Proof. As in Fiorini et al. [Reference Fiorini, Sheahan and Lipsky16], we calculate the moments of the completion time distribution to assess the heaviness of the tail. We assume here that the task size has an exponential tail, which amounts to $\theta_{\text{min}}(S) = \mu$ from Eq. (3.3), and denote $\Delta = \min\{f,r/\alpha\} \gt 0$ and $\Delta^{+} = f + r/\alpha - \Delta = \max\{f,r/\alpha\} \gt 0$. Let U and D be exponentially distributed with failure rate f and repair rate r, respectively. Their distributions are $F_{U}(t) = 1-e^{-ft}$, and $F_{D}(t) = 1-e^{-rt}$.

Notice that the task-size-conditioned Laplace–Stieltjes transform can be expressed as in (D.1)–(D.2)

(D.1)\begin{align} L_{U}(s|t) & = E[e^{-sU}|t \gt U] = \int_{x=0}^{t}e^{-sx}dF_{U}(x|t) = \int_{x=0}^{t}e^{-sx}\dfrac{fe^{-fx}}{F_{U}(t)}dx = \dfrac{f}{s+f}\dfrac{1-e^{-(s+f)t}}{1-e^{-ft}}, \end{align}

(D.2)\begin{align} L_{D}(s|t/\alpha) & = E[e^{-sD}|t \gt D] = \int_{x=0}^{t/\alpha}e^{-sx}dF_{D}(x|t/\alpha) = \int_{x=0}^{t/\alpha}e^{-sx}\dfrac{re^{-rx}}{F_{D}(t/\alpha)}dx = \dfrac{r}{s+r}\dfrac{1-e^{-(s+r)t/\alpha}}{1-e^{-rt/\alpha}}. \end{align}

From Theorem 2.5, the conditional Laplace–Stieltjes transform of the completion time in Eq. (2.23) becomes

\begin{equation*} L_{T}(s|t) = \dfrac{ re^{-(s+f)t} \left[1 + \dfrac{f}{s+r}(1-e^{-(s+r)t/\alpha})\right] + fe^{-(s+r)t/\alpha} \left[1 + \dfrac{r}{s+f}(1-e^{-(s+f)t})\right] }{ (r+f) \left[ 1- \dfrac{rf}{(s+r)(s+f)} (1-e^{-(s+f)t}) (1-e^{-(s+r)t/\alpha}) \right] }, \end{equation*}

where we call

\begin{align*} h_{r}(s,t) & := \dfrac{f}{s+r}(1-e^{-(s+r)t/\alpha}), \\ h_{f}(s,t) & := \dfrac{r}{s+f}(1-e^{-(s+f)t}), \end{align*}

obtaining

(D.3)\begin{equation} L_{T}(s|t) = \dfrac{1}{r+f}\cdot \dfrac{ re^{-(s+f)t}(1 + h_{r}(s,t)) + fe^{-(s+r)t/\alpha}(1 + h_{f}(s,t)) }{1-h_{r}(s,t)h_{f}(s,t)}. \end{equation}

To calculate the mth moment $E[T^{m}]$ of the completion time T, we resort to the derivatives of the Laplace–Stieltjes transform of the completion time distribution evaluated at s = 0 as shown in Eq. (D.4)

(D.4)\begin{equation} E[T^{m}] = \left.(-1)^{m}\dfrac{d^{m}}{ds^{m}}L_{T}(s)\right|_{s=0}. \end{equation}

In particular, however, we calculate the mth partial derivative with respect to s of the conditioned Laplace–Stieltjes transform for the completion time in (D.3), evaluate at s = 0, and uncondition with respect to the task size whenever is possible.

Consider the following function definitions and notation for the partial derivatives of the conditional Laplace Stieljest Transform (LSTm):

1. (1) Main term for the nth derivative:

   \begin{equation*} L_{T}(s;n|t) := \dfrac{1}{r+f}\cdot \dfrac{ re^{-(s+f)t}(1 + h_{r}(s,t)) + \frac{f}{\alpha^{n}}e^{-(s+r)t/\alpha}(1 + h_{f}(s,t)) }{1-h_{r}(s,t)h_{f}(s,t)}, \quad n = 0,1,2,\ldots. \end{equation*}

   Notice $L_{T}(s;0|t) := L_{T}(s|t)$.

2. (2) First recurrent factor of higher-order term for nth derivative:

   (D.5)\begin{equation} \nu_{n}(s,t) := \dfrac{re^{-(s+f)t} + \frac{f}{\alpha^{n-1}}e^{-(s+r)t/\alpha}}{r+f}, \quad n = 1,2,\ldots. \end{equation}

   It holds that $\frac{\partial^{n}}{\partial s^{n}}\nu_{1}(s,t) = (-1)^{n}t^{n}\nu_{n+1}(s,t)$.

3. (3) Recurrent rational exponential functions and terms:

   (D.6)\begin{equation} \theta_i(s,t;a,b) := \dfrac{e^{-i(s+a)t/b}}{1-e^{-i(s+a)t/b}}. \end{equation}

   It can be shown that for $n\geq1$,

   (D.7)\begin{align} \frac{\partial^{n}}{\partial s^{n}}\theta_1(s,t;a,b) & \phantom{:}= (-1)^{n}\left(\frac{t}{b}\right)^{\!\!n} \sum_{i=1}^{n+1}W(n,i)\dfrac{e^{-i(s+a)t/b}}{(1-e^{-(s+a)t/b})^{i}} \notag \\ & := (-1)^{n}\left(\frac{t}{b}\right)^{\!\!n} \sum_{i=1}^{n+1}W(n,i)\theta_{i}(s,t;a,b), \end{align}

   where $W(n,i)$ denotes the nth row and ith column’s Worpitzky number from the Worpitzky triangle [Reference Sloane35], as defined in (D.8)

   (D.8)\begin{equation} W(n,i) = \dfrac{1}{i}\sum_{j=0}^{i} (-1)^{i-j}\binom{i}{j}j^{n}, \quad n \geq 1, \quad 1 \leq i \leq n. \end{equation}

   In order to simplify the formulas, we also give the following notation:

   (D.9)\begin{align} \Theta_1(s,t;a,b) & := \dfrac{t}{b}\cdot\dfrac{e^{-(s+a)t/b}}{1-e^{-(s+a)t/b}}-\dfrac{1}{s+a} \notag \\ & \phantom{:}= \dfrac{t}{b}\theta_1(s,t;a,b)-\dfrac{1}{s+a}. \end{align}

   From here, similarly, for $n\geq0$,

   (D.10)\begin{align} \frac{\partial^{n}}{\partial s^{n}}\Theta_1(s,t;a,b) & \phantom{:}= (-1)^{n} \left( \dfrac{t^{n+1}}{b^{n+1}} \sum_{i=1}^{n+1}W(n,i) \theta_{i}(s,t;a,b) -\dfrac{n!}{(s+a)^{n+1}} \right) \notag \\ & := \Theta_{n+1}(s,t;a,b). \end{align}

   Finally, we call

   (D.11)\begin{equation} \omega_{n}(s,t) := \Theta_{n}(s,t;f,1) + \Theta_{n}(s,t;r,\alpha), \quad n = 1,2,\ldots \end{equation}

   and by definition, $\frac{\partial^{n}}{\partial s^{n}}\omega_{1}(s,t) = \omega_{n+1}(s,t)$.

Now we calculate de first derivative of the conditioned Laplace–Stieltjes transform in Eq. (D.3). Henceforth, functional dependencies of $h_{r}(s,t)$ and $h_{f}(s,t)$ may be omitted for cleaner notation

(D.12)\begin{align} \dfrac{\partial}{\partial s}L_{T}(s|t) = & \phantom{- } \dfrac{\partial}{\partial s}L_{T}(s;0|t) \notag \\ = & -tL_{T}(s;1|t) + \nu_{1}(s,t) \dfrac{ \frac{\partial}{\partial s} (h_{r}h_{f}) }{(1-h_{r}h_{f})^{2}} \notag \\ & + \dfrac{s+f}{r+f} \theta_1(s,t;f,1) \Bigl(\Theta_1(s,t;r,\alpha)+h_{r}h_{f}\Theta_1(s,t;f,1)\Bigr) \dfrac{h_{r}h_{f}}{(1-h_{r}h_{f})^{2}} \notag \\ & + \dfrac{s+r}{r+f} \theta_1(s,t;r,\alpha) \Bigl(\Theta_1(s,t;f,1)+h_{r}h_{f}\Theta_1(s,t;r,\alpha)\Bigr) \dfrac{h_{r}h_{f}}{\bigl(1-h_{r}h_{f}\bigr)^{2}} \notag \\ := & -tL_{T}(s;1|t) + \nu_{1}(s,t) \dfrac{ \frac{\partial}{\partial s} (h_{r}h_{f}) }{(1-h_{r}h_{f})^{2}} \notag \\ & + \left[ \dfrac{s+f}{r+f} \theta_1(s,t;f,1) A(s,t) + \dfrac{s+r}{r+f} \theta_1(s,t;r,\alpha) B(s,t) \right] \dfrac{h_{r}h_{f}}{(1-h_{r}h_{f})^{2}}, \end{align}

with $A(s,t) = \Theta_1(s,t;r,\alpha)+h_{r}h_{f}\Theta_1(s,t;f,1)$ and $B(s,t) = \Theta_1(s,t;f,1)+h_{r}h_{f}\Theta_1(s,t;r,\alpha)$.

It can be shown that

(D.13)\begin{equation} \dfrac{\partial}{\partial s}(h_{r}h_{f}) = h_{r}h_{f}\omega_{1}(s,t). \end{equation}

Then, calling

(D.14)\begin{equation} u_1(s,t) := \dfrac{h_{r}h_{f}}{(1-h_{r}h_{f})^{2}}, \end{equation}

we have from (D.12),

(D.15)\begin{align} \dfrac{\partial}{\partial s}L_{T}(s|t) & = -tL_{T}(s;1|t) + u_1(s,t) \left[ \nu_{1}(s,t)\omega_{1}(s,t) + \dfrac{s+f}{r+f}\theta_1(s,t;f,1)A(s,t)\right.\nonumber\\ &\quad +\left. \dfrac{s+r}{r+f}\theta_1(s,t;r,\alpha)B(s,t) \right]. \end{align}

Moreover,

(D.16)\begin{align} \dfrac{\partial}{\partial s}L_{T}(s;n|t) &= -tL_{T}(s;n+1|t) + u_1(s,t) \left[ \nu_{n+1}(s,t)\omega_{1}(s,t) + \dfrac{s+f}{r+f}\theta_1(s,t;f,1)A(s,t)\right. \nonumber\\ &\quad \left. + \dfrac{s+r}{r+f}\theta_1(s,t;r,\alpha)B(s,t) \right]. \end{align}

To calculate the higher order partial derivatives of $L_{T}(s|t)$ we have to use the general Leibniz rule on the second term in (D.15). For this, we have to study the partial derivatives of the factor $u(s,t)$, and the functions $A(s,t)$ and $B(s,t)$. We start with $u(s,t)$. From (D.13), we observe that

(D.17)\begin{equation} h_{r}(s,t)h_{f}(s,t) = C(t)e^{\int\omega_{1}(s,t)ds}, \end{equation}

where $C(t) := rf\,e^{-t(f+r/\alpha)}$.

Consider now t fixed and the two following one-dimensional functions:

(D.18)\begin{equation} g(x(s)) := C(t)e^{\int x(s)ds}, \quad f(y) := \dfrac{y}{(1-y)^{2}}. \end{equation}

Using the functions defined above Eqs. (D.14) and (D.17) can be rewritten as

(D.19)\begin{align} h_{r}(s,t)h_{f}(s,t) & = g(w_{1}(s,t)), \end{align}

(D.20)\begin{align} u_1(s,t) & = f(g(w_{1}(s,t))), \end{align}

thus, $u_{1}(s,t)$ is a nested composite function.

The derivatives of composite functions can be written using Faà di Bruno’s formula [Reference Faa di Bruno15] as shown in Eq. (D.21):

(D.21)\begin{equation} \dfrac{d^n}{dx^n} f(g(x)) = \sum_{k=1}^{n} f^{(k)}(g(x))\cdot B_{n,k}\Bigr(g'(x),g''(x),\ldots,g^{(n-k+1)}(x)\Bigl), \end{equation}

where $B_{n,k}$ are the incomplete (or partial) exponential Bell polynomials [Reference Bell9, Reference Mihoubi26] defined below

(D.22)\begin{equation} B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1}) = \sum \dfrac{n!}{j_{1}!j_{2}!\cdots j_{n-k+1}!} \left(\dfrac{x_{1}}{1!}\right)^{j_{1}} \left(\dfrac{x_{2}}{2!}\right)^{j_{2}} \cdots \left(\dfrac{x_{n-k+1}}{(n-k+1)!}\right)^{j_{n-k+1}}, \end{equation}

where the summation is carried over all sequences $j_{1},j_{2},j_{3},\ldots,j_{n-k+1}$ of non-negative integers such that conditions below are satisfied

(D.23)\begin{align} j_{1} + j_{2} + \cdots + j_{n-k+1} = k, \end{align}

(D.24)\begin{align} j_{1} + 2j_{2} + 3j_{3} +\cdots + (n-k+1)j_{n-k+1} = n. \end{align}

The nth complete exponential Bell polynomial is given by the sum of the incomplete polynomials as:

(D.25)\begin{equation} B_{n}(x_{1},\ldots,x_{n}) = \sum_{k=1}^{n}B_{n,k}(x_{1},x_{2},\ldots,x_{n-k+1}). \end{equation}

The derivatives of composite functions in Eq. (D.21) are similar to the complete exponential Bell polynomials, of which the variables $x_{1},\ldots,x_{n}$ are replaced by the successive derivatives of the inner function g(x), and the coefficients of the polynomials are factored by the derivatives of the outer function. Since $u(s,t) = f(g(w_{1}(s,t)))$ is a double composite function, its partial derivatives are given by nested Bell polynomials.

From (D.11), we know that the partial derivatives of the innermost function $\omega_{1}(s,t)$ with respect to s are

(D.26)\begin{equation} \dfrac{\partial^{n}}{\partial s^{n}}\omega_{1}(s,t) = \omega_{n+1}(s,t). \end{equation}

The partial derivatives of the first composite function $g(w_{1}(s,t))$ are calculated using (D.21) where the variable coefficients are the derivatives of $g(x(s)) = C(t)e^{\int x(s)ds}$ evaluated in $\omega_{1}(s,t)$, and the incomplete Bell polynomials are evaluated on the partial derivatives of $\omega_{1}(s,t)$ from Eq. (D.26). The general derivative is given below, we omit the functional dependency on $\omega_{i}(s,t)$ to keep the notation uncluttered

\begin{align*} G_{n}(s,t) := \dfrac{\partial^{n}}{\partial s^{n}} \bigl(h_{r}(s,t)h_{f}(s,t)\bigr) = h_{r}(s,t)h_{f}(s,t)B_{n}(\omega_{1},\omega_{2},\ldots,\omega_{n}). \end{align*}

The similarities of the derivatives to the complete Bell polynomials are due to the exponential nature of $g(x(s))$, which makes the successive derivatives of the outer function self-similar, and hence it can be factored out from the coefficients of the partial polynomials. Notice that derivatives of low order for $g(x(s))$ may also be easily calculated recursively following from Eq. (D.13).

The derivatives of the outermost function f(y) are

(D.27)\begin{equation} \dfrac{d^{n}}{dy^{n}}f(y) = \dfrac{d^{n}}{dy^{n}}\dfrac{y}{(1-y)^{2}} = \dfrac{n!}{(1-y)^{2}}\dfrac{y+n}{(1-y)^{n}}. \end{equation}

With this we give below the general form of the partial derivatives of $u(s,t)$ where we drop the functional dependencies of $G_{n}(s,t)$ to keep the notation uncluttered

(D.28)\begin{align} u_{n+1}(s,t) & := \dfrac{\partial^{n}}{\partial s^{n}} u_1(s,t) \notag \\ & = \sum_{k=1}^{n} k!\dfrac{(k + h_{r}h_{f})\phantom{^{k+2}} } {(1 - h_{r}h_{f})^{k+2}} \cdot B_{n,k}\Bigr(G_{1},G_{2},\ldots,G_{n-k+1}\Bigl), \quad n=1,2,\ldots. \end{align}

The derivatives of $A(s,t)$ and $B(s,t)$ are also written in terms of Bell polynomials. This is due to the fact that, from Eq. (D.17) and by definition both functions are composite functions:

\begin{align*} A(s,t) & = \Theta_1(s,t;r,\alpha) + C(t)e^{\int(\Theta_{1}(s,t;f,1) + \Theta_{1}(s,t;r,\alpha))ds}\Theta_1(s,t;f,1), \\ B(s,t) & = \Theta_1(s,t;f,1) + C(t)e^{\int(\Theta_{1}(s,t;f,1) + \Theta_{1}(s,t;r,\alpha))ds}\Theta_1(s,t;r,\alpha). \end{align*}

Denote by $\Theta_{1:k}(s,t;a,b)$ the ordered list $\Theta_{1}(s,t;a,b),\ldots,\Theta_{k}(s,t;a,b)$. It can be shown by induction that

(D.29)\begin{align} \dfrac{\partial^{n}}{\partial s^{n}} A(s,t) & = \Theta_{n+1}(s,t;r,\alpha) + h_{r}(s,t)h_{f}(s,t) \sum _{k=0}^{n}{\binom {n}{k}} B_{n-k+1}\bigl(\Theta_{1:(n-k+1)}(s,t;f,1)\bigr) B_{k}\bigl(\Theta_{1:k}(s,t;r,\alpha)\bigr), \end{align}

(D.30)\begin{align} \dfrac{\partial^{n}}{\partial s^{n}} B(s,t) & = \Theta_{n+1}(s,t;f,1) + h_{r}(s,t)h_{f}(s,t) \sum _{k=0}^{n}{\binom {n}{k}} B_{n-k+1}\bigl(\Theta_{1:(n-k+1)}(s,t;r,\alpha)\bigr) B_{k}\bigl(\Theta_{1:k}(s,t;f,1)\bigr), \end{align}

where the summations are binomial expansions on the indices of $\Theta_{k}(s,t;a,b)$, with the first index shifted to the right by one. Getting back to Eq. (D.15), we had that

(D.31)\begin{align} \dfrac{\partial}{\partial s}L_{T}(s|t)& = -tL_{T}(s;1|t) + u_1(s,t) \left[ \nu_{1}(s,t)\omega_{1}(s,t) + \dfrac{s+f}{r+f}\theta_1(s,t;f,1)A(s,t)\right. \nonumber\\ &\quad \left. + \dfrac{s+r}{r+f}\theta_1(s,t;r,\alpha)B(s,t) \right]. \end{align}

It can be shown from successive differentiation of Eq. (D.31) and considering (D.16) that the nth partial derivative of the conditioned Laplace transform of the completion time is given as:

(D.32)\begin{align} \dfrac{\partial^{n}}{\partial s^{n}}L_{T}(s|t) &= (-t)^{n}L_{T}(s;n|t)\nonumber\\ &\quad + \sum_{k=1}^{n} (-t)^{n-k} \dfrac{\partial^{k-1}}{\partial s^{k-1}} \Biggl( u_1(s,t) \left[ \nu_{n-k+1}(s,t)\omega_{1}(s,t)\right.\nonumber\\ &\quad \left. + A(s,t)\left(\tfrac{s+f}{r+f}\,\theta_1(s,t;f,1)\right) + B(s,t)\left(\tfrac{s+r}{r+f}\,\theta_1(s,t;r,\alpha)\right) \right] \Biggr), \end{align}

where the partial derivatives in the right hand side are calculated with the General Leibniz rule, which is given in (D.33) for the product of three functions $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$:

(D.33)\begin{equation} \dfrac{d^{n}}{dx^{n}}\left(f_{1}(x)f_{2}(x)f_{3}(x)\right) = \sum_{k_{1}+k_{2}+k_{3}=n} \dfrac{n!}{k_{1}!\,k_{2}!\,k_{3}!} \dfrac{d^{k_{1}}}{dx^{k_{1}}}f_{1}(x) \dfrac{d^{k_{2}}}{dx^{k_{2}}}f_{2}(x) \dfrac{d^{k_{3}}}{dx^{k_{3}}}f_{3}(x), \end{equation}

with the summation extending over all triplets $(k_{1},k_{2},k_{3})$ of non-negative integers that are such that $\displaystyle\sum _{t=1}^{3}k_{t}=n$.

Finally, the mth moment of the completion time T is obtained by unconditioning (D.32) with respect to the task size t whenever the improper integral converges

(D.34)\begin{align} E[T^{m}] & = (-1)^{m} \int_{0}^{\infty} \bigl(\left.\tfrac{\partial^{m}}{\partial s^{m}}L_{T}(s|t)\right|_{s=0}\bigr) dF_{S}(t) \notag \\ & = (-1)^{m}\mu \int_{0}^{\infty} \bigl(\left.\tfrac{\partial^{m}}{\partial s^{m}}L_{T}(s|t)\right|_{s=0}\bigr) e^{-\mu t}dt. \end{align}

To study the convergence of the integral in (D.34) we look at the asymptotic behavior of the different terms and factors in (D.32) evaluated at s = 0. We will use the definition of asymptotic equivalence given in (D.35). Recall also that we denote $\Delta = \min\{f,r/\alpha\} \gt 0$ and $\Delta^{+} = f + r/\alpha - \Delta = \max\{f,r/\alpha\} \gt 0$

(D.35)\begin{equation} \psi_{1}(t)\sim\psi_{2}(t) \,\text{if and only if }\, \lim_{t\to\infty}\dfrac{\psi_{1}(t)}{\psi_{2}(t)} = 1. \end{equation}

We classify according to the asymptotic order.

1. (1) Constant factors:

   1. (a) Main factor of the first term: $L_{T}(0;n|t)$

      \begin{equation*} L_{T}(0;n|t) \sim C_{1}(n) := \begin{cases} 1, & \quad f \lt r/\alpha, \\ 1/\alpha^{n}, & \quad f \gt r/\alpha, \\ \tfrac{1}{2}(1+1/\alpha^{n}), & \quad f = r/\alpha \end{cases}, \quad n=0,1,2,\ldots. \end{equation*}

   2. (b) Recurrent factor $\Theta_{n}(0,t;a,b)$:

      \begin{align*} \Theta_{1}(0,t;a,b) & = \dfrac{t}{b}\dfrac{e^{-at/b}}{1-e^{-at/b}}-1/a \sim -1/a, \\ \Theta_{n}(0,t;a,b) & = (-1)^{n-1} \left( \dfrac{t^{n}}{b^{n}} \sum_{i=1}^{n}W(n,i) \dfrac{e^{-i\cdot at/b}}{(1-e^{-at/b})^{i}} -\dfrac{(n-1)!}{a^{n}} \right) \nonumber\\ &\quad \sim (-1)^{n}\dfrac{(n-1)!}{a^{n}}, \quad n = 2,3,\ldots. \end{align*}

   3. (c) Recurrent factor $\omega_{n}(0,t)$:

      \begin{equation*} \omega_{n}(0,t) = \Theta_{n}(0,t;f,1) + \Theta_{n}(0,t;r,\alpha) \sim (-1)^{n}(n-1)!\left(\dfrac{1}{f^{n}}+\dfrac{1}{r^{n}}\right), \quad n = 1,2,\ldots. \end{equation*}

   4. (d) Recurrent factors $A(s,t)$ and $B(s,t)$, and derivatives: from (D.29)–(D.30), and by the continuity of Bell polynomials, we have that for every $n\geq 0$ there exist constants $C_{A}(n) \gt 0$ and $C_{B}(n) \gt 0$ such that

      \begin{align*} \dfrac{\partial^{n}}{\partial s^{n}} A(s,t) & \sim (-1)^{n+1}C_{A}(n), \\ \dfrac{\partial^{n}}{\partial s^{n}} B(s,t) & \sim (-1)^{n+1}C_{B}(n). \end{align*}

2. (2) Polynomial factors with exponential damping:

   1. (a) Recurrent factor $\nu_{n}(0|t)$:

      \begin{equation*} \nu_{n}(0|t) \sim e^{-\Delta t}\cdot C_{\nu}(n) := e^{-\Delta t}\cdot \begin{cases} \tfrac{r}{r+f}, & \quad f \lt r/\alpha, \\ \tfrac{f}{(r+f)}\tfrac{1}{\alpha^{n-1}}, & \quad f \gt r/\alpha, \\ \tfrac{1}{r+f}\left(r+\tfrac{f}{\alpha^{n-1}}\right), & \quad f = r/\alpha \end{cases}, \quad n = 1,2,\ldots. \end{equation*}

   2. (b) Recurrent factors $\theta_{k}(0,t;a,b)$ and derivatives of $\theta_1(0,t;a,b)$:

      \begin{equation*} \theta_{k}(s,t;a,b) \sim e^{-k\cdot at/b}, \end{equation*}

      \begin{equation*} \left.\frac{\partial^{n}}{\partial s^{n}}\theta_1(s,t;a,b)\right|_{s=0} = (-1)^{n}\left(\frac{t}{b}\right)^{\!\!n} \sum_{i=1}^{n+1}W(n,i)\theta_{i}(0,t;a,b) \sim (-1)^{n}\left(\frac{t}{b}\right)^{n}e^{-at/b}, \quad n=1,2,\ldots. \end{equation*}

   3. (c) Accompanying factors of $A(0,t)$ and $B(0,t)$, and derivatives:

      \begin{equation*} \left.\dfrac{\partial^{n}}{\partial s^{n}}\left(\frac{s+a}{r+f}\,\theta_1(s,t;a,b)\right)\right|_{s=0} \sim \dfrac{1}{r+f}\left[n+ a(-1)^{n-1}\dfrac{t^{n-1}}{b^{n-1}}\right]e^{-at/b}, \quad n=1,2,\ldots. \end{equation*}

3. (3) Exponential growth factors: recurrent factor $u_{n}(0,t)$.

   Again due to the continuity of the Bell polynomials, for every $n\in\mathbb{N}$ there exists a constant $C_{G}(n) \gt 0$ such that $G_{n}(0,t) \sim (-1)^{n}C_{G}(n)$, and since the dominant term in (D.28) is for k = n, there exists a constant $C_{B}(n-1) \gt 0$ such that $\left.B_{n-1,n-1}(G_{1})\right|_{s=0} \sim (-1)^{n-1}C_{B}(n-1)$. From here, given that

   \begin{equation*} \dfrac{\bigl(n + h_{r}(0,t)h_{f}(0,t)\bigr)\phantom{^{n+2}}}{\bigl(1 - h_{r}(0,t)h_{f}(0,t)\bigr)^{n+2}} \sim (n+1)e^{(n+2)\Delta t}, \end{equation*}

   we obtain

   \begin{equation*} u_{n}(0,t) \sim (-1)^{n-1}n!\,C_{B}(n-1)\cdot e^{(n+1)\Delta t}, \quad n=1,2,\ldots. \end{equation*}

   Notice that $u_{n}(0,t)$ is the only factor that has an asymptotic behavior that is not decaying nor constant for large t, and also that the asymptotic order increases with n, which is the order of the partial derivative.

It is easy to see from the first derivative in Eq. (D.15) that $\left.\frac{\partial}{\partial s}L_{T}(s|t)\right|_{s=0} \sim -\mathcal{C}e^{\Delta t}$, for some constant $\mathcal{C} \gt 0$, and that the expected completion time $E[T]$ only exists when $\Delta =\min\{f,r/\alpha\} \lt \mu$. For higher order derivatives in Eq. (D.32), we only have to observe the last term in the summation, which will yield the higher order terms overall. Additionally, since the partial derivatives of $u_1(s,t)$ dominate the asymptotic behavior, we only have to look at the term coming from the Leibniz rule for which $u_1(s,t)$ has the partial derivative of order n − 1 and the remaining factors are not differentiated. Then, for $n\geq1$,

(D.36)\begin{align} &\left.\dfrac{\partial^{n}}{\partial s^{n}}L_{T}(s|t)\right|_{s=0} \sim u_{n}(0,t) \left[ \nu_{1}(0,t)\omega_{1}(0,t) + A(0,t)\left(\tfrac{f}{r+f}\,\theta_1(0,t;f,1)\right) \right. \nonumber\\ &\quad\left. + B(0,t)\left(\tfrac{r}{r+f}\,\theta_1(0,t;r,\alpha)\right) \right] \sim \mathcal{C}(n)e^{n\Delta t}, \end{align}

where $\mathcal{C}(n)$ is a constant that depends on n. Then, from (D.34), the mth moment $E[T^{m}]$ is not finite whenever $m\Delta -\mu \geq 0$ which is $m\geq \mu/\Delta = \mu/\min\{f,r/\alpha\} = \varepsilon$, and the completion time distribution is power-tailed [Reference Greiner, Jobmann and Lipsky19], that is, there exists a positive constant c such that

(D.37)\begin{equation} \overline{F_{T}(t)} \sim \dfrac{c}{t^{\varepsilon}}. \end{equation}