Appendix 1: Solving the Probability \(\Pr \left[ {V_2 <U_2} \right] \) in (36)
From (36), we see that the probability \(\Pr \left[ {V_2 <U_2} \right] =0\) if \(v=0\). Hence, we only consider two cases with \(v\ge 1\) as follows.
Case 1: When \(\lambda _1 =\lambda _2 =\lambda \), (36) can be rewritten and solved as
$$\begin{aligned}&\Pr \left[ {V_2 <U_2} \right] \nonumber \\&\quad =v\lambda _3 \lambda \left( {\ln 8} \right) \!\!\int \limits _0^\infty {\left[ {1-\frac{\lambda _3 e^{{-\lambda \theta _x}/\gamma }}{\lambda _3 +\lambda (\theta _x +1)}} \right] ^{u+v-1}\frac{(\theta _x +1)e^{{-\lambda \theta _x}/\gamma }}{\lambda _3 +\lambda (\theta _x +1)}\left[ {\frac{1}{\lambda _3 +\lambda (\theta _x +1)}\!+\!\frac{1}{\gamma }} \right] dx} \nonumber \\&\quad =v\lambda _3 \int \limits _{\lambda _3 +\lambda }^\infty {\left[ {1-\frac{\lambda _3 e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}} \right] ^{u+v-1}\frac{e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =v\lambda _3 \int \limits _{\lambda _3 +\lambda }^\infty {\sum _{p=0}^{v+u-1} {C_{v+u-1}^p \frac{(-\lambda _3)^{p}e^{-p{\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x^{p}}} \frac{e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =v\lambda _3 \sum _{p=0}^{v+u-1} {C_{v+t-1}^p (-\lambda _3)^{p}e^{{\left( {p+1} \right) \left( {\lambda _3 +\lambda } \right) }/\gamma }} \int \limits _{\lambda _3 +\lambda }^\infty {\frac{e^{{-\left( {p+1} \right) x}/\gamma }}{x^{p+1}}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =v\lambda _3 \sum _{p=0}^{v+u-1} {C_{v+t-1}^p (-\lambda _3)^{p}e^{{\left( {p+1} \right) \left( {\lambda _3 +\lambda } \right) }/\gamma }}\nonumber \\&\qquad \times {\left( \,\,{\int \limits _{\lambda _3 +\lambda }^\infty {\frac{e^{{-\left( {p+1} \right) x}/\gamma }}{x^{p+2}}dx} +\frac{1}{\gamma }\int \limits _{\lambda _3 +\lambda }^\infty {\frac{e^{{-\left( {p+1} \right) x}/\gamma }}{x^{p+1}}dx}} \right) } \end{aligned}$$
(75)
Using the \(n\)th order exponential integral function \(E_n \left[ z \right] \) [18, Eq. (1)], and after some manipulations, we obtain the probability \(\Pr \left[ {V_2 <U_2} \right] \) as in (37) when \(\lambda _1 =\lambda _2 =\lambda \) based on (75).
Case 2: When \(\lambda _1 \ne \lambda _2 \), (36) can be rewritten and changed as
$$\begin{aligned}&\Pr \left[ {V_2 <U_2} \right] =v\lambda _3 \int \limits _{\lambda _3 +\lambda _1}^\infty {\left[ {1-\frac{\lambda _3 e^{{-\left( {x-\lambda _3 -\lambda _1} \right) }/\gamma }}{x}} \right] ^{v-1}\frac{e^{{-\left( {x-\lambda _3 -\lambda _1} \right) }/\gamma }}{x}} \left( \frac{1}{x}+\frac{1}{\gamma } \right) \nonumber \\&\qquad \times \left[ {1-\frac{\lambda _3 e^{{-\lambda _2 \left( {x-\lambda _3 -\lambda _1} \right) }/{\left( {\lambda _1 \gamma } \right) }}}{\lambda _3 +\lambda _2 (x-\lambda _3)/\lambda _1}} \right] ^{u}dx \nonumber \\&\quad =v\lambda _3 \int \limits _{\lambda _3 +\lambda }^\infty {\sum _{p=0}^{v-1} {C_{v-1}^p \frac{(-\lambda _3)^{p}e^{-p{\left( {x-\lambda _3 -\lambda _1} \right) }/\gamma }}{x^{p}}} \frac{e^{{-\left( {x-\lambda _3 -\lambda _1} \right) }/\gamma }}{x}} \left( \frac{1}{x}+\frac{1}{\gamma } \right) \nonumber \\&\qquad \times \left[ {1-\frac{\lambda _3 \lambda _1 e^{{\lambda _2 \left( {\lambda _3 +\lambda _1} \right) }/{\left( {\lambda _1 \gamma } \right) }}e^{{-\lambda _2 x}/{\left( {\lambda _1 \gamma } \right) }}}{\lambda _2 x+\lambda _3 (\lambda _1 -\lambda _2)}} \right] ^{u}dx \nonumber \\&\quad =v\lambda _3 \sum _{p=0}^{v-1} {C_{v-1}^p (-\lambda _3 )^{p}e^{{\left( {p+1} \right) \left( {\lambda _3 +\lambda _1} \right) }/\gamma }}\nonumber \\&\qquad \times \left( {\begin{array}{l} \sum _{q=0}^u {C_t^q (-\lambda _3 \lambda _1)^{q}e^{{q\lambda _2 \left( {\lambda _3 +\lambda _1} \right) }/{\left( {\lambda _1 \gamma } \right) }}} \underbrace{\int \limits _{\lambda _3 +\lambda _1}^\infty {\frac{e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _2}{\lambda _1 \gamma }} \right) }}{\left[ {\lambda _2 x+\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{q}x^{p+2}}dx}}_{H_1} \\ +\frac{1}{\gamma }\sum _{q=0}^u {C_t^q (-\lambda _3 \lambda _1)^{q}e^{{q\lambda _2 \left( {\lambda _3 +\lambda _1} \right) }/{\left( {\lambda _1 \gamma } \right) }}\underbrace{\int \limits _{\lambda _3 +\lambda _1}^\infty {\frac{e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _2}{\lambda _1 \gamma }} \right) }}{\left[ {\lambda _2 x+\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{q}x^{p+1}}dx}}_{H_2}} \\ \end{array}} \right) \nonumber \\ \end{aligned}$$
(76)
Using the results in [17, Eq. (47–48)] for the partial expansions of the denominators in the integrals \(H_{1}\) and \(H_{2}\), we have \(H_{1}\) and \(H_{2}\) as follows
$$\begin{aligned} H_1&= \left\{ {\begin{array}{ll} \displaystyle \int \limits _{\lambda _3 +\lambda _1}^\infty {\frac{e^{-x\left( {\frac{p+1}{\gamma }} \right) }}{x^{p+2}}dx} &{},q=0\\ \displaystyle \int \limits _{\lambda _3 +\lambda _1}^\infty {e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _2}{\lambda _1 \gamma }} \right) }\left[ {\sum _{b=1}^q {\frac{m_b^1}{\left[ {\lambda _2 x+\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{b}}+\sum _{b=1}^{p+2} {\frac{n_b^1}{x^{b}}}}} \right] dx} &{},q\ge 1 \\ \end{array}} \right. \end{aligned}$$
(77)
$$\begin{aligned} H_2&= \left\{ {\begin{array}{ll} \displaystyle \int \limits _{\lambda _3 +\lambda _1}^\infty {\frac{e^{-x\left( {\frac{p+1}{\gamma }} \right) }}{x^{p+1}}dx} &{},q=0 \\ \displaystyle \int \limits _{\lambda _3 +\lambda _1}^\infty {e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _2}{\lambda _1 \gamma }} \right) }\left[ {\sum _{b=1}^q {\frac{m_b^2}{\left[ {\lambda _2 x+\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{b}}+\sum _{b=1}^{p+1} {\frac{n_b^2}{x^{b}}}}} \right] dx} &{},q\ge 1 \\ \end{array}} \right. \end{aligned}$$
(78)
where \(m_b^1 =\frac{\left( {-1} \right) ^{p+2}\prod _{c=1}^{q-b} {(p+1+c)\lambda _2^{p+2}}}{(q-b)!\left[ {\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{p+q+2-b}}\), \(n_b^1 =\frac{\left( {-1} \right) ^{p+2-b}\prod _{c=1}^{p+2-b} {(q-1+c)\lambda _2^{p+2-b}} }{(p+2-b)!\left[ {\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{p+q+2-b}}\), \(m_b^2 =\frac{\left( {-1} \right) ^{p+1}\prod _{c=1}^{q-b} {(p+c)\lambda _2^{p+1}} }{(q-b)!\left[ {\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{p+q+1-b}}\), and \(n_b^2 =\frac{\left( {-1} \right) ^{p+1-b}\prod _{c=1}^{p+1-b} {(q-1+c)\lambda _2^{p+1-b}} }{(p+1-b)!\left[ {\lambda _3 (\lambda _1 -\lambda _2)} \right] ^{p+q+1-b}}\).
Substituting (77) and (78) into (76) and after some manipulations, we obtain the probability \(\Pr \left[ {V_2 <U_2} \right] \) as in (38) when \(\lambda _1 \ne \lambda _2\).
Appendix 2: Solving the Probability \(\Pr \left[ {U_2 <V_2 <R} \right] \) in (41)
Similar to “Appendix 1” and from (41), we also see that the probability \(\Pr \left[ {U_2 <V_2 <R} \right] =0\) if \(u=0\). Hence, we only consider two cases with\(u\ge 1\) as follows.
Case 1: When \(\lambda _1 =\lambda _2 =\lambda \), (41) can be rewritten and solved as
$$\begin{aligned}&\Pr \left[ {U_2 <V_2 <R} \right] \nonumber \\&\quad =u\lambda _3 \lambda \left( {\ln 8} \right) \!\!\int \limits _0^R {\left[ {1-\frac{\lambda _3 e^{-\lambda \frac{\theta _x}{\gamma }}}{\lambda _3 +\lambda (\theta _x +1)}} \right] ^{u+v-1}\frac{(\theta _x +1)e^{-\lambda \frac{\theta _x}{\gamma }}}{\lambda _3 +\lambda (\theta _x +1)}\left[ {\frac{1}{\lambda _3 +\lambda (\theta _x +1)}\!+\!\frac{1}{\gamma }} \right] dx} \nonumber \\&\quad =u\lambda _3 \int \limits _{\lambda _3 +\lambda }^{\lambda _3 +\lambda \left( {\theta _R +1} \right) } {\left[ {1-\frac{\lambda _3 e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}} \right] ^{u+v-1}\frac{e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =u\lambda _3 \int \limits _{\lambda _3 +\lambda }^{\lambda _3 +\lambda \left( {\theta _R +1} \right) } {\sum _{p=0}^{v+u-1} {C_{v+u-1}^p \frac{(-\lambda _3)^{p}e^{-p{\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x^{p}}} \frac{e^{{-\left( {x-\lambda _3 -\lambda } \right) }/\gamma }}{x}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =u\lambda _3 \sum _{p=0}^{v+u-1} {C_{v+u-1}^p (-\lambda _3)^{p}e^{{\left( {p+1} \right) \left( {\lambda _3 +\lambda } \right) }/\gamma }} \int \limits _{\lambda _3 +\lambda }^{\lambda _3 +\lambda \left( {\theta _R +1} \right) } {\frac{e^{{-\left( {p+1} \right) x}/\gamma }}{x^{p+1}}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&\quad =u\lambda _3 \sum _{p=0}^{v+u-1} {C_{v+u-1}^p (-\lambda _3)^{p}e^{\frac{\left( {p+1} \right) \left( {\lambda _3 +\lambda } \right) }{\gamma }}}\nonumber \\&\qquad \times {\left( {\int \limits _{\lambda _3 +\lambda }^{\lambda _3 +\lambda \left( {\theta _R +1} \right) } {\frac{e^{\frac{-\left( {p+1} \right) x}{\gamma }}}{x^{p+2}}dx} +\frac{1}{\gamma }\int \limits _{\lambda _3 +\lambda }^{\lambda _3 +\lambda \left( {\theta _R +1} \right) } {\frac{e^{\frac{-\left( {p+1} \right) x}{\gamma }}}{x^{p+1}}dx}} \right) } \end{aligned}$$
(79)
Using the definition of the \(n\)th order exponential integral function [18, Eq. (1)], and after some manipulations, we obtain the probability \(\Pr \left[ {U_2 <V_2 <R} \right] \) as in (42) when \(\lambda _1 =\lambda _2~=~\lambda \).
Case 2: When \(\lambda _1 \ne \lambda _2\), (41) can be rewritten and changed as
$$\begin{aligned}&\Pr \left[ {U_2 <V_2 <R} \right] =u\lambda _3 \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\left[ {1-\frac{\lambda _3 e^{\frac{-\left( {x-\lambda _3 -\lambda _2 } \right) }{\gamma }}}{x}} \right] ^{u-1}\frac{e^{\frac{-\left( {x-\lambda _3 -\lambda _2} \right) }{\gamma }}}{x}} \left( \frac{1}{x}+\frac{1}{\gamma } \right) \nonumber \\&\qquad \times \left[ {1-\frac{\lambda _3 e^{\frac{-\lambda _1 \left( {x-\lambda _3 -\lambda _2} \right) }{\lambda _2 \gamma }}}{\lambda _3 +\frac{\lambda _1 (x-\lambda _3)}{\lambda _2}}} \right] ^{v}dx \nonumber \\&\quad =u\lambda _3 \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\sum _{p=0}^{u-1} {C_{u-1}^p \frac{(-\lambda _3)^{p}e^{\frac{-p\left( {x-\lambda _3 -\lambda _2} \right) }{\gamma }}}{x^{p}}} \frac{e^{\frac{-\left( {x-\lambda _3 -\lambda _2} \right) }{\gamma }}}{x}} \left( \frac{1}{x}+\frac{1}{\gamma } \right) \nonumber \\&\qquad \times \left[ {1-\frac{\lambda _3 \lambda _2 e^{\frac{\lambda _1 \left( {\lambda _3 +\lambda _2} \right) }{\lambda _2 \gamma }}e^{\frac{-\lambda _1 x}{\lambda _2 \gamma }}}{\lambda _1 x+\lambda _3 (\lambda _2 -\lambda _1)}} \right] ^{v}dx \nonumber \\&\quad =u\lambda _3 \sum _{p=0}^{u-1} {C_{u-1}^p (-\lambda _3 )^{p}e^{\frac{\left( {p+1} \right) \left( {\lambda _3 +\lambda _2} \right) }{\gamma }}}\nonumber \\&\qquad \times \left( {\begin{array}{l} \sum _{q=0}^v {C_v^q (-\lambda _3 \lambda _2)^{q}e^{\frac{q\lambda _1 \left( {\lambda _3 +\lambda _2} \right) }{\lambda _2 \gamma }}} \underbrace{\int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\frac{e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _1}{\lambda _2 \gamma }} \right) }}{\left[ {\lambda _1 x+\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{q}x^{p+2}}dx}}_{H_3} \\ +\frac{1}{\gamma }\sum _{q=0}^v {C_v^q (-\lambda _3 \lambda _2)^{q}e^{\frac{q\lambda _1 \left( {\lambda _3 +\lambda _2} \right) }{\lambda _2 \gamma }}\underbrace{\int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\frac{e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _1}{\lambda _2 \gamma }} \right) }}{\left[ {\lambda _1 x+\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{q}x^{p+1}}dx}}_{H_4}} \\ \end{array}} \right) \nonumber \\ \end{aligned}$$
(80)
Similar to solving the integrals \(H_{1}\) and \(H_{2}\), we obtain \(H_{3}\) and \(H_{4}\) as follows
$$\begin{aligned} H_3&= \left\{ {\begin{array}{ll} \displaystyle \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\frac{e^{-x\left( {\frac{p+1}{\gamma }} \right) }}{x^{p+2}}dx} &{},q=0 \\ \displaystyle \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _1}{\lambda _2 \gamma }} \right) }\left[ {\sum _{b=1}^q {\frac{m_b^3}{\left[ {\lambda _1 x+\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{b}}+\sum _{b=1}^{p+2} {\frac{n_b^3}{x^{b}}}}} \right] dx} &{},q\ge 1 \\ \end{array}} \right. \nonumber \\ \end{aligned}$$
(81)
$$\begin{aligned} H_4&= \left\{ {\begin{array}{ll} \displaystyle \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {\frac{e^{-x\left( {\frac{p+1}{\gamma }} \right) }}{x^{p+1}}dx} &{},q=0 \\ \displaystyle \int \limits _{\lambda _3 +\lambda _2}^{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) } {e^{-x\left( {\frac{p+1}{\gamma }+\frac{q\lambda _1}{\lambda _2 \gamma }} \right) }\left[ {\sum _{b=1}^q {\frac{m_b^4}{\left[ {\lambda _1 x+\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{b}}+\sum _{b=1}^{p+1} {\frac{n_b^4}{x^{b}}}}} \right] dx} &{},q\ge 1 \\ \end{array}} \right. \nonumber \\ \end{aligned}$$
(82)
where \(m_b^3 =\frac{\left( {-1} \right) ^{p+2}\prod _{c=1}^{q-b} {(p+1+c)\lambda _1^{p+2}}}{(q-b)!\left[ {\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{p+q+2-b}}, n_b^3 =\frac{\left( {-1} \right) ^{p+2-b}\prod _{c=1}^{p+2-b} {(q-1+c)\lambda _1^{p+2-b}} }{(p+2-b)!\left[ {\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{p+q+2-b}}\), \(m_b^4 =\frac{\left( {-1} \right) ^{p+1}\prod _{c=1}^{q-b} {(p+c)\lambda _1^{p+1}} }{(q-b)!\left[ {\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{p+q+1-b}}\), and \(n_b^4 =\frac{\left( {-1} \right) ^{p+1-b}\prod _{c=1}^{p+1-b} {(q-1+c)\lambda _1^{p+1-b}} }{(p+1-b)!\left[ {\lambda _3 (\lambda _2 -\lambda _1)} \right] ^{p+q+1-b}}\).
Substituting (81) and (82) into (80) and after some manipulations, we obtain the probability \(\Pr \left[ {U_2 <V_2 <R} \right] \) as in (43) when \(\lambda _1 \ne \lambda _2\).
Appendix 3: Solving Formula (52)
Formula (52) is written as
$$\begin{aligned} \Omega&= \frac{\partial \left\{ {\Pr [{ ASR }_{2b}^3 <R]-\Pr [{ ASR }_{2b}^3 <R,\min ({ ASR }_{1b}^3,{ ASR }_{2b}^3)>x]} \right\} }{\partial x} \nonumber \\&= -\frac{\partial \, \Pr [{ ASR }_{2b}^3 <R,\min ({ ASR }_{1b}^3,{ ASR }_{2b}^3)>x]}{\partial x} \end{aligned}$$
(83)
Substituting (25–29) into (83), we have (84) as
$$\begin{aligned} \Omega =-\frac{\partial \, \Pr \left[ \omega _{2b}^3 <\frac{\theta _R }{\gamma }+\left( {\theta _R +1} \right) \omega _{3b}, \omega _{1b}^3 >\frac{\theta _x}{\gamma }+\left( {\theta _x +1} \right) \omega _{3b}, \omega _{2b}^3 >\frac{\theta _x}{\gamma }+\left( {\theta _x +1} \right) \omega _{3b} \right] }{\partial x}\nonumber \\ \end{aligned}$$
(84)
When \(x>R\), then
$$\begin{aligned} \Omega =-\frac{\partial \left\{ 0 \right\} }{\partial x}=0 \end{aligned}$$
(85)
Next, we consider the case in which \(x\le R\), and then \(\Omega \) is obtained as
$$\begin{aligned} \Omega =-\frac{\partial \left\{ {\int \limits _0^\infty {f_{\omega _{3b}} (t)\left\{ {1-F_{\omega _{1b}^3} \left[ {\frac{\theta _x}{\gamma }+\left( {\theta _x +1} \right) t} \right] } \right\} \left\{ {F_{\omega _{2b}^3} \left[ {\frac{\theta _x}{\gamma }+\left( {\theta _x +1} \right) t} \right] -F_{\omega _{2b}^3} \left[ {\frac{\theta _R}{\gamma }+\left( {\theta _R +1} \right) t} \right] } \right\} } dt} \right\} }{\partial x}\nonumber \\ \end{aligned}$$
(86)
where \(f_{\omega _{3b}} (x), F_{\omega _{1b}^3} (x)\) and \(F_{\omega _{2b}^3} (x)\) are the PDF of the exponential random variable \(\omega _{3b}\) and the CDFs of the exponential random variables \(\omega _{1b}^3\) and \(\omega _{2b}^3\), respectively, and are given as
$$\begin{aligned} f_{\omega _{3b}} (x)&= \lambda _3 e^{-\lambda _3 x} \end{aligned}$$
(87)
$$\begin{aligned} F_{\omega _{1b}^3} (x)&= 1-e^{-\lambda _1 x} \end{aligned}$$
(88)
$$\begin{aligned} F_{\omega _{2b}^3} (x)&= 1-e^{-\lambda _2 x} \end{aligned}$$
(89)
Substituting (87–89) into (86), we obtain (90) as
$$\begin{aligned} \Omega =-\frac{\partial \left\{ {\frac{\lambda _3 e^{{-\left( {\lambda _1 +\lambda _2} \right) \theta _x}/\gamma }}{\lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }-\frac{\lambda _3 e^{{-\lambda _1 \theta _x}/\gamma {-\lambda _2 \theta _R}/\gamma }}{\lambda _3 +\lambda _1 \left( {\theta _x +1} \right) +\lambda _2 \left( {\theta _R +1} \right) }} \right\} }{\partial x} \end{aligned}$$
(90)
Substituting (9) into (90) and after some manipulations of the derivation, we obtain \(\Omega \) as in (53).
Appendix 4: Solving (54)
Formula (54) is separated into two parts, called PA and PB, as follows:
$$\begin{aligned} \Pr \left[ {\underbrace{{ ASR }_{2b}^3}_{R_b \in DS}(R_b)<R} \right] =PA{-}PB \end{aligned}$$
(91)
where
$$\begin{aligned} \hbox {P}A&= n\lambda _3 e^{{\left( {\lambda _1 +\lambda _2} \right) }/\gamma }\left( {\lambda _1 +\lambda _2} \right) \left( {\ln 8} \right) \int \limits _0^R {\frac{\left( {\theta _x +1} \right) e^{{-\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }/\gamma }}{\lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }}\nonumber \\&\quad \times \left[ {1-\frac{\lambda _3 e^{{-\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }/\gamma }}{\lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }} \right] ^{n-1}\left( {\frac{1}{\lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }+\frac{1}{\gamma }} \right) dx \end{aligned}$$
(92)
$$\begin{aligned} \hbox {P}B&= n\lambda _3 \lambda _1 e^{{\left( {\lambda _1 -\lambda _2 \theta _R} \right) }/\gamma }\left( {\ln 8} \right) \int \limits _0^R {\frac{\left( {\theta _x +1} \right) e^{{-\lambda _1 \left( {\theta _x +1} \right) }/\gamma }}{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) +\lambda _1 \left( {\theta _x +1} \right) }} \nonumber \\&\quad \times \left[ {1-\frac{\lambda _3 e^{{-\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }/\gamma }}{\lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _x +1} \right) }} \right] ^{n-1}\left( {\frac{1}{\lambda _3 +\lambda _2 \left( {\theta _R +1} \right) +\lambda _1 \left( {\theta _x \!+\!1} \right) }\!+\!\frac{1}{\gamma }} \right) dx\nonumber \\ \end{aligned}$$
(93)
By changing the variable of (92), part PA is obtained as
$$\begin{aligned} PA&= n\lambda _3 e^{\frac{\left( {\lambda _1 +\lambda _2} \right) }{\gamma }}\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-\left( {x-\lambda _3} \right) }{\gamma }}}{x}\left[ {1-\frac{\lambda _3 e^{\frac{-\left( {x-\lambda _1 -\lambda _2 -\lambda _3} \right) }{\gamma }}}{x}} \right] ^{n-1}\left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&= n\lambda _3 e^{\frac{\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-x}{\gamma }}}{x}\sum _{p=0}^{n-1} {\frac{C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{-p\left( {x-\lambda _1 -\lambda _2 -\lambda _3} \right) }{\gamma }}}{x^{p}}} \left( {\frac{1}{x}+\frac{1}{\gamma }} \right) dx} \nonumber \\&= n\lambda _3 e^{\frac{\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}\sum _{p=0}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}} \left\{ {\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-x\left( {p+1} \right) }{\gamma }}}{x^{p+2}}dx} +\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-x\left( {p+1} \right) }{\gamma }}}{\gamma x^{p+1}}dx}} \right\} \nonumber \\&= n\lambda _3 e^{\frac{\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}\sum _{p=0}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}} \left\{ G\left[ {\frac{p+1}{\gamma };a_1 ;a_2 ;p+2} \right] \right. \nonumber \\&\quad \left. +\frac{G\left[ {\frac{p+1}{\gamma };a_1 ;a_2 ;p+1} \right] }{\gamma } \right\} \end{aligned}$$
(94)
Next, by changing the variable of (92), part PB is rewritten as
$$\begin{aligned} PB&= n\lambda _3 \lambda _1 e^{{\left( {\lambda _1 -\lambda _2 \theta _R +\frac{\lambda _1 \lambda _3}{\lambda _1 +\lambda _2}} \right) }/\gamma }\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-\lambda _1 x}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\lambda _1 x+a_3}\left[ {1-\frac{\lambda _3 e^{\frac{-\left( {x-\lambda _1 -\lambda _2 -\lambda _3} \right) }{\gamma }}}{x}} \right] ^{n-1}\left( {\frac{\lambda _1 +\lambda _2}{\lambda _1 x+a_3}+\frac{1}{\gamma }} \right) dx} \nonumber \\&= n\lambda _3 \lambda _1 e^{{\left( {\lambda _1 -\lambda _2 \theta _R +\frac{\lambda _1 \lambda _3}{\lambda _1 +\lambda _2}} \right) }/\gamma }\int \limits _{a_1}^{a_2} {\sum _{p=0}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}} \frac{e^{{-x\left( {\frac{\lambda _1}{\lambda _1 +\lambda _2}+p} \right) }/\gamma }}{x^{p}\left( {\lambda _1 x+a_3} \right) }\left( {\frac{\lambda _1 +\lambda _2}{\lambda _1 x+a_3}+\frac{1}{\gamma }} \right) dx} \nonumber \\&= n\lambda _3 \lambda _1 e^{{\left( {\lambda _1 -\lambda _2 \theta _R +{\lambda _1 \lambda _3}/{\left( {\lambda _1 +\lambda _2} \right) }} \right) }/\gamma } \nonumber \\&\quad \times \left\{ {\begin{array}{l} \underbrace{\left( {\lambda _1 +\lambda _2} \right) \int \limits _{a_1}^{a_2} {\frac{e^{\frac{-\lambda _1 x}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\left( {\lambda _1 x+a_3} \right) ^{2}}dx}}_{H_5}+\underbrace{\left( {\lambda _1 +\lambda _2} \right) \sum _{p=1}^{n-1} {C_{k-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}} \int \limits _{a_1}^{a_2} {\frac{e^{{-x\left( {\frac{\lambda _1}{\lambda _1 +\lambda _2}+p} \right) }/\gamma }}{x^{p}\left( {\lambda _1 x+a_3} \right) ^{2}}dx}}_{H_6} \\ +\underbrace{\int \limits _{a_1}^{a_2} {\frac{e^{\frac{-\lambda _1 x}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\gamma \left( {\lambda _1 x+a_3} \right) }dx}}_{H_7}+\underbrace{\sum _{p=1}^{n-1} {C_{k-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}} \int \limits _{a_1}^{a_2} {\frac{e^{{-x\left( {\frac{\lambda _1}{\lambda _1 +\lambda _2}+p} \right) }/\gamma }}{\gamma x^{p}\left( {\lambda _1 x+a_3} \right) }dx}}_{H_8} \\ \end{array}} \right\} \nonumber \\ \end{aligned}$$
(95)
where the constants \(a_{1}, a_{2}\) and \(a_{3}\) are denoted as
$$\begin{aligned} a_1&= \lambda _1 +\lambda _2 +\lambda _3 \end{aligned}$$
(96)
$$\begin{aligned} a_2&= \lambda _3 +\left( {\lambda _1 +\lambda _2} \right) \left( {\theta _R +1} \right) \end{aligned}$$
(97)
$$\begin{aligned} a_3&= \lambda _2 \lambda _3 +\lambda _2 \left( {\lambda _1 +\lambda _2} \right) \left( {\theta _R +1} \right) \end{aligned}$$
(98)
In (95), part \(H_{5}\) can be calculated as
$$\begin{aligned} H_5&= \left( {\lambda _1 +\lambda _2} \right) \int \limits _{a_1}^{a_2} {\frac{e^{{-\lambda _1 x}/{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\left( {\lambda _1 x+a_3} \right) ^{2}}dx} =\frac{\left( {\lambda _1 +\lambda _2} \right) e^{\frac{a_3}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\lambda _1}\nonumber \\&\times \left\{ {\frac{E_2 \left[ {\frac{a_3 +\lambda _1 a_1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }} \right] }{a_3 +\lambda _1 a_1}-\frac{E_2 \left[ {\frac{a_3 +\lambda _1 a_2}{\gamma \left( {\lambda _1 +\lambda _2} \right) }} \right] }{a_3 +\lambda _1 a_2}} \right\} \nonumber \\&= \frac{\left( {\lambda _1 +\lambda _2} \right) e^{\frac{a_3}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\lambda _1}G\left[ {\frac{1}{\gamma \left( {\lambda _1 +\lambda _2} \right) };a_3 +\lambda _1 a_1 ;a_3 +\lambda _1 a_2 ;2} \right] \end{aligned}$$
(99)
Similar to \(H_{5}\), part \(H_{7}\) is given as
$$\begin{aligned} H_7&= \int \limits _{a_1}^{a_2} {\frac{e^{{-\lambda _1 x}/{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\gamma \left( {\lambda _1 x+a_3} \right) }dx} \nonumber \\&= \frac{e^{\frac{a_3}{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\gamma \lambda _1}\left\{ {E_1 \left[ {\frac{a_3 +\lambda _1 a_1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }} \right] -E_1 \left[ {\frac{a_3 +\lambda _1 a_2}{\gamma \left( {\lambda _1 +\lambda _2} \right) }} \right] } \right\} \nonumber \\&= \frac{e^{{a_3}/{\gamma \left( {\lambda _1 +\lambda _2} \right) }}}{\gamma \lambda _1}G\left[ {\frac{1}{\gamma \left( {\lambda _1 +\lambda _2} \right) };a_3 +\lambda _1 a_1 ;a_3 +\lambda _1 a_2 ;1} \right] \end{aligned}$$
(100)
Using the partial expansion of the denominator the same as in integral \(H_{1}\), part \(H_{6}\) is calculated as
$$\begin{aligned} H_6&= \left( {\lambda _1 +\lambda _2} \right) \sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}\int \limits _{a_1}^{a_2} {e^{\frac{-x\left( {\frac{\lambda _1}{\lambda _1 +\lambda _2}+p} \right) }{\gamma }}}}\nonumber \\&\times {{\left[ {\sum _{b=1}^2 {\frac{m_b^5}{\left[ {\lambda _1 x+a_3)} \right] ^{b}}+\sum _{b=1}^p {\frac{n_b^5}{x^{b}}}}} \right] dx}} \nonumber \\&= \left( {\lambda _1 +\lambda _2} \right) \sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }/\gamma }} \nonumber \\&\quad \times \left\{ {\begin{array}{l} \sum _{b=1}^2 {\frac{m_b^5 e^{\frac{a_3 \left[ {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right] }{\gamma \lambda _1 }}}{\lambda _1}\left\{ {\begin{array}{l} \frac{E_b \left[ {\frac{\left( {a_3 +\lambda _1 a_1} \right) \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma \lambda _1}} \right] }{\left( {a_3 +\lambda _1 a_1} \right) ^{b-1}} \\ -\frac{E_b \left[ {\frac{\left( {a_3 +\lambda _1 a_2} \right) \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma \lambda _1}} \right] }{\left( {a_3 +\lambda _1 a_2} \right) ^{b-1}} \\ \end{array}} \right\} } \\ +\sum _{b=1}^p {n_b^5 \left\{ {\frac{E_b \left[ {\frac{a_1 \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma }} \right] }{a_1^{b-1}}-\frac{E_b \left[ {\frac{a_2 \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma }} \right] }{a_2^{b-1}}} \right\} } \\ \end{array}} \right\} \nonumber \\&= \left( {\lambda _1 +\lambda _2} \right) \sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }/\gamma }}\nonumber \\&\quad \times \left\{ {\begin{array}{l} \sum _{b=1}^2 {\frac{m_b^5 e^{\frac{a_3 \left[ {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right] }{\gamma \lambda _1}}}{\lambda _1}G\left[ {\frac{1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }+\frac{p}{\gamma \lambda _1};a_3 +\lambda _1 a_1 ;a_3 +\lambda _1 a_2 ;b} \right] } \\ +\sum _{b=1}^p {n_b^5 G\left[ {\frac{\lambda _1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }+\frac{p}{\gamma };a_1 ;a_2 ;b} \right] } \\ \end{array}} \right\} \nonumber \\ \end{aligned}$$
(101)
where \(m_b^5 =\frac{\left( {-1} \right) ^{p}\prod _{c=1}^{2-b} {(p-1+c)\lambda _1^p}}{(2-b)!a_3^{p+2-b}}\) and \(n_b^5 =\frac{\left( {-1} \right) ^{p-b}\prod _{c=1}^{p-b} {(1+c)\lambda _1^{p-b}} }{(p-b)!a_3^{p+2-b}}\).
Similar to \(H_{6}\), part \(H_{8}\) can be calculated as
$$\begin{aligned} H_8&= \frac{1}{\gamma }\sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{\frac{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }{\gamma }}\int \limits _{a_1}^{a_2} {e^{\frac{-x\left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) }+p} \right) }{\gamma }}\left[ {\frac{m_1^6}{\lambda _1 x+a_3}+\sum _{b=1}^p {\frac{n_b^6}{x^{b}}}} \right] dx}} \nonumber \\&= \frac{1}{\gamma }\sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }/\gamma }} \nonumber \\&\quad \times \left\{ {\begin{array}{l} \frac{m_1^6 e^{\frac{a_3 \left[ {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right] }{\gamma \lambda _1}}}{\lambda _1}\left\{ {\begin{array}{l} E_1 \left[ {\frac{\left( {a_3 +\lambda _1 a_1} \right) \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma \lambda _1}} \right] \\ -E_1 \left[ {\frac{\left( {a_3 +\lambda _1 a_2} \right) \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma \lambda _1}} \right] \\ \end{array}} \right\} \\ +\sum _{b=1}^p {n_b^6 \left\{ {\frac{E_b \left[ {\frac{a_1 \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma }} \right] }{a_1^{b-1}}-\frac{E_b \left[ {\frac{a_2 \left( {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right) }{\gamma }} \right] }{a_2^{b-1}}} \right\} } \\ \end{array}} \right\} \nonumber \\&= \frac{1}{\gamma }\sum _{p=1}^{n-1} {C_{n-1}^p \left( {-\lambda _3} \right) ^{p}e^{{p\left( {\lambda _1 +\lambda _2 +\lambda _3} \right) }/\gamma }} \nonumber \\&\quad \times \left\{ {\begin{array}{l} \frac{m_1^6 e^{\frac{a_3 \left[ {{\lambda _1}/{\left( {\lambda _1 +\lambda _2} \right) +p}} \right] }{\gamma \lambda _1}}}{\lambda _1}G\left[ {\frac{1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }+\frac{p}{\gamma \lambda _1};a_3 +\lambda _1 a_1 ;a_3 +\lambda _1 a_2 ;1} \right] \\ +\sum _{b=1}^p {n_b^6 G\left[ {\frac{\lambda _1}{\gamma \left( {\lambda _1 +\lambda _2} \right) }+\frac{p}{\gamma };a_3 +\lambda _1 a_1 ;a_3 +\lambda _1 a_2 ;b} \right] } \\ \end{array}} \right\} \end{aligned}$$
(102)
where \(m_1^6 =\left( {-\frac{\lambda _1}{a_3}} \right) ^{p}\) and \(n_b^6 =\frac{\left( {-\lambda _1} \right) ^{p-b}}{a_3^{p+1-b}}\).
Substituting (99–102) into (95), we have part PB, and then combining parts PA and PB based on (91), we obtain \(\Pr \left[ {\underbrace{{ ASR }_{2b}^3}_{R_b \in DS}(R_b)<R} \right] \) as in (55).
