F1000Research F1000Research 2046-1402 F1000 Research Limited London, UK 10.12688/f1000research.172210.1 Research Article Articles R 4- STANDBY- STRENGTH-STRESS MODEL

[version 1; peer review: 1 approved with reservations, 3 not approved]

 Ahmed Isaam Conceptualization Data Curation Formal Analysis Project Administration Software Writing – Original Draft Preparation https://orcid.org/0000-0003-2164-0579 a 1 Abdulrazzaq Humam A Formal Analysis 2 Mohammed M. Jasim Formal Analysis 2 of Mathematics, University of Anbar, Ramadi, Al Anbar Governorate, 009647812524220, Iraq Mathematics, University of Anbar, Ramadi, 009647812524220, Iraq isam_kml@uoanbar.edu.iq

No competing interests were disclosed.

 22 12 2025 2025 14 1423 2 12 2025 Copyright: © 2025 Ahmed I et al. 2025 This is an open access article distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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.

 standby system strength-stress Reliability system maximum likelihood function generalized exponential Lindley distribution. The author(s) declared that no grants were involved in supporting this work. 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 by 6 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 R 4 = ∑ i = 1 4 R ( i )

Where R (i) = P (Y 1 < Ψ 1, Y 2 < 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 R ( i ) = [ ∏ j = 1 i − 1 ∫ Ψ T j ( Ψ ) g i ( Ψ ) dΨ ] [ T ¯ j ( Ψ ) g i ( Ψ ) ]

Where T i ( Ψ ) is the cumulative distribution function and 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.; f 1 ( y ) = P 1 e − ϑ 1 y + ( 1 − P 1 ) e − ϑ 2 y

Therefore; 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 ≥ 0

So, R ( 1 ) = P ( Y 1 ≥ Ψ ) = ∫ 0 ∞ T ¯ 1 ( Ψ ) g 1 ( Ψ ) dΨ = ∫ 0 ∞ [ 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 ∫ 0 ∞ e − ( θ 1 + μ 1 ) Ψ d Ψ + ( 1 − P 1 ) μ 1 ∫ 0 ∞ e − ( θ 2 + μ 1 ) Ψ d Ψ = P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1

Now, R ( 2 ) = P ( Y 1 < Ψ 1 , Y 2 ≥ Ψ 2 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 2 ( Ψ ) g 2 ( Ψ ) dΨ ] = [ 1 − ( P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1 ) ] ∫ 0 ∞ [ P 3 e − ϑ 3 Ψ + ( 1 − P 3 ) e − ϑ 4 Ψ ] μ 2 e − μ 2 Ψ d Ψ = [ 1 − ( P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1 ) ] [ P 3 μ 2 ϑ 3 + μ 2 + ( 1 − P 3 ) μ 2 ϑ 4 + μ 2 ] R ( 3 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 ≥ Ψ 3 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 3 ( Ψ ) g 3 ( Ψ ) dΨ ] = [ 1 − ( P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1 ) ] [ 1 − ( P 3 μ 2 ϑ 3 + μ 2 + ( 1 − P 3 ) μ 2 ϑ 4 + μ 2 ) ] [ ∫ 0 ∞ [ P 5 e − ϑ 5 Ψ + ( 1 − P 5 ) e − ϑ 6 Ψ ] μ 3 e − μ 3 Ψ d Ψ ] [ 1 − ( P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1 ) ] [ 1 − ( P 3 μ 2 ϑ 3 + μ 2 + ( 1 − P 3 ) μ 2 ϑ 4 + μ 2 ) ] [ P 5 μ 3 ϑ 5 + μ 3 + ( 1 − P 5 ) μ 3 ϑ 6 + μ 3 ]

In the same way; R ( 4 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 < Ψ 3 , Y 4 ≥ Ψ 4 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 3 ( Ψ ) g 3 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 4 ( Ψ ) g 4 ( Ψ ) dΨ ] = [ 1 − ( P 1 μ 1 ϑ 1 + μ 1 + ( 1 − P 1 ) μ 1 ϑ 2 + μ 1 ) ] [ 1 − ( P 3 μ 2 ϑ 3 + μ 2 + ( 1 − P 3 ) μ 2 ϑ 4 + μ 2 ) ] [ 1 − ( P 5 μ 3 ϑ 5 + μ 3 + ( 1 − P 5 ) μ 3 ϑ 6 + μ 3 ) ] [ P 7 μ 4 ϑ 7 + μ 4 + ( 1 − P 7 ) μ 4 ϑ 8 + μ 4 ]

By using Equation (1), we get; 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; 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 ) = 1 £ i e − ( Ψ − £ i ) / £ i , Ψ > α > 0 , Ψ ≥ 0

Now, R ( 1 ) = P ( Y 1 ≥ Ψ 1 ) = ∫ 0 ∞ T ¯ 1 ( Ψ ) g 1 ( Ψ ) dΨ = ∫ 0 ∞ [ P 1 e − ϑ 1 Ψ + ( 1 − P 1 ) e − ϑ 2 Ψ ] 1 £ 1 e − ( Ψ − α 1 ) / £ 1 d Ψ = P 1 £ 1 e α 1 £ 1 ∫ α ∞ e − ( ϑ 1 + 1 £ 1 ) Ψ dΨ + ( 1 − P 1 ) £ 1 e α 1 £ 1 ∫ α ∞ e − ( ϑ 2 + 1 £ 1 ) Ψ dΨ = e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] R ( 2 ) = P ( Y 1 < Ψ 1 , Y 2 ≥ Ψ 2 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 2 ( Ψ ) g 2 ( Ψ ) dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] [ P 3 £ 2 e α 2 £ 2 ∫ α ∞ e − ( ϑ 3 + 1 £ 2 ) Ψ dΨ + ( 1 − P 3 ) £ 2 e α 2 £ 2 ∫ α ∞ e − ( ϑ 4 + 1 £ 2 ) Ψ dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] e α 2 £ 2 £ 2 [ P 3 e − ( ϑ 3 + 1 £ 2 ) α 2 ( ϑ 3 + 1 £ 2 ) + ( 1 − P 3 ) e − ( ϑ 4 + 1 £ 2 ) α 2 ( ϑ 4 + 1 £ 2 ) ]

So, R ( 3 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 ≥ Ψ 3 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 3 ( Ψ ) g 3 ( Ψ ) dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] [ 1 − ( e α 2 £ 2 £ 2 [ P 3 e − ( ϑ 3 + 1 £ 2 ) α 2 ( ϑ 3 + 1 £ 2 ) + ( 1 − P 3 ) e − ( ϑ 4 + 1 £ 2 ) α 2 ( ϑ 4 + 1 £ 2 ) ] ) ] [ P 5 £ 3 e α 3 £ 3 ∫ α 3 ∞ e − ( ϑ 5 + 1 £ 3 ) Ψ dΨ + ( 1 − P 5 ) £ 3 e α 3 £ 3 ∫ α 3 ∞ e − ( ϑ 6 + 1 £ 3 ) Ψ dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] [ 1 − ( e α 2 £ 2 £ 2 [ P 3 e − ( ϑ 3 + 1 £ 2 ) α 2 ( ϑ 3 + 1 £ 2 ) + ( 1 − P 3 ) e − ( ϑ 4 + 1 £ 2 ) α 2 ( ϑ 4 + 1 £ 2 ) ] ) ] ( e α 3 £ 3 £ 3 [ P 5 e − ( ϑ 5 + 1 £ 3 ) α 2 ( ϑ 3 + 1 £ 3 ) + ( 1 − P 5 ) e − ( ϑ 6 + 1 £ 3 ) α 3 ( ϑ 4 + 1 £ 3 ) ] ) ]

In the same way; R ( 4 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 < Ψ 3 , Y 4 ≥ Ψ 4 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 3 ( Ψ ) g 3 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 4 ( Ψ ) g 4 ( Ψ ) dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] [ 1 − ( e α 2 £ 2 £ 2 [ P 3 e − ( ϑ 3 + 1 £ 2 ) α 2 ( ϑ 3 + 1 £ 2 ) + ( 1 − P 3 ) e − ( ϑ 4 + 1 £ 2 ) α 2 ( ϑ 4 + 1 £ 2 ) ] ) ] [ P 5 £ 3 e α 3 £ 3 ∫ α 3 ∞ e − ( ϑ 5 + 1 £ 3 ) Ψ dΨ + ( 1 − P 5 ) £ 3 e α 3 £ 3 ∫ α 3 ∞ e − ( ϑ 6 + 1 £ 3 ) Ψ dΨ ] = [ 1 − ( e α 1 £ 1 £ 1 [ P 1 e − ( ϑ 1 + 1 £ 1 ) α 1 ( ϑ 1 + 1 £ 1 ) + ( 1 − P 1 ) e − ( ϑ 2 + 1 £ 1 ) α 1 ( ϑ 2 + 1 £ 1 ) ] ) ] [ 1 − ( e α 2 £ 2 £ 2 [ P 3 e − ( ϑ 3 + 1 £ 2 ) α 2 ( ϑ 3 + 1 £ 2 ) + ( 1 − P 3 ) e − ( ϑ 4 + 1 £ 2 ) α 2 ( ϑ 4 + 1 £ 2 ) ] ) ] [ 1 − ( e α 3 £ 3 £ 3 [ P 5 e − ( ϑ 5 + 1 £ 3 ) α 2 ( ϑ 3 + 1 £ 3 ) + ( 1 − P 5 ) e − ( ϑ 6 + 1 £ 3 ) α 3 ( ϑ 4 + 1 £ 3 ) ] ) ] [ ( e α 4 £ 4 £ 4 [ P 7 e − ( ϑ 7 + 1 £ 4 ) α 4 ( ϑ 7 + 1 £ 4 ) + ( 1 − P 7 ) e − ( ϑ 8 + 1 £ 4 ) α 4 ( ϑ 8 + 1 £ 4 ) ]

By using Equation (1), we get; 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 ) = ω i 2 ( 1 + ω i ) ( 1 + Ψ ) e − ω i Ψ ω i > 0 , Ψ ≥ 0 , i = 1 , … , n R ( 1 ) = P ( Y 1 ≥ Ψ 1 ) = ∫ 0 ∞ T ¯ 1 ( Ψ ) g 1 ( Ψ ) dΨ = ∫ 0 ∞ [ P 1 e − ϑ 1 Ψ + ( 1 − P 1 ) e − ϑ 2 Ψ ] ω 1 2 ( 1 + ω 1 ) ( 1 + Ψ ) e − ω 1 Ψ d Ψ = P 1 ω 1 2 ( 1 + ω 1 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 1 + ω 1 ) Ψ dΨ + ( 1 − P 1 ) ω 1 2 ( 1 + ω 1 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 2 + ω 1 ) Ψ dΨ = P 1 ω 1 2 ( 1 + ω 1 ) [ e − ( ϑ 1 + ω 1 ) Ψ dΨ + ∫ 0 ∞ Ψ e − ( ϑ 1 + ω 1 ) Ψ ] + ( 1 − P 1 ) ω 1 2 ( 1 + ω 1 ) [ e − ( ϑ 2 + ω 1 ) Ψ dΨ + ∫ 0 ∞ Ψ e − ( ϑ 2 + ω 1 ) Ψ ] = ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] R ( 2 ) = P ( Y 1 < Ψ 1 , Y 2 ≥ Ψ 2 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 2 ( Ψ ) g 2 ( Ψ ) dΨ ] = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] P 3 ω 2 2 ( 1 + ω 2 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 3 + ω 2 ) Ψ dΨ + ( 1 − P 3 ) ω 1 2 ( 1 + ω 2 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 4 + ω 2 ) Ψ d Ψ = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] )

So, R ( 3 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 ≥ Ψ 3 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 3 ( Ψ ) g 3 ( Ψ ) dΨ ] = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] [ 1 − ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] ) ] P 5 ω 3 2 ( 1 + ω 3 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 5 + ω 3 ) Ψ dΨ + ( 1 − P 5 ) ω 3 2 ( 1 + ω 3 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 6 + ω 3 ) Ψ d Ψ = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] [ 1 − ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] ) ] ( ω 3 2 ( 1 + ω 3 ) [ P 5 ϑ 5 + ω 3 + 1 ( ϑ 5 + ω 3 ) 2 + ( 1 − P 5 ) ϑ 6 + ω 3 + 1 ( ϑ 6 + ω 3 ) 2 ] )

In the same way; R ( 4 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 < Ψ 3 , Y 4 ≥ Ψ 4 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 3 ( Ψ ) g 3 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 4 ( Ψ ) g 4 ( Ψ ) dΨ ] = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] [ 1 − ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] ) ] [ 1 − ( ω 3 2 ( 1 + ω 3 ) [ P 5 ϑ 5 + ω 3 + 1 ( ϑ 5 + ω 3 ) 2 + ( 1 − P 5 ) ϑ 6 + ω 3 + 1 ( ϑ 6 + ω 3 ) 2 ] ) ] P 7 ω 4 2 ( 1 + ω 4 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 7 + ω 4 ) Ψ dΨ + ( 1 − P 7 ) ω 4 2 ( 1 + ω 4 ) ∫ 0 ∞ ( 1 + Ψ ) e − ( ϑ 8 + ω 4 ) Ψ d z = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] [ 1 − ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] ) ] [ 1 − ( ω 3 2 ( 1 + ω 3 ) [ P 5 ϑ 5 + ω 3 + 1 ( ϑ 5 + ω 3 ) 2 + ( 1 − P 5 ) ϑ 6 + ω 3 + 1 ( ϑ 6 + ω 3 ) 2 ] ) ] ( ω 4 2 ( 1 + ω 4 ) [ P 7 ϑ 7 + ω 4 + 1 ( ϑ 7 + ω 4 ) 2 + ( 1 − P 7 ) ϑ 8 + ω 4 + 1 ( ϑ 8 + ω 4 ) 2 ] )

So, R ( 4 ) = [ 1 − ( ω 1 2 ( 1 + ω 1 ) [ P 1 ϑ 1 + ω 1 + 1 ( ϑ 1 + ω 1 ) 2 + ( 1 − P 1 ) ϑ 2 + ω 1 + 1 ( ϑ 2 + ω 1 ) 2 ] ) ] [ 1 − ( ω 2 2 ( 1 + ω 2 ) [ P 3 ϑ 3 + ω 2 + 1 ( ϑ 3 + ω 2 ) 2 + ( 1 − P 3 ) ϑ 4 + ω 2 + 1 ( ϑ 4 + ω 2 ) 2 ] ) ] [ 1 − ( ω 3 2 ( 1 + ω 3 ) [ P 5 ϑ 5 + ω 3 + 1 ( ϑ 5 + ω 3 ) 2 + ( 1 − P 5 ) ϑ 6 + ω 3 + 1 ( ϑ 6 + ω 3 ) 2 ] ) ] ( ω 4 2 ( 1 + ω 4 ) [ P 7 ϑ 7 + ω 4 + 1 ( ϑ 7 + ω 4 ) 2 + ( 1 − P 7 ) ϑ 8 + ω 4 + 1 ( ϑ 8 + ω 4 ) 2 ] )

By using Equation (1), we have; 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 ≥ 0 , i = 1 , … , n R ( 1 ) = P ( Y 1 ≥ Ψ 1 ) = ∫ 0 ∞ T ¯ 1 ( Ψ ) g 1 ( Ψ ) dΨ = ∫ 0 ∞ [ 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 ∫ 0 ∞ 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 ) ∫ 0 ∞ e − θ 2 Ψ e − τ 1 Ψ ( 1 − e − τ 1 Ψ ) α 1 − 1 dΨ = α 1 P 1 ∫ 0 1 Ψ α 1 − 1 ( 1 − Ψ ) ϑ 1 τ 1 d Ψ + α 1 ( 1 − P 1 ) ∫ 0 1 Ψ α 1 − 1 ( 1 − Ψ ) ϑ 2 τ 1 d Ψ = α 1 P 1 Beta ( α 1 , 1 + ϑ 1 τ 1 ) + α 1 Beta ( α 1 , 1 + ϑ 2 τ 1 ) = α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] R ( 2 ) = P ( Y 1 < Ψ 1 , Y 2 ≥ Ψ 2 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 2 ( Ψ ) g 2 ( Ψ ) dΨ ] = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] α 2 P 3 ∫ 0 1 Ψ α 2 − 1 ( 1 − Ψ ) ϑ 3 τ 1 d Ψ + α 2 ( 1 − P 3 ) ∫ 0 1 Ψ α 2 − 1 ( 1 − Ψ ) ϑ 4 τ 2 d Ψ = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] α 2 Г ( α 2 ) [ P 3 Г ( 1 + ϑ 3 τ 2 ) Г ( α 2 + 1 + ϑ 3 τ 2 ) + ( 1 − P 3 ) Г ( 1 + ϑ 4 τ 2 ) Г ( α 2 + 1 + ϑ 4 τ 2 ) ]

So, R ( 3 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 ≥ Ψ 3 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 3 ( Ψ ) g 3 ( Ψ ) dΨ ] = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] [ 1 − ( α 2 Г ( α 2 ) [ P 3 Г ( 1 + ϑ 3 τ 2 ) Г ( α 2 + 1 + ϑ 3 τ 2 ) + ( 1 − P 3 ) Г ( 1 + ϑ 4 τ 2 ) Г ( α 2 + 1 + ϑ 4 τ 2 ) ] ) ] α 2 P 3 ∫ 0 1 Ψ α 2 − 1 ( 1 − Ψ ) ϑ 3 τ 1 d Ψ + α 2 ( 1 − P 3 ) ∫ 0 1 Ψ α 2 − 1 ( 1 − Ψ ) ϑ 4 τ 2 d Ψ = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] [ 1 − ( α 2 Г ( α 2 ) [ P 3 Г ( 1 + ϑ 3 τ 2 ) Г ( α 2 + 1 + ϑ 3 τ 2 ) + ( 1 − P 3 ) Г ( 1 + ϑ 4 τ 2 ) Г ( α 2 + 1 + ϑ 4 τ 2 ) ] ) ] α 3 Г ( α 3 ) [ P 5 Г ( 1 + ϑ 5 τ 2 ) Г ( α 3 + 1 + ϑ 5 τ 3 ) + ( 1 − P 5 ) Г ( 1 + ϑ 6 τ 3 ) Г ( α 3 + 1 + ϑ 6 τ 3 ) ]

In the same way: R ( 4 ) = P ( Y 1 < Ψ 1 , Y 2 < Ψ 2 , Y 3 < Ψ 3 , Y 4 ≥ Ψ 4 ) = [ ∫ 0 ∞ T 1 ( Ψ ) g 1 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 2 ( Ψ ) g 2 ( Ψ ) dΨ ] [ ∫ 0 ∞ T 3 ( Ψ ) g 3 ( Ψ ) dΨ ] [ ∫ 0 ∞ T ¯ 4 ( Ψ ) g 4 ( Ψ ) dΨ ] = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] = [ 1 − ( α 1 Г ( α 1 ) [ P 1 Г ( 1 + ϑ 1 τ 1 ) Г ( α 1 + 1 + ϑ 1 τ 1 ) + ( 1 − P 1 ) Г ( 1 + ϑ 2 τ 1 ) Г ( α 1 + 1 + ϑ 2 τ 1 ) ] ) ] [ 1 − ( α 2 Г ( α 2 ) [ P 3 Г ( 1 + ϑ 3 τ 2 ) Г ( α 2 + 1 + ϑ 3 τ 2 ) + ( 1 − P 3 ) Г ( 1 + ϑ 4 τ 2 ) Г ( α 2 + 1 + ϑ 4 τ 2 ) ] ) ] [ 1 − ( α 3 Г ( α 3 ) [ P 5 Г ( 1 + ϑ 5 τ 2 ) Г ( α 3 + 1 + ϑ 5 τ 3 ) + ( 1 − P 5 ) Г ( 1 + ϑ 6 τ 3 ) Г ( α 3 + 1 + ϑ 6 τ 3 ) ] ) ] α 4 Г ( α 4 ) [ P 7 Г ( 1 + ϑ 7 τ 4 ) Г ( α 4 + 1 + ϑ 7 τ 4 ) + ( 1 − P 7 ) Г ( 1 + ϑ 8 τ 4 ) Г ( α 4 + 1 + ϑ 8 τ 4 ) ]

Therefore, we have; R 4 = R ( 1 ) + R ( 2 ) + R ( 3 ) + R ( 4 )

3. Estimation of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> </mml:math> </inline-formula> <sub>4</sub>

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 f 1(y) and f 2(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 by 8 and 9 L ( ϑ 1 , ϑ 2 , P , ʎ i / Ψ ij , y ij ) = ∏ i = 1 n [ ∑ j = 1 2 P j f ( y j , ϑ j ) ] ∏ i = 1 n ∏ 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 [ − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i ] μ i k i e − μ i ∑ j = 1 k i Ψ ij Ln ( L ) = Ln ( n ! n 1 ! n 2 ! ) + ( n 1 . Ln ( P ) ) + ( n 2 . Ln ( 1 − P ) ) + ( n 1 . Ln ϑ 1 ) + ( n 2 Ln ϑ 2 ) − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i + k i ln μ i − μ i ∑ j = 1 k i Ψ ij

We use; ( ∂ ln L ∂ P ) = 0 , ( ∂ ln L ∂ ϑ 1 ) = 0 , ( ∂ ln L ∂ ϑ 2 ) = 0 and ∂ ln L ∂ μ i = 0 .

Therefore, we get; P ` = n 1 / ( n 1 + n 2 ) , ϑ ` 1 = n 1 / ∑ j = 1 n 1 y 1 j , ϑ ` 2 = n 2 / ∑ j = 1 n 2 y 2 j and μ i ̂ = k i ∑ j = 1 k i Ψ ij

So, we have; R ̂ 4 = R ̂ ( 1 ) + R ̂ ( 2 ) + R ̂ ( 3 ) + 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; L ( ϑ 1 , ϑ 2 , P , £ i , α i / Ψ ij , y ij ) = ∏ i = 1 n [ ∑ j = 1 2 P j f ( y j , ϑ j ) ] ∏ i = 1 n ∏ j = 1 k i [ 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 [ − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i ] ( 1 £ i ) k i e − 1 £ i ∑ j = 1 k i ( Ψ ij − α i ) Ln ( L ) = Ln ( n ! n 1 ! n 2 ! ) + ( n 1 . Ln ( P ) ) + ( n 2 . Ln ( 1 − P ) ) + ( n 1 . Ln ϑ 1 ) + ( n 2 Ln ϑ 2 ) − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i − k i ln £ i − 1 £ i ∑ j = 1 k i ln ( Ψ ij − α i )

We use; ( ∂ ln L ∂ P ) = 0 , ( ∂ ln L ∂ ϑ 1 ) = 0 , ( ∂ ln L ∂ ϑ 2 ) = 0 , ∂ ln L ∂ α i = 0 and ∂ ln L ∂ £ i = 0 . P ` = n 1 / ( n 1 + n 2 ) , ϑ ` 1 = n 1 / ∑ j = 1 n 1 y 1 j , ϑ ` 2 = n 2 / ∑ j = 1 n 2 y 2 j and £ i ̂ = ∑ j = 1 k i ln ( Ψ ij − α i ) k i

Where; α i = min ( Ψ ij ) , So, we have; R ̂ 4 = R ̂ ( 1 ) + R ̂ ( 2 ) + R ̂ ( 3 ) + 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 by 10 ; L ( ϑ 1 , ϑ 2 , P , ω i / Ψ ij , y ij ) = ∏ i = 1 n [ ∑ j = 1 2 P j f ( y j , ϑ j ) ] ∏ i = 1 n ∏ j = 1 k i 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 [ − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i ] ω i 2 k i ( 1 + ω i ) k i [ ∏ j = 1 k i ( 1 + Ψ ij ) ] e − ω i ∑ j = 1 k i ( Ψ ij ) Ln ( L ) = Ln ( n ! n 1 ! n 2 ! ) + ( n 1 . Ln ( P ) ) + ( n 2 . Ln ( 1 − P ) ) + ( n 1 . Ln ϑ 1 ) + ( n 2 Ln ϑ 2 ) − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i + 2 k i ln ω i − ω i ∑ j = 1 k i Ψ ij − k i ln ( 1 + ω i ) + ∑ j = 1 k i ln ( 1 + Ψ ij )

We use; ( ∂ ln L ∂ P ) = 0 , ( ∂ ln L ∂ ϑ 1 ) = 0 , ( ∂ ln L ∂ ϑ 2 ) = 0 and ∂ ln L ∂ ω i = 0 .

Therefore, we get; P ` = ( n 1 n 1 + n 2 ) , ϑ ` 1 = n 1 / ∑ j = 1 n 1 y 1 j , ϑ ` 2 = n 2 ∑ j = 1 n 2 y 2 j and ω i ̂ = ( 1 − Ψ ¯ i ) + Ψ ¯ i + 6 Ψ ¯ i + 1 2 Ψ ¯ i

Where; Ψ ¯ i = ∑ j = 1 k i ( Ψ ij ) k i . So, we have; R ̂ 4 = R ̂ ( 1 ) + R ̂ ( 2 ) + R ̂ ( 3 ) + 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; L ( ϑ 1 , ϑ 2 , P , τ i , α i / Ψ ij , y ij ) = ∏ i = 1 n [ ∑ j = 1 2 P j f ( y j , ϑ j ) ] ∏ i = 1 n ∏ j = 1 k i [ α i τ i e − τ i Ψ ij ( 1 − e − τ i Ψ ij ) α i − 1 ] = ( n ! n 1 ! ∗ n 2 ! ) . P n 1 ( 1 − P ) n 2 . ϑ 1 n 1 . ϑ 2 n 2 e [ − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i ] α i k i τ i k i e − τ i ∑ j = 1 k i ( Ψ ij ) ( ∏ j = 1 k i ( 1 − e − τ i Ψ ij ) α i − 1 ) Ln ( L ) = Ln ( n ! n 1 ! n 2 ! ) + ( n 1 . Ln ( P ) ) + ( n 2 . Ln ( 1 − P ) ) + ( n 1 . Ln ϑ 1 ) + ( n 2 Ln ϑ 2 ) − ∑ i = 1 n 1 ϑ 1 y 1 i − ∑ i = 1 n 2 ϑ 2 y 2 i + k i Ln α i + k i Ln τ i − τ i ∑ j = 1 k i ( Ψ ij ) + ( α i − 1 ) ∑ j = 1 k i Ln ( 1 − e − τ i Ψ ij )

We use; ( ∂ ln L ∂ P ) = 0 , ( ∂ ln L ∂ ϑ 1 ) = 0 , ( ∂ ln L ∂ ϑ 2 ) = 0 , ∂ ln L ∂ α i = 0 and ∂ ln L ∂ τ i = 0 . P ` = n 1 / ( n 1 + n 2 ) , ϑ ` 1 = n 1 / ∑ j = 1 n 1 y 1 j , ϑ ` 2 = n 2 / ∑ j = 1 n 2 y 2 j , α i ̂ = k i ∑ j = 1 k i ln ( 1 − e − τ i Ψ ij )

And τ i ̂ = [ ∑ j = 1 k i ( ( Ψ ij e − τ i Ψ ij ) / ( 1 − e − τ i Ψ ij ) ) ∑ j = 1 k i ln ( 1 − e − τ i Ψ ij ) + 1 k i ∑ j = 1 k i Ψ ij ( 1 − e − τ i Ψ ij ) ] − 1

So, we have; R ̂ 4 = R ̂ ( 1 ) + R ̂ ( 2 ) + R ̂ ( 3 ) + 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:

Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>.</mml:mo> </mml:math> </inline-formula>
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>λ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>6</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>λ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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:

Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mo>.</mml:mo> </mml:math> </inline-formula>
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:msub> <mml:mi>β</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.3</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>6</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:msub> <mml:mi>β</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.3</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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:

Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>.</mml:mo> </mml:math> </inline-formula>
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>γ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>6</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>γ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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 <inline-formula> <mml:math display="inline"> <mml:mspace width="0.25em"/> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mtext>Stress</mml:mtext> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mspace width="0.25em"/> </mml:math> </inline-formula>follows a combination of the two (exp.-dis.)

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

Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>8</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math> </inline-formula>, <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>p</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.2</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>5</mml:mn> <mml:mo>,</mml:mo> <mml:mn>7</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>λ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.3</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula>.
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
Values of <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>1</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo mathvariant="fraktur">,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>2</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo mathvariant="fraktur">,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>3</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo mathvariant="fraktur">,</mml:mo> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> </mml:math> </inline-formula> and <inline-formula> <mml:math display="inline"> <mml:mi mathvariant="fraktur">R</mml:mi> <mml:mn>4</mml:mn> </mml:math>,</inline-formula> <inline-formula> <mml:math display="inline"> <mml:msub> <mml:mi>θ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>…</mml:mo> <mml:mo>,</mml:mo> <mml:mn>6</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>α</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>1.5</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>;</mml:mo> <mml:msub> <mml:mi>λ</mml:mi> <mml:mi>i</mml:mi> </mml:msub> <mml:mo>=</mml:mo> <mml:mn>0.3</mml:mn> <mml:mrow> <mml:mo stretchy="true">(</mml:mo> <mml:mi>i</mml:mi> <mml:mo>=</mml:mo> <mml:mn>1</mml:mn> <mml:mo>,</mml:mo> <mml:mn>2</mml:mn> <mml:mo>,</mml:mo> <mml:mn>3</mml:mn> <mml:mo>,</mml:mo> <mml:mn>4</mml:mn> <mml:mo stretchy="true">)</mml:mo> </mml:mrow> <mml:mo>.</mml:mo> </mml:math> </inline-formula>
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 … , 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 … , 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 … , 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 … , 8 ) α i ( i = 1 , … , 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 , … , 4 ) with varying values ϑ i ( i = 1 , 2 … , 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 … , 8 ) and α i , £ i ( i = 1 , … , 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 … , 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 … , 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 … , 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 … , 8 ) α i ( i = 1 , … , 4 ) with varying values of ʎ i ( i = 1 , 2 , 3 , 4 ) , we observed that values of R (1), R (2) and R 4 have increased whereas R (3) and R (4) are decreased. Also if fixed values α i , ʎ i ( i = 1 , … , 4 ) with varying values ϑ i ( i = 1 , 2 … , 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 … , 8 ) and α i , ʎ i ( i = 1 , … , 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 Beg MA : On the estimation of Pr(Y < X) for the Two parameter Exponential Distribution. Metrika. 1980;27:29–34. 10.1007/BF01893574 Doloi C Borah M Gogoi J : Cascade System with Pr(X < Y < Z). Journal of Informatics and Mathematical Sciences. 2013;5(1):37–47. Doloi C Borah M : Cascade System with Mixture of Distributions. International Journal of Statistics and Systems. 2012;7(1):11–24. Dutta Bhowal : cascade reliability model for n -warm standby. Assam stat. Rev. 1999;12(2):66–72. 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. GoPalan MN Venkateswarlu P : Reliability analysis of time-dependent cascade system with deterministic cycle times. Microelectron. Reliab. 1982;22(4):841–872. 10.1016/S0026-2714(82)80198-4 Gogoi J Borah M : Interference on Reliability for Cascade Model. Journal of Informatics and Mathematical Sciences. 2012;4(1):77–83. Kamel I : Single stage shrinkage estimation methods for reliability function of exponential - Shanker stress - Strength models. AIP Conf. Proc. 2023;2834:080cs027. Kamel I Anwar T Najim A : Estimation of (S-S) reliability for inverted exponential distribution. AIp Conference proceedings. 2023;2591:050006. 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. Pandit SN Sriwastav GL : Studies in Cascade Reliability I. IEEE Trans. on Reliability. 1975;24(1): pp.53–56. Sumathi T Maheswari U Swathi N : Cascade Reliability of Stress- Strength system when Strength follows mixed Exponential distribution. 2013;27–31. 10.5256/f1000research.189925.r456041 Reviewer response for version 1 Bansal Shikha 1 Referee https://orcid.org/0000-0002-9586-2177 SRM Institute of Science and Technology, Ghaziabad, Uttar Pradesh, India

Competing interests: No competing interests were disclosed.

 7 5 2026 Copyright: © 2026 Bansal S 2026 This is an open access peer review report distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. recommendation reject

After a thorough review of the manuscript titled  “R₄ Standby–Strength–Stress Model”, the paper has been evaluated carefully. B ased on the current version, the decision is Not approved, as the manuscript requires major revisions before it can be considered suitable for indexing. The detailed comments are provided below:

Introduction Section:

The introduction is not structured appropriately. While the basic concept of the standby system is briefly explained, the section lacks sufficient background discussion. The manuscript does not clearly present the research motivation, research gap, novelty, or specific contributions of the study. These aspects must be clearly articulated.

Literature Review:

The literature review is insufficient and limited in scope. The manuscript does not adequately cite recent and relevant research articles related to standby systems and strength–stress models. A more comprehensive and updated review of the existing literature is required.

Inconsistent Notation:

Different notations for reliability have been used in the abstract and throughout the manuscript. The notation must be consistent throughout the paper.

Definition of Notations:

Several symbols and notations are used without proper definition prior to their first appearance. All mathematical symbols and parameters must be clearly defined before use.

Justification of Distribution Assumption:

The manuscript uses the exponential distribution; however, no justification for this assumption is provided. A clear explanation or theoretical/practical justification for selecting the exponential distribution is required.

Conclusion Section:

The conclusion is not presented appropriately. It is written as a general paragraph and does not effectively summarize the research's key findings, contributions, and implications. The conclusion should clearly highlight the main results and insights derived from the study.

Future Research Directions:

The manuscript does not include a section on future research directions. This should be added to indicate possible extensions or improvements of the proposed model.

Language and Grammar:

The manuscript contains multiple grammatical and language errors.

 Final Decision:

Based on the above comments, the manuscript in its present form is not suitable for indexing.

Is the work clearly and accurately presented and does it cite the current literature?

No

If applicable, is the statistical analysis and its interpretation appropriate?

No

Are all the source data underlying the results available to ensure full reproducibility?

No source data required

Is the study design appropriate and is the work technically sound?

No

Are the conclusions drawn adequately supported by the results?

No

Are sufficient details of methods and analysis provided to allow replication by others?

No

Reviewer Expertise:

Mathematical Modelling , ReliabilityTheory, Optimization

I confirm that I have read this submission and believe that I have an appropriate level of expertise to state that I do not consider it to be of an acceptable scientific standard, for reasons outlined above.

 10.5256/f1000research.189925.r444718 Reviewer response for version 1 Rathi Puran 1 Referee M.D. University, Rohtak, Haryana, India

Competing interests: No competing interests were disclosed.

 25 4 2026 Copyright: © 2026 Rathi P 2026 This is an open access peer review report distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. recommendation reject

1. Mathematical Notation and Exposition Mathematical notation in many formulas are presented without adequate explanation. Key variables and parameters Ψ, Y, θ, μ etc. are introduced without clear definitions. Some expressions also appear to be duplicated. Greater care is needed to ensure clarity, consistency, and completeness of mathematical presentation.

2. Justification of Distributional Choices and Novelty The manuscript does not adequately justify the choice of the mixture of exponential, Lindley, and generalized exponential distributions. In addition, the novelty of the proposed model relative to existing work is not clearly established. The literature review should be expanded to better position the contribution and clearly state how this work advances the current state of the art.

3. Simulation Study and Interpretation The simulation design is insufficiently described, and important details regarding parameter settings and experimental design are missing. Although several tables are presented, their interpretation is weak and key findings are not clearly highlighted. The MSE-based comparisons are mentioned but not analyzed in sufficient depth. Moreover, the manuscript lacks a discussion of practical or applied implications of the results.

4. Organization and Structure The paper is lengthy and contains repetitive material. Several extended algebraic derivations would be more appropriate for an appendix. The Results and Conclusions sections are relatively weak and should be substantially revised to provide clearer synthesis, interpretation, and contribution.

Is the work clearly and accurately presented and does it cite the current literature?

Partly

If applicable, is the statistical analysis and its interpretation appropriate?

No

Are all the source data underlying the results available to ensure full reproducibility?

Partly

Is the study design appropriate and is the work technically sound?

Partly

Are the conclusions drawn adequately supported by the results?

Partly

Are sufficient details of methods and analysis provided to allow replication by others?

Partly

Reviewer Expertise:

Reliability Theory and Modeling, Optimization Technique, Time Series and SQC

I confirm that I have read this submission and believe that I have an appropriate level of expertise to state that I do not consider it to be of an acceptable scientific standard, for reasons outlined above.

 10.5256/f1000research.189925.r456037 Reviewer response for version 1 Sharma Sunita 1 Referee https://orcid.org/0009-0002-3940-3758 Manipal Institute of Technology, Manipal, Karnataka, India

Competing interests: No competing interests were disclosed.

 25 4 2026 Copyright: © 2026 Sharma S 2026 This is an open access peer review report distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. recommendation reject

The manuscript studies a multi-component standby strength–stress reliability model and derives reliability expressions under different assumed lifetime distributions (exponential, Lindley and generalized exponential).

However, the paper in its current form has several major limitations. The work lacks a clear discussion of the research gap, motivation and significance of the proposed study. The literature review is weak, with insufficient recent references and limited connection to current research in strength–stress and standby system reliability.

Additionally, the manuscript does not provide practical implications or real-life applications to justify the usefulness of the derived formulas. The selection of distributions is not properly justified. The overall structure and presentation of the paper are not well organized and there are several inconsistencies in writing, notation and grammar, which affect readability.

Recommendation: Not acceptable in the present form. The authors should address the above issues and significantly revise the manuscript before it can be considered for indexing.

Is the work clearly and accurately presented and does it cite the current literature?

No

If applicable, is the statistical analysis and its interpretation appropriate?

Partly

Are all the source data underlying the results available to ensure full reproducibility?

No source data required

Is the study design appropriate and is the work technically sound?

No

Are the conclusions drawn adequately supported by the results?

Partly

Are sufficient details of methods and analysis provided to allow replication by others?

Partly

Reviewer Expertise:

Bayesian Estimation, Reliability Theory

I confirm that I have read this submission and believe that I have an appropriate level of expertise to state that I do not consider it to be of an acceptable scientific standard, for reasons outlined above.

 10.5256/f1000research.189925.r444720 Reviewer response for version 1 Susam Selim Orhun 1 Referee Munzur University, Tunceli, Turkey

Competing interests: No competing interests were disclosed.

 25 4 2026 Copyright: © 2026 Susam SO 2026 This is an open access peer review report distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. recommendation approve-with-reservations

The overall probabilistic structure of the model and the decomposition of system reliability are correct and consistent with standard standby system formulations. Nevertheless, the manuscript contains several technical inaccuracies at the level of intermediate derivations.

In particular, some probability density functions for exponential and mixture exponential distributions are written with incorrect signs in the exponential terms and, in a few cases, without proper normalization. In addition, the distribution function T_i(\cdot) is defined as a cumulative distribution function, but its use in the integral expressions is not always consistent with this definition, leading to confusion between T_i and 1-T_i. Although the final closed-form reliability expressions appear to be based on the correct underlying distributions, the intermediate equations should be carefully revised to ensure mathematical consistency and transparency.

Is the work clearly and accurately presented and does it cite the current literature?

Yes

If applicable, is the statistical analysis and its interpretation appropriate?

Partly

Are all the source data underlying the results available to ensure full reproducibility?

Yes

Is the study design appropriate and is the work technically sound?

Partly

Are the conclusions drawn adequately supported by the results?

Yes

Are sufficient details of methods and analysis provided to allow replication by others?

Partly

Reviewer Expertise:

Multivariate distribution functions and copulas

I confirm that I have read this submission and believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard, however I have significant reservations, as outlined above.