Keywords
standby system, strength-stress, Reliability system, maximum likelihood function, generalized exponential, Lindley distribution.
This article is included in the Fallujah Multidisciplinary Science and Innovation gateway.
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.
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.
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.
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 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 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
Consider that Yi (i = 1,2,..,n) is the strength, which is an independent random variable (r.v.) arranged in the order of activation. In addition, i (i = 1,2,…,n) is the stress that is independent r.v. so, 4 is given by2,3
Where (i) = P (Y1 < 1, Y2 < 2,…, Yi-1 < i-1, Yi i) is reliability.
Consider fi(y), gi( ) as the p.d.f. (i = 1,2,…,n); therefore, we have
Where is the cumulative distribution function and . In this study, we examine four different distributions.
Considering that the strength follows a mixture of tow (exp-dis.) with p.d.f.;
Therefore;
Also the stress follows one parameter (exp-dis.) with p.d.f;
By using Equation (1), we get;
Let the strength follow a combination of the tow (exp.-dis.) with p.d.f.;1
Therefore;
Also the stress follows two parameter (exp-dis.) with p.d.f.;
By using Equation (1), we get;
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
By using Equation (1), we have;
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.;
In this section, we will estimate 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 and .
Let ij be r.s. from one parameter (exp-dis.) with parameter .
Where, i = 1,…,n, j = 1,…, . The L.f. is given by8 and9
We use; ( and .
Therefore, we get;
Let ij be r.s. from two parameter (exp-dis.) with parameter Where, i = 1,…,n, j = 1,…, . The L.f. is given by;
Let ij be r.s. from one parameter (exp-dis.) with parameter . Where, i = 1,…,n, j = 1,…, . The L.f. is given by10;
We use; ( and .
Therefore, we get;
In this section, the numerical results are presented to compare the performance of the different reliability values obtained for four different distributions.
The numerical results can be seen in the Tables 1, 2, 3 and 4:
The numerical results can be seen in the Tables 5, 6, 7 and 8:
The numerical results can be seen in the Tables 9, 10, 11 and 12:
The numerical results can be seen in the Tables 13, 14, 15 and 16:
5.1 From Tables 1, 2, 3 and 4 for one parameter exponential; for fixed values with varying values of , we observed that values of (1) and 4 have increased whereas (2), (3) and (4) are decreased. Also if fixed values with varying values it is observed that values of (1) and 4 have decreased whereas (2), (3) and (4) are increased. But if the fixed values and with varying values , it is observed that there is a consistency of values (1), (2), (3), (4) and 4.
5.2 From Tables 5, 6, 7 and 8 for two parameter generalized exponential distribution; for fixed values with varying values of we observed that values of (3) and (4) have increased whereas (1), (2) and 4 are decreased. In addition, if fixed values with varying values it is observed that the values of (1) and 4 decrease, whereas (2), (3), and (4) increase. But if the fixed values and with varying values , it is observed that there is a consistency of values (1), (2), (3), (4) and 4.
5.3 From Tables 9, 10, 11 and 12 for one parameter Lindley distribution; for fixed values with varying values of , we observed that values of (1), (2) and 4 have increased where (3) and (4) are decreased. Also if fixed values with varying values it is observed that values of (1) and 4 have decreased whereas (2), (3) and (4) are increased. But if the fixed values and with varying values , it is observed that there is a consistency of values (1), (2), (3), (4) and 4.
5.4 From the Tables 13, 14, 15 and 16 for two parameter exponential; for following fixed values with varying values of , we observed that values of (1), (2) and R4 have increased whereas (3) and (4) are decreased. Also if fixed values with varying values it is observed that values of (1) and 4 have decreased where as (2), (3) and (4) show increasing. But if the fixed values and with varying values , it is observed that there is a consistency of values (1), (2), (3), (4) and 4.
All data used in the research was generated using simulation and does not belong to any specific entity. No data associated with this article.
| Views | Downloads | |
|---|---|---|
| F1000Research | - | - |
|
PubMed Central
Data from PMC are received and updated monthly.
|
- | - |
Is the work clearly and accurately presented and does it cite the current literature?
No
Is the study design appropriate and is the work technically sound?
No
Are sufficient details of methods and analysis provided to allow replication by others?
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
Are the conclusions drawn adequately supported by the results?
No
Competing Interests: No competing interests were disclosed.
Reviewer Expertise: Mathematical Modelling , ReliabilityTheory, Optimization
Is the work clearly and accurately presented and does it cite the current literature?
Partly
Is the study design appropriate and is the work technically sound?
Partly
Are sufficient details of methods and analysis provided to allow replication by others?
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
Are the conclusions drawn adequately supported by the results?
Partly
Competing Interests: No competing interests were disclosed.
Reviewer Expertise: Reliability Theory and Modeling, Optimization Technique, Time Series and SQC
Is the work clearly and accurately presented and does it cite the current literature?
No
Is the study design appropriate and is the work technically sound?
No
Are sufficient details of methods and analysis provided to allow replication by others?
Partly
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
Are the conclusions drawn adequately supported by the results?
Partly
Competing Interests: No competing interests were disclosed.
Reviewer Expertise: Bayesian Estimation, Reliability Theory
Is the work clearly and accurately presented and does it cite the current literature?
Yes
Is the study design appropriate and is the work technically sound?
Partly
Are sufficient details of methods and analysis provided to allow replication by others?
Partly
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
Are the conclusions drawn adequately supported by the results?
Yes
Competing Interests: No competing interests were disclosed.
Reviewer Expertise: Multivariate distribution functions and copulas
Alongside their report, reviewers assign a status to the article:
| Invited Reviewers | ||||
|---|---|---|---|---|
| 1 | 2 | 3 | 4 | |
|
Version 1 22 Dec 25 |
read | read | read | read |
Provide sufficient details of any financial or non-financial competing interests to enable users to assess whether your comments might lead a reasonable person to question your impartiality. Consider the following examples, but note that this is not an exhaustive list:
Sign up for content alerts and receive a weekly or monthly email with all newly published articles
Already registered? Sign in
The email address should be the one you originally registered with F1000.
You registered with F1000 via Google, so we cannot reset your password.
To sign in, please click here.
If you still need help with your Google account password, please click here.
You registered with F1000 via Facebook, so we cannot reset your password.
To sign in, please click here.
If you still need help with your Facebook account password, please click here.
If your email address is registered with us, we will email you instructions to reset your password.
If you think you should have received this email but it has not arrived, please check your spam filters and/or contact for further assistance.
Comments on this article Comments (0)