An M/PH/K queue with constant impatient time

Abstract

This paper is concerned with an M/PH/K queue with customer abandonment, constant impatient time, and many servers. By combining the method developed in Choi et al. (Math Oper Res 29:309–325, 2004) and Kim and Kim (Perform Eval 83–84:1–15, 2015) and the state space reduction method introduced in Ramaswami (Stoch Models 1:393–417, 1985), the paper develops an efficient algorithm for computing performance measures for the queueing system of interest. The paper shows a number of properties associated with matrices used in the development of the algorithm, which make it possible for the algorithm, under certain conditions, to handle systems with up to one hundred servers. The paper also obtains analytical properties of performance measures that are useful in gaining insight into the queueing system of interest.

Appendices

Appendix A: transition blocks for the count-server-for-phase approach

In this appendix, we construct the transition blocks in (3) explicitly. First, we need to specify how states in $\varOmega (0)\cup \varOmega (1)\cup {\ldots }\cup \varOmega (\textit{K})$ are organized. In general, we define, for k = 0, 1, ..., K, and m = 1, 2, ..., K,

$$\begin{aligned} \varOmega (k,m)=\left\{ (n_{1} ,\ldots ,n_{m} ):\; \; \text{ integer } n_{i} \ge 0,\; i=1,2,\ldots ,m,\; \sum _{i=1}^{m}n_{i} =k\right\} .\qquad \end{aligned}$$

(49)

Note that $\varOmega$(k) = $\varOmega$(k, m ${}_{s}$), for k = 0, 1, 2, ..., K. We organize the states in $\varOmega$(k, m) lexicographically. Then we have

$$\begin{aligned} \varOmega (\textit{k}, \textit{m}) = \cup _{i=0}^{k} (\varOmega (k-i, m-1)\times \{i\}). \end{aligned}$$

(50)

It is easy to see that, for m = 1, we have $\varOmega$(k, 1) = $\{$ k $\}$, and for k = 0, $\varOmega (0, m) = \{(0, {\ldots }, 0)\}$.

We begin with $\{$ S ${}^{+}$(k, m ${}_{s}$), k = 0, 1, ..., K–1$\}$. The basic components to construct those matrices are $\{$ $\lambda$, ${\varvec{\beta }}$ = ($\beta$ ${}_{1}$, ..., $\beta _{m_{s} }$)$\}$, since the corresponding transitions are triggered by the arrival of a customer. The vector ${\varvec{\beta }}$ has to be utilized to specify the service phase of the arriving customer. An effective way to construct those matrices is to generate them iteratively. To that end, we need to construct matrices $\{$ S ${}^{+}$(k, m), k = 0, 1, ..., K–1, m = 1, 2, ..., m ${}_{s}\}$ for transitions from $\varOmega$(k, m) to $\varOmega$(k+1, m). We further decompose the transitions into transitions from $\{\varOmega$(k, m–1)$\times \{$0$\}$, $\varOmega$(k–1, m–1)$\times \{$1$\}$, ..., $\varOmega$(0, m–1)$\times \{$ k $\} \}$ to $\{ \varOmega$(k+1, m–1)$\times \{$0$\}$, $\varOmega$(k, m–1)$\times \{$1$\}$, ..., $\varOmega$(0, m–1)$\times \{$ k+1$\} \}$, respectively. Specifically, for S ${}^{+}$(k, m), the construction components are $\{$ $\lambda \beta$ ${}_{1}$, ..., $\lambda \beta {}_{m}$ $\}$, and $S^+(k,m)$ is given by

$$\begin{aligned} \begin{array}{l} {\begin{array}{ccccc} {\ }_{{\varOmega (k+1,m-1)\times \{ 0\} } } &{} {\mathbf{\; }_{\varOmega (k,m-1)\times \{ 1\} } } &{} {\mathbf{\; \; \; \; }\cdots } &{} {\mathbf{\; \; \; \; \; }_{\varOmega (1,m-1)\times \{ k\} } } &{} {_{\varOmega (0,m-1)\times \{ k+1\} } } \end{array}\; } \\ \quad \quad \left( \begin{array}{cccccc} _{{S^{+} (k,m-1)}} &{} \ \ \ \ _{{\lambda \beta _{m} I}} &{} {} &{} {} &{} {} &{} {} \\ {} &{} _{{S^{+} (k-1,m-1)}} &{} \ \ _{{\lambda \beta _{m} I}} &{} {} &{} {} &{} {} \\ {} &{} {\ddots } &{} {\ddots } &{} {\ddots } &{} {} &{} {} \\ {} &{} {} &{} {} &{} \ \ \ _{{S^{+} (1,m-1)}} &{} _{{\lambda \beta _{m} I}} &{} {} \\ {} &{} {} &{} {} &{} {} &{} _{{S^{+} (0,m-1)}} &{} \ \ \ \ \ \ _{{\lambda \beta _{m} }} \end{array}\right) \end{array} \end{aligned}$$

(51)

and S ${}^{+}$(0, m) = $\lambda$($\beta$ ${}_{1}$, ..., $\beta {}_{m}$), for m = 1, 2, ..., m ${}_{s}$, and S ${}^{+}$(k, 1) = $\lambda \beta$ ${}_{1}$, for k = 0, 1, ..., K–1.

Now, we construct $\{$ S ${}^{-}$(k, m), k = 1, 2, ..., K, m = 1, 2, ..., m ${}_{s} \}$, which are for transitions from $\varOmega$(k, m) to $\varOmega$(k–1, m). The corresponding transitions are triggered by a service completion. Thus, the construction is based on $\{ t_{1}^{0} ,\; t_{2}^{0}$, ..., $t_{m_{s} }^{0} \}$ (Note: Recall that T ${}^{0}$ = $\left( t_{j}^{0} \right) _{m_{s} \times 1}$), and we obtain $S^-(k,m)$ as

$$\begin{aligned} \begin{array}{l} {\begin{array}{cccc} {_{\varOmega (k-1,m-1)\times \{ 0\} } } &{} {_{\mathbf{\; \; \; }\varOmega (k-2,m-1)\times \{ 1\} } } &{} {\ }_{\cdots } \mathbf{\ \; \; }_{\varOmega (1,m-1)\times \{ k-2\} } &{} {_{\varOmega (0,m-1)\times \{ k-1\} } } \ \end{array}} \\ \left( \begin{array}{ccccc} _{{S^{^{\_ } } (k,m-1)}} &{} {} &{} {} &{} {} &{} {} \\ _{{t_{m}^{0} I}} &{} {\ \ \ \ \ \ \ \ _{S^{^{\_ } } (k-1,m-1)}} &{} {} &{} {} &{} {} \\ {} &{} {\ \ \ddots } &{} {\ \ \ \ \ddots } &{} {} &{} {} \\ {} &{} {} &{} {\ \ \ \ddots } &{} {\ddots } &{} {} \\ {} &{} {} &{} {} &{} {\ \ \ \ \ _{(k-1)t_{m}^{0} I}} &{} \ \ \ \ \ _{{S^{^{\_ } } (1,m-1)}} \\ {} &{} {} &{} {} &{} {} &{} {\ \ \ \ \ \ \ \ _{kt_{m}^{0} }} \end{array}\right) \end{array} \end{aligned}$$

(52)

and S ${}^{-}$(1, m) = ($t_{1}^{0} ,\; t_{2}^{0}$, ..., $t_{m}^{0}$)$\prime$, for m = 1, 2, ..., m ${}_{s}$, and S ${}^{-}$(k, 1) = $kt_{1}^{0}$, for k = 1, 2, ..., K.

Finally, we construct $\{$ S(k, m), k = 0, 1, ..., K, m = 1, 2, ..., m ${}_{s} \}$, which are for transitions from $\varOmega$(k, m) to $\varOmega$(k, m). We decompose the transitions into three types, based on the decomposition of the states in $\varOmega$(k, m):

• (i) Type “+”: transitions from $\varOmega$(j, m–1)$\times \{$ k–j $\}$ to $\varOmega$(j+1, m–1)$\times \{$ k–j–1$\}$;

• (ii) Type “–”: transitions from $\varOmega$(j, m–1)$\times \{$ k–j $\}$ to $\varOmega$(j–1, m–1)$\times \{$ k–j+1$\}$; and

• iii) Transitions from $\varOmega$(j, m–1)$\times \{$ k–j $\}$ to $\varOmega$(j, m–1)$\times \{$ k–j $\}$.

Then the transition matrix S(k, m) from $\varOmega$(k, m) to $\varOmega$(k, m) can be written as

$$\begin{aligned} \begin{array}{l} {\begin{array}{ccccc} {\mathbf{\; }_{\varOmega (k,m-1)\times \{ 0\} } } &{} {\mathbf{\; \; }_{\varOmega (k-1,m-1)\times \{ 1\} } } &{} {\mathbf{\; \; \; \; \; }\cdots } &{} {\mathbf{\; \; \; \; }_{\varOmega (1,m-1)\times \{ k-1\} } } &{} {{}_{\varOmega (0,m-1)\times \{ k\} } } \end{array}} \\ \left( \begin{array}{ccccc} _{{S(k,m-1)}} &{} \ \ _{{{\hat{S}}^{-} (k,m-1)}} &{} {} &{} {} &{} {} \\ _{{{\hat{S}}^{+} (k-1,m-1)}} &{} _{{S(k-1,m-1)}} &{} \ \ \ _{{{\hat{S}}^{-} (k-1,m-1)}} &{} {} &{} {} \\ {} &{} {\ddots } &{} {\ddots } &{} {\ddots } &{} {} \\ {} &{} {} &{} _{{(k-1){\hat{S}}^{+} (1,m-1)}} &{} \ _{S(1,m-1)} &{} \ \ _{{\hat{S}}^{-} (1,m-1)} \\ {} &{} {} &{} {} &{} \ \ _{k{\hat{S}}^{+} (0,m-1)} &{} \ \ \ _{S(0,m-1)} \end{array}\right) \\ {\mathbf{\; \; \; \; \; \; \; \; \; }+\; \left( \begin{array}{ccccc} {0} &{} {} &{} {} &{} {} &{} {} \\ {} &{} _{t_{m,m} I} &{} {} &{} {} &{} {} \\ {} &{} {} &{} {\ddots } &{} {} &{} {} \\ {} &{} {} &{} {} &{} _{(k-1)t_{m,m} I} &{} {} \\ {} &{} {} &{} {} &{} {} &{} _{kt_{m,m} } \end{array}\right) .} \end{array} \end{aligned}$$

(53)

If m = 1, we have S(k, 1) = kt ${}_{1,1}$, for k = 0, 1, 2, ..., K. We need to construct two sets of matrices $\{ {\hat{S}}^{+} (k,m)$, k = 1, 2, ..., K, m = 1, 2, ..., m ${}_{s}$–1$\}$ and $\{ {\hat{S}}^{-} (k,m)$, k = 0, 1, ..., K–1, m = 1, 2, ..., m ${}_{s}$–1$\}$. Note that $\{ {\hat{S}}^{+} (k,m)$, k = 1, 2, ..., K, m = 1, 2, ..., m ${}_{s}$–1$\}$ are for the transitions from phase m to phases $\{$1, 2, ..., m–1$\}$, and $\{ {\hat{S}}^{-} (k,m)$, k = 1, 2, ..., K, m = 1, 2, ..., m ${}_{s}$–1$\}$ are for the transitions from phases $\{$1, 2, ..., m–1$\}$ to phase m. We use the construction methods for $\{$ S ${}^{+}$(k, m), k = 0, 1, ..., K–1, m = 1, 2, ..., m ${}_{s} \}$ and $\{$ S ${}^{-}$(k, m), k = 1, 2, ..., K, m = 1, 2, ..., m ${}_{s} \}$ in this construction.

• (i) The construction of $\{ {\hat{S}}^{+} (k,m), \textit{k} = 1, 2, {\ldots }, \textit{K}, \textit{m} = 1, 2, {\ldots }, \textit{m}{}_{s}-1\}$ is similar to that of $\{\textit{S}{}^{+}(\textit{k}, \textit{m}), \textit{k} = 0, 1, {\ldots }, \textit{K--}1, \textit{m} = 2, 3, {\ldots }, \textit{m}{}_{s} \}$, except that $\{{\lambda \beta }{}_{1}, {\ldots }, {\lambda \beta {}_{m}}\}$ is replaced with $\{\textit{t}{}_{m} {}_{+}{}_{1} {}_{,1}, \textit{t}{}_{m} {}_{+}{}_{1,2}, {\ldots }, \textit{t}{}_{m}{}_{+1} {}_{,}{{}_{m}}\}$. In addition, we have ${\hat{S}}^{+} (0,m)=(t_{m+1,1} ,\; \cdots ,\; t_{m+1,m} )$, for $\textit{m} = 1, 2, {\ldots }, \textit{m}{}_{s}$–1, and ${\hat{S}}^{+} (k,1)=t_{2,1}$, for k = 0, 1, ..., K–1.

• (ii) The construction of $\{ {\hat{S}}^{-} (k,m)$, k = 0, 1, ..., K–1, m = 2, ..., m ${}_{s}$–1$\}$ is similar to that of $\{$ S ${}^{-}$(k, m), k = 0, 1, ..., K–1, m = 2, 3, ..., m ${}_{s} \}$, except that $\{ t_{1}^{0} ,\; t_{2}^{0}$, ..., $t_{m_{s} -1}^{0} \}$ is replaced with $\{\textit{t}{}_{1,}{{}_{m}}{}_{+1}, \textit{t}{}_{2,}{{}_{m}}{}_{+1}, {\ldots }, \textit{t}{}_{m}{}_{,}{{}_{m}}{}_{+1} \}$. In addition, we have ${\hat{S}}^{-} (1,m)=(t_{1,m+1} ,\; \ldots ,\; t_{m,m+1} )'$, for m = 1, 2, ..., m ${}_{s}$–1, and ${\hat{S}}^{-} (k,1)=kt_{1,2}$, for k = 1, 2, ..., K.

Finally, we summarize the above construction methods to outline the steps to construct $\{$ S ${}^{+}$(k, m ${}_{s}$), k = 0, 1, ..., K-1$\}$, $\{$ S ${}^{-}$(k, m ${}_{s}$), k = 1, 2, ..., K $\}$, and $\{$ S(k, m ${}_{s}$), k = 0, 1, 2, ..., K $\}$.

Algorithm A.1

Construction of transition blocks for the count-server-for-phase approach

• A.I.1 Compute $\{$ S ${}^{+}$(k, m ${}_{s}$), k = 0, 1, ..., K–1$\}$:

  • (i) S ${}^{+}$(k, 1) = $\lambda \beta$ ${}_{1}$, for k = 0, 1, ..., K;

  • (ii) Use Eq. (51) to construct $\{$ S ${}^{+}$(k, m), for k = 0, 1, ..., K $\}$, for m = 2, 3, ..., m ${}_{s}$.

• A.I.2 Compute $\{$ S ${}^{-}$(k, m ${}_{s}$), k = 1, 2, ..., K $\}$:

  • (i) S ${}^{-}$(k, 1) = $kt_{1}^{0}$, for k = 1, 2, ..., K;

  • (ii) Use equation (52) to construct $\{$ S ${}^{-}$(k, m), for k = 0, 1, ..., K $\}$, for m = 2, 3, ..., m ${}_{s}$.

• A.I.3 Compute $\{$ S(k, m ${}_{s}$), k = 1, 2, ..., K $\}$:

  • If m = 1, we have S(k, 1) = kt ${}_{1,1}$, for k = 0, 1, 2, ..., K.

  • For m = 2, 3, ..., m ${}_{s}$,

    • (i) Construct $\{ {\hat{S}}^{+} (k,j), \textit{k} = 1, 2, {\ldots }, \textit{K}, \textit{j} = 1, 2, {\ldots }, \textit{m--}1\}$ using Eq. (51) with ($\textit{t}{}_{m}{}_{+1} {}_{,1}, \textit{t}{}_{m} {}_{+}{}_{1,2}, {\ldots }, \textit{t}{}_{m}{}_{+1} {}_{,}{}_{m}$) in place of $\lambda$($\beta$ ${}_{1}$, ..., $\beta {}_{m}$).

    • (ii) Construct $\{ {\hat{S}}^{-} (k,m), \textit{k} = 0, 1, {\ldots }, \textit{K--}1, \textit{m} = 2, {\ldots }, \textit{m}{}_{s}-1\}$ using Eq. (52) with (t ${}_{1} {}_{,}$ ${}_{m}$ ${}_{+1}$, t ${}_{2,}$ ${}_{m}$ ${}_{+1}, {\ldots }, \textit{t}{}_{m}{}_{,}$ ${}_{m}$ ${}_{+1}$)$\prime$ in place of ($t_{1}^{0} ,\; t_{2}^{0}$, ..., $t_{m}^{0}$)$\prime$.

    • (iii) Construct $\{\textit{S}(\textit{k}, \textit{m}), \textit{k} = 0, 1, {\ldots }, \textit{K}\}$.

  end

Appendix B. Matrices R and G, and Vectors u${}_{1}$ and u${}_{2}$

The following solution approach was introduced in Choi et al. (2004), and used in solving the M/PH/1 case in Kim and Kim (2015). The theoretical basis for this solution approach is the following theorem (Theorem 2.15 and 2.16 in Gohberg et al. 1982).

Theorem B.1

(Gohberg et al. (1982)) Consider second order matrix differential equation

$$\begin{aligned} \frac{\text{ d }^{2} }{\text{ d }x^{2} } \mathbf{u}(x)+\frac{\text{ d }}{\text{ d }x} \mathbf{u}(x)B_{1} +\mathbf{u}(x)B_{2} = \mathbf{0}, \end{aligned}$$

(54)

where u(x) is the row vector function to be found, and B ${}_{1}$ and B ${}_{2}$ are matrices. Suppose that X ${}_{1}$ and X ${}_{2}$ are matrices that are solutions of the auxiliary equation

$$\begin{aligned} X^{2} +XB_{1} +B_{2} = \mathbf{0}. \end{aligned}$$

(55)

If X ${}_{1}$ and X ${}_{2}$ have no common eigenvalues, then the general solution of Eq. (54) is given by

$$\begin{aligned} \mathbf{u}(x)=\mathbf{u}_{1} \exp \{ X_{1} x\} +\mathbf{u}_{2} \exp \{ X_{2} x\} \end{aligned}$$

(56)

where u ${}_{1}$ and u ${}_{2}$ are two constant vectors.

For our problem (15), we have

   B ${}_{1}$ = –($\lambda$I + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$)) and

   B ${}_{2}$ = $\lambda$Q$\otimes$ I + $\lambda$M$\otimes$(S(K, m ${}_{s}$)+S ${}^{-}$(K, m ${}_{s}$)S ${}^{+}$(K–1, m ${}_{s}$) /$\lambda$).

Let X ${}_{1}$ = $\lambda$(I – R) and X ${}_{2} {}_{ }$= $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$), where R and G are defined in Sect. 3. It can be shown that the two matrices are solutions to Eq. (55). Then one can use Eq. (56) to find a solution to Eq. (15), which satisfies all boundary conditions. For that purpose, we need that exp$\{$ X ${}_{1}$ x $\}$ and exp$\{$ X ${}_{2}$ x $\}$ provides 2m ${}_{e}$ m ${}_{s}$ ${}^{K}$ independent solutions to (15). To ensure that, we need the following results.

Proposition B.2

If $\rho$ $\ne$ 1, matrices $\lambda$(I – R) and $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$) have no common eigenvalues. If $\rho$ = 1, matrices $\lambda$(I – R) and $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$) have one common eigenvalue zero with algebraic multiplicity one.

Proof

First, we consider the case with $\rho$ < 1. Since sp(R) = 1, it is clear that the real parts of all eigenvalues of $\lambda$(I – R) are nonnegative. By Proposition 4.2, the maximal real part of all eigenvalues of $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$) is negative. Note that the eigenvalue of $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$) with the maximal real part has to be real. Thus, the two matrices have no common eigenvalues.

If $\rho$ > 1, we have sp(R) = sp(G) < 1. Then all eigenvalues of $\lambda$(I – R) have a positive real part. On the other hand, by Proposition 4.2, all eigenvalues of $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$) have a nonpositive real part. Therefore, the two matrices have no common eigenvalue.

If $\rho$ = 1, zero is an eigenvalue of both $\lambda$(I – R) and $\lambda$G + Q $\otimes$ I + M $\otimes$ S(K, m ${}_{s}$). Similar to Choi et al. (2004), it can be shown that the algebraic multiplicity of the two matrices is one. This completes the proof of Proposition B.2. $\square$

By Proposition B.2, if $\rho$ $\ne$ 1, it can be shown that Eq. (56) gives 2m ${}_{e}$ m ${}_{s}$ ${}^{K}$ independent solutions to (15). Then all we need to do is to find $\{$ u ${}_{1}$, u ${}_{2} \}$ for a solution to (10). To do so, we use the function dp ${}_{K}$ ${}_{+1}$(x)/dx, which can be found in two ways: using Eqs. (10) and (16). Equalizing the resulted expressions at $x=0$ and $x=\tau -$ leads to the following linear system for $\{$ u ${}_{1}$, u ${}_{2} \}$:

$$\begin{aligned}&\displaystyle {\mathbf{0}=\mathbf{u}_{1} e^{\lambda (R-I)\tau } \left( D_{K} (I\otimes S^{+} (K-1,m_{s} ))\frac{1}{\lambda } -R\right) } \nonumber \\&\displaystyle \mathbf{\; \; \; \; \; }+\mathbf{u}_{2} \left( G-I+\frac{1}{\lambda } \left( Q\otimes I+M\otimes S(K,m_{s})\right) \right) \nonumber \\&\displaystyle \mathbf{\; \ \ \ \ } + \mathbf{u}_2\frac{1}{\lambda } \left( D_{K} (I\otimes S^{+} (K-1,m_{s} ))\right) ; \nonumber \\&\displaystyle \mathbf{0}=\mathbf{u}_{1} \left( Q\otimes I+M\otimes S(K,m_{s} )+\lambda R\right) \nonumber \\&\displaystyle \ \ \ \ \ \ \ +\mathbf{u}_{2} \lambda e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) \tau } \left( I-G\right) . \end{aligned}$$

(57)

Finally, we use the law of total probability to normalize $\{$ u ${}_{1}$, u ${}_{2} \}$, i.e., $\sum _{k=0}^{K}{} \mathbf{p}_{k} \mathbf{e} +\int _{0}^{\tau }{} \mathbf{p}_{K+1} (x)\text{ d }x\mathbf{e} =1$. By routine calculations, we obtain

$$\begin{aligned} \displaystyle \sum _{k=0}^{K}{} \mathbf{p}_{k} \mathbf{e}= & {} \mathbf{p}_{K} \left( I+\sum _{k=0}^{K-1}\left( \prod _{j=0}^{K-(k+1)}D_{K-j} \right) \right) \mathbf{e} \nonumber \\ \displaystyle= & {} \left( \mathbf{u}_{1} \exp \{ \lambda (R-I)\tau \} +\mathbf{u}_{2} \right) \frac{1}{\lambda } \left( I+\sum _{k=0}^{K-1}\left( \prod _{j=0}^{K-(k+1)}D_{K-j} \right) \right) \mathbf{e},\qquad \end{aligned}$$

(58)

and

$$\begin{aligned}&\displaystyle \int _{0}^{\tau }{} \mathbf{p}_{K+1} (x)\text{ d }x \nonumber \\&\quad \displaystyle =\mathbf{u}_{1} \int _{0}^{\tau }e^{\{ \lambda (R-I)(\tau -x)\}} \text{ d }x +\mathbf{u}_{2} \int _{0}^{\tau }\lambda e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) x} \text{ d }x . \end{aligned}$$

(59)

Using properties given in Sect. 4, the integrals in Eq. (59) can be obtained.

Proposition B.3

We have, for $\rho$ < 1,

$$\begin{aligned}&\displaystyle \int _{0}^{\tau }\exp \{ \lambda (R-I)(\tau -x)\} \text{ d }x \nonumber \\&\quad \displaystyle =\frac{1}{\lambda } \left( e^{\lambda (R-I)\tau } -I+\lambda \tau {\xi ({\varvec{\pi }} \otimes {\varvec{\phi }} )}\right) \left( R-I+ {{{\varvec{\xi }} ({\varvec{\pi }} \otimes {\varvec{\phi }} )}} \right) ^{-1} ; \end{aligned}$$

(60)

$$\begin{aligned}&\displaystyle \int _{0}^{\tau }e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) x} \text{ d }x \nonumber \\&\quad \displaystyle \ \ =\left( e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) x} -I\right) \left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) ^{-1},\qquad \end{aligned}$$

(61)

for $\rho$ > 1,

$$\begin{aligned} \displaystyle {\int _{0}^{\tau }\exp \{ \lambda (R-I)(\tau -x)\} \text{ d }x =\frac{1}{\lambda } \left( e^{(R-I)\tau } -I\right) \left( R-I\right) ^{-1} ;}\end{aligned}$$

(62)

$$\begin{aligned} \begin{array}{l} \displaystyle {\int _{0}^{\tau }e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) x} \text{ d }x } \\ \displaystyle \ \ \ \ \ \ \ =\left( e^{\left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} )\right) \tau } -I+\tau {{{\varvec{\zeta }} ({\varvec{\pi }} \otimes {\varvec{\phi }})}} \right) \\ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \times \left( \lambda G+Q\otimes I+M\otimes S(K,m_{s} +)+ {{{\varvec{\zeta }} ({\varvec{\pi }} \otimes {\varvec{\phi }} )}} \right) ^{-1}, \end{array} \end{aligned}$$

(63)

and, for $\rho$ = 1, Eqs. (60) and (63) hold.

Proof

The proof is based on Propositions 4.3 and  4.4. Details are omitted.

By Proposition B.3, the linear system with (19), (20), and (21) for $\{$ u ${}_{1}$, u ${}_{2} \}$ is obtained for the case with $\rho$ $\ne$ 1.

If $\rho$ = 1, by Proposition B.2, (56) gives 2m ${}_{e}$ m ${}_{s}$ ${}^{K}$ – 1 independent solutions to (15). We need to find one more solution to (15). By Proposition 4.2, it is easy to verify that

$$\begin{aligned} \mathbf{v}(x)=({\varvec{\pi }} \otimes {\varvec{\phi }} )\left( \lambda x I + (A_{0} -A_{2} )\left( A_{0} +A_{1} +A_{2} +\mathbf{e}{{\otimes ({\varvec{\pi }} \otimes {\varvec{\phi }})}}\right) ^{-1} \right) \end{aligned}$$

(64)

is another independent solution to (15). Then the solution to (15) can be expressed as

$$\begin{aligned} \mathbf{u}(x)=\mathbf{u}_{1} \exp \{ X_{1} x\} +\mathbf{u}_{2} \exp \{ X_{2} x\} + u_{3} \mathbf{v}(x), \end{aligned}$$

(65)

where $u_3$ is a constant. Similar to the case with $\rho$ $\ne$ 1, a linear system for $\{$ u ${}_{1}$, u ${}_{2}$, $u_{3}\}$ can be established by using two boundary conditions (i.e., conditions at $x=0$ and $x=\tau -$) and the law of total probability. Once $\{$ u ${}_{1}$, u ${}_{2}$, $u_3\}$ is obtained, a solution to (15) can be obtained. Details are omitted.