Analysis of energy saving in user equipment in LTE-A using stochastic modelling

Abstract

Energy saving in User Equipment (UE) is one of the important issues for limited sources of power in the device. It is critical for the UE to maximize its energy efficiency. In this paper, we have presented two stochastic models, namely the Markov model and semi-Markov model, for the UE based on the states of discontinuous reception (DRX) mechanism, i.e., a power saving method in mobile communication networks. Explicit expressions are derived for transient and steady-state system size probabilities for the Markov model. For the semi-Markov model, steady-state probabilities are computed. Further, the performance measures such as mean and variance are computed for both models. Using these models, based on the states of DRX mechanism, energy saving in the UE is calculated. Finally, sensitivity analysis is performed in which the results obtained are compared for both models. Numerical results obtained in this paper ensure that energy saving can be maximized in the UE using the Markov modelling of DRX mechanism rather than semi-Markov modelling. The energy saving using the Markov model is atleast 33.19$\%$ more than the semi-Markov model. Also, for energy saving in the UE, the semi-Markov model for DRX mechanism is compared with the Markov model. The semi-Markov models for the DRX mechanism are available in the literature without considering the packet arrivals. Our analysis of DRX mechanism and conclusion on its performance can be designed and implemented to an extension for the existing DRX mechanism. We believe that, these models can also be extended to study the energy saving of hardware and other components of the system.

Appendices

Appendix

Appendix 1

Let $f_{i,j}(s)$ represents the Laplace transform of $P_{i,j}(t)$. Taking Laplace transform of (4), we get

$$\begin{aligned} (s+ \alpha )f_{0,1}(s) = \beta f_{0,0}(s). \end{aligned}$$

(15)

On the inversion of the above equation, we have

$$\begin{aligned} P_{0,1}(t) = \beta e^{-\alpha t}* P_{0,0}(t), \end{aligned}$$

where ’$*$’ represents the convolution of functions. Taking Laplace transform on Equation (5), we have

$$\begin{aligned} f_{0,j}(s) = \frac{ (\alpha - \lambda )}{(s+ \alpha )}f_{0,j-1}(s),~ j = 2,3,\ldots ,N \end{aligned}$$

which recursively yields,

$$\begin{aligned} f_{0,j}(s) = \frac{ (\alpha - \lambda )^{j-1}}{(s+ \alpha )^{j-1}}f_{0,1}(s), ~j = 2,3,\ldots ,N. \end{aligned}$$

Using equation (15) in above equation, we obtain

$$\begin{aligned} f_{0,j}(s) = \frac{\beta (\alpha - \lambda )^{j-1}}{(s+ \alpha )^{j}}f_{0,0}(s), ~j = 1,2,\ldots ,N. \end{aligned}$$

(16)

On inversion of equation (16), we obtain the required transient probabilities $P_{0,j}(t)$, $j = 1,2,\ldots ,N.$ Now, taking Laplace transform on (6), we get

$$\begin{aligned} f_{0,N+1}(s) = \frac{(\alpha - \lambda )}{(s+\nu )}f_{0,N}(s). \end{aligned}$$

Using (16) for $j = N$ in above equation, we obtain

$$\begin{aligned} f_{0,N+1}(s) = \frac{\beta (\alpha - \lambda )^{N}}{(s+\nu )(s+ \alpha )^{N}}f_{0,0}(s). \end{aligned}$$

(17)

Inversion yields,

$$\begin{aligned} P_{0,N+1}(t) = \beta (\alpha - \lambda )^{N} e^{-\nu t}*e^{-\alpha t}\frac{t^{N-1}}{(N-1)! }* P_{0,0}(t). \end{aligned}$$

Taking Laplace transform of (7), we get

$$\begin{aligned} f_{i,N+1}(s) = \frac{\lambda }{(s+ \nu )}f_{i-1,N+1}(s), ~i = 1, 2, \ldots \end{aligned}$$

which recursively yields,

$$\begin{aligned} f_{i,N+1}(s) = (\frac{\lambda }{s+ \nu })^i f_{0,N+1}(s), ~i = 1, 2, \ldots . \end{aligned}$$

Using (17) in above equation, we obtain

$$\begin{aligned} f_{i,N+1}(s) = \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{(s+\nu )^{i+1}(s+ \alpha )^{N}}f_{0,0}(s), ~i = 0, 1, \ldots .\nonumber \\ \end{aligned}$$

(18)

Taking inverse Laplace transform of (18) yields the required transient probabilities $P_{i,N+1}(t)$, $i = 0,1,\ldots$.

Appendix 2

Define a generating function with coefficients as the transient probabilities of working states,

$$\begin{aligned} G(z,t) = P_{0,0} (t) +\sum _{i=1}^{\infty } P_{i,0} (t) z^i. \end{aligned}$$

Then,

$$\begin{aligned} \frac{\partial G(z,t)}{\partial z} = P_{0,0}^{'} (t) +\sum _{i=1}^{\infty } P_{i,0}^{'} (t) z^i. \end{aligned}$$

Substituting (1), (2) and (3) in the above equation, we get

$$\begin{aligned} \frac{\partial G(z,t)}{\partial z}&- \bigg (\lambda z + \frac{\mu }{z} - (\mu + \lambda )\bigg )G(z,t) \\&=-\bigg (\beta + \frac{\mu }{z} - \mu \bigg ) P_{0,0} (t)\\&+ (\nu - \lambda )\sum _{i=0}^{\infty } P_{i,N+1} (t) z^i + \lambda z \sum _{i=1}^{N} P_{0,i} (t). \end{aligned}$$

Solving the above partial differential equation, we obtain

$$\begin{aligned}&G(z,t) = -\bigg (\beta + \frac{\mu }{z} - \mu \bigg )\int _{0}^{t} P_{0,0} (y) e^{\big (\lambda z + \frac{\mu }{z} - (\mu + \lambda )\big )(t-y)} dy\\&\quad +(\nu - \lambda )\int _{0}^{t}\sum _{i=0}^{\infty } P_{i,N+1} (y) z^i e^{\big (\lambda z + \frac{\mu }{z} - (\mu + \lambda )\big )(t-y)} dy \\&+\lambda z \int _{0}^{t}\sum _{i=1}^{N} P_{0,i} (y) e^{\big (\lambda z + \frac{\mu }{z} - (\mu + \lambda )\big )(t-y)} dy + e^{-\big (\mu + \lambda - (\lambda z + \frac{\mu }{z})\big )t}. \end{aligned}$$

If $a = 2\sqrt{\lambda \mu }$ and $b = \sqrt{\frac{\lambda }{\mu }}$ then

$$\begin{aligned} e^{\big (\lambda z + \frac{\mu }{z}\big )(t-y)} = \sum _{n= - \infty }^{\infty } (bz)^n I_{n}(a(t-y)), \end{aligned}$$

where $I_n(.)$ is the modified bessel function of first kind. Then,

$$\begin{aligned}&G(z,t) = -\bigg (\beta + \frac{\mu }{z} - \mu \bigg )\int _{0}^{t} P_{0,0} (y) e^{ - (\mu + \lambda )(t-y)} \\&\bigg [\sum _{n= - \infty }^{\infty } (bz)^n I_{n}(a(t-y))\bigg ] dy +(\nu - \lambda )\int _{0}^{t}\sum _{i=0}^{\infty } P_{i,N+1} (y) z^i\\&\bigg [e^{ - (\mu + \lambda )(t-y)} \sum _{n= - \infty }^{\infty } (bz)^n I_{n}(a(t-y))\bigg ] dy\\&\quad +\lambda z \int _{0}^{t} \sum _{i=1}^{N} P_{0,i} (y) e^{- (\mu + \lambda )(t-y)} \sum _{n= - \infty }^{\infty } (bz)^n I_{n}(a(t-y)) dy \\&\quad + e^{ - (\mu + \lambda )t} \sum _{n= - \infty }^{\infty } (bz)^n I_{n}(a(t)). \end{aligned}$$

Further comparing the coefficient of $z^n$ for $n = 1,2,\ldots$ from both sides in above equation, we find

$$\begin{aligned}&P_{n,0}(t)=-\int _{0}^{t} P_{0,0} (y) e^{ - (\mu + \lambda )(t-y)} b^n \nonumber \\&\quad [(\beta - \mu )I_{n}(.)+\mu b I_{n+1}(.)] dy\nonumber \\&\quad +(\nu - \lambda )\int _{0}^{t}\sum _{i=0}^{n} P_{i,N+1} (y) e^{ - (\mu + \lambda )(t-y)} b^{n-i} I_{n-i}(.) dy \nonumber \\&\quad +(\nu - \lambda )\int _{0}^{t}\sum _{i=1}^{\infty } P_{n+i,N+1} (y) e^{ - (\mu + \lambda )(t-y)} b^{-i} I_{-i}(.) dy \nonumber \\&\quad +\lambda \int _{0}^{t} \sum _{i=1}^{N} P_{0,i} (y) e^{- (\mu + \lambda )(t-y)} b^{n-1} I_{n-1}(.) dy \nonumber \\&\qquad + e^{ - (\mu + \lambda )t} b^n I_{n}(at), \end{aligned}$$

(19)

where$I_n(.) = I_n(a(t-y))$.

And comparing the coefficient of $z^{-n}$ for n = 1,2,3,.. on both sides, we have,

$$\begin{aligned}&0 = -\int _{0}^{t} P_{0,0} (y) e^{ - (\mu + \lambda )(t-y)} b^{-n} [(\beta - \mu )I_{-n}(.)\nonumber \\&\quad +\mu b I_{-n+1}(.)] dy +(\nu - \lambda )\int _{0}^{t}\sum _{i=0}^{\infty } P_{i,N+1} (y) e^{ - (\mu + \lambda )(t-y)}\nonumber \\&\quad \times b^{-n-i} I_{-n-i}(.) dy +\lambda \int _{0}^{t} \sum _{i=1}^{N} P_{0,i} (y) e^{- (\mu + \lambda )(t-y)}\nonumber \\&\quad \times b^{-n-1} I_{-n-1}(.) dy + e^{ - (\mu + \lambda )t} b^{-n} I_{-n}(at). \end{aligned}$$

(20)

Using the property of modified Bessel function that $I_n(.) = I_{-n}(.)$ and Eqs. (19) and (20) [(19) - $b^{2n} \times (20)]$, we have for $n = 1, 2, \ldots$

$$\begin{aligned} P_{n,0}(t)= & {} -\int _{0}^{t} P_{0,0} (y) e^{ - (\mu + \lambda )(t-y)} \mu b^{n+1}\nonumber \\&\times [I_{n+1}(.) - I_{n-1}(.)] dy +(\nu - \lambda )\nonumber \\&\qquad \qquad \qquad \int _{0}^{t}\sum _{i=0}^{n} P_{i,N+1} (y) e^{ - (\mu + \lambda )(t-y)}\nonumber \\&\times b^{n-i} [I_{n-i}(.)- I_{n+i}(.)] dy +(\nu - \lambda )\nonumber \\&\qquad \qquad \qquad \int _{0}^{t}\sum _{i=1}^{\infty } P_{n+i,N+1} (y)\nonumber \\&\times e^{ - (\mu + \lambda )(t-y)}b^{-i} [I_{i}(.) - I_{2n+i}(.)] dy +\lambda \nonumber \\&\qquad \qquad \qquad \int _{0}^{t} \sum _{i=1}^{N} P_{0,i} (y)\nonumber \\&\times e^{- (\mu + \lambda )(t-y)} b^{n-1}[ I_{n-1}(.)-I_{n+1}(.)] dy . \end{aligned}$$

(21)

After doing some simple manipulation on the above equation, we get

$$\begin{aligned} P_{n,0}(t) =&-\mu b^{n+1} P_{0,0} (t)\\&\times [I_{n+1}(at) - I_{n-1}(at)]e^{ - (\mu + \lambda )(t)}\\&+\sum _{i=0}^{n} (\nu - \lambda )b^{n-i} P_{i,N+1} (t) \times [I_{n-i}(at)\\&- I_{n+i}(at)] e^{ - (\mu + \lambda )(t)}\\&+\sum _{i=1}^{\infty } (\nu - \lambda )b^{-i} P_{n+i,N+1} (t) \times [I_{i}(at)\\&- I_{2n+i}(at)] e^{ - (\mu + \lambda )(t)}\\&+ \sum _{i=1}^{N} \lambda b^{n-1} P_{0,i} (t)\times [ I_{n-1}(at)\\&-I_{n+1}(at)] e^{- (\mu + \lambda )(t)} . \end{aligned}$$

Using $I_{n-1}(\alpha x) - I_{n+1}(\alpha x) = \frac{2 n}{\alpha x} I_n (\alpha x)$ in (21) yields the required transient probabilities $P_{n,0}(t)$, $n \ge 1$.

Appendix 3

In Eq. (19), substituting n = 0, we have

$$\begin{aligned}&P_{0,0}(t) = -\int _{0}^{t} P_{0,0} (y) e^{ - (\mu + \lambda )(t-y)} [(\beta - \mu )I_{0}(.)+\mu b I_{1}(.)] dy \nonumber \\&+(\nu - \lambda )\int _{0}^{t}\sum _{i=0}^{\infty } P_{i,N+1} (y) e^{ - (\mu + \lambda )(t-y)} b^{-i} I_{i}(.) dy +\frac{\lambda }{b} \nonumber \\&\times \int _{0}^{t} \sum _{i=1}^{N} P_{0,i} (y) e^{- (\mu + \lambda )(t-y)} I_{1}(.) dy + e^{ - (\mu + \lambda )t} I_{0}(at). \end{aligned}$$

(22)

As

$$\begin{aligned} \sum _{n = 0}^{\infty } P_{n,0}(t) + \sum _{j = 1}^{N} P_{0,j}(t)+ \sum _{i = 0}^{\infty } P_{i,N+1}(t) = 1 . \end{aligned}$$

Taking Laplace transform of above equation, we get

$$\begin{aligned} \sum _{n = 0}^{\infty } f_{n,0}(s) + \sum _{j = 1}^{N} f_{0,j}(s)+ \sum _{i = 0}^{\infty } f_{i,N+1}(t) = \frac{1}{s}. \end{aligned}$$

(23)

Taking Laplace transform of transient probabilities obtained in Theorem 2 and using $L[I_n(\alpha t)] = \frac{1}{\sqrt{s^2 - {\alpha }^2} }\bigg (\frac{s-\sqrt{s^2 - {\alpha }^2 }}{\alpha }\bigg )^n$ and $L\bigg [\frac{2 n}{\alpha t} I_n(\alpha t)\bigg ] = \frac{2{\alpha }^{n-1}}{(s+\sqrt{s^2 - {\alpha }^2 })^n}$, we get

$$\begin{aligned}&f_{n,0}(s) = \mu b^{n+1}f_{0,0} (s) \frac{2{\alpha }^{n-1}}{(d+\sqrt{d^2 - {\alpha }^2 })^n}\nonumber \\&\quad +(\nu - \lambda )\sum _{i=0}^{n} b^{n-i} f_{i,N+1} (s) \frac{1}{\sqrt{d^2 - {\alpha }^2} }\nonumber \\&\quad \times \Bigg [\left( \frac{d-\sqrt{d^2 - {\alpha }^2 }}{\alpha }\right) ^{n-i}- \left( \frac{d-\sqrt{d^2 - {\alpha }^2 }}{\alpha }\right) ^{n+i}\Bigg ]\nonumber \\&\quad +(\nu - \lambda )\sum _{i=1}^{\infty } b^{-i}f_{n+i,N+1} (s) \frac{1}{\sqrt{d^2 - {\alpha }^2} }\nonumber \\&\quad \times \Bigg [\left( \frac{d-\sqrt{d^2 - {\alpha }^2 }}{\alpha }\right) ^{i}- \left( \frac{d-\sqrt{d^2 - {\alpha }^2 }}{\alpha }\right) ^{2n+i}\Bigg ]\nonumber \\&\quad +\lambda b^{n-1} \sum _{i=1}^{N} f_{0,i} (s)\frac{2{\alpha }^{n-1}}{(d+\sqrt{d^2 - {\alpha }^2 })^n} \end{aligned}$$

(24)

where $d= s+\lambda +\mu$ .

Using (16), (18) and (24) in equation (23), we obtain the required Laplace transform of transient probability of idle state (0,0). Inverse Laplace transform of the above equation yields the idle state probability $P_{0,0}(t)$.

Appendix 4

From Eq. (11), we get

$$\begin{aligned} \pi _{0,1} = \frac{\beta }{\alpha } \pi _{0,0}. \end{aligned}$$

(25)

From Eq. (12), we get

$$\begin{aligned} \alpha \pi _{0,j} = (\alpha - \lambda )\pi _{0,j-1}., j = 2,3,\ldots ,N \end{aligned}$$

which recursively yields,

$$\begin{aligned} \pi _{0,j} =\frac{ (\alpha - \lambda )^{j-1}}{\alpha ^{j-1}}\pi _{0,1}, j = 2,3,\ldots ,N . \end{aligned}$$

Using the value of $\pi _{0,1}$ obtained in equation (25), we get

$$\begin{aligned} \pi _{0,j} =\frac{\beta (\alpha - \lambda )^{j-1}}{\alpha ^{j}}\pi _{0,0}, j = 1,2,\ldots ,N . \end{aligned}$$

(26)

From Eq. (13), we have

$$\begin{aligned} \pi _{0,N+1} = \frac{(\alpha - \lambda )}{\nu }\pi _{0,N} \end{aligned}$$

and using Eq. (26) we obtain, for $j = N$

$$\begin{aligned} \pi _{0,N+1} = \frac{\beta (\alpha - \lambda )^{N}}{\nu \alpha ^{N}}\pi _{0,0}. \end{aligned}$$

(27)

From Eq. (14), we get

$$\begin{aligned} \pi _{i,N+1} = \frac{\lambda }{\nu } \pi _{i-1,N+1}, i = 1, 2, \ldots \end{aligned}$$

which recursively yields

$$\begin{aligned} \pi _{i,N+1} = \frac{\lambda ^i}{\nu ^i } \pi _{0,N+1}, i = 1, 2,\ldots \end{aligned}$$

and using value of $\pi _{0,N+1}$ obtained in equation (27), we have

$$\begin{aligned} \pi _{i,N+1} = \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{\nu ^{i+1}\alpha ^{N} } \pi _{0,0}, i = 0, 1,\ldots . \end{aligned}$$

(28)

Taking Laplace transform of (21) for $n = 1,2,\ldots$ we have

$$\begin{aligned}&f_{n,0}(s) = f_{0,0} (s) \mu b^{n+1} \Bigg [ \frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+1}\\&-\frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-1} \Bigg ]\\&+(\nu - \lambda )\sum _{i=0}^{n} f_{i,N+1} (s) b^{n-i} \Bigg [\frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-i}\\&- \frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+i} \Bigg ]\\&+(\nu - \lambda )\sum _{i=1}^{\infty } f_{n+i,N+1} (s) b^{-i} \Bigg [\frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{i}\\&- \frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{2n+i}\Bigg ]\\&+\lambda \sum _{i=1}^{N} f_{0,i} (s) b^{n-1}\Bigg [ \frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-1}\\&- \frac{1}{\sqrt{d^2-a^2}}\left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+1}\Bigg ] \end{aligned}$$

where $d = s +\lambda +\mu$ , $a = 2\sqrt{\lambda \mu }$ and $b = \sqrt{\frac{\lambda }{\mu }}$.

Using Eqs. (16) and (18) in above equation, we get

$$\begin{aligned}&f_{n,0}(s) = \frac{\mu b^{n+1}}{\sqrt{d^2-a^2}}\Bigg [ \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+1}\\&- \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-1}\Bigg ] f_{0,0} (s)\\&+(\nu - \lambda )\sum _{i=0}^{n} \frac{ b^{n-i}}{\sqrt{d^2-a^2}} \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{(s+\nu )^{i+1}(s+ \alpha )^{N}}\\&\times \left[ \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-i}- \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+i}\right] f_{0,0} (s)\\&+(\nu - \lambda )\sum _{i=1}^{\infty } \frac{ b^{-i}}{\sqrt{d^2-a^2}}\frac{\beta \lambda ^{n+i} (\alpha - \lambda )^{N}}{(s+\nu )^{n+i+1}(s+ \alpha )^{N}}\\&\times \left[ \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{i} - \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{2n+i}\right] f_{0,0} (s) \\&+\lambda \sum _{i=1}^{N} \frac{b^{n-1}}{\sqrt{d^2-a^2}} \frac{\beta (\alpha - \lambda )^{i-1}}{(s+ \alpha )^{i}}\Bigg [ \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n-1}\\&- \left( \frac{d - \sqrt{d^2-a^2} }{a}\right) ^{n+1}\Bigg ] f_{0,0} (s) . \end{aligned}$$

The steady-state probability $\pi _{n,0}$ for $n= 1,2,\ldots$ can be found using the well known properties of the Laplace transform. It is well known that for steady-state

$$\begin{aligned} \frac{\lambda }{\mu }<1 \end{aligned}$$

and

$$\begin{aligned} \lim _{t \rightarrow \infty } P_{n,0}(t) = \pi _{n,0} = \lim _{s \rightarrow 0} s f_{n,0} (s), \end{aligned}$$

where $n= 1,2,\ldots$ . Hence,

$$\begin{aligned}&\pi _{n,0} =\Bigg \{\frac{\mu b^{n+1}}{(\mu - \lambda )}\left[ b^{n+1}- b^{n-1}\right] +\sum _{i=0}^{n} \frac{(\nu - \lambda ) b^{n-i} }{(\mu - \lambda )}\nonumber \\&\times \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{\nu ^{i+1} \alpha ^{N}} \left[ b^{n-i}- b^{n+i}\right] +(\nu - \lambda )\sum _{i=1}^{\infty } \frac{ b^{-i}}{(\mu - \lambda )} \nonumber \\&\times \frac{\beta \lambda ^{n+i} (\alpha - \lambda )^{N}}{\nu ^{n+i+1}\alpha ^{N}} \left[ b^{i} - b^{2n+i}\right] +\lambda \sum _{i=1}^{N} \frac{b^{n-1}}{(\mu - \lambda )} \nonumber \\&\times \frac{\beta (\alpha - \lambda )^{i-1}}{\alpha ^{i}}\left[ b^{n-1}- b^{n+1} \right] \Bigg \} \pi _{0,0}.\nonumber \\&\pi _{n,0} =\Bigg \{ \left[ \frac{\mu b^{n+1}}{(\mu - \lambda )} + \frac{ \beta b^{n-1} ((\alpha - \lambda )^N -\alpha ^N)}{ \alpha ^N (\mu - \lambda )} \right] \nonumber \\&\times \left[ b^{n+1}- b^{n-1} \right] + \frac{\beta b^{n} (\alpha - \lambda )^{N}}{(\mu - \lambda ) \alpha ^{N}} \nonumber \\&\times \Bigg [ \frac{(\nu - \lambda ) b^{n+2}(1-(\frac{\lambda }{\nu b^2})^{n+1})}{(\nu b^2 -\lambda )}- b^n \left( 1-\left( \frac{\lambda }{\nu }\right) ^{n+1}\right) \nonumber \\&+\left( \frac{\lambda }{\nu }\right) ^{n+1} (b^{-n} - b^n)\Bigg ] \Bigg \}\pi _{0,0}.\nonumber \\&\pi _{n,0} = A_n \pi _{0,0}, \end{aligned}$$

(29)

where

$$\begin{aligned}&A_n = \left[ \frac{\mu b^{n+1}}{(\mu - \lambda )} + \frac{ \beta b^{n-1} ((\alpha - \lambda )^N -\alpha ^N)}{ \alpha ^N (\mu - \lambda )} \right] \left[ b^{n+1}- b^{n-1} \right] \\&+ \frac{\beta b^{n} (\alpha - \lambda )^{N}}{(\mu - \lambda ) \alpha ^{N}} \Bigg [ \frac{(\nu - \lambda ) b^{n+2}(1-(\frac{\lambda }{\nu b^2})^{n+1})}{(\nu b^2 -\lambda )}\\&- b^n \left( 1-\left( \frac{\lambda }{\nu }\right) ^{n+1}\right) +\left( \frac{\lambda }{\nu }\right) ^{n+1} (b^{-n} - b^n)\Bigg ]. \end{aligned}$$

And as

$$\begin{aligned} \sum _{i = 0}^{\infty } \pi _{i,0} + \sum _{j = 1}^{N} \pi _{0,j}+ \sum _{i = 0}^{\infty } \pi _{i,N+1} = 1. \end{aligned}$$

Substituting (26), (28) and (29) in above equation, we get

$$\begin{aligned}&\pi _{0,0}+\sum _{i = 1}^{\infty }A_i \pi _{0,0} + \sum _{j = 1}^{N}\frac{\beta (\alpha - \lambda )^{j-1}}{\alpha ^{j}}\pi _{0,0}\\&+ \sum _{i = 0}^{\infty } \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{\nu ^{i+1}\alpha ^{N} } \pi _{0,0} = 1 . \end{aligned}$$

Hence,

$$\begin{aligned} \pi _{0,0} = \frac{1}{1+\sum _{i = 1}^{\infty }A_i + \sum _{j = 1}^{N}\frac{\beta (\alpha - \lambda )^{j-1}}{\alpha ^{j}}+ \sum _{i = 0}^{\infty } \frac{\beta \lambda ^i (\alpha - \lambda )^{N}}{\nu ^{i+1}\alpha ^{N} } }. \end{aligned}$$

Here, $\pi _{0,0}$ exist as $\lambda \le \alpha$, $\lambda < \mu$ and $\lambda < \nu$.

Appendix 5

The two stage method is used to solve the SMP model which is described by its kernel matrix [5]. Let $K_{m,n}(t)$ denotes the elements of kernel matrix K(t), where $K_{m,n}(t)$ is the probability that the system has just entered the state m and in the next transition is going the enter the state n within time t. In this Section, we present the steady-state analysis for the semi-Markov model. The non-zero elements for kernel matrix are given as:

$$\begin{aligned}&K_{I_{0},S_1}(t) = \int _{0}^{t} (1-F_{I_{0}, W_1}(x)) dF_{I_{0}, S_1}(x)\\&K_{I_{0},S_1}(t) = {\left\{ \begin{array}{ll} 0 &{} \,\hbox {if}\, t< \frac{1}{\beta } \\ e^{-\frac{\lambda }{\beta }} &{} \,\hbox {if}\, t \ge \frac{1}{\beta } \end{array}\right. }\\&K_{I_{0},W_1}(t) = \int _{0}^{t} (1-F_{I_{0}, S_1}(x)) dF_{I_{0}, W_1}(x)\\&K_{I_{0},W_1}(t) = {\left\{ \begin{array}{ll} 1 - e^{-\lambda t} &{} \,\hbox {if}\, t< \frac{1}{\beta }\\ 1 - e^{-\frac{\lambda }{\beta }} &{} \,\hbox {if}\, t \ge \frac{1}{\beta } \end{array}\right. }\\&K_{S_j,S_{j+1}}(t) = \int _{0}^{t} (1-F_{S_j, W_1}(x)) dF_{S_j, S_{j+1}}(x)\\&K_{S_j,S_{j+1}}(t) = {\left\{ \begin{array}{ll} 0 &{} \,\hbox {if}\,t< \frac{1}{\alpha }\\ e^{-\frac{\lambda }{\alpha }} &{} \,\hbox {if}\, t \ge \frac{1}{\alpha } \end{array}\right. } , j= 1,2,\ldots ,N-1\\&K_{S_N,L_0}(t) = \int _{0}^{t} (1-F_{S_N, W_1}(x)) dF_{S_N, L_0}(x)\\&K_{S_N,L_0}(t) = {\left\{ \begin{array}{ll} 0 &{} \,\hbox {if}\, t< \frac{1}{\nu }\\ e^{-\frac{\lambda }{\alpha }} &{} \,\hbox {if}\, t \ge \frac{1}{\alpha } \end{array}\right. }\\&K_{W_k,W_{k+1}}(t) = \int _{0}^{t} (1-F_{W_k, W_{k-1}}(x)) dF_{W_k, W_{k+1}}(x)\\&K_{W_k,W_{k+1}}(t) = \frac{\lambda }{\lambda +\mu }(1-e^{-(\lambda +\mu )t}), k = 1,2,\ldots \\&K_{W_1,I_{0}}(t) = \int _{0}^{t} (1-F_{W_1, W_{2}}(x)) dF_{W_1, I_{0}}(x)\\&K_{W_1,I_{0}}(t) = \frac{\mu }{\lambda +\mu }(1-e^{-(\lambda +\mu )t})\\&K_{W_k,W_{k-1}}(t) = \int _{0}^{t} (1-F_{W_k, W_{k+1}}(x)) dF_{W_k, W_{k-1}}(x)\\&K_{W_k,W_{k-1}}(t) = \frac{\mu }{\lambda +\mu }(1-e^{-(\lambda +\mu )t}), k = 2,3,\ldots \\&K_{S_j,W_1}(t) = \int _{0}^{t}(1-F_{S_j, S_{j+1}}(x)) dF_{S_j, W_1}(x)\\&K_{S_j,W_1}(t) = {\left\{ \begin{array}{ll} 1 - e^{-\lambda t} &{} \,\hbox {if}\, t< \frac{1}{\alpha }\\ 1 - e^{-\frac{\lambda }{\alpha }} &{} \,\hbox {if}\, t \ge \frac{1}{\alpha } \end{array}\right. }, j = 1,2,\ldots ,N\\&K_{L_i,L_{i+1}}(t) = \int _{0}^{t} (1-F_{L_i, W_{k}}(x)) dF_{L_i, L_{i+1}}(x)\\&K_{L_i,L_{i+1}}(t) = {\left\{ \begin{array}{ll} 1 - e^{-\lambda t} &{} \,\hbox {if}\, t< \frac{1}{\nu }\\ 1 - e^{-\frac{\lambda }{\nu }} &{} \,\hbox {if}\, t \ge \frac{1}{\nu } \end{array}\right. }, i = 0,1,\ldots \\&K_{L_0,I_{0}}(t) = \int _{0}^{t} (1-F_{L_0, L_{1}}(x)) dF_{L_0, I_{0}}(x)\\&K_{L_0,I_{0}}(t) = {\left\{ \begin{array}{ll} 0 &{} \,\hbox {if}\, t< \frac{1}{\nu }\\ e^{-\frac{\lambda }{\nu }} &{} \,\hbox {if}\, t \ge \frac{1}{\nu } \end{array}\right. }\\&K_{L_i,W_{i}}(t) = \int _{0}^{t} (1-F_{L_i, L_{i+1}}(x)) dF_{L_i, W_i}(x)\\&K_{L_i,W_{i}}(t) = {\left\{ \begin{array}{ll} 0 &{} \,\hbox {if}\, t < \frac{1}{\nu }\\ e^{-\frac{\lambda }{\nu }} &{} \,\hbox {if}\, t \ge \frac{1}{\nu } \end{array}\right. }, i = 1,2,\ldots \end{aligned}$$

By the two stage analysis of semi-Markov model,

$K_{m,n}(\infty ) = \lim _{t \rightarrow \infty } K_{m,n}(t)$, where $m,n \in {\varOmega }$, where matrix $K(\infty )$ with entries $K_{m,n}(\infty )$ gives the one step transition probabilities for the EMC(Embedded Markov Chain) of the SMP model and row sum of matrix $K(\infty )$ is 1. Hence, the one step transition probabilities for the model are given as:

$K_{I_{0},S_1}(\infty ) = e^{-\frac{\lambda }{\beta }}$

$K_{I_{0},W_1}(\infty ) = 1-e^{-\frac{\lambda }{\beta }}$

$K_{S_j,S_{j+1}}(\infty ) = e^{-\frac{\lambda }{\alpha }}, j = 1,2,\ldots ,N-1$

$K_{S_N,L_0}(\infty ) = e^{-\frac{\lambda }{\alpha }}$

$K_{W_k,W_{k+1}}(\infty ) = \frac{\lambda }{\lambda +\mu }, k = 1,2,\ldots$

$K_{W_1,I_{0}}(\infty ) = \frac{\mu }{\lambda +\mu }$

$K_{W_k,W_{k-1}}(\infty ) = \frac{\mu }{\lambda +\mu }, k = 2,3,\ldots$

$K_{S_j,W_1}(\infty ) = 1-e^{-\frac{\lambda }{\alpha }}, j = 1,2,\ldots ,N$

$K_{L_i,L_{i+1}}(\infty ) = 1-e^{-\frac{\lambda }{\nu }}, i = 0,1,\ldots$

$K_{L_0,I_{0}}(t) = e^{-\frac{\lambda }{\nu }}$

$K_{L_i,W_{i}}(\infty ) = e^{-\frac{\lambda }{\nu }}, i = 1,2,\ldots$

The steady-state probabilities for states of EMC, given by vector $\mathbf{H }$ and is defined as

$(H_{I},H_{W_1},\ldots ,H_{S_1},\ldots ,H_{S_N},H_{L_0},H_{L_1},\ldots )$ can be obtained by solving [35]:

$$\begin{aligned} H = HK(\infty ), \sum _{m \in {\varOmega }} H_m = 1 . \end{aligned}$$

Hence, using $H = HK(\infty )$ and the entries of matrix $K(\infty )$, we get the system of linear equations, i.e.,

$H_{I_{0}} = \frac{\mu }{\lambda +\mu }H_{W_1}+ e^{-\frac{\lambda }{\nu }}H_{L_0}$

$H_{W_1} = \big (1-e^{-\frac{\lambda }{\beta }}\big )H_{I_0}+ \frac{\mu }{\lambda +\mu }H_{W_2}+ e^{-\frac{\lambda }{\nu }}H_{L_1}+\big (1-e^{-\frac{\lambda }{\alpha }}\big )(H_{S_1}+H_{S_2}+\ldots +H_{S_N})$

$H_{W_k} = \frac{\lambda }{\lambda +\mu }H_{W_{k-1}}+ \frac{\mu }{\lambda +\mu }H_{W_{k+1}}+ e^{-\frac{\lambda }{\nu }}H_{L_i}, k = 2,\ldots$

$H_{S_1} = e^{-\frac{\lambda }{\beta }}H_{I_0}$

$H_{S_{j+1}} = e^{-\frac{\lambda }{\alpha }}H_{S_j}, j = 1,2,\ldots ,N-1$

$H_{L_0} = e^{-\frac{\lambda }{\alpha }}H_{S_N}$

$H_{L_{i+1}} =\big (1-e^{-\frac{\lambda }{\nu }}\big )H_{L_{i}} , i= 0,1,\ldots$ .

Solving this system of equations, we get the steady-state probabilities of each state, $H_m, m \in {\varOmega }$ in terms of $H_{I_0}$, i.e.,

$H_{S_j} = e^{-\frac{\lambda }{\beta }}\big (e^{-\frac{\lambda }{\alpha }}\big )^{j-1} H_{I_{0}}$, $j = 1,2,\ldots ,N$

$H_{L_i} = e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }\big )}\big (1- e^{-\frac{\lambda }{\nu }}\big )^i H_{I_{0}}$, $i = 0,1,\ldots$

$H_{W_1} = \big (\frac{\lambda +\mu }{\mu }\big )\big (1-e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}\big ) H_{I_{0}}$

$H_{W_k} = \big (\frac{\lambda +\mu }{\mu }\big )\bigg [\big (\frac{\lambda }{\mu }\big )^{k-1}\big (1-e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}\big )+$

$\sum _{j=2}^{k} e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }\big )}\bigg [\big (\frac{\lambda }{\mu }\big )^{k-j}(1-e^{-\frac{\lambda }{\nu }})^j\bigg ] \bigg ]H_{I_{0}}, k = 2,\ldots$

Using equation $\sum _{m \in {\varOmega }} H_m = 1$ and the probabilities, $H_m, m \in {\varOmega }$ in terms of $H_{I_{0}}$, we obtain

$$\begin{aligned} H_{I_{0}} = \frac{1}{D}, \end{aligned}$$

where,

$$\begin{aligned}&D = \frac{2\mu }{\mu -\lambda } - \frac{\lambda +\mu }{\mu -\lambda } e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}\\&\quad + \frac{e^{-\frac{\lambda }{\beta }}-e^{-\big (\frac{\lambda N}{\alpha }+\frac{\lambda }{\beta }\big )}}{1- e^{-\frac{\lambda }{\alpha }} } + \frac{e^{-\big (\frac{\lambda N}{\alpha }+\frac{\lambda }{\beta }\big )}}{e^{-\frac{\lambda }{\nu }}}\\&\quad - \frac{\lambda (\lambda +\mu )}{\mu }\frac{e^{-\big (\frac{\lambda N}{\alpha }+\frac{\lambda }{\beta }\big )}(1- e^{-\frac{\lambda }{\nu }})^2}{\big (\frac{\lambda }{\mu }-1+e^{-\frac{\lambda }{\nu }}\big )}\bigg [\frac{e^{\frac{\lambda }{\nu }}}{\mu }-\frac{1}{\mu -\lambda }\bigg ]. \end{aligned}$$

Let $T_m$ denotes the sojourn time in state $m \in {\varOmega }$. Since, the EMC obtained for this model is irreducible, aperiodic and positive recurrent, i.e, ergodic, hence the steady-state probabilities of each state $m \in {\varOmega }$ of semi-Markov model [35] can be obtained as:

$$\begin{aligned} \pi _m = \frac{H_m E[T_m]}{\sum _{n \in {\varOmega }}H_n E[T_n] }, m \in {\varOmega }, \end{aligned}$$

where $E[T_m]$ denotes the expected sojourn time in state $m \in {\varOmega }$. Then the expected sojourn time spent in each state can be obtained as:

$$\begin{aligned}&E[T_{I_{0}}] = \int _{0}^{\infty } (1-F_{I_{0}, S_1}(t)) (1-F_{I_{0}, W_1}(t)) dt\\&E[T_{I_{0}}] = \frac{1 - e^{\frac{-\lambda }{\beta }}}{\lambda }\\&E[T_{W_k}] = \int _{0}^{\infty } (1-F_{W_k, W_{k-1}}(t)) (1-F_{W_k, W_{k+1}}(t)) dt\\&E[T_{W_k}] = \frac{1}{\lambda +\mu } , k = 1,2,\ldots \\&E[T_{S_j} ]= \int _{0}^{\infty } (1-F_{S_j, W_1}(t)) (1-F_{S_j, S_{j+1}}(t)) dt\\&E[T_{S_j}] = \frac{1 - e^{\frac{-\lambda }{\alpha }}}{\lambda }, j = 1,2,\ldots ,N-1\\&E[T_{S_N} ]= \int _{0}^{\infty } (1-F_{S_N, W_1}(t)) (1-F_{S_N, L_{0}}(t)) dt\\&E[T_{S_N}] = \frac{1 - e^{\frac{-\lambda }{\alpha }}}{\lambda }\\&E[T_{L_i}] = \int _{0}^{\infty } (1-F_{L_i, L_{i+1}}(t)) (1-F_{L_i, W_k}(t)) dt\\&E[T_{L_i}] = \frac{1 - e^{\frac{-\lambda }{\nu }}}{\lambda }, i = 0,1,\ldots . \end{aligned}$$

The stability conditions for the existence of steady-state probabilities of the semi-Markov model are given as $\lambda < \mu$, $\lambda \le \alpha$ and $\lambda < \nu$. Hence, under stability conditions using expected sojourn times and the steady-state probabilities of EMC, we get

$$\begin{aligned} \sum _{n \in {\varOmega }}H_n E[T_n] = M H_{I_0}, \end{aligned}$$

where

$$\begin{aligned}&M = \frac{\mu }{\lambda (\mu -\lambda )} - \frac{e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}}{\mu -\lambda } - \frac{e^{-\big (\frac{\lambda N}{\alpha }+\frac{\lambda }{\beta }\big )}}{\lambda }\big [2-e^{\frac{\lambda }{\nu }} \big ]\\&\quad - \frac{\lambda }{\mu }\frac{e^{-\big (\frac{\lambda N}{\alpha }+\frac{\lambda }{\beta }\big )}\big (1- e^{-\frac{\lambda }{\nu }}\big )^2}{\big (\frac{\lambda }{\mu }-1+e^{-\frac{\lambda }{\nu }}\big )}\bigg [\frac{e^{\frac{\lambda }{\nu }}}{\mu }-\frac{1}{\mu -\lambda }\bigg ]. \end{aligned}$$

Hence, the steady-state probabilities for the semi-Markov model are given as:

$$\begin{aligned}&\pi _{I_{0}} = \frac{1 - e^{-\frac{\lambda }{\beta }}}{\lambda M },\\&\pi _{W_1} = \frac{1-e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}}{\mu M },\\&\pi _{W_k} = \frac{1}{\mu M }\bigg [\big (\frac{\lambda }{\mu }\big )^{k-1}\big (1-e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }+\frac{\lambda }{\nu }\big )}\big ) \\&\qquad +\frac{\big (\frac{\lambda }{\mu }\big )\big (1-e^{-\frac{\lambda }{\nu }}\big )^2}{\bigg (\frac{\lambda }{\mu }-1+e^{-\frac{\lambda }{\nu }}\bigg )} e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }\big )}\\&\qquad \left[ \left( \frac{\lambda }{\mu }\right) ^{k-2}-\left( 1-e^{-\frac{\lambda }{\nu }}\right) ^{k-2}\right] \bigg ], ~k = 2,3,\ldots \\&\pi _{S_j} = \frac{e^{-\frac{\lambda }{\beta }}\big (e^{-\frac{\lambda }{\alpha }}\big )^{j-1}\big (1-e^{-\frac{\lambda }{\alpha }}\big )}{\lambda M}, ~j= 1,2,\ldots ,N \\&\pi _{L_i} = \frac{e^{-\big (\frac{\lambda }{\beta }+\frac{\lambda N}{\alpha }\big )}\big (1- e^{-\frac{\lambda }{\nu }}\big )^{i+1}}{\lambda M }, ~i= 0, 1,\ldots \end{aligned}$$

The steady-state probabilities $\pi _m \forall$ $m \in S$ exist as $\lambda < \mu$, $\lambda \le \alpha$ and $\lambda < \nu$.