**Lemma 1.**

*(a) Let Ln*(*γ*) *be the matrix the entries of which define the intensities of transitions when some customer leaves the buffer due to impatience.*

*The matrices Ln*(*γ*), *n* = 1, *N*, *can be computed by the following way:*

*1. Calculate the matrices L*(*l*) *n* (*γ*) *using the recursive formulas:*

$$L\_n^{(0)}(\gamma) = n\gamma\_{R\*}$$

$$L\_{n}^{(l)}(\boldsymbol{\eta}) = \begin{pmatrix} n\gamma\_{\mathbb{R}-l}I & O & \cdots & O \\ L\_{1}^{(l-1)}(\boldsymbol{\eta}) & (n-1)\gamma\_{\mathbb{R}-l}I & \cdots & O \\ & O & L\_{2}^{(l-1)}(\boldsymbol{\eta}) & \cdots & O \\ & \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & \gamma\_{\mathbb{R}-l}I \\ & O & O & \cdots & L\_{n}^{(l-1)}(\boldsymbol{\eta}) \end{pmatrix}, l = \overline{1, R-1}.$$

*Here and after, I is the identity matrix and O is a zero matrix of an appropriate dimension;*


$$H = \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 0\\ a\_2 & 0 & 0 & \cdots & 0 & 0\\ p\_{3,1}a\_3 & p\_{3,2}a\_3 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ p\_{R-1,1}a\_{R-1} & p\_{R-1,2}a\_{R-1} & p\_{R-1,3}a\_{R-1} & \cdots & 0 & 0\\ p\_{R,1}a\_R & p\_{R,2}a\_R & p\_{R,3}a\_R & \cdots & p\_{R,R-1}a\_R & 0 \end{pmatrix}.$$

*Calculation of the matrices Yn*(*H*), *n* = 1, *N*, *can be performed as follows:*


 . . .

$$Z\_{\pi}^{(0)}(H\_j) = nh\_{r\_j,1}^j, n = \overline{1,N}, j = \overline{1,R-2},$$

$$Z\_{\pi}^{(l)}(H\_j) = \begin{pmatrix} nh\_{r\_j-l,1}^jI & O & \cdots & O \\ Z\_1^{(l-1)}(H\_j) & (n-1)h\_{r\_j-l,1}^jI & \cdots & O \\ & O & Z\_2^{(l-1)}(H\_j) & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & h\_{r\_j-l,1}^jI \\ & O & O & \cdots & Z\_n^{(l-1)}(H\_j) \\ \end{pmatrix}$$

$$l = \overline{1,...,r\_j-2}, \ n = \overline{1,N}, j = \overline{1,R-2},$$

,

*where hja*,*b is the* (*a*, *b*)*th entry of the matrix Hj and rj is the number of rows of the matrix Hj. 3. Calculate the matrices X*(*l*) *n* (*Hj*) *using the recursive formulas:*

 −

 −

$$\mathbf{X}\_{\mathbf{n}}^{(0)}(H\_{\hat{j}}) = h\_{1,r\_{j}}^{\hat{i}}, \; n = \overline{0, N-1}, \; j = \overline{1, R-2},$$

$$\mathbf{X}\_{\mathbf{n}}^{(l)}(H\_{\hat{j}}) = \begin{pmatrix} h\_{1,r\_{j}-l}^{\hat{i}}I & \mathbf{X}\_{0}^{(l-1)}(H\_{\hat{j}}) & O & \cdots & O & O\\ O & h\_{1,r\_{j}-l}^{\hat{i}}I & \mathbf{X}\_{1}^{(l-1)}(H\_{\hat{j}}) & \cdots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \cdots & h\_{1,r\_{j}-l}^{\hat{i}}I & \mathbf{X}\_{\mathbf{n}}^{(l-1)}(H\_{\hat{j}}) \end{pmatrix},$$

$$l = \overline{1, r\_j - 2}, \text{ } n = \overline{0, N - 1}, j = \overline{1, R - 2}.$$

*4. Calculate the matrices Zn*(*Hj*) = *Z*(*rj*−<sup>2</sup>) *n* (*Hj*), *n* = 1, *N*, *and Xn*(*Hj*) = *X*(*rj*−<sup>2</sup>) *n* (*Hj*), *n* = 0, *N* − 1, *j* = 1, *R* − 2.

*5. Calculate the matrices Y*(*j*) *n* , *n* = 1, *N*, *using the recursive formulas:*

$$Y\_{n}^{(0)} = \begin{pmatrix} 0 & nH\_{M-1,M} & 0 & \cdots & 0 & 0\\ H\_{M,M-1} & 0 & (n-1)H\_{M-1,M} & \cdots & 0 & 0\\ 0 & 2H\_{M,M-1} & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & H\_{M-1,M}\\ 0 & 0 & 0 & \cdots & nH\_{M,M-1} & 0 \end{pmatrix},$$

$$Y\_{n}^{(j)} = \begin{pmatrix} O & nX\_{0}(H\_{j}) & O & \cdots & O & O\\ Z\_{1}(H\_{j}) & Y\_{1}^{(j-1)} & (n-1)X\_{1}(H\_{j}) & \cdots & O & O\\ O & Z\_{2}(H\_{j}) & Y\_{2}^{(j-1)} & \cdots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & O & O & \cdots & Y\_{n-1}^{(j-1)} & 1X\_{n-1}(H\_{j})\\ O & O & O & \cdots & Z\_{n}(H\_{j}) & Y\_{n}^{(j-1)} \end{pmatrix},$$

$$j = \overline{1, \mathcal{R} - 2}.$$

*6. Calculate the matrices Yn*(*H*) *as Yn*(*H*) = *Y*(*<sup>R</sup>*−<sup>2</sup>) *n* , *n* = 1, *N*.

*(c) Let An*(**h**), *n* = 0, *N* − 1, *be the matrix the entries of which define the transition probabilities at the moment when a new customer arrives to the system and the system capacity is not exhausted (there are n*, 0 ≤ *n* < *N*, *customers in the buffer). Here, the row vector* **h** *has the following form* **h** = (*h*1, *h*2,..., *hR*) *where hr is the probability that the arrived to the system customer has type-r*, *r* = 1, *R*.

*Computation of the matrices An*(**h**) *can be performed as follows:*

*<sup>A</sup>*0(**h**) = **h** *and An*(**h**) = *A*(*<sup>R</sup>*−<sup>2</sup>) *n* (**h**) *where the matrices A*(*l*) *n* (**h**) *of block size* (*n* + 1) × (*n* + <sup>2</sup>), *n* = 1, *N* − 1, *are recursively computed as*

$$A\_{n}^{(0)}(\mathbf{h}) = \begin{pmatrix} h\_{R-1} & h\_{R} & 0 & \cdots & 0 & 0 \\ 0 & h\_{R-1} & h\_{R} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & h\_{R-1} & h\_{R} \end{pmatrix},$$

$$A\_{n}^{(l)}(\mathbf{h}) = \begin{pmatrix} h\_{R-l-1} & \hat{\mathbf{h}}^{(l)} & \mathbf{0} & \mathbf{0} & \cdots & \mathbf{0} & \mathbf{0} \\ \mathbf{0}^{T} & h\_{R-l-1}I & A\_{1}^{(l-1)} & O & \cdots & O & O \\ \mathbf{0}^{T} & O & h\_{R-l-1}I & A\_{2}^{(l-1)} & \cdots & O & O \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ \mathbf{0}^{T} & O & O & O & \cdots & h\_{R-l-1}I & A\_{n}^{(l-1)} \end{pmatrix},$$

$$l = \overline{1, R-2},$$

*where the vectors* **h ¯** (*l*) *are defined as* **h¯** (*l*) = (*hR*−*l*, *hR*−*l*+1,..., *hR*), *l* = 1, *R* − 2.

*(d)Let E*<sup>−</sup>*n* , *n* = 1, *N*, *be the matrix the entries of which define the transition probabilities at the moment when a customer with the maximal (among currently presenting in the system) priority is chosen for service.*

*The matrices E*<sup>−</sup>*n can be computed as*

$$E\_1^- = (\underbrace{1, 1, \dots, 1}\_{R})^T \prime$$

$$E\_{\mathfrak{n}}^{-} = \begin{pmatrix} I\_{K\_{\mathbb{R}}^{\langle n\rangle}} \\ O\_{K\_{\mathbb{R}-1}^{\langle n\rangle} \times (K\_{\mathbb{R}}^{\langle n\rangle} - K\_{\mathbb{R}-1}^{\langle n\rangle})} & I\_{K\_{\mathbb{R}-1}^{\langle n\rangle}} \\ \vdots & \\ O\_{K\_{\mathbb{R}}^{\langle n\rangle} \times (K\_{\mathbb{R}}^{\langle n\rangle} - K\_{\mathbb{2}}^{\langle n\rangle})} & I\_{K\_{\mathbb{2}}^{\langle n\rangle}} \\ O\_{K\_{1}^{\langle n\rangle} \times (K\_{\mathbb{R}}^{\langle n\rangle} - K\_{1}^{\langle n\rangle})} & I\_{K\_{1}^{\langle n\rangle}} \end{pmatrix}, n = \overline{2, N},$$
 
$$K\_{r}^{\langle n\rangle} = \binom{n + r - 2}{r - 1}, r = \overline{1, R}.$$

*where*

$$Here, \ (^{n+r-2}\_{r-1}) = \mathbb{C}^{r-1}\_{n+r-2} \text{ is the binomial coefficient.}$$

*(e) Let the entries of the square matrix <sup>E</sup>*<sup>ˆ</sup>*r*, *r* = 1, *R*, *of size* (*<sup>N</sup>*+*R*−<sup>1</sup> *R*−1 ) *define the transition probabilities at the moment when a type-r customer arrives at the system when there are N customers in the buffer and the arriving customer tries to force out a customer with a lower priority from the buffer. All entries in each row of this matrix are equal to zero except one entry which is equal to 1. We assume that each row and column of the matrix E*ˆ*r correspond to some state* {*η*1, *η*2, ... , *ηR*} *of the process ζ<sup>t</sup>*, *t* ≥ 0. *Note, that all states of the process ζ<sup>t</sup>*, *t* ≥ 0, *are enumerated in the reverse lexicographical order of components η*(1) *t* , ... , *η*(*R*) *t . For example, the first row and column of the matrix E*ˆ*r correspond to the state* {*<sup>N</sup>*, 0, 0, ... , <sup>0</sup>}*, the second row and column correspond to the state* {*N* − 1, 1, 0, ... , <sup>0</sup>}*,*...*, the last row and column correspond to the state* {0, 0, 0, ... , *<sup>N</sup>*}. *In the row of the matrix E*ˆ*r that corresponds to the state* {*η*1, *η*2, ... , *ηR*}*, the entry 1 is located in the column that corresponds to the same state* {*η*1, *η*2, ... , *ηR*} *only in the case if ηl* = 0 *for all l*, *R* ≥ *l* > *r*. *In this case, the arriving type-r customer is lost, because the customers with lower priority are absent in the buffer. If ηl* > 0 *for some l*, *R* ≥ *l* > *r and r*∗ *is a maximum of such values l*, *then the entry 1 is located in the column that corresponds to the state* {*η*1, ... , *η<sup>r</sup>*−1, *ηr* + 1, *ηr*+1, ... , *ηr*∗−1, *ηr*∗ − 1, 0, ... , <sup>0</sup>}. *In this case, the customer of type-r*∗ *has the lowest priority among the customers presenting in the system and an arriving type-r customer forces out one type-r*∗ *customer which departs from the system (is lost).*

**Proof.** The derivation of the form of the matrices that describe the transitions of the process *ζ*(*n*) *t* , *t* ≥ 0, is quite complicated and cumbersome. In derivations, we used some ideas of the paper [29]. To explain the scheme of the derivation of the form of the presented matrices, we show here how to compute the matrices *Ln*(*γ*), *n* = 1, *R*, the entries of which define the intensities of transitions of the components of the process *ζ*(*n*) *t* , *t* ≥ 0, when some customer leaves the buffer due to impatience. The rest of the matrices that define the intensities of transition of the components of the process *ζ*(*n*) *t* , *t* ≥ 0, can be obtained by the same way based on the careful account of possible transitions.

Computation of the matrices *Ln*(*γ*) can be performed as follows. Let us introduce the matrices *L*(*l*) *n* (*γ*) of the transition intensities of the components *n*(*R*) *t* , ... , *n*(*<sup>R</sup>*−*l*) *t* at the moment when there are *n* customers in the buffer and one of the customers leaves it due to impatience conditional on the fact that all customers have types *R*, *R* − 1, . . . , *R* − *l*, where *l* = 0, *R* − 1.

It is clear, that for *l* = 0, the matrices *L*(0) *n* (*γ*) have the scalar form *L*(0) *n* (*γ*) = *<sup>n</sup>γR*, because all *n* customers are of type-*R* in this situation.

Let us consider the matrix *L*(1) *n* . This matrix defines the transition intensities of the components *n*(*R*) *t* , *n*(*<sup>R</sup>*−<sup>1</sup>) *t* at the moment when there are *n* customers in the buffer and one of the customers leaves it due to impatience conditional on the fact that all customers have types *R* or *R* − 1. Taking into account the reverse lexicographic order of components, by definition the first row of the matrix *L*(1) *n* (*γ*) corresponds to the state where all *n* customers are of type-(*R* − <sup>1</sup>), the second row corresponds to the state where *n* − 1 customers are of type-(*R* − 1) and one customer is of type-*R*, etc., the last row corresponds to the state where all *n* customers are of type-*R*. After the customer leaves the system, the number of customers in the buffer decreases by 1. Thus, the first column of the matrix *L*(1) *n* (*γ*)

corresponds to the state where all *n* − 1 customers are of type-(*R* − <sup>1</sup>), the second column corresponds to the state where *n* − 2 customers are of type-(*R* − 1) and one customer is of type-*R*, etc., the last column corresponds to the state where all *n* − 1 customers are of type-*R*. Taking into account these considerations, it is easy to verify that the matrix *L*(1) *n* (*γ*) of size (*n* + 1) × *n* has the form

$$L\_n^{(1)}(\\\gamma) = \begin{pmatrix} n\gamma\_{R-1} & 0 & \cdots & 0\\ \gamma\_R & (n-1)\gamma\_{R-1} & \cdots & 0\\ 0 & 2\gamma\_R & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \gamma\_{R-1}\\ 0 & 0 & \cdots & n\gamma\_R \end{pmatrix},$$

or

$$L\_n^{(1)}(\\\gamma) = \begin{pmatrix} n\gamma\_{R-1} & 0 & \cdots & 0\\ L\_1^{(0)}(\gamma) & (n-1)\gamma\_{R-1} & \cdots & 0\\ 0 & L\_2^{(0)}(\gamma) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \gamma\_{R-1}\\ 0 & 0 & \cdots & L\_n^{(0)}(\gamma) \end{pmatrix}$$

.

Using the same reasonings, it can be shown that the matrix *L*(*l*) *n* (*γ*) of block size (*n* + 1) × *n* has the following form

$$L\_{n}^{(l)}(\\\chi) = \begin{pmatrix} n\gamma\_{R-l}I & O & \cdots & O\\ L\_{1}^{(l-1)}(\gamma) & (n-1)\gamma\_{R-l}I & \cdots & O\\ O & L\_{2}^{(l-1)}(\gamma) & \cdots & O\\ \vdots & \vdots & \ddots & \vdots\\ O & O & \cdots & \gamma\_{R-l}I\\ O & O & \cdots & L\_{n}^{(l-1)}(\gamma) \end{pmatrix}, \; l = \overline{2, R-1}.$$

It is clear that the required matrices *Ln*(*γ*) can be computed as *Ln*(*γ*) = *L*(*<sup>R</sup>*−<sup>1</sup>) *n* (*γ*), *n* = 1, *N*. This proves the proposed formulas for computation of the matrices *Ln*(*γ*).

**Remark 1.** *Derivation of the form of the matrices defined in Lemma 1 creates an opportunity to analyze not only the system under study in this paper but also many other queueing systems with a finite buffer and many types of customers having different priorities.*

Let us introduce the following notation:


By analyzing all possible transitions of the Markov chain *ξ<sup>t</sup>*, *t* ≥ 0, during an interval of infinitesimal length and rewriting the intensities of these transitions in the block matrix form, we obtain the following result.

**Theorem 1.** *The infinitesimal generator Q of the Markov chain ξ<sup>t</sup>*, *t* ≥ 0, *has the following block-tridiagonal structure*

$$Q = \begin{pmatrix} Q\_{0,0} & Q\_{0,1} & O & O & \dots & O & O\\ Q\_{1,0} & Q\_{1,1} & Q\_{1,2} & O & \dots & O & O\\ O & Q\_{2,1} & Q\_{2,2} & Q\_{2,3} & \dots & O & O\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & O & \dots & Q\_{N+1,N} & Q\_{N+1,N+1} \end{pmatrix}.$$

*The non-zero blocks are defined as follows:*

$$\begin{aligned} \mathcal{Q}\_{0,0} &= D\_{0\prime} \\\\ \mathcal{Q}\_{1,1} &= D\_0 \oplus S\_{\prime} \end{aligned}$$

$$Q\_{n,n} = D\_0 \oplus S \otimes I\_{K\_{n-1}} + I\_{WM} \otimes (Y\_{n-1} + I\_{n-1}), \ n = \overline{\mathbf{1}\_r} \mathbf{N}\_r$$

$$Q\_{N+1,N+1} = (D\_0 \oplus S) \otimes I\_{K\_N} + I\_{WM} \otimes (Y\_N + I\_N) + (1-q) \sum\_{r=1}^R D\_r \otimes I\_{MK\_N} + \overline{\mathbf{1}\_r} \mathbf{N}\_r$$

$$q \sum\_{r=1}^R D\_r \otimes I\_M \otimes \hat{E}\_r,$$

$$Q\_{0,1} = \sum\_{r=1}^R D\_r \otimes \mathbf{J}\_r$$

$$Q\_{n,n+1} = \sum\_{r=1}^R D\_r \otimes I\_M \otimes A\_{n-1}(\mathbf{h}\_r), \ n = \overline{\mathbf{1}\_r N\_r}$$

$$Q\_{1,0} = I\_W \otimes \mathbf{S}\_{\mathbf{0}},$$

$$Q\_{n,n-1} = I\_W \otimes \mathbf{S}\_{\mathbf{0}} \mathbf{f} \otimes \mathbf{E}\_{n-1}^- + I\_{WM} \otimes L\_{n-1}(\gamma), \ n = \overline{\mathbf{1}\_r N + 1}.$$

The Markov chain *ξ<sup>t</sup>*, *t* ≥ 0, is an irreducible and has a finite state space. Therefore, the stationary probabilities of the system states

$$\pi(n\_{\prime\prime}\nu\_{\prime\prime}m\_{\prime\prime}\eta^{(1)},\ldots,\eta^{(R)}) = $$

$$=\lim\_{t\to\infty}P\{n\_t=n\_{\prime\prime}\nu\_t=\nu\_{\prime\prime}m\_t=m\_{\prime\prime}\eta^{(1)}\_t=\eta^{(1)},\ldots,\eta^{(R)}\_t=\eta^{(R)}\},$$

always exist.

Let us form the row vectors *πn*, *n* = 0, *N* + 1, of these probabilities which are enumerated in the reverse lexicographic order of the components *η*(1) *t* , ... , *η*(*R*) *t* and the direct lexicographic order of the components *νt* and *mt*.

It is well known that the probability vectors *πn*, *n* = 0, *N* + 1, satisfy the following system of linear algebraic equations:

> (*<sup>π</sup>*0, *π*1,..., *<sup>π</sup>N*+<sup>1</sup>)*Q* = **0**, (1) (*<sup>π</sup>*0, *π*1,..., *<sup>π</sup>N*+<sup>1</sup>)**<sup>e</sup>** = 1

where *Q* is the infinitesimal generator of the Markov chain *ξ<sup>t</sup>*, *t* ≥ 0.

To compute the steady-state distribution of this Markov chain, it is necessary to solve system (1). The matrix of this system has the block-tridiagonal structure. Markov chains having the structure of the generator similar to the one defined in Theorem 1 are sometimes called in the existing literature as the Level-Dependent Quasi-Birth-and-Death processes; see, e.g., [31]. System (1) is finite and can be directly solved via the use of the variety of the standard computer programs. However, the number of equations of the finite system (1) for queueing model under study can be large especially when the buffer capacity *N* or the number of priority classes is large. Therefore, to effectively solve this system, it is desirable to apply an algorithm that exploits the sparse block-tridiagonal structure of the generator *Q*. In particular, the algorithm given in [32] can be recommended.

## **4. Performance Measures**

The average number of customers in the buffer is

$$N\_{buffr} = \sum\_{n=2}^{N+1} (n-1)\pi\_n \mathbf{e}\_n$$

The average number *N*(*r*) *buffer*of type-*<sup>r</sup>*, *r* = 1, *R*, customers in the buffer can be computed as

$$N\_{bnffer}^{(r)} = \sum\_{n=2}^{N+1} \pi\_n(I\_{WM} \odot L\_{n-1}(\mathbf{h}\_r)) \mathbf{e}\_n$$

The intensity of the output flow of successfully serviced customers is

$$\lambda\_{\text{out}} = \sum\_{n=1}^{N+1} \pi\_n \left( I\_W \otimes \mathbf{S\_0} \otimes I\_{K\_{n-1}} \right) \mathbf{e}\_n$$

The intensity of the output flow of customers who leave the buffer due to impatience is

$$\lambda\_{imp} = \sum\_{n=2}^{N+1} \pi\_n(I\_{WM} \odot L\_{n-1}(\gamma)) \mathbf{e}.$$

The probability *Ploss* of loss of an arbitrary customer is computed

$$P\_{\rm loss} = 1 - \frac{\lambda\_{\rm out}}{\lambda}.$$

The probability *Pimp*−*loss* of loss of an arbitrary customer due to impatience is computed

$$P\_{imp-loss} = \frac{\lambda\_{imp}}{\lambda}.$$

The intensity *λ*(*r*) *imp* of the output flow of the type-*<sup>r</sup>*, *r* = 1, *R*, customers who leave the buffer due to impatience is

$$\lambda\_{imp}^{(r)} = \sum\_{n=2}^{N+1} \pi\_n(I\_{WM} \otimes L\_{n-1}(\gamma\_r)) \mathbf{e}\_n$$

where *γr* is the row vector of size *R* with all zero entries except the *r*-th entry which is equal to *γ<sup>r</sup>*.

The average intensity *λ* ˜ (*r*) of the type-*l*, *l* = *r* + 1, *R*, customers transformation to the type-*<sup>r</sup>*, *r* = 1, *R* − 1, customers is computed as

$$\tilde{\lambda}^{(r)} = \sum\_{l=r+1}^{R} \kappa\_l \mathcal{N}\_{buffer}^{(l)} p\_{l,r} \dots$$

The probability *P*(*r*) *imp*−*loss*, *r* = 1, *R*, of loss of an arbitrary type-*r* customer due to impatience can be computed

$$P\_{imp-loss}^{(r)} = \frac{\lambda\_{imp}^{(r)}}{\lambda\_r + \bar{\lambda}^{(r)}}.$$

Here, we assume that *λ* ˜ (*R*) = 0.

The probability of an arbitrary type-*r* customer loss upon arrival without trying to force out a customer with lower priority is

$$P\_{ent-loss-without-force-out}^{(r)} = (1 - q)\lambda\_r^{-1}\pi\_{N+1}(D\_r \otimes I\_{MK\_N})\mathbf{e}\_r \; r = \overline{1, \overline{\mathcal{R}}}.$$

The probability of an arbitrary type-*r* customer loss upon arrival despite an attempt to force out a customer with lower priority is

$$P\_{\textrm{ent-loss-width-force-out}}^{(r)} = q\lambda\_r^{-1}\mathfrak{r}\_{N+1}(D\_r \otimes I\_M \otimes \mathcal{E}\_r)\mathbf{e}\_r \, r = \overline{1,\overline{\mathcal{R}}\_r}$$

where the matrix *E* ˜ *r* has all zero entries except the diagonal entries which are equal to the diagonal entries of the matrix *E* ˆ *r*.

The probability of an arbitrary customer loss upon arrival is

$$P\_{ent-loss} = \frac{\sum\_{r=1}^{R} \left( (1 - q)\pi \mathbf{r}\_{N+1} (D\_r \otimes I\_{MK\_N}) \mathbf{e} + q \pi \mathbf{r}\_{N+1} (D\_r \otimes I\_M \otimes E\_r) \mathbf{e} \right)}{\lambda}$$

.

The probability of an arbitrary type-*r* customer loss upon arrival is

$$P\_{ent-loss}^{(r)} = P\_{ent-loss-width-force-out}^{(r)} + P\_{ent-loss-width-force-out}^{(r)}r = \overline{1,\overline{R}}.$$

The probability that an arbitrary type-*r* customer meets the full buffer upon arrival and forces out a customer with lower priority is

$$P\_{force-out}^{(r)} = q\lambda\_r^{-1}\pi\_{N+1}(D\_r \otimes I\_M \otimes E\_r) \mathbf{e}\_r \ \mathbf{r} = \overline{1, R}\_r$$

where the matrix *E* ¯ *r* = *E* ˆ *r* − *E* ˜ *r*.

Let the square matrix *E* ˆ *<sup>r</sup>*,*l*, *r* = 1, *R* − 1, *l* = *r* + 1, *R*, of size (*<sup>N</sup>*+*R*−<sup>1</sup> *R*−1 ) define the transition probabilities of the process *ζ*(*N*) *t* , *t* ≥ 0, at the moment when a type-*r* customer arrives to the system when there are *N* customers in the buffer and the arriving customer forces out a type-*l* customer from the buffer. This matrix is defined by analogy with the matrix *E* ˆ *r* defined above. All entries in each row of this matrix are equal to zero except one entry which can be equal to 1. We assume that each row and column of the matrix *E* ˆ *<sup>r</sup>*,*l* correspond to some state {*η*1, *η*2, ... , *ηR*} of the process *ζ*(*N*) *t* , *t* ≥ 0. In the row of the matrix *E* ˆ *<sup>r</sup>*,*l* that corresponds to the state {*η*1, *η*2, ... , *ηR*}, the entry 1 is located in the column that corresponds to the state {*η*1, ... , *η<sup>r</sup>*−1, *ηr* + 1, *ηr*+1, ... , *ηl*−1, *ηl* − 1, 0, ... , 0} only in the case if *ηm* = 0 for all *m*, *R* ≥ *m* > *l*, and *ηl* > 0. If this condition is false, all entries of this row are zero entries.

The intensity *λ*(*r*) *f orce*−*out* of forcing out from the buffer type-*<sup>r</sup>*, *r* = 2, *R*, customers is

$$
\lambda\_{forrc-out}^{(r)} = q \sum\_{l=1}^{r-1} \pi\_{N+1}(D\_l \otimes I\_M \otimes \mathcal{E}\_{l,r}) \mathbf{e}.
$$

The probability *Pf orce*−*loss* of the loss of an arbitrary customer due to forcing out is

$$P\_{force-loss} = \frac{\sum\_{r=2}^{R} \lambda\_{force-out}^{(r)}}{\lambda}.$$

The probability *P*(*r*) *f orce*−*loss* of the loss of an arbitrary type-*<sup>r</sup>*, *r* = 2, *R*, customer due to forcing out is

$$P\_{force-loss}^{(r)} = \frac{\lambda\_{force-out}^{(r)}}{\lambda\_I + \tilde{\lambda}^{(r)}}.$$
