standby Reliability system R 4 , based on four different distributions the above mentioned, the standby Reliability functions R ( 2 ) for distributions are given by adding Marginal Reliability R ( 1 ) and R ( 2 ) , R ( 3 ) given by adding R ( 1 ) , R ( 2 ) and R ( 3 ) . Also R 4 given by adding R ( 1 ) , R ( 2 ) , R ( 3 ) and R ( 4 ) , and parameters estimator by maximum likelihood.

Conclusions

a simulation study will be conducted to see the behavior of { R ( 1 ) , R ( 2 ) , R ( 3 ) , R ( 4 ) , R 4 } of a standby system for four different distributions, in this simulation study will be conducted to see the estimator for all distributions and compare the results by using one important statistical criteria mean square error (MSE). Then results will be discussed to see which one of the estimators is the best for each one of the distributions separately.

Keywords

standby system, strength-stress, Reliability system, maximum likelihood function, generalized exponential, Lindley distribution.

Corresponding author: Isaam Ahmed

Competing interests: No competing interests were disclosed.

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

Copyright: © 2025 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.

How to cite: Ahmed I, Abdulrazzaq HA and Mohammed MJ. R ₄- STANDBY- STRENGTH-STRESS MODEL [version 1; peer review: awaiting peer review]. F1000Research 2025, 14:1423 (https://doi.org/10.12688/f1000research.172210.1) First published: 22 Dec 2025, 14:1423 (https://doi.org/10.12688/f1000research.172210.1) Latest published: 22 Dec 2025, 14:1423 (https://doi.org/10.12688/f1000research.172210.1)

1. Introduction

The estimation of reliability in stress and stress systems has become more important to researchers, particularly in the last century, when the distributions are independent. A standby system is a system that contains several components that are operating, and one of the components is active, while the remaining components are in standby mode. The system fails when the stress is greater than the strength of the active component, as well as other components in the standby mode. Consider a standby system that contains (n) identical components.⁷ Assume that one component in the system is operational and performs its assigned tasks, whereas the remaining (n-1) components are in standby.⁴ We assumed that these components are operational and idle. This system is known as the standby system. When the stress is greater than the $(s$ trength), it leads to failure of the component in the working position. There are two types of components: one in the active position and the other in the standby position. In the component standby system, there are two failure distributions: the first failure distribution occurs when the function is in the active mode and the second failure distribution when the component is in the standby mode. When the failure rate of the component in standby mode is such that the component is in active mode, we say that the vehicle has (hot standby).¹² If the failure rate of the component in standby mode is zero, then we call the system (gold standby), which we will adopt in this research. The estimation of the reliability function in the standby system was based on a combination of exponential distribution(exp.-dis.). This study was conducted by⁶ on the estimation of the reliability in the strength and stress systems for a system that contains multiple components and uses the standby system for different distributions.¹¹

2. Discussion of reliability

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^2,3

(1)

R_{4} = \sum_{i = 1}^{4} R (i)

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

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

(2)

R (i) = [\prod_{j = 1}^{i - 1} \int_{Ψ} T_{j} (Ψ) g_{i} (Ψ) dΨ] [{\bar{T}}_{j} (Ψ) g_{i} (Ψ)]

Where $T_{i} (Ψ)$ is the cumulative distribution function and ${\bar{T}}_{i} (Ψ) = 1 - T_{i} (Ψ)$ . In this study, we examine four different distributions.

2.1 (Stress) follows one parameter (exp.-dis.) and (Strength) follows a combination of the two (exp.-dis.)

Considering that the strength follows a mixture of tow (exp-dis.) with p.d.f.;

(3)

f_{1} (y) = P_{1} e^{- ϑ_{1} y} + (1 - P_{1}) e^{- ϑ_{2} y}

Therefore; ${\bar{T}}_{1} (y) = P_{1} e^{- ϑ_{1} y} + (1 - P_{1}) e^{- ϑ_{2} y}$

Also the stress follows one parameter (exp-dis.) with p.d.f;

g_{i} (Ψ) = μ_{i} e^{- μ_{i} Ψ} μ_{i} > 0, Ψ \geq 0

So,

\begin{matrix} R (1) = P (Y_{1} \geq Ψ) = \int_{0}^{\infty} {\bar{T}}_{1} (Ψ) g_{1} (Ψ) dΨ = \int_{0}^{\infty} [P_{1} e^{- ϑ_{1} Ψ} + (1 - P_{1}) e^{- ϑ_{2} Ψ}] μ_{i} e^{- μ_{i} Ψ} d Ψ \\ = P_{1} μ_{1} \int_{0}^{\infty} e^{- (θ_{1} + μ_{1}) Ψ} d Ψ + (1 - P_{1}) μ_{1} \int_{0}^{\infty} e^{- (θ_{2} + μ_{1}) Ψ} d Ψ \\ = \frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}} \end{matrix}

Now,

\begin{matrix} R (2) = P (Y_{1} < Ψ_{1}, Y_{2} \geq Ψ_{2}) = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{2} (Ψ) g_{2} (Ψ) dΨ] \\ = [1 - (\frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}})] \int_{0}^{\infty} [P_{3} e^{- ϑ_{3} Ψ} + (1 - P_{3}) e^{- ϑ_{4} Ψ}] μ_{2} e^{- μ_{2} Ψ} d Ψ \\ = [1 - (\frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}})] [\frac{P_{3} μ_{2}}{ϑ_{3} + μ_{2}} + \frac{(1 - P_{3}) μ_{2}}{ϑ_{4} + μ_{2}}] \end{matrix}

\begin{matrix} R (3) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} \geq Ψ_{3}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{3} (Ψ) g_{3} (Ψ) dΨ] \\ = [1 - (\frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}})] [1 - (\frac{P_{3} μ_{2}}{ϑ_{3} + μ_{2}} + \frac{(1 - P_{3}) μ_{2}}{ϑ_{4} + μ_{2}})] [\int_{0}^{\infty} [P_{5} e^{- ϑ_{5} Ψ} + (1 - P_{5}) e^{- ϑ_{6} Ψ}] μ_{3} e^{- μ_{3} Ψ} d Ψ] \\ [1 - (\frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}})] [1 - (\frac{P_{3} μ_{2}}{ϑ_{3} + μ_{2}} + \frac{(1 - P_{3}) μ_{2}}{ϑ_{4} + μ_{2}})] [\frac{P_{5} μ_{3}}{ϑ_{5} + μ_{3}} + \frac{(1 - P_{5}) μ_{3}}{ϑ_{6} + μ_{3}}] \end{matrix}

In the same way;

\begin{matrix} R (4) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} < Ψ_{3}, Y_{4} \geq Ψ_{4}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} T_{3} (Ψ) g_{3} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{4} (Ψ) g_{4} (Ψ) dΨ] \\ = [1 - (\frac{P_{1} μ_{1}}{ϑ_{1} + μ_{1}} + \frac{(1 - P_{1}) μ_{1}}{ϑ_{2} + μ_{1}})] [1 - (\frac{P_{3} μ_{2}}{ϑ_{3} + μ_{2}} + \frac{(1 - P_{3}) μ_{2}}{ϑ_{4} + μ_{2}})] [1 - (\frac{P_{5} μ_{3}}{ϑ_{5} + μ_{3}} + \frac{(1 - P_{5}) μ_{3}}{ϑ_{6} + μ_{3}})] \\ [\frac{P_{7} μ_{4}}{ϑ_{7} + μ_{4}} + \frac{(1 - P_{7}) μ_{4}}{ϑ_{8} + μ_{4}}] \end{matrix}

By using Equation (1), we get;

(4)

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

2.2 (Stress) follows two parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

Let the strength follow a combination of the tow (exp.-dis.) with p.d.f.;¹

f_{i} (y) = P_{2 i - 1} ϑ_{2 i - 1} e^{- ϑ_{2 i - 1} y} + (1 - P_{2 i - 1}) ϑ_{2 i} e^{- ϑ_{2 i} y}

Therefore; ${\bar{T}}_{1} (y) = P_{1} e^{- ϑ_{1} y} + (1 - P_{1}) e^{- ϑ_{2} y}$

Also the stress follows two parameter (exp-dis.) with p.d.f.;

g_{i} (Ψ) = \frac{1}{£_{i}} e^{- (Ψ - £_{i}) / £_{i}}, Ψ > α > 0, Ψ \geq 0

Now,

\begin{matrix} R (1) = P (Y_{1} \geq Ψ_{1}) = \int_{0}^{\infty} {\bar{T}}_{1} (Ψ) g_{1} (Ψ) dΨ \\ = \int_{0}^{\infty} [P_{1} e^{- ϑ_{1} Ψ} + (1 - P_{1}) e^{- ϑ_{2} Ψ}] \frac{1}{£_{1}} e^{- (Ψ - α_{1}) / £_{1}} d Ψ \\ = \frac{P_{1}}{£_{1}} e^{\frac{α_{1}}{£_{1}}} \int_{α}^{\infty} e^{- (ϑ_{1} + \frac{1}{£_{1}}) Ψ} dΨ + \frac{(1 - P_{1})}{£_{1}} e^{\frac{α_{1}}{£_{1}}} \int_{α}^{\infty} e^{- (ϑ_{2} + \frac{1}{£_{1}}) Ψ} dΨ \\ = \frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}] \end{matrix}

\begin{matrix} R (2) = P (Y_{1} < Ψ_{1}, Y_{2} \geq Ψ_{2}) = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{2} (Ψ) g_{2} (Ψ) dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] [\frac{P_{3}}{£_{2}} e^{\frac{α_{2}}{£_{2}}} \int_{α}^{\infty} e^{- (ϑ_{3} + \frac{1}{£_{2}}) Ψ} dΨ + \frac{(1 - P_{3})}{£_{2}} e^{\frac{α_{2}}{£_{2}}} \int_{α}^{\infty} e^{- (ϑ_{4} + \frac{1}{£_{2}}) Ψ} dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] \frac{e^{\frac{α_{2}}{£_{2}}}}{£_{2}} [\frac{P_{3} e^{- (ϑ_{3} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{2}})} + \frac{(1 - P_{3}) e^{- (ϑ_{4} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{4} + \frac{1}{£_{2}})}] \end{matrix}

So,

\begin{matrix} R (3) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} \geq Ψ_{3}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{3} (Ψ) g_{3} (Ψ) dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] [1 - (\frac{e^{\frac{α_{2}}{£_{2}}}}{£_{2}} [\frac{P_{3} e^{- (ϑ_{3} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{2}})} + \frac{(1 - P_{3}) e^{- (ϑ_{4} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{4} + \frac{1}{£_{2}})}])] \\ [\frac{P_{5}}{£_{3}} e^{\frac{α_{3}}{£_{3}}} \int_{α_{3}}^{\infty} e^{- (ϑ_{5} + \frac{1}{£_{3}}) Ψ} dΨ + \frac{(1 - P_{5})}{£_{3}} e^{\frac{α_{3}}{£_{3}}} \int_{α_{3}}^{\infty} e^{- (ϑ_{6} + \frac{1}{£_{3}}) Ψ} dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] [1 - (\frac{e^{\frac{α_{2}}{£_{2}}}}{£_{2}} [\frac{P_{3} e^{- (ϑ_{3} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{2}})} + \\ \frac{(1 - P_{3}) e^{- (ϑ_{4} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{4} + \frac{1}{£_{2}})}])] \\ (\frac{e^{\frac{α_{3}}{£_{3}}}}{£_{3}} [\frac{P_{5} e^{- (ϑ_{5} + \frac{1}{£_{3}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{3}})} + \frac{(1 - P_{5}) e^{- (ϑ_{6} + \frac{1}{£_{3}}) α_{3}}}{(ϑ_{4} + \frac{1}{£_{3}})}])] \end{matrix}

In the same way;

\begin{matrix} R (4) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} < Ψ_{3}, Y_{4} \geq Ψ_{4}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} T_{3} (Ψ) g_{3} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{4} (Ψ) g_{4} (Ψ) dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] [1 - (\frac{e^{\frac{α_{2}}{£_{2}}}}{£_{2}} [\frac{P_{3} e^{- (ϑ_{3} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{2}})} + \frac{(1 - P_{3}) e^{- (ϑ_{4} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{4} + \frac{1}{£_{2}})}])] \\ [\frac{P_{5}}{£_{3}} e^{\frac{α_{3}}{£_{3}}} \int_{α_{3}}^{\infty} e^{- (ϑ_{5} + \frac{1}{£_{3}}) Ψ} dΨ + \frac{(1 - P_{5})}{£_{3}} e^{\frac{α_{3}}{£_{3}}} \int_{α_{3}}^{\infty} e^{- (ϑ_{6} + \frac{1}{£_{3}}) Ψ} dΨ] \\ = [1 - (\frac{e^{\frac{α_{1}}{£_{1}}}}{£_{1}} [\frac{P_{1} e^{- (ϑ_{1} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{1} + \frac{1}{£_{1}})} + \frac{(1 - P_{1}) e^{- (ϑ_{2} + \frac{1}{£_{1}}) α_{1}}}{(ϑ_{2} + \frac{1}{£_{1}})}])] [1 - (\frac{e^{\frac{α_{2}}{£_{2}}}}{£_{2}} [\frac{P_{3} e^{- (ϑ_{3} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{2}})} + \\ \frac{(1 - P_{3}) e^{- (ϑ_{4} + \frac{1}{£_{2}}) α_{2}}}{(ϑ_{4} + \frac{1}{£_{2}})}])] \\ [1 - (\frac{e^{\frac{α_{3}}{£_{3}}}}{£_{3}} [\frac{P_{5} e^{- (ϑ_{5} + \frac{1}{£_{3}}) α_{2}}}{(ϑ_{3} + \frac{1}{£_{3}})} + \frac{(1 - P_{5}) e^{- (ϑ_{6} + \frac{1}{£_{3}}) α_{3}}}{(ϑ_{4} + \frac{1}{£_{3}})}])] [(\frac{e^{\frac{α_{4}}{£_{4}}}}{£_{4}} [\frac{P_{7} e^{- (ϑ_{7} + \frac{1}{£_{4}}) α_{4}}}{(ϑ_{7} + \frac{1}{£_{4}})} + \frac{(1 - P_{7}) e^{- (ϑ_{8} + \frac{1}{£_{4}}) α_{4}}}{(ϑ_{8} + \frac{1}{£_{4}})}] \end{matrix}

By using Equation (1), we get;

(5)

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

2.3 (Stress) follows one parameter Lindley distribution and (Stress) follows a combination of two (exp.-dis.)

Let the strength follow a combination of the tow (exp.-dis.) which is given in Equation (3), and the stress follows a one-parameter Lindley distribution with p.d.f.⁵

g_{i} (Ψ) = \frac{ω_{i}^{2}}{(1 + ω_{i})} (1 + Ψ) e^{- ω_{i} Ψ} ω_{i} > 0, Ψ \geq 0, i = 1, \dots, n

\begin{matrix} R (1) = P (Y_{1} \geq Ψ_{1}) \\ = \int_{0}^{\infty} {\bar{T}}_{1} (Ψ) g_{1} (Ψ) dΨ = \int_{0}^{\infty} [P_{1} e^{- ϑ_{1} Ψ} + (1 - P_{1}) e^{- ϑ_{2} Ψ}] \frac{ω_{1}^{2}}{(1 + ω_{1})} (1 + Ψ) e^{- ω_{1} Ψ} d Ψ \\ = \frac{P_{1} ω_{1}^{2}}{(1 + ω_{1})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{1} + ω_{1}) Ψ} dΨ + \frac{(1 - P_{1}) ω_{1}^{2}}{(1 + ω_{1})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{2} + ω_{1}) Ψ} dΨ \\ = \frac{P_{1} ω_{1}^{2}}{(1 + ω_{1})} [e^{- (ϑ_{1} + ω_{1}) Ψ} dΨ + \int_{0}^{\infty} Ψ e^{- (ϑ_{1} + ω_{1}) Ψ}] + \frac{(1 - P_{1}) ω_{1}^{2}}{(1 + ω_{1})} [e^{- (ϑ_{2} + ω_{1}) Ψ} dΨ + \int_{0}^{\infty} Ψ e^{- (ϑ_{2} + ω_{1}) Ψ}] \\ = \frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}] \end{matrix}

\begin{matrix} R (2) = P (Y_{1} < Ψ_{1}, Y_{2} \geq Ψ_{2}) = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{2} (Ψ) g_{2} (Ψ) dΨ] \\ = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ \frac{P_{3} ω_{2}^{2}}{(1 + ω_{2})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{3} + ω_{2}) Ψ} dΨ + \frac{(1 - P_{3}) ω_{1}^{2}}{(1 + ω_{2})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{4} + ω_{2}) Ψ} d Ψ \\ = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}]) \end{matrix}

So,

\begin{matrix} R (3) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} \geq Ψ_{3}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{3} (Ψ) g_{3} (Ψ) dΨ] \\ = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ [1 - (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}])] \frac{P_{5} ω_{3}^{2}}{(1 + ω_{3})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{5} + ω_{3}) Ψ} dΨ \\ + \frac{(1 - P_{5}) ω_{3}^{2}}{(1 + ω_{3})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{6} + ω_{3}) Ψ} d Ψ \\ = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ [1 - (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}])] \\ (\frac{ω_{3}^{2}}{(1 + ω_{3})} [P_{5} \frac{ϑ_{5} + ω_{3} + 1}{{(ϑ_{5} + ω_{3})}^{2}} + (1 - P_{5}) \frac{ϑ_{6} + ω_{3} + 1}{{(ϑ_{6} + ω_{3})}^{2}}]) \end{matrix}

In the same way;

\begin{matrix} R (4) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} < Ψ_{3}, Y_{4} \geq Ψ_{4}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} T_{3} (Ψ) g_{3} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{4} (Ψ) g_{4} (Ψ) dΨ] \\ = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ [1 - (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}])] \\ [1 - (\frac{ω_{3}^{2}}{(1 + ω_{3})} [P_{5} \frac{ϑ_{5} + ω_{3} + 1}{{(ϑ_{5} + ω_{3})}^{2}} + (1 - P_{5}) \frac{ϑ_{6} + ω_{3} + 1}{{(ϑ_{6} + ω_{3})}^{2}}])] \frac{P_{7} ω_{4}^{2}}{(1 + ω_{4})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{7} + ω_{4}) Ψ} dΨ \\ + \frac{(1 - P_{7}) ω_{4}^{2}}{(1 + ω_{4})} \int_{0}^{\infty} (1 + Ψ) e^{- (ϑ_{8} + ω_{4}) Ψ} d z = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ [1 - (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}])] \\ [1 - (\frac{ω_{3}^{2}}{(1 + ω_{3})} [P_{5} \frac{ϑ_{5} + ω_{3} + 1}{{(ϑ_{5} + ω_{3})}^{2}} + (1 - P_{5}) \frac{ϑ_{6} + ω_{3} + 1}{{(ϑ_{6} + ω_{3})}^{2}}])] \\ (\frac{ω_{4}^{2}}{(1 + ω_{4})} [P_{7} \frac{ϑ_{7} + ω_{4} + 1}{{(ϑ_{7} + ω_{4})}^{2}} + (1 - P_{7}) \frac{ϑ_{8} + ω_{4} + 1}{{(ϑ_{8} + ω_{4})}^{2}}]) \end{matrix}

So,

\begin{matrix} R (4) = [1 - (\frac{ω_{1}^{2}}{(1 + ω_{1})} [P_{1} \frac{ϑ_{1} + ω_{1} + 1}{{(ϑ_{1} + ω_{1})}^{2}} + (1 - P_{1}) \frac{ϑ_{2} + ω_{1} + 1}{{(ϑ_{2} + ω_{1})}^{2}}])] \\ [1 - (\frac{ω_{2}^{2}}{(1 + ω_{2})} [P_{3} \frac{ϑ_{3} + ω_{2} + 1}{{(ϑ_{3} + ω_{2})}^{2}} + (1 - P_{3}) \frac{ϑ_{4} + ω_{2} + 1}{{(ϑ_{4} + ω_{2})}^{2}}])] \\ [1 - (\frac{ω_{3}^{2}}{(1 + ω_{3})} [P_{5} \frac{ϑ_{5} + ω_{3} + 1}{{(ϑ_{5} + ω_{3})}^{2}} + (1 - P_{5}) \frac{ϑ_{6} + ω_{3} + 1}{{(ϑ_{6} + ω_{3})}^{2}}])] \\ (\frac{ω_{4}^{2}}{(1 + ω_{4})} [P_{7} \frac{ϑ_{7} + ω_{4} + 1}{{(ϑ_{7} + ω_{4})}^{2}} + (1 - P_{7}) \frac{ϑ_{8} + ω_{4} + 1}{{(ϑ_{8} + ω_{4})}^{2}}]) \end{matrix}

By using Equation (1), we have;

(6)

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

2.4 (Stress) follows two parameter generalized (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

Let the strength follow a combination of the tow (exp-dis.) which is given in Equation (3), and the stress follows a two-parameter generalized (exp-dis.) with p.d.f.;

g_{i} (Ψ) = α_{i} τ_{i} e^{- τ_{i} Ψ} {(1 - e^{- τ_{i} Ψ})}^{α_{i} - 1} α_{i}, τ_{i} > 0, Ψ \geq 0, i = 1, \dots, n

\begin{matrix} R (1) = P (Y_{1} \geq Ψ_{1}) = \int_{0}^{\infty} {\bar{T}}_{1} (Ψ) g_{1} (Ψ) dΨ \\ = \int_{0}^{\infty} [P_{1} e^{- ϑ_{1} Ψ} + (1 - P_{1}) e^{- ϑ_{2} Ψ}] α_{1} τ_{1} e^{- τ_{1} Ψ} {(1 - e^{- τ_{1} Ψ})}^{α_{1} - 1} d Ψ \\ = α_{1} τ_{1} P_{1} \int_{0}^{\infty} e^{- θ_{1} Ψ} e^{- τ_{1} Ψ} {(1 - e^{- τ_{1} Ψ})}^{α_{1} - 1} d Ψ + α_{1} τ_{1} (1 - P_{1}) \int_{0}^{\infty} e^{- θ_{2} Ψ} e^{- τ_{1} Ψ} {(1 - e^{- τ_{1} Ψ})}^{α_{1} - 1} dΨ \\ = α_{1} P_{1} \int_{0}^{1} Ψ^{α_{1} - 1} {(1 - Ψ)}^{\frac{ϑ_{1}}{τ_{1}}} d Ψ + α_{1} (1 - P_{1}) \int_{0}^{1} Ψ^{α_{1} - 1} {(1 - Ψ)}^{\frac{ϑ_{2}}{τ_{1}}} d Ψ \\ = α_{1} P_{1} Beta (α_{1}, 1 + \frac{ϑ_{1}}{τ_{1}}) + α_{1} Beta (α_{1}, 1 + \frac{ϑ_{2}}{τ_{1}}) = α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}] \end{matrix}

\begin{matrix} R (2) = P (Y_{1} < Ψ_{1}, Y_{2} \geq Ψ_{2}) = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{2} (Ψ) g_{2} (Ψ) dΨ] \\ = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ α_{2} P_{3} \int_{0}^{1} Ψ^{α_{2} - 1} {(1 - Ψ)}^{\frac{ϑ_{3}}{τ_{1}}} d Ψ + α_{2} (1 - P_{3}) \int_{0}^{1} Ψ^{α_{2} - 1} {(1 - Ψ)}^{\frac{ϑ_{4}}{τ_{2}}} d Ψ \\ = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ α_{2} Г (α_{2}) [P_{3} \frac{Г (1 + \frac{ϑ_{3}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{3}}{τ_{2}})} + (1 - P_{3}) \frac{Г (1 + \frac{ϑ_{4}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{4}}{τ_{2}})}] \end{matrix}

So,

\begin{matrix} R (3) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} \geq Ψ_{3}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{3} (Ψ) g_{3} (Ψ) dΨ] \\ = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ [1 - (α_{2} Г (α_{2}) [P_{3} \frac{Г (1 + \frac{ϑ_{3}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{3}}{τ_{2}})} + (1 - P_{3}) \frac{Г (1 + \frac{ϑ_{4}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{4}}{τ_{2}})}])] \\ α_{2} P_{3} \int_{0}^{1} Ψ^{α_{2} - 1} {(1 - Ψ)}^{\frac{ϑ_{3}}{τ_{1}}} d Ψ + α_{2} (1 - P_{3}) \int_{0}^{1} Ψ^{α_{2} - 1} {(1 - Ψ)}^{\frac{ϑ_{4}}{τ_{2}}} d Ψ \\ = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ [1 - (α_{2} Г (α_{2}) [P_{3} \frac{Г (1 + \frac{ϑ_{3}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{3}}{τ_{2}})} + (1 - P_{3}) \frac{Г (1 + \frac{ϑ_{4}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{4}}{τ_{2}})}])] \\ α_{3} Г (α_{3}) [P_{5} \frac{Г (1 + \frac{ϑ_{5}}{τ_{2}})}{Г (α_{3} + 1 + \frac{ϑ_{5}}{τ_{3}})} + (1 - P_{5}) \frac{Г (1 + \frac{ϑ_{6}}{τ_{3}})}{Г (α_{3} + 1 + \frac{ϑ_{6}}{τ_{3}})}] \end{matrix}

In the same way:

\begin{matrix} R (4) = P (Y_{1} < Ψ_{1}, Y_{2} < Ψ_{2}, Y_{3} < Ψ_{3}, Y_{4} \geq Ψ_{4}) \\ = [\int_{0}^{\infty} T_{1} (Ψ) g_{1} (Ψ) dΨ] [\int_{0}^{\infty} T_{2} (Ψ) g_{2} (Ψ) dΨ] [\int_{0}^{\infty} T_{3} (Ψ) g_{3} (Ψ) dΨ] [\int_{0}^{\infty} {\bar{T}}_{4} (Ψ) g_{4} (Ψ) dΨ] = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ = [1 - (α_{1} Г (α_{1}) [P_{1} \frac{Г (1 + \frac{ϑ_{1}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{1}}{τ_{1}})} + (1 - P_{1}) \frac{Г (1 + \frac{ϑ_{2}}{τ_{1}})}{Г (α_{1} + 1 + \frac{ϑ_{2}}{τ_{1}})}])] \\ [1 - (α_{2} Г (α_{2}) [P_{3} \frac{Г (1 + \frac{ϑ_{3}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{3}}{τ_{2}})} + (1 - P_{3}) \frac{Г (1 + \frac{ϑ_{4}}{τ_{2}})}{Г (α_{2} + 1 + \frac{ϑ_{4}}{τ_{2}})}])] \\ [1 - (α_{3} Г (α_{3}) [P_{5} \frac{Г (1 + \frac{ϑ_{5}}{τ_{2}})}{Г (α_{3} + 1 + \frac{ϑ_{5}}{τ_{3}})} + (1 - P_{5}) \frac{Г (1 + \frac{ϑ_{6}}{τ_{3}})}{Г (α_{3} + 1 + \frac{ϑ_{6}}{τ_{3}})}])] \\ α_{4} Г (α_{4}) [P_{7} \frac{Г (1 + \frac{ϑ_{7}}{τ_{4}})}{Г (α_{4} + 1 + \frac{ϑ_{7}}{τ_{4}})} + (1 - P_{7}) \frac{Г (1 + \frac{ϑ_{8}}{τ_{4}})}{Г (α_{4} + 1 + \frac{ϑ_{8}}{τ_{4}})}] \end{matrix}

Therefore, we have;

(7)

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

3. Estimation of $R$ 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 $ϑ_{1}$ and $ϑ_{2}$ .

3.1 (Stress) follows one parameter (exp.-dis.)

Let $Ψ$ _ij be r.s. from one parameter (exp-dis.) with parameter $ʎ_{i}$ .

Where, i = 1,…,n, j = 1,…, $k_{i}$ . The L.f. is given by⁸ and⁹

\begin{matrix} L (ϑ_{1}, ϑ_{2}, P, ʎ_{i} / Ψ_{ij}, y_{ij}) = \prod_{i = 1}^{n} [\sum_{j = 1}^{2} P_{j} f (y_{j}, ϑ_{j})] \prod_{i = 1}^{n} \prod_{j = 1}^{k_{i}} [μ_{i} e^{- μ_{i} Ψ_{ij}}] \\ = (n! / n_{1}! * n_{2}!) . P^{n_{1}} {(1 - P)}^{n_{2}} . ϑ_{1}^{n_{1}} . ϑ_{2}^{n_{2}} e^{[- \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i}]} {μ_{i}}^{k_{i}} e^{- μ_{i} \sum_{j = 1}^{k_{i}} Ψ_{ij}} \end{matrix}

Ln (L) = Ln (\frac{n!}{n_{1}! n_{2}!}) + (n_{1} . Ln (P)) + (n_{2} . Ln (1 - P)) + (n_{1} . Ln ϑ_{1}) + (n_{2} Ln ϑ_{2}) - \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i} + k_{i} ln μ_{i} - μ_{i} \sum_{j = 1}^{k_{i}} Ψ_{ij}

We use; ( $\frac{\partial ln L}{\partial P}) = 0, (\frac{\partial ln L}{\partial ϑ_{1}}) = 0, ($ $\frac{\partial ln L}{\partial ϑ_{2}}) = 0$ and $\frac{\partial ln L}{\partial μ_{i}} = 0$ .

Therefore, we get; $\overset{`}{P} = n_{1} / (n_{1} + n_{2}), {\overset{`}{ϑ}}_{1} = n_{1} / \sum_{j = 1}^{n_{1}} y_{1 j}, {\overset{`}{ϑ}}_{2} = n_{2} / \sum_{j = 1}^{n_{2}} y_{2 j} and \hat{μ_{i}} = \frac{k_{i}}{\sum_{j = 1}^{k_{i}} Ψ_{ij}}$

So, we have;

(8)

\hat{R} 4 = \hat{R} (1) + \hat{R} (2) + \hat{R} (3) + \hat{R} (4)

3.2 (Stress) follows two parameter (exp.-dis.)

Let $Ψ$ _ij be r.s. from two parameter (exp-dis.) with parameter $α_{i}, £_{i} [1];$ Where, i = 1,…,n, j = 1,…, $k_{i}$ . The L.f. is given by;

\begin{matrix} L (ϑ_{1}, ϑ_{2}, P, £_{i}, α_{i} / Ψ_{ij}, y_{ij}) = \prod_{i = 1}^{n} [\sum_{j = 1}^{2} P_{j} f (y_{j}, ϑ_{j})] \prod_{i = 1}^{n} \prod_{j = 1}^{k_{i}} [\frac{1}{£_{i}} e^{- (Ψ_{ij} - α_{i}) / £_{i}}] \\ = (n! / n_{1}! * n_{2}!) . P^{n_{1}} {(1 - P)}^{n_{2}} . ϑ_{1}^{n_{1}} . ϑ_{2}^{n_{2}} e^{[- \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i}]} {(\frac{1}{£_{i}})}^{k_{i}} e^{- \frac{1}{£_{i}} \sum_{j = 1}^{k_{i}} (Ψ_{ij - α_{i}})} \end{matrix}

Ln (L) = Ln (\frac{n!}{n_{1}! n_{2}!}) + (n_{1} . Ln (P)) + (n_{2} . Ln (1 - P)) + (n_{1} . Ln ϑ_{1}) + (n_{2} Ln ϑ_{2}) - \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i} - k_{i} ln £_{i} - \frac{1}{£_{i}} \sum_{j = 1}^{k_{i}} ln (Ψ_{ij} - α_{i})

We use; ( $\frac{\partial ln L}{\partial P}) = 0, (\frac{\partial ln L}{\partial ϑ_{1}}) = 0, ($ $\frac{\partial ln L}{\partial ϑ_{2}}) = 0, \frac{\partial ln L}{\partial α_{i}} = 0$ and $\frac{\partial ln L}{\partial £_{i}} = 0 .$

\overset{`}{P} = n_{1} / (n_{1} + n_{2}), {\overset{`}{ϑ}}_{1} = n_{1} / \sum_{j = 1}^{n_{1}} y_{1 j}, {\overset{`}{ϑ}}_{2} = n_{2} / \sum_{j = 1}^{n_{2}} y_{2 j} and \hat{£_{i}} = \frac{\sum_{j = 1}^{k_{i}} ln (Ψ_{ij} - α_{i})}{k_{i}}

Where; $α_{i} =$ min $(Ψ_{ij})$ , So, we have;

(9)

\hat{R} 4 = \hat{R} (1) + \hat{R} (2) + \hat{R} (3) + \hat{R} (4)

3.3 (Stress) follows one parameter Lindley distribution

Let $Ψ$ _ij be r.s. from one parameter (exp-dis.) with parameter $ω_{i}$ . Where, i = 1,…,n, j = 1,…, $k_{i}$ . The L.f. is given by¹⁰;

\begin{matrix} L (ϑ_{1}, ϑ_{2}, P, ω_{i} / Ψ_{ij}, y_{ij}) = \prod_{i = 1}^{n} [\sum_{j = 1}^{2} P_{j} f (y_{j}, ϑ_{j})] \prod_{i = 1}^{n} \prod_{j = 1}^{k_{i}} \frac{ω_{i}^{2}}{(1 + ω_{i})} (1 + Ψ_{ij}) e^{- ω_{i} Ψ_{ij}} \\ = (n! / n_{1}! * n_{2}!) . P^{n_{1}} {(1 - P)}^{n_{2}} . ϑ_{1}^{n_{1}} . ϑ_{2}^{n_{2}} e^{[- \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i}]} \frac{ω_{i}^{2 k_{i}}}{{(1 + ω_{i})}^{k_{i}}} [\prod_{j = 1}^{k_{i}} (1 + Ψ_{ij})] e^{- ω_{i} \sum_{j = 1}^{k_{i}} (Ψ_{ij})} \end{matrix}

Ln (L) = Ln (\frac{n!}{n_{1}! n_{2}!}) + (n_{1} . Ln (P)) + (n_{2} . Ln (1 - P)) + (n_{1} . Ln ϑ_{1}) + (n_{2} Ln ϑ_{2}) - \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i} + 2 k_{i} ln ω_{i} - ω_{i} \sum_{j = 1}^{k_{i}} Ψ_{ij} - k_{i} ln (1 + ω_{i}) + \sum_{j = 1}^{k_{i}} ln (1 + Ψ_{ij})

We use; ( $\frac{\partial ln L}{\partial P}) = 0, (\frac{\partial ln L}{\partial ϑ_{1}}) = 0, ($ $\frac{\partial ln L}{\partial ϑ_{2}}) = 0$ and $\frac{\partial ln L}{\partial ω_{i}} = 0$ .

Therefore, we get; $\overset{`}{P} = (\frac{n_{1}}{n_{1}} + n_{2}), {\overset{`}{ϑ}}_{1} = n_{1} / \sum_{j = 1}^{n_{1}} y_{1 j}, {\overset{`}{ϑ}}_{2} = \frac{n_{2}}{\sum_{j = 1}^{n_{2}}} y_{2 j} and \hat{ω_{i}} = \frac{(1 - {\bar{Ψ}}_{i}) + \sqrt{{\bar{Ψ}}_{i} + 6 {\bar{Ψ}}_{i} + 1}}{2 {\bar{Ψ}}_{i}}$

Where; ${\bar{Ψ}}_{i} = \frac{\sum_{j = 1}^{k_{i}} (Ψ_{ij})}{k_{i}}$ . So, we have;

(10)

\hat{R} 4 = \hat{R} (1) + \hat{R} (2) + \hat{R} (3) + \hat{R} (4)

3.4 (Stress) follows two parameter generalized (exp.-dis.)

Let $Ψ$ _ij be r.s. from two parameter (exp.-dis.) with parameter $α_{i}, τ_{i}$ . Where, i = 1,…,n, j = 1,…, $k_{i}$ . The L.f is given by;

\begin{matrix} L (ϑ_{1}, ϑ_{2}, P, τ_{i}, α_{i} / Ψ_{ij}, y_{ij}) = \prod_{i = 1}^{n} [\sum_{j = 1}^{2} P_{j} f (y_{j}, ϑ_{j})] \prod_{i = 1}^{n} \prod_{j = 1}^{k_{i}} [α_{i} τ_{i} e^{- τ_{i} Ψ ij} {(1 - e^{- τ_{i} Ψ ij})}^{α_{i} - 1}] \\ = (\frac{n!}{n_{1}!} * n_{2}!) . P^{n_{1}} (1 - P)^{n_{2}} . ϑ_{1}^{n_{1}} . ϑ_{2}^{n_{2}} e^{[- \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i}]} {α_{i}}^{k_{i}} {τ_{i}}^{k_{i}} e^{- τ_{i} \sum_{j = 1}^{k_{i}} (Ψ_{ij})} (\prod_{j = 1}^{k_{i}} {(1 - e^{- τ_{i} Ψ ij})}^{α_{i} - 1}) \end{matrix}

Ln (L) = Ln (\frac{n!}{n_{1}! n_{2}!}) + (n_{1} . Ln (P)) + (n_{2} . Ln (1 - P)) + (n_{1} . Ln ϑ_{1}) + (n_{2} Ln ϑ_{2}) - \sum_{i = 1}^{n_{1}} ϑ_{1} y_{1 i} - \sum_{i = 1}^{n_{2}} ϑ_{2} y_{2 i} + k_{i} Ln α_{i} + k_{i} Ln τ_{i} - τ_{i} \sum_{j = 1}^{k_{i}} (Ψ_{ij}) + (α_{i} - 1) \sum_{j = 1}^{k_{i}} Ln (1 - e^{- τ_{i} Ψ_{ij}})

We use; ( $\frac{\partial ln L}{\partial P}) = 0, (\frac{\partial ln L}{\partial ϑ_{1}}) = 0, ($ $\frac{\partial ln L}{\partial ϑ_{2}}) = 0, \frac{\partial ln L}{\partial α_{i}} = 0$ and $\frac{\partial ln L}{\partial τ_{i}} = 0 .$

\overset{`}{P} = n_{1} / (n_{1} + n_{2}), {\overset{`}{ϑ}}_{1} = n_{1} / \sum_{j = 1}^{n_{1}} y_{1 j}, {\overset{`}{ϑ}}_{2} = n_{2} / \sum_{j = 1}^{n_{2}} y_{2 j}, \hat{α_{i}} = \frac{k_{i}}{\sum_{j = 1}^{k_{i}} ln (1 - e^{- τ_{i} Ψ_{ij}})}

And $\hat{τ_{i}} = {[\frac{\sum_{j = 1}^{k_{i}} ((Ψ_{ij} e^{- τ_{i} Ψ_{ij}}) / (1 - e^{- τ_{i} Ψ_{ij}}))}{\sum_{j = 1}^{k_{i}} ln (1 - e^{- τ_{i} Ψ_{ij}})} + \frac{1}{k_{i}} \sum_{j = 1}^{k_{i}} \frac{Ψ_{ij}}{(1 - e^{- τ_{i} Ψ_{ij}})}]}^{- 1}$

So, we have;

(11)

\hat{R} 4 = \hat{R} (1) + \hat{R} (2) + \hat{R} (3) + \hat{R} (4)

4. The simulation manner

In this section, the numerical results are presented to compare the performance of the different reliability values obtained for four different distributions.

4.1 (Stress) follows one parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

The numerical results can be seen in the Tables 1, 2, 3 and 4:

Table 1. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8) .$

$λ_{1}$	$λ_{2}$	$λ_{3}$	$λ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.3	0.3	0.3	0.3750	0.2344	0.1465	0.0916	0.8474
0.5	0.5	0.5	0.5	0.5000	0.2500	0.1250	0.0625	0.9375
0.7	0.7	0.7	0.7	0.5833	0.2431	0.1013	0.0422	0.9699
0.9	0.9	0.9	0.9	0.6429	0.2296	0.0820	0.0293	0.9837
1.1	1.1	1.1	1.1	0.6875	0.2148	0.0671	0.0210	0.9905
1.5	1.5	1.5	1.5	0.7500	0.1875	0.0469	0.0117	0.9961

Table 2. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8)$ .

$λ_{1}$	$λ_{2}$	$λ_{3}$	$λ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.5	0.7	0.9	0.3750	0.3125	0.1823	0.0837	0.9535
0.8	1.0	1.2	1.4	0.6154	0.2564	0.0905	0.0278	0.9901
1.2	1.4	1.6	1.8	0.7059	0.2167	0.0590	0.0144	0.9960
1.8	2.0	2.2	2.4	0.5958	0.3233	0.0659	0.0124	0.9974
2.2	2.4	2.6	2.8	0.8148	0.1533	0.0268	0.0044	0.9992

Table 3. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); λ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$θ_{1}$	$θ_{2}$	$θ_{3}$	$θ_{4}$	$θ_{5}$	$θ_{6}$	$θ_{7}$	$θ_{8}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.8934	0.0851	0.0155	0.0039	0.9980
0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	0.8431	0.8431	0.1189	0.0075	0.9956
0.3	0.4	0.5	0.6	0.7	0.8	0.9	1	0.7982	0.1456	0.0370	0.0116	0.9924
0.5	0.6	0.7	0.8	0.9	1.0	1.1	1.2	0.7214	0.1833	0.0576	0.0211	0.9834

Table 4. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); λ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$p_{1}$	$p_{3}$	$p_{5}$	$p_{7}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.7500	0.1875	0.0469	0.0117	0.9961
0.3	0.4	0.5	0.6	0.7500	0.1875	0.0469	0.0117	0.9961
0.5	0.6	0.7	0.8	0.7500	0.1875	0.0469	0.0078	0.9922
0.7	0.8	0.9	1	0.7500	0.1875	0.0469	0.0117	0.9961

4.2 (Stress) follows two parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

The numerical results can be seen in the Tables 5, 6, 7 and 8:

Table 5. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8), α_{i} = 1.5 .$

$β_{1}$	$β_{2}$	$β_{3}$	$β_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	R (1)
0.3	0.3	0.3	0.3	0.4108	0.2420	0.1426	0.0840	0.8794
0.5	0.5	0.5	0.5	0.3779	0.2351	0.1463	0.0910	0.8502
0.7	0.7	0.7	0.7	0.3499	0.2275	0.1479	0.0961	0.8214
0.9	0.9	0.9	0.9	0.3258	0.2196	0.1481	0.0998	0.7934
1.1	1.1	1.1	1.1	0.3048	0.2119	0.1473	0.1024	0.7664
1.5	1.5	1.5	1.5	0.2699	0.1971	0.1439	0.1050	0.7159

Table 6. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8), α_{i} = 1.5$ .

$β_{1}$	$β_{2}$	$β_{3}$	$β_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.5	0.7	0.9	0.4108	0.2227	0.1283	0.0776	0.8393
0.8	1.0	1.2	1.4	0.3374	0.2087	0.1340	0.0889	0.7690
1.2	1.4	1.6	1.8	0.2952	0.1958	0.1336	0.0933	0.7179
1.8	2.0	2.2	2.4	0.2486	0.1775	0.1291	0.0955	0.6507
2.2	2.4	2.6	2.8	0.2249	0.1664	0.1250	0.0952	0.6115

Table 7. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); α_{i} = 1.5 (i = 1, 2, 3, 4), β_{i} = 0.3 (i = 1, 2, 3, 4)$ .

$θ_{1}$	$θ_{2}$	$θ_{3}$	$θ_{4}$	$θ_{5}$	$θ_{6}$	$θ_{7}$	$θ_{8}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.7262	0.1403	0.0478	0.0216	0.9359
0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	0.6078	0.1684	0.0672	0.0332	0.8766
0.3	0.4	0.5	0.6	0.7	0.8	0.9	1	0.5090	0.1768	0.0792	0.0419	0.8069
0.5	0.6	0.7	0.8	0.9	1.0	1.1	1.2	0.3578	0.1629	0.0854	0.0497	0.6557

Table 8. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); α_{i} = 1.5 (i = 1, 2, 3, 4), β_{i} = 0.3 (i = 1, 2, 3, 4)$ .

$p_{1}$	$p_{3}$	$p_{5}$	$p_{7}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.4108	0.2420	0.1426	0.0840	0.8794
0.3	0.4	0.5	0.6	0.4108	0.2420	0.1426	0.0840	0.8794
0.5	0.6	0.7	0.8	0.4108	0.2420	0.1426	0.0840	0.8794
0.7	0.8	0.9	1	0.4108	0.2420	0.1426	0.0840	0.8794

4.3 (Stress) follows a one-parameter Lindley distribution, and (Stress) follows a combination of two (exp.-dis.)

The numerical results can be seen in the Tables 9, 10, 11 and 12:

Table 9. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8) .$

$γ_{1}$	$γ_{2}$	$γ_{3}$	$γ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.3	0.3	0.3	0.1947	0.1568	0.1263	0.1017	0.5795
0.5	0.5	0.5	0.5	0.3333	0.2222	0.1481	0.0988	0.8025
0.7	0.7	0.7	0.7	0.4404	0.2464	0.1379	0.0772	0.9019
0.9	0.9	0.9	0.9	0.5220	0.2495	0.1193	0.0570	0.9478
1.1	1.1	1.1	1.1	0.5852	0.2427	0.1007	0.0418	0.9704
1.5	1.5	1.5	1.5	0.6750	0.2194	0.0713	0.0232	0.9888

Table 10. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8)$ .

$γ_{1}$	$γ_{2}$	$γ_{3}$	$γ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.5	0.7	0.9	0.1947	0.2684	0.2364	0.1568	0.8564
0.8	1.0	1.2	1.4	0.4839	0.2867	0.1403	0.0585	0.9693
1.2	1.4	1.6	1.8	0.6115	0.2549	0.0925	0.0297	0.9886
1.8	2.0	2.2	2.4	0.7218	0.2077	0.0541	0.0129	0.9965
2.2	2.4	2.6	2.8	0.7677	0.1825	0.0399	0.0081	0.9982

Table 11. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); γ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$θ_{1}$	$θ_{2}$	$θ_{3}$	$θ_{4}$	$θ_{5}$	$θ_{6}$	$θ_{7}$	$θ_{8}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.8555	0.1061	0.0247	0.0078	0.9940
0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	0.7904	0.1435	0.0398	0.0141	0.9878
0.3	0.4	0.5	0.6	0.7	0.8	0.9	1	0.7339	0.1706	0.0542	0.0210	0.9798
0.5	0.6	0.7	0.8	0.9	1.0	1.1	1.2	0.6411	0.2039	0.0790	0.0364	0.9604

Table 12. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); γ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$p_{1}$	$p_{3}$	$p_{5}$	$p_{7}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.6750	0.2194	0.0713	0.0232	0.9888
0.3	0.4	0.5	0.6	0.6750	0.2194	0.0713	0.0232	0.9888
0.5	0.6	0.7	0.8	0.6750	0.2194	0.0713	0.0232	0.9888
0.7	0.8	0.9	1	0.6750	0.2194	0.0713	0.0232	0.9888

4.4 (Stress) follows two parameter generalized (exp.-dis.) and $(Stress)$ follows a combination of the two (exp.-dis.)

The numerical results can be seen in the Tables 13, 14, 15 and 16:

Table 13. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8); α_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$λ_{1}$	$λ_{2}$	$λ_{3}$	$λ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.3	0.3	0.3	0.2693	0.1968	0.1438	0.1051	0.7150
0.5	0.5	0.5	0.5	0.4000	0.2400	0.1440	0.0864	0.8704
0.7	0.7	0.7	0.7	0.4927	0.2499	0.1268	0.0643	0.9338
0.9	0.9	0.9	0.9	0.5612	0.2463	0.1081	0.0474	0.9629
1.1	1.1	1.1	1.1	0.6136	0.2371	0.0916	0.0354	0.9777
1.5	1.5	1.5	1.5	0.6883	0.2145	0.0669	0.0208	0.9906

Table 14. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8); α_{i} = 1.5 (i = 1, 2, 3, 4)$ .

$λ_{1}$	$λ_{2}$	$λ_{3}$	$λ_{4}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.3	0.5	0.7	0.9	0.2693	0.2923	0.2160	0.1248	0.9024
0.8	1.0	1.2	1.4	0.5294	0.2772	0.1229	0.0474	0.9769
1.2	1.4	1.6	1.8	0.6354	0.2452	0.0839	0.0258	0.9903
1.8	2.0	2.2	2.4	0.7279	0.2038	0.0524	0.0124	0.9966
2.2	2.4	2.6	2.8	0.7673	0.1822	0.0402	0.0083	0.9980

Table 15. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); α_{i} = 1.5 (i = 1, 2, 3, 4); λ_{i} = 0.3 (i = 1, 2, 3, 4)$ .

$ϑ_{1}$	$ϑ_{2}$	$ϑ_{3}$	$ϑ_{4}$	$ϑ_{5}$	$ϑ_{6}$	$ϑ_{7}$	$ϑ_{8}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.5471	0.1536	0.0709	0.0405	0.8120
0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	0.4223	0.1619	0.0846	0.0518	0.7206
0.3	0.4	0.5	0.6	0.7	0.8	0.9	1	0.3391	0.1564	0.0895	0.0578	0.6428
0.5	0.6	0.7	0.8	0.9	1.0	1.1	1.2	0.2367	0.1354	0.0874	0.0611	0.5207

Table 16. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); α_{i} = 1.5 (i = 1, 2, 3, 4); λ_{i} = 0.3 (i = 1, 2, 3, 4) .$

$p_{1}$	$p_{3}$	$p_{5}$	$p_{7}$	$R (1)$	$R$ (2)	$R$ (3)	$R$ (4)	$R$ 4
0.1	0.2	0.3	0.4	0.2693	0.1968	0.1438	0.1051	0.7150
0.3	0.4	0.5	0.6	0.2693	0.1968	0.1438	0.1051	0.7150
0.5	0.6	0.7	0.8	0.2693	0.1968	0.1438	0.1051	0.7150
0.7	0.8	0.9	1	0.2693	0.1968	0.1438	0.1051	0.7150

5. Conclusions

5.1 From Tables 1, 2, 3 and 4 for one parameter exponential; for fixed values $P_{i} = 0.2 (i = 1, 3, 5, 7);$ $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ with varying values of $λ_{i} (i = 1, 2, 3, 4)$ , we observed that values of $R$ (1) and $R$ ₄ have increased whereas $R$ (2), $R$ (3) and $R$ (4) are decreased. Also if fixed values $λ_{i} = 1.5 (i = 1, 2, 3, 4)$ with varying values $ϑ_{i} (i = 1, 2 \dots, 8)$ it is observed that values of $R$ (1) and $R$ ₄ have decreased whereas $R$ (2), $R$ (3) and $R$ (4) are increased. But if the fixed values $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ and $λ_{i} = 1.5 (i = 1, 2, 3, 4)$ with varying values $P_{i} (i = 1, 3, 5, 7)$ , it is observed that there is a consistency of values $R$ (1), $R$ (2), $R$ (3), $R$ (4) and $R$ 4.

5.2 From Tables 5, 6, 7 and 8 for two parameter generalized exponential distribution; for fixed values $P_{i} = 0.2 (i = 1, 3, 5, 7); ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ $α_{i} (i = 1, \dots, 4)$ with varying values of $£_{i} (i = 1, 2, 3, 4),$ we observed that values of $R$ (3) and $R$ (4) have increased whereas $R$ (1), $R$ (2) and $R$ ₄ are decreased. In addition, if fixed values $α_{i}, £_{i} (i = 1, \dots, 4)$ with varying values $ϑ_{i} (i = 1, 2 \dots, 8)$ it is observed that the values of $R$ (1) and $R$ ₄ decrease, whereas $R$ (2), $R$ (3), and $R$ (4) increase. But if the fixed values $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ and $α_{i}, £_{i} (i = 1, \dots, 4)$ with varying values $P_{i} (i = 1, 3, 5, 7)$ , it is observed that there is a consistency of values $R$ (1), $R$ (2), $R$ (3), $R$ (4) and $R$ 4.

5.3 From Tables 9, 10, 11 and 12 for one parameter Lindley distribution; for fixed values $P_{i} = 0.2 (i = 1, 3, 5, 7);$ $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ with varying values of $τ_{i} (i = 1, 2, 3, 4)$ , we observed that values of $R$ (1), $R$ (2) and $R$ ₄ have increased where $R$ (3) and $R$ (4) are decreased. Also if fixed values $τ_{i} = 1.5 (i = 1, 2, 3, 4)$ with varying values $ϑ_{i} (i = 1, 2 \dots, 8)$ it is observed that values of $R$ (1) and $R$ ₄ have decreased whereas $R$ (2), $R$ (3) and $R$ (4) are increased. But if the fixed values $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ and $λ_{i} = 1.5 (i = 1, 2, 3, 4)$ with varying values $P_{i} (i = 1, 3, 5, 7)$ , it is observed that there is a consistency of values $R$ (1), $R$ (2), $R$ (3), $R$ (4) and $R$ 4.

5.4 From the Tables 13, 14, 15 and 16 for two parameter exponential; for following fixed values $P_{i} = 0.2 (i = 1, 3, 5, 7); ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ $α_{i} (i = 1, \dots, 4)$ with varying values of $ʎ_{i} (i = 1, 2, 3, 4)$ , we observed that values of $R$ (1), $R$ (2) and R₄ have increased whereas $R$ (3) and $R$ (4) are decreased. Also if fixed values $α_{i}, ʎ_{i} (i = 1, \dots, 4)$ with varying values $ϑ_{i} (i = 1, 2 \dots, 8)$ it is observed that values of $R$ (1) and $R$ ₄ have decreased where as $R$ (2), $R$ (3) and $R$ (4) show increasing. But if the fixed values $ϑ_{i} = 0.5 (i = 1, 2 \dots, 8)$ and $α_{i}, ʎ_{i} (i = 1, \dots, 4)$ with varying values $P_{i} (i = 1, 3, 5, 7)$ , it is observed that there is a consistency of values $R$ (1), $R$ (2), $R$ (3), $R$ (4) and $R$ 4.

Data availability

All data used in the research was generated using simulation and does not belong to any specific entity. No data associated with this article.

References

1. Beg MA: On the estimation of Pr(Y < X) for the Two parameter Exponential Distribution. Metrika. 1980; 27: 29–34. Publisher Full Text
2. Doloi C, Borah M, Gogoi J: Cascade System with Pr(X < Y < Z). Journal of Informatics and Mathematical Sciences. 2013; 5(1): 37–47.
3. Doloi C, Borah M: Cascade System with Mixture of Distributions. International Journal of Statistics and Systems. 2012; 7(1): 11–24.
4. Dutta, Bhowal: cascade reliability model for n -warm standby. Assam stat. Rev. 1999; 12(2): 66–72.
5. Gogoi J, Borah M: Estimation of reliability for multi component systems using exponential, gamma and Lindley stress-strength distributions. Journal of Reliability and Statistical Studies. 2012; 5(1): 33–41.
6. GoPalan MN, Venkateswarlu P: Reliability analysis of time-dependent cascade system with deterministic cycle times. Microelectron. Reliab. 1982; 22(4): 841–872. Publisher Full Text
7. Gogoi J, Borah M: Interference on Reliability for Cascade Model. Journal of Informatics and Mathematical Sciences. 2012; 4(1): 77–83.
8. Kamel I: Single stage shrinkage estimation methods for reliability function of exponential - Shanker stress - Strength models. AIP Conf. Proc. 2023; 2834: 080cs027.
9. Kamel I, Anwar T, Najim A: Estimation of (S-S) reliability for inverted exponential distribution. AIp Conference proceedings. 2023; 2591: 050006.
10. Kamel I, Shahoodh K, Ali K: On Bayesian estimation in parallel and series system stress-strength model. pak. J. Statist. 2025; 41(2): 129–137.
11. Pandit SN, Sriwastav GL: Studies in Cascade Reliability I. IEEE Trans. on Reliability. 1975; 24(1): pp. 53–56.
12. Sumathi T, Maheswari U, Swathi N: Cascade Reliability of Stress- Strength system when Strength follows mixed Exponential distribution.2013; 27–31.

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¹ of Mathematics, University of Anbar, Ramadi, Al Anbar Governorate, 009647812524220, Iraq
² Mathematics, University of Anbar, Ramadi, 009647812524220, Iraq

Isaam Ahmed
Roles: Conceptualization, Data Curation, Formal Analysis, Project Administration, Software, Writing – Original Draft Preparation

Humam A Abdulrazzaq
Roles: Formal Analysis

M. Jasim Mohammed
Roles: Formal Analysis

Competing interests

No competing interests were disclosed.

Grant information

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

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Ahmed I, Abdulrazzaq HA and Mohammed MJ. R ₄- STANDBY- STRENGTH-STRESS MODEL [version 1; peer review: awaiting peer review]. F1000Research 2025, 14:1423 (https://doi.org/10.12688/f1000research.172210.1)

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[1] 1. Beg MA: On the estimation of Pr(Y < X) for the Two parameter Exponential Distribution. Metrika. 1980; 27: 29–34. Publisher Full Text

[2] 2. Doloi C, Borah M, Gogoi J: Cascade System with Pr(X < Y < Z). Journal of Informatics and Mathematical Sciences. 2013; 5(1): 37–47.

[3] 3. Doloi C, Borah M: Cascade System with Mixture of Distributions. International Journal of Statistics and Systems. 2012; 7(1): 11–24.

[4] 4. Dutta, Bhowal: cascade reliability model for n -warm standby. Assam stat. Rev. 1999; 12(2): 66–72.

[5] 5. Gogoi J, Borah M: Estimation of reliability for multi component systems using exponential, gamma and Lindley stress-strength distributions. Journal of Reliability and Statistical Studies. 2012; 5(1): 33–41.

[6] 6. GoPalan MN, Venkateswarlu P: Reliability analysis of time-dependent cascade system with deterministic cycle times. Microelectron. Reliab. 1982; 22(4): 841–872. Publisher Full Text

[7] 7. Gogoi J, Borah M: Interference on Reliability for Cascade Model. Journal of Informatics and Mathematical Sciences. 2012; 4(1): 77–83.

[8] 8. Kamel I: Single stage shrinkage estimation methods for reliability function of exponential - Shanker stress - Strength models. AIP Conf. Proc. 2023; 2834: 080cs027.

[9] 9. Kamel I, Anwar T, Najim A: Estimation of (S-S) reliability for inverted exponential distribution. AIp Conference proceedings. 2023; 2591: 050006.

[10] 10. Kamel I, Shahoodh K, Ali K: On Bayesian estimation in parallel and series system stress-strength model. pak. J. Statist. 2025; 41(2): 129–137.

[11] 11. Pandit SN, Sriwastav GL: Studies in Cascade Reliability I. IEEE Trans. on Reliability. 1975; 24(1): pp. 53–56.

[12] 12. Sumathi T, Maheswari U, Swathi N: Cascade Reliability of Stress- Strength system when Strength follows mixed Exponential distribution.2013; 27–31.

R 4- STANDBY- STRENGTH-STRESS MODEL

Abstract

Background

Methods

Results

Conclusions

Keywords

1. Introduction

2. Discussion of reliability

(1)

(2)

2.1 (Stress) follows one parameter (exp.-dis.) and (Strength) follows a combination of the two (exp.-dis.)

(3)

(4)

2.2 (Stress) follows two parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

(5)

2.3 (Stress) follows one parameter Lindley distribution and (Stress) follows a combination of two (exp.-dis.)

(6)

2.4 (Stress) follows two parameter generalized (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

(7)

3. Estimation of R 4

3.1 (Stress) follows one parameter (exp.-dis.)

(8)

3.2 (Stress) follows two parameter (exp.-dis.)

(9)

3.3 (Stress) follows one parameter Lindley distribution

(10)

3.4 (Stress) follows two parameter generalized (exp.-dis.)

(11)

4. The simulation manner

4.1 (Stress) follows one parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

Table 1. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8).

Table 2. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8) .

Table 3. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);λi=1.5(i=1,2,3,4) .

Table 4. Values of R(1),R(2),R(3),R(4) and R4 , θi=0.5(i=1,2…,6);λi=1.5(i=1,2,3,4) .

4.2 (Stress) follows two parameter (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

Table 5. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8),αi=1.5.

Table 6. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8),αi=1.5 .

Table 7. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);αi=1.5(i=1,2,3,4),βi=0.3(i=1,2,3,4) .

Table 8. Values of R(1),R(2),R(3),R(4) and R4 , θi=0.5(i=1,2…,6);αi=1.5(i=1,2,3,4),βi=0.3(i=1,2,3,4) .

4.3 (Stress) follows a one-parameter Lindley distribution, and (Stress) follows a combination of two (exp.-dis.)

Table 9. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8).

Table 10. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8) .

Table 11. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);γi=1.5(i=1,2,3,4) .

Table 12. Values of R(1),R(2),R(3),R(4) and R4 , θi=0.5(i=1,2…,6);γi=1.5(i=1,2,3,4) .

4.4 (Stress) follows two parameter generalized (exp.-dis.) and (Stress) follows a combination of the two (exp.-dis.)

Table 13. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8);αi=1.5(i=1,2,3,4) .

Table 14. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);θi=0.5(i=1,2…,8);αi=1.5(i=1,2,3,4) .

Table 15. Values of R(1),R(2),R(3),R(4) and R4 , pi=0.2(i=1,3,5,7);αi=1.5(i=1,2,3,4);λi=0.3(i=1,2,3,4) .

Table 16. Values of R(1),R(2),R(3),R(4) and R4, θi=0.5(i=1,2…,6);αi=1.5(i=1,2,3,4);λi=0.3(i=1,2,3,4).

5. Conclusions

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References

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R ₄- STANDBY- STRENGTH-STRESS MODEL

3. Estimation of $R$ ₄

Table 1. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8) .$

Table 2. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8)$ .

Table 3. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); λ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

Table 4. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); λ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

Table 5. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8), α_{i} = 1.5 .$

Table 6. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8), α_{i} = 1.5$ .

Table 7. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); α_{i} = 1.5 (i = 1, 2, 3, 4), β_{i} = 0.3 (i = 1, 2, 3, 4)$ .

Table 8. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); α_{i} = 1.5 (i = 1, 2, 3, 4), β_{i} = 0.3 (i = 1, 2, 3, 4)$ .

Table 9. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8) .$

Table 10. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8)$ .

Table 11. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); γ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

Table 12. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); γ_{i} = 1.5 (i = 1, 2, 3, 4)$ .

4.4 (Stress) follows two parameter generalized (exp.-dis.) and $(Stress)$ follows a combination of the two (exp.-dis.)

Table 13. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8); α_{i} = 1.5 (i = 1, 2, 3, 4)$ .

Table 14. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); θ_{i} = 0.5 (i = 1, 2 \dots, 8); α_{i} = 1.5 (i = 1, 2, 3, 4)$ .

Table 15. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $p_{i} = 0.2 (i = 1, 3, 5, 7); α_{i} = 1.5 (i = 1, 2, 3, 4); λ_{i} = 0.3 (i = 1, 2, 3, 4)$ .

Table 16. Values of $R (1), R (2), R (3), R (4)$ and $R 4$ , $θ_{i} = 0.5 (i = 1, 2 \dots, 6); α_{i} = 1.5 (i = 1, 2, 3, 4); λ_{i} = 0.3 (i = 1, 2, 3, 4) .$