Exact Reliability and Signature Formulas for Linear m-Consecutive-k-out-of-n: F Systems

Abstract

An m-consecutive-k-out-of-n: F ($m/C/k/n: F$) system consists of n linearly ordered components such that the system fails if and only if there are at least m nonoverlapping runs of k consecutive failed components. Our motivation in this work is to obtain efficient formulas for the signature and reliability of the $m/C/k/n: F$ system with independent and identical (i.i.d) components that are easy to implement and have a low computational time. We demonstrate that the reliability formula derived for this system requires less computational time than the $m/C/k/n: F$ system formula currently in use. For the minimal and maximal signatures of the $m/C/k/n: F$ system, we provide precise equations. In addition, the average number of faulty components at the time of an $m/C/k/n: F$ system failure and mean time to failure (MTTF) of an $m/C/k/n: F$ system are analyzed through the system signature.

1. Introduction

A lot of time and energy has been spent studying the reliability characteristics of consecutive-type systems throughout the last three decades. This is explained by the fact that these systems have been used to model and determine the best designs for a variety of systems, including relay stations for spacecrafts, oil pipeline systems, vacuum systems in accelerators, and communication networks. The linear consecutive-$k$-out-of-$n$: $F$ system is the first consecutive-type structure that was documented in reliability literature. It fails if and only if at least $k$ consecutive components fail. It is composed of $n$ linearly ordered components. In order to allow more flexible operation principles, a number of modifications and generalizations have been proposed more recently (see, e.g., Huang et al. [1], Kuo and Zuo [2], Saenz-de-Cabezon and Wynn [3], and Zuo and Tian [4]). A logical extension of the conventional $m$-out-of-$n$: $F$ and the consecutive-$k$-out-of-$n$: $F$ systems, the $m/C/k/n: F$ system consists of $n$ linearly ordered components, and it fails if and only if there are at least $m$ ($m>1$) nonoverlapping runs of $k$ ($k>1$) consecutively failed components ($n \ge mk$). An $m/C/k/n: F$ system was first introduced and studied by Griffith [5]. Then, Papastavridis [6] offered recursive and exact failure probability formulas for this system for the general case of unequal component failure distributions and i.i.d. cases, respectively.

For the i.i.d. scenario, Makri and Philippou [7] have provided a precise formula for the related system’s reliability. Agarwal et al. [8] presented a clear, closed-form formula for assessing this system’s reliability using a graphical evaluation and review technique (GERT). Additionally, they demonstrated that the formula produced by GERT is significantly more efficient than the current formulae of $m/C/k/n: F$ systems for i.i.d. components because of its short computing time. Eryilmaz et al. [9] investigated the reliability properties of $m/C/k/n: F$ systems with replaceable components. They derived precise formulae and recurrence relations for the signature of the system. This system has been the subject of continuous research until recently (a survey of $m/C/k/n: F$ systems and their generalizations may be found in Triantafyllov and Ioannis [10], Özbey [11], and Kan [12]). The $m/C/k/n: F$ system transforms to a consecutive-$k$-out-of-$n$: $F$ system and $m$-out-of-$n$: $F$ system when $m=1$ and $k=1$, respectively. For the consecutive-$k$-out-of-$n$ system, Gökdere [13] proposed a new method for determining the time-dependent component reliability of the system under a stress-strength setup. Also, a repairable consecutive-$k$-out-of-$n$: $F$ system with $n$ linearly and circularly arranged components is studied by Gökdere and Ng [14]. They developed a novel and simple method for determining the time-dependent transition performance probability and time-dependent reliability of the system when the state of the system is known within a certain amount of time. In Özbey [15], an age replacement policy for the linear and circular consecutive-$k$-out-of-$n$: $F$ systems has been discussed.

The concept of the signature of a coherent system was introduced by Samaniego [16]. Numerous related applications have shown that the concept of a signature is emerging as a highly natural and potent tool for the analysis of coherent systems. There is a wealth of literature on this topic that concentrates on both conceptual and computational issues. For a reliability structure with n components, let $T_1, T_2,\dots ,T_n$ represent the component lifetimes and let $T_{1:n}\le T_{2:n}\le \dots \le T_{n:n}$ represent the order statistics of $T_1, T_2,\dots ,T_n$. If $T$ denotes the system’s lifetime, the signature vector of the system is defined as the probability vector $\left(p_1\left(n\right), p_2\left(n\right),\dots ,p_n\left(n\right)\right)$ with entries $p_i\left(n\right)=P\left(T=T_{i:n}\right)(i=1, 2,\dots ,n)$, where $T_{i:n}$ denotes the smallest $i$th among $T_1, T_2,\dots ,T_n$. Samaniego [16] proved that the reliability of any coherent system with i.i.d. components can be computed using the system signature as

$$P\left(T>t\right)=\sum _{i=1}^n{p}_i\left(n\right)P\left(T_{i:n}>t\right)$$

(1)

The signature vector’s $i$th element can be calculated as

$$p_i\left(n\right)={\displaystyle \frac{r_{n-i+1}\left(n\right)}{\left(\begin{array}{c}n\\ n-i+1\end{array}\right)}}-{\displaystyle \frac{r_{n-i}\left(n\right)}{\left(\begin{array}{c}n\\ n-i\end{array}\right)}} , i=1, 2,\dots ,n.$$

(2)

The symbol $r_i\left(n\right)$ $(i=\mathrm{1,2},\dots ,n)$ indicates the number of path sets of the structure that have exactly $i$ working components [17].

In this research, we examine the $m$/C/$k$/$n$: $F$ system, which is composed of i.i.d. components. We deduce exact formulas for both system reliability and system signature. Mathematica is utilized to perform numerical calculations related to failure probability. We also show that formulas obtained in this paper have a low computational time. The minimal and maximal signatures of the $m$/C/$k$/$n$: $F$ system are presented using exact formulas. Furthermore, the mean time to failure (MTTF) of the $m$/C/$k$/$n$: $F$ system and the average number of defective components at the time of a system failure are examined using the system signature.

2. Linear $\mathit{m}/\mathit{C}/\mathit{k}/\mathit{n}: \mathit{F}$ System

Based on the number of failed components in the $m/C/k/n: F$ system, we can divide it into stages as follows:

$$\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)=\left\{\begin{array}{cc}0,& \mathrm{a}\mathrm{l}\mathrm{l} \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{a}\mathrm{r}\mathrm{e} \mathrm{w}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ 1,& \mathrm{i}\mathrm{f} \mathrm{o}\mathrm{n}\mathrm{e} \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{s}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ ⋮& ⋮\\ mk-1,& \mathrm{i}\mathrm{f} mk-1 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ mk,& \mathrm{i}\mathrm{f} mk \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{y} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e} \mathrm{o}\mathrm{r} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\\ ⋮& ⋮\\ d-1& \mathrm{i}\mathrm{f} d-1 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{y} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e} \mathrm{o}\mathrm{r} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\\ d,& \mathrm{i}\mathrm{f} d \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\end{array}\right.$$

where $i=0, 1,\dots ,d$; $d=n+m-\left[n/k\right]$ and the greatest integer smaller than or equal to value $x$ are $\left[x\right]$. Note that if $m=\left[n/k\right]$, then $d=n$.

At each stage we described above, there are cases that cause and do not cause system failures. For system reliability, we need to find the number of different cases that do not cause a system failure in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$. Also, for the signature of the system, we need to find the number of cases that cause a system failure in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$.

In order to obtain the $m/C/k/n: F$ system reliability denoted by $R_m\left(k,n\right)$, we have to first calculate the system reliability in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$ denoted by $R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\right)$. For a system consisting of i.i.d. components with component reliability, $p$, in Stage ($i$), the reliability can be calculated as follows:

$$R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\right)=\left(\begin{array}{c}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t} \mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{s} \mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\\ \mathrm{d}\mathrm{o} \mathrm{n}\mathrm{o}\mathrm{t} \mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{u}\mathrm{r}\mathrm{e} \mathrm{i}\mathrm{n} \mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\end{array}\right)p^{n-i}{\left(1-p\right)}^i$$

(3)

where $i=0, 1,\dots ,d-1$. Using (3), the reliability $R_m\left(k,n\right)$ can be computed as

$$R_m\left(k,n\right)=\sum _{i=0}^{d-1}R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\right)$$

(4)

For the signature of the system denoted by $\mathbf{p}=\left(p_1\left(m,k,n\right), p_2\left(m,k,n\right),\dots ,p_n\left(m,k,n\right)\right)$, we have to first obtain the system signature in the stages where the system may not work. In other words, in order to obtain the signature of the system, it is sufficient to find the system signature obtained between $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(mk\right)$ and $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(d\right)$ denoted by $p_{mk}\left(m,k,n\right)$ and $p_d\left(m,k,n\right)$, respectively. That is,

$$p_1\left(m,k,n\right)=\cdots =p_{mk-1}\left(m,k,n\right)=0 \mathrm{a}\mathrm{n}\mathrm{d} p_{d+1}\left(m,k,n\right)=\cdots =p_n\left(m,k,n\right)=0.$$

(5)

Note that, if $d=n$, then $p_1\left(m,k,n\right)=\cdots =p_{mk-1}\left(m,k,n\right)=0$ and $p_n\left(m,k,n\right)=1$.

The signature of the $m/C/k/n: F$ system consisting of i.i.d. components is presented by

$$p_i\left(m,k,n\right)={\displaystyle \frac{\left(\begin{array}{c}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{s} \mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\\ \mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{u}\mathrm{r}\mathrm{e} \mathrm{i}\mathrm{n} \mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\end{array}\right)\times \left(\begin{array}{c}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{w}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}\\ \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{i}\mathrm{n} \mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\end{array}\right)!}{n!}},$$

(6)

where $i=mk,mk+1,\dots ,d$.

In order to explain exactly what we want to express above, we provide the following example.

Example 1.

Let $m=3$, $k=2$ and $n=10$. Then, $d=10+3-\left[10/2\right]=8$. Counting the number of malfunctioning components in the $3/C/2/10: F$ system, we have

$$\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)=\left\{\begin{array}{cc}0,& \mathrm{a}\mathrm{l}\mathrm{l} \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{a}\mathrm{r}\mathrm{e} \mathrm{w}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ 1,& \mathrm{i}\mathrm{f} \mathrm{o}\mathrm{n}\mathrm{e} \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{s}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ ⋮& ⋮\\ 5& \mathrm{i}\mathrm{f} 5 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\\ 6,& \mathrm{i}\mathrm{f} 6 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{y} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e} \mathrm{o}\mathrm{r} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\\ 7& \mathrm{i}\mathrm{f} 7 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{y} \mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e} \mathrm{o}\mathrm{r} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\\ 8,& \mathrm{i}\mathrm{f} 8 \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s} \mathrm{f}\mathrm{a}\mathrm{i}\mathrm{l}, \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m} \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\end{array}\right.$$

where $i=0, 1,\dots ,8$.

In Stage$\left(0\right)$, there is a possible value: 0000000000, where 0 represents the working state of a component. In this stage, all components work, therefore the system works. From (3), the system’s reliability in Stage$\left(0\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(0\right)\right)=p}^{10}{\left(1-p\right)}^0.$$

In Stage$\left(1\right)$, there are 10 different cases depending on the previous stage, 0000000001, 0000000010, 0000000100, 0000001000, 0000010000, 0000100000, 0001000000, 0010000000, 0100000000, and 1000000000, where 1 represents the failed state of a component. Because, in the previous stage, there is one possible case consisting of 10 working components, and in order to move from this stage to the next stage, any of the working components must fail. In this stage, one component fails, but the system works and there are no cases that cause a system failure. From (3), the system’s reliability in Stage$\left(1\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(1\right)\right)=10p}^9{\left(1-p\right)}^1.$$

In Stage$\left(2\right)$, there are 45 different cases depending on the previous stage, and each of these different cases repeats (number of failed components in this stage = 2)! = 2 times. At this stage, there are 90 cases where two components are faulty, and in each of these cases, the system operates. From (3), the system’s reliability in Stage$\left(2\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(2\right)\right)=45p}^8{\left(1-p\right)}^2.$$

In Stage$\left(3\right)$, there are 120 different cases depending on the previous stage, and each of these different cases repeats (number of failed components in this stage = 3)! = 6 times. At this stage, there are 720 cases where three components are faulty, and in each of these cases, the system operates. From (3), the system’s reliability in the Stage$\left(3\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(3\right)\right)=120p}^7{\left(1-p\right)}^3.$$

In Stage$\left(4\right)$, there are 210 different cases depending on the previous stage, and each of these different cases repeats (number of failed components in this stage = 4)! = 24 times. At this stage, there are 5040 cases where four components are faulty, and in each of these cases, the system operates. From (3), the system’s reliability in Stage$\left(4\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(4\right)\right)=210p}^6{\left(1-p\right)}^4.$$

In Stage$\left(5\right)$, there are 252 different cases depending on the previous stage, and each of these different cases repeats (number of failed components in this stage = 5)! = 120 times. At this stage, there are 30,240 cases where five components are faulty, and in each of these cases, the system operates. From (3), the system’s reliability in Stage$\left(5\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(5\right)\right)=252p}^5{\left(1-p\right)}^5.$$

In Stage$\left(6\right)$, there are 210 different cases depending on the previous stage, and each of these different cases repeats (number of failed components in this stage = 6)! = 720 times. In this stage, there are 151,200 cases in which six components are failed, and in some of these cases, the system works and in some of these cases the system fails. By performing hand calculations (Theorem 1 provides the systematic expression for these calculations), we can say that 175 different cases in this stage do not fail the system and each of these different cases repeats 720 times. Therefore, the number of cases that cause a system failure is 25,200. Cases that fail the system are not transferred to the next stage and are used to calculate the system signature vector to be obtained in this stage. As a result, 126,000 cases remain to be used in the next stage. From (3), the system’s reliability in Stage$\left(6\right)$ can be written as:

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(6\right)\right)=175p}^4{\left(1-p\right)}^6.$$

Using (6) for the signature vector in this stage, we obtain

$$p_6\left(m,k,n\right)={\displaystyle \frac{\left(25200\right)\left(4\right)!}{10!}}=0.1667$$

In Stage$\left(7\right)$, there are 504,000 cases in which seven components are failed depending on the previous stage. In some of these cases the system works, and in some of these cases the system fails. By performing hand calculations, we can say that 40 different cases in this stage do not fail the system and each of these different cases repeats 5040 times. Thus, the number of cases causing system failures becomes 302,400, and 201,600 cases remain to be used in the next stage.

For the system’s reliability in the Stage$\left(7\right)$, we can write

$${R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(7\right)\right)=40p}^3{\left(1-p\right)}^7.$$

Using (6) for the signature vector in this stage, we obtain

$$p_7\left(m,k,n\right)={\displaystyle \frac{\left(302400\right)\left(3\right)!}{10!}}=0.5$$

In Stage$\left(8\right)$, there are 604,800 cases in which eight components are failed depending on the previous stage. By performing hand calculations, we can say that all cases in this stage fail the system. Thus, the number of cases causing system failures becomes 604,800 and there are no cases to be used in the next stage. Using (6) for the signature vector in this stage, we obtain

$$p_8\left(m,k,n\right)={\displaystyle \frac{\left(604800\right)\left(2\right)!}{10!}}=0.3333$$

For this example, from (4), the reliability of the $R_3\left(\mathrm{2,10}\right)$ system can be obtained as

$$\begin{array}{c}{R}_3\left(\mathrm{2,10}\right)={\displaystyle \sum _{i=0}^7}R\left(\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)\right)\hfill \\ {=p}^{10}{q}^0+{10p}^9{q}^1+{45p}^8{q}^2+{120p}^7{q}^3+{210p}^6{q}^4+{252p}^5{q}^5+{175p}^4{q}^6+{40p}^3{q}^7.\hfill \end{array}$$

If $p=0.75$, then $R_3\left(\mathrm{2,10}\right)=0.994821$. This is in agreement with the result from Agarwal et al. ([4], p. 32). Similarly, for this example, the system’s signature is determined from (5) and (6) to be

$$\mathbf{p}=\left(\mathrm{0,0},\mathrm{0,0},\mathrm{0,0.1667,0.5,0.3333,0},0\right).$$

As a result, the signature we obtained for the $3/C/2/10: F$ system was the same as that reported by Eryılmaz et al. ([9], p.347). To obtain precise formulas for the signature and dependability of an $m/C/k/n: F$ system, we must establish a formula for the number of distinct scenarios, including the $i$ failed components designated by $M_i^L$, $0\le i\le d$ that do not result in the system failing. In this case, Theorem 1 is borrowed from [18] to meet the need.

Theorem 1.

The number of different cases that do not cause the failure of the $m/C/k/n: F$ system, including $i$ failed components, $M_i^L$, for $0\le i\le d$

$$M_i^L=\sum _{s=0}^{m-1}\left(\begin{array}{c}n-i+s\\ s\end{array}\right)\sum _{r=0}^{m^{*}}{\left(-1\right)}^r\left(\begin{array}{c}n-i+1\\ r\end{array}\right)\left(\begin{array}{c}n-\left(s+r\right)k\\ n-i\end{array}\right),$$

(7)

where $m^{*}=\mathit{min}\left(n-i+1,\left[i/k\right]-s\right)$.

The following Theorem, which provides an accurate formula for the reliability of the $m/C/k/n: F$ system, is now ready to be presented.

Theorem 2.

The reliability, $R_m\left(k,n\right)$, for the $m/C/k/n: F$ system consisting of i.i.d. components with component reliability, $p$, is presented by

$$R_m\left(k,n\right)=\sum _{i=0}^{d-1}{M}_i^L{p}^{n-i}{\left(1-p\right)}^i,$$

(8)

where $d=n+m-\left[n/k\right]$.

Proof. 

To find the system’s reliability, we divided it into different stages depending on the number of failed components in the system. Then, we found the reliability that can be obtained in each stage using (3), and then we obtained the system’s reliability using (4). It is worth mentioning that, for the exact reliability formula of the system, it is very important to find the number of different cases that do not cause a system failure in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$. We can say that this quantity is equal to $M_i^L$, which is presented in Theorem 1. Consequently, the Theorem’s proof is now complete. □

In

Table 1. Reliability of the $m/C/k/n: F$ system for different values of $n$, $k$, and $m$ when $p=0.75$.

$\mathit{n}$	$\mathit{m}$	$\mathit{k}$	Reliability	Computational Time (secs) of Our Proposed Method	Computational Time (Secs) of Agarval [4].
100	3	3	0.890793	0.016	0.031
200			0.577539	0.047	0.094
300			0.303943	0.094	0.219
400			0.140993	0.171	0.406
500			0.060229	0.297	0.672
100	3	5	0.999957	0.016	0.016
200			0.999603	0.031	0.062
300			0.998649	0.078	0.141
400			0.996870	0.141	0.265
500			0.994101	0.250	0.438
100	5	3	0.994730	0.015	0.031
200			0.914469	0.063	0.125
300			0.718185	0.156	0.328
400			0.483287	0.296	0.609
500			0.287115	0.485	0.985

, we compute the reliability of the $m/C/k/n: F$ system for several values of $n$, $k$, and $m$, where $p=0.75$. For computations, we utilized Mathematica on a Windows computer. The Mathematica codes we used for these computations can be accessed from Reliability Code I (https://drive.google.com/file/d/1KessvmONSexFvmIRaqhkMliTAccfdQU0/view (accessed on 7 August 2024)) and Reliability Code II (https://drive.google.com/file/d/1gtZrSFSrOCPXGgPMm6Cv7PVildRGH-aO/view (accessed on 7 August 2024)) links. From the table, we can observe that the corresponding computational time for our proposed method is less than that of Agarwal et al. [8].

The precise formula for the signature of the $m/C/k/n: F$ system is presented in the Theorem that follows.

Theorem 3.

For $mk\le i\le d$, $d=n+m-\left[n/k\right]$, the signature $p_i\left(n,k,m\right)$ of an $m/C/k/n: F$ system consisting of i.i.d. components is presented by

$$p_i\left(n,k,m\right)=\left[M_{i-1}^L\left(n-i+1\right)\left(i-1\right)!-M_i^L\left(i\right)!\right]{\displaystyle \frac{\left(n-i\right)!}{\left(n\right)!}},$$

(9)

where $p_1\left(m,k,n\right)=\cdots =p_{mk-1}\left(m,k,n\right)=0 and p_{d+1}\left(m,k,n\right)=\cdots =p_n\left(m,k,n\right)=0.$

Proof. 

By the definition of the system we have

$$p_1\left(m,k,n\right)=\cdots =p_{mk-1}\left(m,k,n\right)=0 \mathrm{a}\mathrm{n}\mathrm{d} p_{d+1}\left(m,k,n\right)=\cdots =p_n\left(m,k,n\right)=0 \mathrm{i}\mathrm{f} d<n ,$$

or

$$p_1\left(m,k,n\right)=\cdots =p_{mk-1}\left(m,k,n\right)=0 \mathrm{i}\mathrm{f} d=n.$$

In order to obtain the system’s signature, it is sufficient to know the number of cases that cause the system’s failure in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$, $mk\le i\le d$. We can say that, in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$, there are $M_i^L$ different cases that do not cause the system’s failure, and each of these different cases repeats (number of the failed components)! times, that is, $i!$ times. Furthermore, we can say that the number of all cases in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$ is proportional to the number of working components, $\left(n-i+1\right)$ in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i-1\right)$. As a result, the number of cases that cause the system’s failure in $\mathrm{S}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\left(i\right)$ can be obtained as

$$M_{i-1}^L\left(n-i+1\right)\left(i-1\right)!-M_i^L\left(i\right)!.$$

Thus, the proof is completed using (6). □

One can use the system’s signature to find the mean number of failed components that fail at the time of the system’s failure (see e.g., Eryilmaz [19]). Let $Z_m\left(k,n\right)$ denote the number of failed components at the moment of an $m/C/k/n: F$ system failure. The mean number of malfunctioning components at the time of the system’s failure can be calculated from

$$E\left(Z_m\left(k,n\right)\right)=\sum _{i=mk}^{d}i p_i\left(n,k,m\right),$$

(10)

using the predicted value of the signature of an $m/C/k/n: F$ system. Using Equation (9) in Equation (10), we calculated the mean number of failed components at the system failure moment for a range of values of $n$, $k$, and $m$ in

Table 2. Mean number of failed components at the moment of the $m/C/k/n: F$ system’s failure for different values of $n$, $k$, and $m$.

$\mathit{n}$	$\mathit{m}$	$\mathit{k}$	$\mathit{E}\left({\mathit{Z}}_{\mathit{m}}\left(\mathit{k},\mathit{n}\right)\right)$	Computational Time (Secs)
100	3	3	34.3288	0.015
300			67.9265	0.172
500			93.9803	0.625
100	3	5	56.9951	0.016
300			130.4946	0.141
500			193.1706	0.531
100	5	3	42.5738	0.015
300			83.2840	0.282
500			114.8181	1.031

. Also, in Table 2, we present the corresponding times for the desired numerical results for the proposed system. For computations, we utilized Mathematica on a Windows computer. The Mathematica code we used for these computations can be accessed from the Mean Number of Failed Components Code link (https://drive.google.com/file/d/1dQ7dWVE-6o4VOQ-llGedXyrrvuKO8zJP/view (accessed on 7 August 2024)).

3. Additional Results for $\mathit{m}/\mathit{C}/\mathit{k}/\mathit{n}:\mathit{F}$ Systems

A generalized mixture of series or parallel systems can be used to represent any coherent system [20]. If $T_1,T_2,\dots ,T_n$ denote the i.i.d. distributed lifetimes of the components of a coherent system and $T$ denotes the lifetime of a coherent system, then the system’s reliability can be expressed as either

$$P\left(T>t\right)=\sum _{i=1}^n{\alpha }_{i}P\left(T_{1:i}>t\right),$$

(11)

or

$$P\left(T>t\right)=\sum _{i=1}^n{\beta }_{i}P\left(T_{i:i}>t\right),$$

(12)

where ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ and ${\beta }_1,{\beta }_2,\dots ,{\beta }_n$ satisfy the conditions ${\sum }_{i=1}^n{\alpha }_i=1$, ${\sum }_{i=1}^n{\beta }_i=1$. Also, where

$$P\left(T_{1:i}>t\right)=P\left(T_1>t,T_2>t,\dots ,T_i>t\right),$$

and

$$P\left(T_{i:i}>t\right)=1-P\left(T_1\le t,T_2\le t,\dots ,T_i\le t\right),$$

for $1\le i\le n$. The vectors $\left({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\right)$ and $\left({\beta }_1,{\beta }_2,\dots ,{\beta }_n\right)$ are, respectively, called the minimal and maximal signatures of the system. Eryilmaz [21] proved that the minimal and maximal signatures of any coherent system with i.i.d. components can be presented as

$${\alpha }_i=\sum _{j=n-i}^{n_{max}}{\left(-1\right)}^{j-\left(n-i\right)}\left(\begin{array}{c}j\\ n-i\end{array}\right)r_{n-j}\left(n\right),$$

(13)

$${\beta }_i=-\sum _{j=0}^{\mathrm{m}\mathrm{i}\mathrm{n}\left(i,n_{max}\right)}{\left(-1\right)}^{i-j}\left(\begin{array}{c}n-j\\ n-i\end{array}\right)r_{n-j}\left(n\right),$$

(14)

where $n_{max}$ is the number of failed components in the system such that it can still work. For the minimal and maximal signatures of the $m/C/k/n: F$ system, the following Lemma provides precise equations.

Lemma 1.

The minimal and maximal signatures of an $m/C/k/n: F$ system consisting of i.i.d. components are presented by

$${\alpha }_i\left(n,k,m\right)=\sum _{j=d-i}^{d-1}{\left(-1\right)}^{j+i-n}\left(\begin{array}{c}j\\ n-i\end{array}\right)M_j^L,$$

(15)

where $1\le i\le n$ and ${\alpha }_1\left(n,k,m\right)={\alpha }_2\left(n,k,m\right)=\cdots ={\alpha }_{n-d}\left(n,k,m\right)=0$ if $d<n$,

$${\beta }_i\left(n,k,m\right)=\sum _{j=0}^i{\left(-1\right)}^{i-j+1}\left(\begin{array}{c}n-j\\ n-i\end{array}\right)M_j^L,$$

(16)

where $mk\le i\le n$ and ${\beta }_i\left(n,k,m\right)=0$ if $1\le i<mk$.

Proof. 

The proof shows right away that $r_{n-j}\left(n\right)=M_j^L$ in (12). □

Reliability functions and other reliability properties of coherent systems can be easily calculated using Equations (11) and (12) for a given joint cumulative distribution function or joint survival function.

Assume for the moment that the multivariate Pareto distribution with a survival function belongs to the random vector $\left(T_1,T_2,\dots ,T_n\right)$.

$${\overline{F}}_a\left(t_1,t_2,\dots ,t_n\right)={\left(\sum _{j=1}^n{t}_j-n+1\right)}^{-a},$$

where $a>0$ and $t_j>1$ for $j=1, 2,\dots ,n$ (see, e.g., [22], p. 600).

For the multivariate Pareto model, Eryilmaz [9] showed that the MTTF of an $m/C/k/n: F$ system can be computed from

$$\mathrm{M}\mathrm{T}\mathrm{T}\mathrm{F}=\sum _{i=1}^n{\alpha }_i\left(n,k,m\right){\displaystyle \frac{1}{i\left(a-1\right)}},$$

(17)

where $a>1$

In

Table 3. MTTF of the $m/C/k/n: F$ system for different values of $n$, $k$, and $m$.

n	m	k	MTTF	Computational Time (Secs)
10	3	3	12.6448	$\approx 0$
	3	2	5.8254	$\approx 0$
	2	3	7.0000	$\approx 0$
20	3	3	5.5391	$\approx 0$
	3	2	3.0259	0.015
	2	3	4.0372	0.016
30	3	3	4.1180	0.015
	3	2	2.2189	0.016
	2	3	3.1472	0.016
40	3	3	3.4361	0.047
	3	2	1.8118	0.031
	2	3	2.6831	0.031
50	3	3	3.0197	0.063
	3	2	1.5596	0.078
	2	3	2.3880	0.062

, we compute the MTTF of the $m/C/k/n: F$ system using Equation (15) in Equation (17) for several values of $n$, $m$, and $k$ when $a=1.2$. Also, in Table 3, we present the corresponding times for desired numerical results for the proposed system. For computations, we utilized Mathematica on a Windows computer. The Mathematica code we used for these computations can be accessed from the MTTF code link (https://drive.google.com/file/d/1XlFyT_IcgK3QyexfqnozhVl3iVurE5tZ/view (accessed on 7 August 2024)).

4. Conclusions

Without seriously affecting the system’s ability to function, the $m/C/k/n: F$ system provides a workable model for situations in which the failure of $k$ successive components is seen as proof of the system’s decline. The deterioration level rises by 1 for each instance of a strand of $k$ successively failing components, and the system breaks down when the deterioration level reaches $m$. It is evident that such a system is more dependable than the conventional $m$-out-of-$n$: $F$ and consecutive-$k$-out-of-$n$: $F$ systems for $m>1$ and $k>1$, respectively. In this paper, we obtain exact formulas for computing the reliability and the signature of $m/C/k/n: F$ systems with i.i.d. components. We show that the reliability formula deduced for this system has a lower computational time compared with the existing formula of the $m/C/k/n: F$ system. In addition, the system signature is used to analyze the mean number of malfunctioning components at the time of the system’s failure as well as the mean time to system failure. The reliability, mean number of failed components, and MTTF for the m/C/k/n: F system are computed on a PC with the Windows XP operating system using Wolfram Mathematica 7 for different combinations of n, k, and m. Some results are presented in Table 1, Table 2 and Table 3. The computing time is presented in seconds. From our perspective in this article, it is very significant and practical to study the $m/C/k/n: F$ system and some new systems. Our findings are useful for directing the dependability of microwave-signal transmitting systems and the stability of network nodes.