Asymptotic Mean Time To Failure and Higher Moments for Large, Recursive Networks

As explained in many textbooks [1, 2, 3]), if $R(t)$ is the system’s reliability, the MTTF is defined by

The higher moments $\mu_{m}$ are

For instance, the standard deviation $\sigma$ is given by $\sigma^{2}=\langle\,t^{2}\,\rangle-\langle\,t\,\rangle^{2}$. The moment generating function $f(z)$ is a formal series defined by

by construction, $\langle\,t^{m}\,\rangle=f^{(m)}(0)$ (the $m$th derivative of $f(z)$ taken at $z=0$). Since $f(0)=1$, the cumulant generating function $g(z)$ is subsequently defined by

where $\kappa_{m}$ is the cumulant of order $m$; $\kappa_{2}=\mu_{2}-\mu_{1}^{2}$ is the variance, $\kappa_{3}=\mu_{3}-3\,\mu_{1}\,\mu_{2}+2\,\mu_{1}^{3}$, and so on. Since $g(z)=\ln\langle\,e^{z\,t}\,\rangle$, $\kappa_{m}=g^{(m)}(0)$. Using the same integration by parts as in eq. (1) leads to

where $\widetilde{R}$ is the Laplace transform of the reliability.

Calculations are often performed by replacing $R(t)$ with an exponential function, $R_{\rm exp}(t)=\exp(-\lambda\,t)$. We have then

and $\displaystyle\langle\,e^{z\,t}\,\rangle=\left(1-\frac{z}{\lambda}\right)^{-1}$. Consequently,

from which

When $p(t)=\exp(-\lambda\,t)$ is the common reliability of the system’s elements, the total reliability is

where ${\mathcal{R}}(p)$ is the reliability polynomial; using the change of variable $\displaystyle t=-\frac{\ln p}{\lambda}$, we get

and

We make full use of eqs. (10) and (11) in the next sections. The moment generating function also reads

• •

  For $n$ elements in series, ${\mathcal{R}}_{n}(p)=p^{n}$, so that $R_{n}(t)=\exp(-n\,\lambda\,t)$, and $\displaystyle{\rm MTTF}^{(\rm series)}_{n}=\frac{1}{n\,\lambda}$ [1, 3]. Equations( 6) and ( 8) become

  	$\displaystyle\mu^{\rm(series)}_{m}$	$\displaystyle=$	$\displaystyle\frac{m!}{n^{m}\,\lambda^{m}}$		(13)
  	$\displaystyle\kappa^{\rm(series)}_{m}$	$\displaystyle=$	$\displaystyle\frac{(m-1)!}{n^{m}\,\lambda^{m}}\,.$		(14)

• •

  If the $n$ elements are in parallel, then $R_{n}(t)=1-(1-\exp(-\lambda\,t))^{n}$, and [1, 3]

  $${\rm MTTF}^{(\rm parallel)}_{n}=\frac{1}{\lambda}\,\sum_{i=1}^{n}\,\frac{1}{i}=\frac{1}{\lambda}\,(\psi(n+1)-\psi(1))\rightarrow\frac{1}{\lambda}\,\left(\ln n+\mathbf{C}+\frac{1}{2\,n}+\cdots\right)$$

  (15)

  for $n$ large, where $\displaystyle\psi(z)=\frac{d\ln\Gamma(z)}{dz}$ is the digamma function and $\mathbf{C}\approx 0.577216$ is the Euler gamma constant (see eq. 8.362.1 of [9]). In this case, the MTTF diverges as $n$ goes to infinity.

• •

  For $k$-out-of-$n$:G systems, the reliability polynomial is [1, 3]

  $${\mathcal{R}}_{k,n}(p)=\sum_{i=k}^{n}\,\frac{n!}{(n-i)!\,i!}\,p^{i}\,(1-p)^{n-i}\,.$$

  (16)

  All the moments could be obtained from eq. (11). However, the formula for the cumulants being eventually extremely simple, we focus on these parameters. Starting from eq. (12), we find

  	$\displaystyle\langle\,e^{z\,t}\,\rangle$	$\displaystyle=$	$\displaystyle 1+\frac{z}{\lambda}\,\int_{0}^{1}\,\frac{dp}{p^{1+z/\lambda}}\,\sum_{i=k}^{n}\,\frac{n!}{(n-i)!\,i!}\,p^{i}\,(1-p)^{n-i}$		(17)
  		$\displaystyle=$	$\displaystyle 1+\frac{z}{\lambda}\,\sum_{i=k}^{n}\,\frac{n!}{(n-i)!\,i!}\,\frac{\Gamma\left(i-\frac{z}{\lambda}\right)\,\Gamma(n+1-i)}{\Gamma\left(n+1-\frac{z}{\lambda}\right)}$
  		$\displaystyle=$	$\displaystyle 1+\frac{z}{\lambda}\,\frac{n!}{\Gamma\left(n+1-\frac{z}{\lambda}\right)}\,\sum_{i=k}^{n}\,\frac{\Gamma\left(i-\frac{z}{\lambda}\right)}{i!}$
  		$\displaystyle=$	$\displaystyle\frac{n!}{\Gamma\left(n+1-\frac{z}{\lambda}\right)}\,\frac{\Gamma\left(k-\frac{z}{\lambda}\right)}{(k-1)!}\,;$

  the last equality is proven by induction. Consequently,

  $$g(z)=\ln\frac{\Gamma\left(k-\frac{z}{\lambda}\right)}{\Gamma(k)}\,\frac{\Gamma(n+1)}{\Gamma\left(n+1-\frac{z}{\lambda}\right)}\,,$$

  (18)

  whence (see eq. 8.363.8 of [9])

  	$\displaystyle\kappa_{m}$	$\displaystyle=$	$\displaystyle g^{(m)}(0)=\left(\frac{-1}{\lambda}\right)^{m}\,\left(\psi^{(m-1)}(k)-\psi^{(m-1)}(n+1)\right)$		(19)
  		$\displaystyle=$	$\displaystyle\frac{(m-1)!}{\lambda^{m}}\,\sum_{i=k}^{n}\,\frac{1}{i^{m}}\,;$

  Equation (19) generalizes the expressions obtained for the series ($k=n$) and parallel ($k=1$) cases, and a similar expression for the variance [10]. When $m>1$, $\kappa_{m}$ is bounded by the finite value $\displaystyle\frac{(m-1)!}{\lambda^{m}}\,\zeta(m)$, where $\zeta(m)$ is the Zeta function.

Let us now investigate more complex systems.

This architecture, displayed in Fig. 1, is constituted by the repetition of perfect graphs $K_{4}$ (the fully-connected graph with four nodes). This configuration is exactly solvable even when edges and nodes have distinct reliabilities [6]. For perfect nodes, and with all edge reliabilities equal to $p$, the two-terminal reliability has the simple form

The two eigenvalues are displayed in Fig. 2 as functions of $p$. As $n$ increases, the contribution of $\zeta_{+}$ prevails over that of $\zeta_{-}$. Therefore, when $n$ is large,

because the contribution of $\zeta_{-}^{n}(p)$ vanishes exponentially. The approximation given by eq. (24) lies at the heart of our method for deriving the MTTF’s asymptotic expansion. Inserting it into eq. (10) gives

Here, the lower bound (zero) does not play a significant role because $\zeta_{\pm}(0)=0$. As $n$ increases, $\zeta_{+}^{n}(p)$ is negligible except close to $p=1$, as illustrated in Fig. 3.

The gist of the asymptotic expansion is therefore to consider $\alpha_{+}$ and $\zeta_{+}$ in the vicinity of unity. Setting $q=1-p$, we have from eqs. (21)–(22)

Note that $\zeta_{+}(1)=\alpha_{+}(1)=1$ because ${\mathcal{R}}_{n}(p=1)=1$. We can write

At this point, we have to rescale the variable $q$ in order to extract the asymptotic behavior of the integral. Equation (27) gives

this suggests setting $\tau=n\,q^{4}$, or equivalently $q=\tau^{1/4}\,n^{-1/4}$, so that

In the last exponential, the argument is $-2\,\tau^{5/4}\,n^{-1/4}+4\,\tau^{7/4}\,n^{-3/4}+\cdots$. Because of the $\exp(-\tau)$ factor, we can neglect the contribution of large $\tau$’s, so that when $n$ is large, we only need to expand the second exponential and all other factors in the limit $\tau\to 0$:

The error made by replacing the upper bound of the integral by $+\infty$ vanishes exponentially as $n$ goes to infinity. Keeping only the prevailing terms in each factor of eq. (31) leads to

For the leading-order term (and this term only), $\alpha_{+}$ does not play any role since it may safely be replaced with 1. The asymptotic $n$-dependence is not $1/n$ or $\ln n$ anymore as in the series and parallel cases, but a power-law, $n^{-1/4}$, which slowly decreases with $n$. The following terms of the expansion may be derived easily by expanding all the factors in eq. (31):

The exact MTTF is obtained straightforwardly by using eq. (10) and the value ${\mathcal{R}}_{n}(p)$ deduced from the three-term recursion relation at the origin of eq. (20):

with ${\mathcal{R}}_{0}(p)=1$ and ${\mathcal{R}}_{1}(p)=p\,(1+2\,p-7\,p^{3}+7\,p^{4}-2\,p^{5})$. The exact values of the MTTF are then obtained by a simple integration of ${\mathcal{R}}_{n}(p)$. The exact and the asymptotic (limited to the first three terms of the expansion) results for the MTTF are plotted in Fig. 4. Even for moderate values of $n$, the agreement between the two is good.

Following this method, we also find the asymptotic expansion of $\langle\,t^{2}\,\rangle$ by adding the factor $\displaystyle-2\,\ln\left(1-\frac{\tau^{1/4}}{n^{1/4}}\right)$ and another $1/\lambda$ in eq. (31). As regards the leading term of the expansion, a mere factor $2\,\frac{\tau^{1/4}}{n^{1/4}}\,\lambda^{-1}$ is added in the integral. Finally

Further terms can be routinely obtained using mathematical software such as Mathematica.

After simplification, the variance of the distribution is therefore deduced to behave as

We could perform similar calculations for higher moments or cumulants, and again would find asymptotic power-law behaviors.

This architecture, displayed in Fig. 5, has been considered in previous studies [11, 12, 13] and recently solved for the two-terminal reliability ${\mathcal{R}}_{n}$ between $S_{0}$ and $S_{n}$ [5]. For perfect nodes,

When $n\to\infty$, ${\mathcal{R}}_{n}$ clearly tends to the constant $\displaystyle{\mathcal{R}}_{\infty}=\frac{p^{2}}{\big{(}1-p\,(1-p)\big{)}^{2}}\neq 1$ ($p\,(1-p)$ is always less than 1/4, so the last contribution in eq. (37) decreases faster than $4^{-n}$). This stems from the existence of one path with a finite number of hops, namely $S_{0}\to T\to S_{n}$. The MTTF’s asymptotic behavior is therefore different from that of the preceding section: it does not vary with $n$. This is also true for higher moments, with

where $\langle\,t^{m}\,\rangle_{\infty}=\lim_{n\to\infty}\,\langle\,t^{m}\,\rangle_{n}$. The first of these integrals are

where $\psi^{\prime}$ is the derivative of the digamma function $\psi$. From the first values of $\langle\,t^{m}\,\rangle$, we can infer the general result

Depending on the parity of $m$, the difference $\psi^{(m-1)}(1/3)-\psi^{(m-1)}(2/3)$ may actually be further simplified (leaving only $\psi^{(m-1)}(1/3)$ for $m$ even, or powers of $\pi$ for $m$ odd). It is easy to prove that, asymptotically,

In that limit, it looks as if only the $S_{0}\to T\to S_{n}$ connection exists.

This configuration, displayed in Fig. 6, is a slight generalization of $n$ double links in parallel. As $n\to\infty$, there is an infinity of paths of finite length connecting $S$ to $T$; for perfect nodes, the associated two-terminal reliability ${\mathcal{R}}_{n}$ is [7]

with

Here again — if we forget that 1 is a third eigenvalue – we have two eigenvalues $\zeta_{\pm}$. However, the situation is different from that of the $K_{4}$ ladder, because $\zeta_{\pm}\to 0$ when $p\to 1$, while $\zeta_{+}\to 1$ and $\alpha_{+}\to 1$ when $p\to 0$. We also expect that

the right-hand side of eq. (46) corresponding to $n$ elements of reliability $p^{2}$ in parallel.

As $n$ increases, the contribution of $\zeta_{+}$ prevails over that of $\zeta_{-}$ (see Fig. 7), so that ${\mathcal{R}}_{n}\approx 1-\alpha_{+}\,\zeta_{+}^{n}$. Consequently,

Because of the $1/p$ factor, the asymptotic expansion of $\langle\,t\,\rangle_{n}$ is now controlled by the behaviors of $\zeta_{+}$ and $\alpha_{+}$ for $p\to 0$:

We can write

Each term on the right-hand side of eq. (50) vanishes for $p\to 0$, so that the $1/p$ factor does not lead to a diverging integral in eq. (47). The first term of eq. (50) gives the prevailing contribution, namely

This contribution is — unsurprisingly — half the usual result for $n$ elements in parallel, because the reliability $p^{2}$ translates into a $2\,\lambda$ failure rate. For the two other contributions, the change of variable $\tau=n\,p^{2}$ gives a factor $\exp(-\tau)$; the remaining factors in eq. (50) must be expanded in the vicinity of $\tau\to 0$, as in section 3.1. Summing the three contributions gives

A comparison of the exact results with the asymptotic expansion in which we have kept the first three terms of eq. (52) is plotted in Fig. 8. The agreement is good, even for moderate values of $n$.

In the preceding subsections, we have considered architectures for which ${\mathcal{R}}_{n}$ is exactly known through the analytic expressions of two eigenvalues $\zeta_{\pm}$. In complex systems, more than two eigenvalues may coexist, and be known only as roots of polynomial equations. However, the MTTF’s asymptotic expression can still be derived from the knowledge of the generating function ${\mathcal{G}}(z)$ [14] of the ${\mathcal{R}}_{n}$’s, namely

Such is the case of the Street $3\times n$, displayed in Fig. 9. This configuration has been studied for the two-terminal reliability ${\mathcal{R}}_{n}$ between $S_{0}$ and $U_{n}$ [6, 15, 16, 17, 18, 19, 20, 21] (there is actually an offset of 1 between our $n$ and these references’ $n$ because our source is $S_{0}$). For perfect nodes, ${\mathcal{G}}$ is given by ${\mathcal{N}}/({\mathcal{D}}_{1}\,{\mathcal{D}}_{2})$, with [6]

${\mathcal{N}}$, ${\mathcal{D}}_{1}$ and ${\mathcal{D}}_{2}$ are polynomials in both $z$ and $p$.

The eigenvalue of greatest modulus, named $\zeta_{+}$ again, actually obeys ${\mathcal{D}}_{2}=0$ for $z=1/\zeta_{+}$ (in the limit $p\to 1$, ${\mathcal{D}}_{2}\to 1-z$ and $\zeta_{+}\to 1$); all other eigenvalues tend to zero in that limit. Even though it is not possible to get an analytic expression for $\zeta_{+}$ as a function of $p$ (${\mathcal{D}}_{2}$ is of degree 6 in $z$), we can readily compute it numerically. We can also deduce from the constraint ${\mathcal{D}}_{2}(z=1/\zeta_{+})=0$ the expansion of $\zeta_{+}$ as a function of $q$ for small $q$’s, starting with $\zeta_{+}=1$:

The scaling variable $\tau$ should be $n$ times the leading term of eq. (58), namely $\tau=n\,q^{3}$, so that

$\alpha_{+}$ is deduced from $p$ and the numerical value of $\zeta_{+}$ through the residue of ${\mathcal{G}}$ at $z=1/\zeta_{+}$. The general result is