Appendix 1: proof of Proposition 1
In fixed DTR approach, the secondary user periodically keeps sensing for fixed detection duration D and then keeps transmitting for fixed transmission duration T in the duration of idle period.
The expectation access time of the (n + 1)th transmission duration (i.e., \( \left[ {\left( {n + 1} \right)D + nT,\left( {n + 1} \right)\left( {D + T} \right)} \right] \)) can be given as:
$$ \int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {\left[ {t - \left( {n + 1} \right)D - nT} \right]f\left( t \right)dt} + T\int_{{\left( {n + 1} \right)\left( {D + T} \right)}}^{\infty } {f\left( t \right)dt} . $$
(31)
Then, the expectation access time of the first M transmission durations can be obtained as follows:
$$ \sum\limits_{n = 0}^M {\left\{ {\int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {\left[ {t - \left( {n + 1} \right)D - nT} \right]f\left( t \right)dt} + T\int_{{\left( {n + 1} \right)\left( {D + T} \right)}}^{\infty } {f\left( t \right)dt} } \right\}} $$
(32)
and the expectation idle time of the first M detection and transmission durations can be given as:
$$ \sum\limits_{n = 0}^M {\left\{ {\int_{{n\left( {D + T} \right)}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {tf\left( t \right)dt} } \right\}} . $$
(33)
Thus, according to the definition of spectrum holes utilization, we can obtain spectrum holes utilization using fixed DTR approach as follows:
$$ \begin{array}{*{20}c} {\eta_{\text{SH}} = \mathop {\lim }\limits_{{M \to \infty }} \frac{{\sum\limits_{n = 0}^M {\left\{ {\int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {\left[ {t - \left( {n + 1} \right)D - nT} \right]f\left( t \right)dt} } \right\}} }}{{\sum\limits_{n = 0}^M {\left( {\int_{{n\left( {D + T} \right)}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {tf\left( t \right)dt} } \right)} }}} \\ { + \mathop {\lim }\limits_{{M \to \infty }} \frac{{\sum\limits_{n = 0}^M {\left\{ {T\int_{{\left( {n + 1} \right)\left( {D + T} \right)}}^{\infty } {f\left( t \right)dt} } \right\}} }}{{\sum\limits_{n = 0}^M {\left( {\int_{{n\left( {D + T} \right)}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {tf\left( t \right)dt} } \right)} }}} \\ { = \mathop {\lim }\limits_{{M \to \infty }} \frac{{\sum\limits_{n = 0}^M {\left\{ {\int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {tue^{- ut} dt + \int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {\left[ {\left( {n + 1} \right)D + nT} \right]ue^{- ut} dt} } } \right\}} }}{{\sum\limits_{n = 0}^M {\left( {\int_{{n\left( {D + T} \right)}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {ute^{- ut} dt} } \right)} }}} \\ { + \mathop {\lim }\limits_{{M \to \infty }} \frac{{\sum\limits_{n = 0}^M {\left\{ {T\int_{{\left( {n + 1} \right)\left( {D + T} \right)}}^{\infty } {ue^{- ut} dt} } \right\}} }}{{\sum\limits_{n = 0}^M {\left( {\int_{{n\left( {D + T} \right)}}^{{\left( {n + 1} \right)\left( {D + T} \right)}} {ute^{- ut} dt} } \right)} }}} \\ { = u\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\frac{{\partial \left[ {e^{{ - u\left( {n + 1} \right)\left( {D + T} \right)}} - e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} } \right]}}{{\partial u}}} } \\ { + u\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\left[ {e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} - e^{{ - u\left( {n + 1} \right)\left( {D + T} \right)}} } \right]} } \\ { + u\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\left\{ {e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} \frac{{\partial \left( {e^{- uT} - 1} \right)}}{{\partial u}} + \left( {e^{- uT} - 1} \right)\frac{{\partial e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} }}{{\partial u}}} \right\}} } \\ { = u\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\left[ {e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} - e^{{ - u\left( {n + 1} \right)\left( {D + T} \right)}} } \right]} } \\ { = \frac{{e^{- uD} - e^{{ - u\left( {D + T} \right)}} }}{{1 - e^{{ - u\left( {D + T} \right)}} }} = \frac{{e^{uT} - 1}}{{e^{{u\left( {D + T} \right)}} - 1}}} \\ \end{array} . $$
(34)
We know that the probability that spectrum hole terminates in \( \left[ {\left( {n + 1} \right)D + nT,\left( {n + 1} \right)D + \left( {n + 1} \right)T} \right] \) is
$$ P\left\{ {\left( {n + 1} \right)D + nT \le t \le \left( {n + 1} \right)\left( {D + T} \right)} \right\}. $$
(35)
Thus, we can obtain that the collision probability is:
$$ \begin{array}{*{20}c} {P^c = P_{{\max }}^c \mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 1}^M {P\left\{ {\left( {n + 1} \right)D + nT \le t \le \left( {n + 1} \right)\left( {D + T} \right)} \right\}} } \hfill \\ { = v\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\int_{{\left( {n + 1} \right)D + nT}}^{{\left( {n + 1} \right)D + \left( {n + 1} \right)T}} {ue^{- ut} dt} } } \hfill \\ { = v\mathop {\lim }\limits_{{M \to \infty }} \sum\limits_{n = 0}^M {\left( {e^{{ - u\left[ {\left( {n + 1} \right)D + nT} \right]}} - e^{{ - u\left[ {\left( {n + 1} \right)D + \left( {n + 1} \right)T} \right]}} } \right)} } \hfill \\ { = v\left[ {e^{- uD} - e^{{ - u\left( {D + T} \right)}} } \right]\mathop {\lim }\limits_{{M \to \infty }} \left[ {1 + e^{{ - u\left( {D + T} \right)}} + \cdots e^{{ - uM\left( {D + T} \right)}} } \right]} \hfill \\ { = v\frac{{e^{- uD} - e^{{ - u\left( {D + T} \right)}} }}{{1 - e^{{ - u\left( {D + T} \right)}} }}} \hfill \\ { = v\frac{{e^{uT} - 1}}{{e^{{u\left( {D + T} \right)}} - 1}}} \hfill \\ \end{array} . $$
(36)
Appendix 2: proof of Proposition 2
The probability density function of exponential distribution is given by Eq. 1.
-
1.
\( \varsigma \ge P_{{\max }}^c \)
Now, the secondary user can start to transmit at the first time slot in the idle period and keep transmitting until collision occurs. That is to say, the optimal access time slot is the first time slot and available transmission duration is limitless, i.e.,
$$ x^{\text{opt}} = 0{\text{ and }}T^a = \infty . $$
(37)
Therefore, according to Eq. 37, we can obtain that
$$ \eta_{{{\text{SH}},\max }} = {{E\left( {0,\infty } \right)} \mathord{\left/ {\vphantom {{E\left( {0,\infty } \right)} {E\left( {0,\infty } \right)}}} \right. } {E\left( {0,\infty } \right)}} = 1. $$
-
2.
\( \varsigma < P_{{\max }}^c \)
According to Eq. 14, we can obtain:
$$ \begin{array}{*{20}c} {P^C = P_{{\max }}^C \int_x^{x + T} {f\left( t \right)dt} \le \varsigma } \hfill \\ { \Rightarrow v\int_x^{x + T} {ue^{- ut} dt} \le \varsigma } \hfill \\ { \Rightarrow e^{- ux} - e^{{ - u\left( {x + T} \right)}} \le {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { \Rightarrow T \le {{ - \ln \left( {1 - {{\varsigma e^{ux} } \mathord{\left/ {\vphantom {{\varsigma e^{ux} } v}} \right. } v}} \right)} \mathord{\left/ {\vphantom {{ - \ln \left( {1 - {{\varsigma e^{ux} } \mathord{\left/ {\vphantom {{\varsigma e^{ux} } v}} \right. } v}} \right)} u}} \right. } u}} \hfill \\ \end{array} $$
(38)
According to Eq. 37, we can obtain:
$$ T^a = \max \left\{ T \right\} = {{ - 1} \mathord{\left/ {\vphantom {{ - 1} u}} \right. } u}\ln \left( {e^{- ux} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right) - x. $$
(39)
The transmission duration expectation of exponential distribution can be given as:
$$ \begin{array}{*{20}c} {E\left( {x,T} \right) = \int_x^{x + T} {\left( {t - x} \right)} f\left( t \right)dt + T\int_{x + T}^{\infty } {f\left( t \right)dt} } \hfill \\ { = \int_x^{x + T} {ute^{- ut} dt} - x\int_x^{x + T} {ue^{- ut} dt} + T\int_{x + T}^{\infty } {ue^{- ut} dt} } \hfill \\ { = \frac{1}{u}e^{- ux} - Te^{{ - u\left( {x + T} \right)}} - \frac{1}{u}e^{{ - u\left( {x + T} \right)}} + Te^{{ - u\left( {x + T} \right)}} } \hfill \\ { = \frac{1}{u}e^{- ux} - \frac{1}{u}e^{{ - u\left( {x + T} \right)}} } \hfill \\ { \le \frac{\varsigma }{uv}} \hfill \\ \end{array} . $$
(40)
It can easily obtained that
$$ E\left( {x,T} \right)_{{\max }} = \frac{\varsigma }{uv}\,{\text{when}}\;\,T = T^a \;{\text{and}}\;x \le {1 \mathord{\left/ {\vphantom {1 u}} \right. } u}\ln \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right). $$
(41)
It can be seen from Eq. 41 that E(x, T)max is constant when the channel-usage and collision tolerable level are certain.
$$ \begin{array}{*{20}c} {E\left( {0,\infty } \right) = \int_0^{\infty } {f\left( t \right)dt} } \hfill \\ { = \int_0^{\infty } {ue^{- ut} dt} = {1 \mathord{\left/ {\vphantom {1 u}} \right. } u}} \hfill \\ \end{array} $$
(42)
According to Eqs. 40 and 41, we can obtain that x
opt is arbitrary in \( \left[ {0,{1 \mathord{\left/ {\vphantom {1 u}} \right. } u}\ln \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)} \right] \) and the maximum spectrum holes utilization is:
$$ \eta_{{{\text{SH}},\max }} = {{E\left( {x,T} \right)_{{\max }} } \mathord{\left/ {\vphantom {{E\left( {x,T} \right)_{{\max }} } {E\left( {0,\infty } \right)}}} \right. } {E\left( {0,\infty } \right)}} = {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}. $$
Appendix 3: proof of Proposition 3
The probability density function of Pareto distribution is:
$$ f\left( {\left. t \right|k,\sigma } \right) = k\sigma^k t^{- 1 - k} \;\left( {t \ge \sigma } \right). $$
(43)
We analyze Proposition 3 in the following two cases:
-
1.
\( \varsigma \ge P_{{\max }}^c . \)
Now, the secondary user can start to transmit at the first time slot in the idle period and keep transmitting until collision occurs. That is to say, the optimal access time slot is the first time slot and available transmission duration is limitless, i.e.,
$$ \begin{array}{*{20}c} {x^{\text{opt}} = 0} \hfill \\ {T^a = \infty } \hfill \\ \end{array} . $$
(44)
Therefore, according to Eq. 43, we can obtain that
$$ \eta_{{{\text{SH}},\max }} = {{E\left( {0,\infty } \right)} \mathord{\left/ {\vphantom {{E\left( {0,\infty } \right)} {E\left( {0,\infty } \right)}}} \right. } {E\left( {0,\infty } \right)}} = 1. $$
-
2.
\( \varsigma < P_{{\max }}^c \)
According to Eq. 14, we can obtain:
$$ \begin{array}{*{20}c} {P^C = P_{{\max }}^C \int_x^{x + T} {f\left( t \right)dt} \le \varsigma } \hfill \\ { \Rightarrow \sigma^k x^{- k} - \sigma^k \left( {x + T} \right)^{- k} \le {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { \Rightarrow T \le \left( {x^{- k} - \frac{\varsigma }{{v\sigma^k }}} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - x} \hfill \\ \end{array} . $$
(45)
According to Eq. 44, we can obtain:
$$ T^a = \max \left\{ T \right\} = \left( {x^{- k} - \frac{\varsigma }{{v\sigma^k }}} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - x, $$
(46)
and P
c = ς when T = T
a, i.e.,
$$ \int_x^{{x + T^a }} {f\left( t \right)dt} = {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v} $$
(47)
We compute partial derivative of both sides of Eq. 46 and can obtain:
$$ \left( {\frac{{dT^a }}{dx} + 1} \right)f\left( {x + T^a } \right) = f\left( x \right). $$
(48)
Then, we compute partial derivative of transmission duration expectation with respect to x as follows:
$$ \begin{array}{*{20}c} {\frac{{\partial E\left( {x,T^a } \right)}}{{\partial x}}} \hfill \\ { = \frac{\partial }{{\partial x}}\left[ {\int_x^{{x + T^a }} {tf\left( t \right)dt} - x\int_x^{{x + T^a }} {f\left( t \right)dt} + T^a \int_{{x + T^a }}^{\infty } {f\left( t \right)dt} } \right]} \hfill \\ { = \frac{\partial }{{\partial x}}\left[ {\int_x^{{x + T^a }} {tf\left( t \right)dt} - {{x\varsigma } \mathord{\left/ {\vphantom {{x\varsigma } v}} \right. } v} + T^a \int_{{x + T^a }}^{\infty } {f\left( t \right)dt} } \right]\,\;\left( {{\text{see}}\;{\text{Eq}}.\,47} \right)} \hfill \\ { = \left( {1 + \frac{{dT^a }}{dx}} \right)\left( {x + T^a } \right)f\left( {x + T^a } \right) - xf\left( x \right) + \frac{{dT^a }}{dx}\int_{x + T}^{\infty } {f\left( t \right)dt} } \hfill \\ { - T^a \left( {1 + \frac{{dT^a }}{dx}} \right)f\left( {x + T^a } \right) - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = x\left( {1 + \frac{{dT^a }}{dx}} \right)f\left( {x + T^a } \right) - xf\left( x \right) + \frac{{dT^a }}{dx}\int_{{x + T^a }}^{\infty } {f\left( t \right)dt} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \frac{{dT^a }}{dx}\int_{{x + T^a }}^{\infty } {f\left( t \right)dt} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}\,\left( {{\text{see}}\;{\text{Eq}}{\text{. 48}}} \right)} \hfill \\ \end{array} $$
(49)
Note that the probability density function is variable in Eq. 49. Then, the partial derivative of transmission duration expectation of Pareto distribution with respect to x can be given as:
$$ \begin{array}{*{20}c} {\frac{{\partial E\left( {x,T^a } \right)}}{{\partial x}} = \frac{{dT^a }}{dx}\int_{{x + T^a }}^{\infty } {f\left( t \right)dt} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \left[ {\left( {\frac{x}{{x + T^a }}} \right)^{- 1 - k} - 1} \right]\sigma^k \left( {x + T^a } \right)^{- k} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \sigma^k \left[ {T^a x^{- 1 - k} + x^{- k} - \left( {x + T^a } \right)^{- k} } \right] - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \sigma^k T^a x^{- 1 - k} } \hfill \\ { > 0\,\left( {k > 0} \right)} \hfill \\ \end{array} $$
(50)
Thus, E(x, T) monotonically increase with the increase of x when \( x < \sigma \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)^{{{1 \mathord{\left/ {\vphantom {1 k}} \right. } k}}} \).
On the other hand, transmission duration is limitless, i.e., T = ∞ when \( x > \sigma \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)^{{{1 \mathord{\left/ {\vphantom {1 k}} \right. } k}}} \), and we have
$$ \begin{array}{*{20}c} {E\left( {x,\infty } \right) = \int_x^{\infty } {\left( {t - x} \right)} f\left( t \right)dt} \hfill \\ { = \int_x^{\infty } {k\sigma t^{- k} dt} - x\int_x^{\infty } {k\sigma t^{- 1 - k} dt} } \hfill \\ { = - \sigma x^{1 - k} - \frac{{k\sigma }}{1 - k}x^{1 - k} } \hfill \\ { = \frac{\sigma }{k - 1}x^{1 - k} \left( {k > 1} \right)} \hfill \\ \end{array} . $$
(51)
Thus, E(x, ∞) monotonically decrease with the increase of x for \( x > \sigma \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)^{{{1 \mathord{\left/ {\vphantom {1 k}} \right. } k}}} \).
Moreover,
$$ E\left( {x,T} \right) = \frac{{\sigma^{2 - k} }}{k - 1}\left( {\frac{v}{\varsigma }} \right)^{{{{\left( {1 - k} \right)} \mathord{\left/ {\vphantom {{\left( {1 - k} \right)} k}} \right. } k}}} \,{\text{for}}\,x = \sigma \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)^{{{1 \mathord{\left/ {\vphantom {1 k}} \right. } k}}} $$
(52)
Therefore, we can obtain that maximum transmission duration expectation is:
$$ E\left( {x,T} \right)_{{\max }} = \frac{{\sigma^{2 - k} }}{k - 1}\left( {\frac{v}{\varsigma }} \right)^{{{{\left( {1 - k} \right)} \mathord{\left/ {\vphantom {{\left( {1 - k} \right)} k}} \right. } k}}}, $$
(53)
$$ \begin{array}{*{20}c} {E\left( {0,\infty } \right) = \int_0^{\infty } t f\left( t \right)dt} \hfill \\ { = \int_0^{\infty } {k\sigma^k t^{- k} dt} } \hfill \\ { = \frac{{k\sigma }}{k - 1}\,\left( {k > 1} \right)} \hfill \\ \end{array} . $$
(54)
According to Eqs. 53 and 54, we can obtain that the maximum spectrum holes utilization is:
$$ \eta_{{{\text{SH}},\max }} = \frac{{E\left( {x,T} \right)_{{\max }} }}{{E\left( {0,\infty } \right)}} = \frac{{\sigma^{1 - k} }}{k}\left( {\frac{v}{\varsigma }} \right)^{{{{\left( {1 - k} \right)} \mathord{\left/ {\vphantom {{\left( {1 - k} \right)} k}} \right. } k}}}, $$
(55)
where
$$ x = x^{\text{opt}} = \sigma \left( {{v \mathord{\left/ {\vphantom {v \varsigma }} \right. } \varsigma }} \right)^{{{1 \mathord{\left/ {\vphantom {1 k}} \right. } k}}} . $$
Appendix 4: proof of Proposition 4
The probability density function of generalized Pareto distribution is
$$ f\left( {t;k,\sigma } \right) = \frac{1}{\sigma }\left( {1 + k\frac{t}{\sigma }} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} . $$
(56)
We analyze Proposition 4 in the following two cases:
-
1.
\( \zeta \ge P_{{\max }}^c \)
Now, the secondary user can start to transmit at the first time slot in the idle period and keep transmitting until collision occurs. That is to say, the optimal access time slot is the first time slot and available transmission duration is limitless, i.e.,
$$ x^{\text{opt}} = 0{\text{ and }}T^a = \infty . $$
(57)
Therefore, according to Eq. 57, we can obtain that
$$ \eta_{{{\text{SH}},\max }} = {{E\left( {0,\infty } \right)} \mathord{\left/ {\vphantom {{E\left( {0,\infty } \right)} {E\left( {0,\infty } \right)}}} \right. } {E\left( {0,\infty } \right)}} = 1. $$
-
2.
\( \zeta < P_{{\max }}^c \)
According to Eq. 14, we can obtain:
$$ \begin{array}{*{20}c} {P^C = P_{{\max }}^C \int_x^{x + T} {f\left( t \right)dt} \le \varsigma } \hfill \\ { \Rightarrow \left( {1 + k\frac{x}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - \left( {1 + k\frac{x + T}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} \le {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { \Rightarrow T \le \frac{\sigma }{k}\left[ {\left( {\left( {1 + k\frac{x}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} {{ - \varsigma } \mathord{\left/ {\vphantom {{ - \varsigma } v}} \right. } v}} \right)^{- k} - 1} \right] - x} \hfill \\ \end{array} \;. $$
(58)
According to Eq. 58, we can obtain:
$$ T^a = \max \left\{ T \right\} = \frac{\sigma }{k}\left[ {\left( {\left( {1 + k\frac{x}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} {{ - \varsigma } \mathord{\left/ {\vphantom {{ - \varsigma } v}} \right. } v}} \right)^{- k} - 1} \right] - x\;. $$
(59)
According to Eq. 49, the partial derivative of transmission duration expectation of generalized Pareto distribution with respect to x can be given as:
$$ \begin{array}{*{20}c} {\frac{{\partial E\left( {x,T^a } \right)}}{{\partial x}} = \frac{{dT^a }}{dx}\int_{{x + T^a }}^{\infty } {f\left( t \right)dt} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \left[ {\left( {\frac{{1 + k\frac{x}{\sigma }}}{{1 + k\frac{{x + T^a }}{\sigma }}}} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} - 1} \right]\left( {1 + k\frac{{x + T^a }}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \left[ {\left( {1 + \frac{{{{kT^a } \mathord{\left/ {\vphantom {{kT^a } \sigma }} \right. } \sigma }}}{{1 + {{kx} \mathord{\left/ {\vphantom {{kx} \sigma }} \right. } \sigma }}}} \right)\left( {1 + k\frac{x}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - \left( {1 + k\frac{{x + T^a }}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} } \right] - {\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \hfill \\ { = \left( {\frac{{kT^a }}{\sigma }} \right)\left( {1 + k\frac{x}{\sigma }} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} \ge 0\;\quad \left( {{\text{see}}\;{\text{Eq}}.\;58} \right)} \hfill \\ \end{array} $$
(60)
Therefore, we can obtain that E(x, T) is constant and is identical with exponential distribution if k = 0, and E(x, T) monotonically increase with the increase of x for
$$ 0 \le x < \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right],\,{\text{if}}\,k > 0. $$
On the other hand, transmission duration is limitless, i.e., T = ∞ when \( x > \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right] \), and we have
$$ \begin{array}{*{20}c} {E\left( {x,\infty } \right) = \int_x^{\infty } {\left( {t - x} \right)} f\left( t \right)dt} \hfill \\ { = \int_x^{\infty } {\frac{t}{\sigma }\left( {1 + \frac{kt}{\sigma }} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} dt} - x\int_x^{\infty } {\frac{1}{\sigma }\left( {1 + \frac{kt}{\sigma }} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} dt} } \hfill \\ { = x\left( {1 + \frac{kx}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} - \frac{\sigma }{k - 1}\left( {1 + \frac{kx}{\sigma }} \right)^{{{{1 - 1} \mathord{\left/ {\vphantom {{1 - 1} k}} \right. } k}}} - x\left( {1 + \frac{kx}{\sigma }} \right)^{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} k}} \right. } k}}} } \hfill \\ { = \frac{\sigma }{1 - k}\left( {1 + \frac{kx}{\sigma }} \right)^{{{{1 - 1} \mathord{\left/ {\vphantom {{1 - 1} k}} \right. } k}}} } \hfill \\ \end{array} $$
Thus, E(x, ∞) monotonically decrease with the increase of x for
$$ x > \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right]. $$
Moreover,
$$ E\left( {x,\infty } \right) = \frac{\sigma }{1 - k}\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{1 - k} . $$
where
$$ x = \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right]. $$
Therefore, we can obtain that the optimal access time slot is
$$ x^{\text{opt}} = \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right] $$
and the maximum transmission duration expectation is:
$$ E\left( {x,T} \right)_{{\max }} = E\left( {x,\infty } \right) = \frac{\sigma }{1 - k}\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{1 - k} . $$
(61)
$$ \begin{array}{*{20}c} {E\left( {0,\infty } \right) = \int_0^{\infty } t f(t)dt} \hfill \\ { = \int_0^{\infty } {\frac{t}{\sigma }\left( {1 + k\frac{t}{\sigma }} \right)^{{{{ - 1 - 1} \mathord{\left/ {\vphantom {{ - 1 - 1} k}} \right. } k}}} dt} } \hfill \\ { = \frac{\sigma }{1 - k}\,\left( {0 \le k < 1} \right)} \hfill \\ \end{array} $$
(62)
According to Eqs. 61 and 62, we can obtain that the maximum spectrum holes utilization is
$$ \eta_{{{\text{SH}},\max }} = {{E\left( {x,T} \right)_{{\max }} } \mathord{\left/ {\vphantom {{E\left( {x,T} \right)_{{\max }} } {E\left( {0,\infty } \right)}}} \right. } {E\left( {0,\infty } \right)}} = \left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{1 - k}, $$
where
$$ x = x^{\text{opt}} = \frac{\sigma }{k}\left[ {\left( {{\varsigma \mathord{\left/ {\vphantom {\varsigma v}} \right. } v}} \right)^{- k} - 1} \right]. $$
