The fluid model discussed in the previous section is investigated when it has the background process as an M/M/1/N queue with mean arrival and service rates to be \( {\lambda}_j=\frac{\lambda }{j+1} \) and μj = μ, respectively.
Taking Laplace transform of Eqs. (8–10), with respect to z, we have
$$ s\;{F_0}^{\ast }(s)-{F}_0(0)=-\frac{\lambda_0}{r_0}{F_0}^{\ast }(s)+\frac{\mu_1}{r_0}{F^{\ast}}_1(s), $$
(15)
$$ s\;{F_j}^{\ast }(s)-{F}_j(0)=-\frac{\left({\lambda}_j+{\mu}_j\right)}{r_j}{F_j}^{\ast }(s)+\frac{\lambda_{j-1}}{r_j}{F_{j-1}}^{\ast }(s)+\frac{\mu_{j+1}}{r_j}{F_{j+1}}^{\ast }(s),s\ge 0,j\in \varOmega -\left\{0,N\right\} $$
(16)
$$ s\;{F_N}^{\ast }(s)-{F}_N(0)=\frac{\lambda_{N-1}}{r_N}{F_{N-1}}^{\ast }(s)-\frac{\mu_N}{r_N}{F_N}^{\ast }(s), $$
(17)
where
$$ {F_j}^{\ast }(s)=\underset{0}{\overset{\infty }{\int }}{e}^{- sz}\;{F}_j(z)\; dz\kern0.37em \mathrm{and}\kern0.37em {F}_0(0)={d}_0 $$
(18)
Matrix notation (15–17) can be given as:
$$ A(s)\;{\mathbbm{F}}^{\ast }(s)=\mathbbm{F}(0) $$
(19)
where \( {\mathbbm{F}}^{\ast }(s)={\left[{F^{\ast}}_0(s),{F^{\ast}}_1(s),...,{F^{\ast}}_N(s)\right]}^T \), \( \mathbbm{F}(0)={\left[{F}_0(0),{F}_1(0),...,{F}_N(0)\right]}^T \)and
$$ A(s)=\left(\begin{array}{l}s+\frac{\lambda_0}{r_0}\kern2.20em -\frac{\mu_1}{r_0}\kern3.759999em \mathrm{0...}\kern7.50em 0\kern3.70em 0\\ {}-\frac{\lambda_0}{r}\kern1.72em s+\frac{\lambda_1+{\mu}_1}{r}\kern1.54em -\frac{\mu_2}{r}...\kern6.70em 0\kern3.70em 0\\ {}\kern1.10em 0\kern3.68em -\frac{\lambda_1}{r}\kern1.8em s+\frac{\lambda_2+{\mu}_2}{r}\kern0.88em -\frac{\mu_3}{r}\;...\kern1.90em 0\kern3.70em 0\\ {}\kern1.25em .\kern4.75em .\kern4.679998em .\kern4.10em .\kern3.4em 0\kern3.70em 0\\ {}\kern1.25em .\kern4.75em .\kern4.66em .\kern4.10em .\kern3.4em 0\kern3.70em 0\\ {}\kern1.25em .\kern4.75em .\kern4.66em .\kern4.10em .\kern3.4em 0\kern3.70em 0\\ {}\kern1.10em 0\kern4.60em 0\kern4.359999em 0\kern3.90em 0\;...\kern1.68em -\frac{\lambda_{N-1}}{r}\kern1.08em s+\frac{\mu_N}{r}\end{array}\right) $$
The matrix A(s) transforms into a symmetric tridiagonal matrix by diagonal matrix
$$ \varPhi =\operatorname{diag}\left[{\phi}_0,{\phi}_1,...,{\phi}_N\right] $$
(20)
with
$$ {\phi}_0=1,\kern0.84em {\phi}_i=\sqrt{\frac{i!\kern0.36em r}{r_0}{\left(\frac{\mu }{\lambda}\right)}^i}\kern1.5em ,i=1,2,...,N $$
(21)
and we get
$$ sI+B=\varPhi\;A{\varPhi}^{-1} $$
(22)
where the symmetric tridiagonal matrix B with elements are the same that of matrix A(s).
$$ B=\left(\begin{array}{l}\frac{\lambda_0}{r_0}\kern2.70em \sqrt{\frac{\lambda \mu}{r_0r}}\kern3.80em 0\kern4.20em \mathrm{0...}\kern3.819998em 0\kern5.20em 0\\ {}\sqrt{\frac{\lambda \mu}{r_0r}}\kern1.24em \frac{\lambda }{2r}+\frac{\mu }{r}\kern2.30em \sqrt{\frac{\lambda \mu}{2{r}^2}}\kern3.40em 0\kern0.2em ...\kern3.799998em 0\kern4.99em 0\\ {}\kern0.72em 0\kern2.20em \sqrt{\frac{\lambda \mu}{2{r}^2}}\kern3.479999em \frac{\lambda }{3r}+\frac{\mu }{r}\kern1.2em \sqrt{\frac{\lambda \mu}{3{r}^2}}\;...\kern2.999999em 0\kern5.10em 0\\ {}\kern0.84em .\kern3.85em .\kern4.779998em .\kern4.30em .\kern4.90em 0\kern5.10em 0\\ {}\kern0.84em .\kern3.85em .\kern4.78em .\kern4.30em .\kern4.90em 0\kern5.10em 0\\ {}\kern0.84em .\kern3.85em .\kern4.78em .\kern4.30em .\kern4.90em 0\kern5.10em 0\\ {}\kern0.72em 0\kern3.70em 0\kern4.50em 0\kern3.98em \mathrm{0...}.\kern1.7em \frac{\lambda }{\left(N-1\right)r}+\frac{\mu }{r}\kern1.34em \sqrt{\frac{\lambda \mu}{N{r}^2}}\kern0.24em \\ {}\kern0.84em 0\kern3.60em 0\kern4.50em 0\kern3.98em 0\;...\kern2.8em \sqrt{\frac{\lambda \mu}{N{r}^2}}\kern3.75em \frac{\mu }{r}\kern0.24em \end{array}\right) $$
If we take the determinant θn(s)of the bottom right square submatrices of the matrix A(s) and βn(s) is the determinant of the top left submatrices of the matrix A(s), then θn(s) and βn(s) satisfy the following difference equations.
$$ {\theta}_n(s)-\left(s+\left(\frac{\lambda }{N-n+2\Big)}+\mu \right)\frac{1}{r}\right)\;{\theta}_{n-1}(s)+\frac{\lambda \mu}{r^2\left(N-n+2\right)}\;{\theta}_{n-2}(s)=0 $$
(23)
$$ {\beta}_n(s)-\left(s+\left(\frac{\lambda }{n}+\mu \right)\frac{1}{r}\right)\;{\beta}_{n-1}(s)+\frac{\lambda \mu}{\left(n-1\right){r}^2}\;{\beta}_{n-2}(s)=0,\kern1.22em n=2,3,...,N. $$
(24)
with initial conditions
$$ {\theta}_0(s)=1={\beta}_0(s)\kern0.24em \mathrm{and}\kern0.24em {\theta}_1(s)=s+\frac{\mu }{r},{\beta}_1(s)=s+\frac{\lambda }{r_0} $$
(25)
Can be represent the elements of the inverse of the matrix sI + Bas following
$$ {c}_{ij}(s)=\left\{{\displaystyle \begin{array}{l}\sqrt{\frac{i!{\left(\lambda \mu \right)}^{j-i}}{j!{r}^{2\left(j-i\right)}}}\frac{\theta_{N-j}(s){\beta}_i(s)}{\mid sI+B\mid}\kern0.9000001em ,i<j\\ {}\frac{\theta_{N-i}(s){\beta}_i(s)}{\mid sI+B\mid}\kern6.699996em ,i=j\\ {}\sqrt{\frac{j!{\left(\lambda \mu \right)}^{i-j}}{i!{r}^{2\left(i-j\right)}}}\frac{\theta_{N-i}(s){\beta}_j(s)}{\mid sI+B\mid}\kern0.6em ,i>j\end{array}}\right) $$
(26)
From Eq. (19), we see that:
$$ {\displaystyle \begin{array}{l}\;\mathbbm{F}(s)=A{(s)}^{-1}\mathbbm{F}(0)\\ {}\kern1.32em ={\varPhi}^{-1}{\left( sI+B\right)}^{-1}\varPhi\;\mathbbm{F}(0)\end{array}} $$
(27)
$$ {\displaystyle \begin{array}{l}{F_n}^{\ast }(s)=\sum \limits_{k=0}^N{\phi_n}^{-1}{\phi}_k\;{c}_{nk}{F}_k(0)\\ {}\kern1.68em ={\phi_n}^{-1}\;{\phi}_0\;{c}_{n0}\;{F}_0(0)\end{array}} $$
(28)
where
$$ {\phi_n}^{-1}{\phi}_0=\left\{{\displaystyle \begin{array}{l}1\kern6.999996em ,n=0\\ {}\sqrt{\frac{r_0}{n!r}\;{\left(\frac{\lambda }{\mu}\right)}^n}\kern1.6em ,n=1,2,...,N\end{array}}\right) $$
(29)
$$ {c}_{n0}=\left\{{\displaystyle \begin{array}{l}\frac{\theta_N(s)\;{\beta}_0(s)}{\mid sI+B\mid}\kern6.099997em ,n=0\\ {}\sqrt{\frac{{\left(\lambda \mu \right)}^n}{n!{r}_0\;{r}^{2n-1}}}\kern0.24em \frac{\theta_N(s){\beta}_0(s)\;}{\mid sI+B\mid },n=1,2,...,N\end{array}}\right) $$
(30)
Let ξm (m = 0, 1, ..., N)are the roots of polynomial of ∣A(s)∣. These roots are the negative eigenvalues of the matrix M(0). Since the matrix Mis positive definite, real and symmetric, the eigenvalues of Mare real, distinct, and positive. Hence the roots of ∣A(s)∣ are real, distinct, and negative. The determinant ∣sI + B ∣ = s ∣ M(s)∣is equal to \( s\prod \limits_{m=1}^N\left(s-{\xi}_m\right) \), where M(s) is given by:
$$ M(s)=\left(\begin{array}{l}s+\frac{\lambda_0}{r_0}+\frac{\mu }{r}\kern1.10em -\sqrt{\frac{\lambda \mu}{2{r}^2}}\kern3.059999em 0\kern4.399999em \mathrm{0...}\kern3.759999em 0\kern6.959998em 0\\ {}-\sqrt{\frac{\lambda \mu}{2{r}^2}}\kern2.14em s+\frac{\lambda }{2r}+\frac{\mu }{r}\kern0.96em -\sqrt{\frac{\lambda \mu}{3{r}^2}}\kern3.00em \mathrm{0...}\kern3.759999em 0\kern6.959998em 0\\ {}\kern0.72em 0\kern3.74em -\sqrt{\frac{\lambda \mu}{3{r}^2}}\kern1.94em s+\frac{\lambda }{3r}+\frac{\mu }{r}\kern0.72em \sqrt{\frac{\lambda \mu}{4{r}^2}}\;...\kern3.28em 0\kern6.959998em 0\\ {}\kern0.84em .\kern4.759999em .\kern5.70em .\kern4.20em .\kern4.879999em 0\kern6.959998em 0\\ {}\kern0.84em .\kern4.759999em .\kern5.70em .\kern4.20em .\kern4.879999em 0\kern6.959998em 0\\ {}\kern0.84em .\kern4.759999em .\kern5.70em .\kern4.20em .\kern4.879999em 0\kern6.959998em 0\\ {}\kern0.72em 0\kern4.639999em 0\kern5.40em 0\kern4.00em \mathrm{0...}.\kern0.72em s+\frac{\lambda }{\left(N-2\right)r}+\frac{\mu }{r}\kern1.26em -\sqrt{\frac{\lambda \mu}{\left(N-1\right){r}^2}}\kern0.24em \\ {}\kern0.84em 0\kern4.52em 0\kern5.40em 0\kern4.00em 0\;...\kern1.32em -\sqrt{\frac{\lambda \mu}{\left(N-1\right){r}^2}}\kern1.56em s+\frac{\lambda }{\left(N-1\right)r}+\frac{\mu }{r}\kern0.24em \end{array}\right) $$
$$ {F_n}^{\ast }(s)=\left\{{\displaystyle \begin{array}{l}\frac{\theta_N(s)\;{\beta}_0(s){d}_o}{\mid sI+B\mid}\kern13.40001em ,\kern1.32em n=0\\ {}\sqrt{\frac{r_0}{n!r}\;{\left(\frac{\lambda }{\mu}\right)}^n}\sqrt{\frac{{\left(\lambda \mu \right)}^n}{r_0\;{r}^{2n-1}\;n!}}\kern0.24em \frac{\theta_{N-n}(s){\beta}_0(s)\;{d}_0\;}{\mid sI+B\mid}\kern1.3em ,n=1,2,...,N\end{array}}\right) $$
(31)
or
$$ {F_n}^{\ast }(s)=\left\{{\displaystyle \begin{array}{l}\frac{\theta_N(s)\;{d}_o}{s\prod \limits_{i=1}^N\left(s-{\xi}_i\right)}\kern5.599997em ,\kern1.32em n=0\\ {}\frac{1}{n!}{\left(\frac{\lambda }{r}\right)}^n\kern0.24em \frac{\theta_{N-n}(s)\;{d}_0\;}{s\prod \limits_{i=1}^N\left(s-{\xi}_i\right)}\kern1.4em ,n=1,2,...,N\end{array}}\right) $$
(32)
$$ {F_n}^{\ast }(s)=\frac{p_n}{s}+\sum \limits_{k=1}^N\frac{d_{n,k}}{s-{\xi}_k} $$
(33)
where
$$ {d}_{n,k}=\left\{{\displaystyle \begin{array}{l}\frac{\theta_N\left({\xi}_k\right)\;{d}_o}{\xi_k\prod \limits_{i=1,i\ne k}^N\left({\xi}_k-{\xi}_i\right)},\kern1.32em n=0\\ {}\frac{1}{n!}\;{\left(\frac{\lambda }{r}\right)}^n\kern0.24em \frac{\theta_{N-n}\left({\xi}_k\right)\;{d}_0\;}{\xi_k\prod \limits_{i=1,i\ne k}^N\left({\xi}_k-{\xi}_i\right)},n=1,2,...,N\end{array}}\right) $$
(34)
Applying Laplace inverse transform on Eq. (33), we get
$$ {F}_n(z)={p}_n+\sum \limits_{k=1}^N{d}_{n,k}\;{e}^{-{\xi}_kz},n=0,1,...,N;z\ge 0 $$
(35)
Similarly, the closed form expressions for Fn(z) of both models as given by (34) and (35) are obtained analytically. Therefore, the stationary distribution of the buffer content is given as follows:
$$ F(z)=\underset{t\to \infty }{\lim}\Pr \left(Z(t)\le z\right)=\sum \limits_{j=o}^N{F}_j(z) $$
(36)
Or
$$ F(z)=1+\sum \limits_{j=0}^N\sum \limits_{k=1}^N{d}_{j,k}{e}^{-{\xi}_kz} $$
(37)
Finally, the constants pnmust satisfy the conditions (14). Also,
$$ {p}_n=\underset{s\to 0}{\lim }s\;{F_n}^{\ast }(s), and\kern0.36em {\theta}_n(0)={\left(\frac{\mu }{r}\right)}^n $$
(38)
Then,
$$ {p}_n=\frac{\rho^n}{n!}\;{p}_0,n=0,1,...,N $$
(39)
$$ {p}_0={\left[1+\sum \limits_{n=1}^N\frac{\rho^n}{n!}\right]}^{-1} $$
(40)
Where \( \rho =\frac{\lambda }{\mu }. \)