=Paper=
{{Paper
|id=Vol-2792/paper12
|storemode=property
|title=Simulation a Modified Erlang Loss System with Multi-type Servers and Multi-type Customers
|pdfUrl=https://ceur-ws.org/Vol-2792/paper12.pdf
|volume=Vol-2792
|authors=Stepan Rogozin
}}
==Simulation a Modified Erlang Loss System with Multi-type Servers and Multi-type Customers==
https://ceur-ws.org/Vol-2792/paper12.pdf
Simulation a Modified Erlang Loss System with
Multi-type Servers and Multi-type Customers?
Stepan Rogozin
Institute of Applied Mathematical Research Karelian Research Centre RAS,
Petrozavodsk, Russia
ppexa@mail.ru
Abstract. We consider a modified multiserver Erlang loss system with
two-priority classes of customers: the first priority customers (class-1) are
lost if find all servers busy, while the second priority customers (class-2)
join an infinite capacity queue if find all servers busy. A new feature of
this system is that a multi-type servers assignment for the first priority
customers is allowed. We assume Poisson inputs and general service times
for both classes. We show how the product form of class-1 stationary
probabilities can be used to obtain the stability condition of the whole
system. Also we perform simulation to confirm theoretical results.
Keywords: Modified Erlang System · Two-Priority Customers · Multi-
type Customers · Multi-type Servers · Simulation
1 Introduction
Internet traffic has been explosively increased in recent years and the most likely
it will grow in a future. Increased using of the smartphones, tablet and laptop
computers, smart watches etc. causes a spectrum shortage problems in wire-
less networks. One of the way to solve this problem is to use cognitive wireless
networks [3]. There can be two classes of users in these networks: primary (li-
censed) users and secondary (un-licensed) users. Secondary users may use the
bandwidths only if they do not interfere primary users and, in particular, if
primary users are not present. Primary users can interrupt transmissions of sec-
ondary users, if there are no free channels. In this case secondary user evacuates
to the head of the buffer. When one of the channels becomes available, secondary
users can resume transmissions. We also assume that different channels in gen-
eral have different transmission rates and can accept a limited set of classes of
the primary users. This assumption is related to the notion of flexible servers
[9,10]. This notion means that some service capacity can be transferred from one
pool of servers to another to satisfy different requirements. We also mention the
concept of cross-trained servers [11,12,13], where one pool of servers can serve
a limited set of customer types, while a second pool has been trained to serve
?
The research is supported by Russian Foundation for Basic Research, project No.
18-07-00156.
Copyright © 2020 for this paper by its authors. Use permitted under
Creative Commons License Attribution 4.0 International (CC BY 4.0).
all types of customers. Models with flexible servers and multiclass multiserver
setting often use for the purpose of finding an optimal allocation by minimizing
a cost function [14,9,15].
Based on this motivation we will study the following queueing system. First,
following work [1], we consider a modified Erlang loss system with c identical
servers and two classes of customers: the first priority customers (class-1) are
lost if find all servers busy, while the second priority customers (class-2) join an
infinite capacity queue if find all servers busy. Class-1 customers have absolute
priority over class-2 customers meaning that the transmission of a class-2 cus-
tomer might be interrupted by a class-1 customer. Interrupted class-1 customers
evacuate to the head of the buffer and resume their transmission as soon as a
server is available.
In the present research we study an extension of this system which in fact is a
combination of described modified Erlang system and another multi-class multi-
server Erlang-type system introduced and studied in the paper [2]. In this system
each server is served by only a limited set of customer classes and different classes
of customers in general have different arrival rates. Also each server in general
has different service rates. The customer assignment probabilities appear in this
system: for each customer class and for each set of idle servers the assignment
probability of this class customers to each available server must be given. The
analysis performed in the work [2] shows that one can choose the assignment
probabilities in such a way that the system becomes reversible (that is described
by a reversible Markov process) and in this case the stationary probabilities {Pi }
can be found in an explicit form. These stationary probabilities will appear in
stability condition of our system (1) and mean the probabilities that the system
has i busy servers by class-1 customers.
In this work we combine these systems in such a way that the first priority
customers (class-1) considered in the work [1] are divided into I subclasses which
are the classes of customers in the paper [2]. We note that the stationary proba-
bilities in this system are the same as in [2], since the second priority customers
do not affect the assignment and the service of the first priority customers. We
show that stability condition of the system described in [1] is also the stability
condition of our extended system provided the service rates are the same for
all servers, as in classical multi-server systems (identical servers). In this case
one can use found stationary probabilities {Pi } to obtain stability condition (1)
(see below) of our extended system. We note that this condition is correct if the
assignment probabilities of class-1 customers satisfy system equations (12) (see
below). In turn system (12) is adapted from [2].
A similar model was studied by Akutsu and Phung-Duc [3] where primary
customer first sense the channels before occupying them. In this work Poisson
arrivals and exponential service times of all classes of users are assumed. The
stability condition is suggested and verified by simulation in [3]. Other queueing
models of cognitive radio networks could be found in the papers [4,6,7,8].
The rest of paper is the following. In Section 2 we describe an extended sys-
tem and show the stability condition of this system. In Section 3 we present
some preliminary results from the work [2]. In Section 4 we construct an ex-
ample of the extended system to show in detail how to find the assignment
probabilities (which must satisfy some balance equations) and then calculate
the required stationary probabilities {Pi }. In Section 5 we perform simulation
to demonstrate the stability/instability of the system depending on whether the
conditions on the assignment probabilities are met or violated the mentioned
balance equations.
2 Description of the System
In this work we study the extension of the following system with J identical
servers and two-priority classes of customers: the first priority customers (class-
1) are lost if find all servers busy, while the second priority customers (class-
2) join an infinite capacity queue if find all servers busy. Customers of class-i
follow Poisson input with rate λ(i) . The service discipline for class-2 customers
is assumed to be FCFS (first-come-first-served). Also we assume that class-i
(i)
customers have independent identically distributed (iid) service times {Sn }
with generic element S (i) . Denote by ρ2 = λ(2) ES (2) the traffic intensity of class-
2 customers and let Pi be the stationary probability that i servers are occupied by
class-1 customers. Such a modified Erlang loss system is motivated and studied
in the paper [1] in which in particular the following stability criterion of this
system is obtained:
J
X
ρ2 + iPi < J. (1)
i=1
A new distinctive feature of the modified system is that a multi-type server
assignment for the priority customers is assumed. Each server can serve only a
limited set of subclasses of class-1 customers and different subclasses of customers
in general have different arrival rates. We assume Poisson inputs and server-
dependent service rates for the sets of priority subclasses. At the same time,
class-2 customers belong to only one type and can be served by any server.
More exactly, we assume J numbered servers {1, ..., J} and the first priority
(class-1) customers are multi-class divided on subclasses {1, ..., I}. Subclass cus-
tomer i can be served by servers from a set S(i), and server j can serve customers
from subclasses C(j), for each i = 1, ..., I and j = 1, ..., J. Customers of subclass
i arrive at rate λi , and in general, server j has rate µj .
Now we show that condition (1) is also the stability condition of this extended
system provided the service rates are the same for all servers. We apply the proof
from [1] (see Section 2) and show that relations (20),(21) in [1] are correct for
the extended system. Specifically, we show that the completed work (i.e., the
total service time) of class-1 customers in interval [0, t) satisfies:
J
Z tX
i1(Q1 (u) = i)du, t ≥ 0, (2)
0 i=1
where 1 is the indicator function and Q1 (t) is the number of servers occupied by
class-1 customers at instant t. Indeed, we can split indicators in (2) into 2J − 1
non intersection subsets of indicators:
J
X
1(Q1 (t) = 1) = 1({only the i-th server is busy at instant t}),
i=1
X
1(Q1 (t) = 2) = 1({only the i-th and the j-th servers
i,j∈{1,..,J}, i6=j
are busy at instant t}),
...
1(Q1 (t) = J) = 1({all servers are busy at instant t}).
Then we can obtain the corresponding stationary probabilities as in relations (21)
in the paper [1]. We denote by V̂1 (t) the work of class-1 customers accepted by
the system in time interval [0, t]. Note that V̂1 (t) does not include the lost work.
Let W1 (t) be the workload (the remaining work to be done) of class-i customers
at instant t− . It is easy to see that,
J
Z tX
V̂1 (t) = [i1(Q1 (u) = i)]du + W1 (t), t ≥ 0. (3)
0 i=1
Because the number of class-1 customers is upper bounded as Q1 (t) ≤ J, then
it is easy to show that the class-1 customer queue is a positive recurrent regen-
erative process [5] and the following limit exists:
Z J
EV̂1 (t) 1 tX
lim = lim [iP(Q1 (u) = i)]du =
t→∞ t t→∞ t 0
i=1
J J
1 t
X Z X
= i lim P(Q1 (u) = i)du = iPi . (4)
t→∞ t 0
i=1 i=1
The key observation is that the stationary probability Pi , that i servers are
occupied by class-1 customers, can be represented as the limiting fraction of
the corresponding busy time. This observation implies the last equality in (4).
The rest of the proof is the same as in [1]. We note that if the service rates
are different then (2) does not express the completed work of class-1 customers,
and the equality (3) does not hold. Therefore we further assume that the service
rates are the same for all servers.
3 Preliminary Results
Now we present some results from paper [2]. We denotes by X(t) the set of
numbers of servers which are not busy by class-1 customers at time t. The state
space of the process {X(t)} is S ⊆ {1, ..., J}, and is all possible combinations
of the numbers of servers. When the system is in state S an arriving class-
1 customer of subclass i selects a server j ∈ S with the probability Pi,j (S).
These assignment probabilities are the control parameters that we can choose to
obtain the stationary distribution of the system. If i ∈ / C(j) then Pi,j (S) = 0. If
S = {k} and i ∈ C(k) then Pi,k (S) = 1 for all k ∈ {1, ..., J}. We note that class-
2 customers do not affect the assignment and the service of class-1 customers.
Therefore all results obtained in the paper [2] for the loss system with multi-class
customers and multi-class servers and briefly presented below are applicable for
our extended system.
We assume Poisson input and exponential service times for all of customer
subclasses. Under this assumption the process {X(t)} is a continuous-time Markov
chain. If we have arbitrary service time distribution, we obtain the same station-
ary distribution. See the proof of this in Proposition 7 in the paper [2]. Now we
show that one can choose the assignment probabilities in such a way that one
can obtain the stationary distribution of the process {X(t)} (as in the paper [2])
and also obtain probabilities Pi in (1).
Now we assume that the process {X(t)} is reversible. Actually, in the paper
[2] this statement is claimed but not shown in details (see Section 3 in [2]),
and we check it for our example by conducting simulation in Section 5. Because
{X(t)} is reversible, then the following detailed balance equations hold:
π(S)νj (S) = π(S\{j})µj for all subset S ⊆ {1, ..., J} and j ∈ S, (5)
where π(S) is the stationary probabilities that servers S are not busy by class-1
customers, νj (S) is the rate at which server j ∈ S becomes busy, when the system
is in state S. From these equations one can obtain the stationary distribution
for S = {j1 , ..., jm }, j1 , ..., jm ∈ {1, ..., J} and m = 1, ..., J:
µj1 µj2 µj3 µjm
π(S) = π(∅) ··· , (6)
νj1 ({j1 }) νj2 ({j1 , j2 }) νj3 ({j1 , j2 , j3 }) νjm (S)
where π(∅) normalizes the sum to 1. Further we denote
π = π(S), π (k) = π(S\{k}), π ({j,k}) = π(S\{j, k}),
(k)
νj = νj (S) and νj = νj (S\{k})
for all S ⊆ {1, ..., J} and j, k ∈ S.
Proposition 1. If detailed balance equations (5) hold, then the following recur-
sion holds:
(k)
νj νj
= (j) . (7)
νk ν k
Proof. From the detailed balance equations we have:
πνj = π (j) µj ,
πνk = π (k) µk .
Then we obtain:
(k)
π (k) νj = π ({j,k}) µj , j 6= k,
(j)
π (j) νk = π ({j,k}) µk , j 6= k,
and finally we obtain:
(k) π ({j,k}) µj µj µk
νj = (k)
= π ({j,k}) , (8)
π πνk
(j) π ({j,k}) µk µj µk
νk = = π ({j,k}) . (9)
π (j) πνj
Now recursion (7) follows from (8) and (9) .
We denote by ν(S) the rate at which one of the idle servers becomes busy
when the system is in state S. It is clear that the following relations holds for
all S ⊆ {1, ..., J}: X X
ν(S) = νj (S) = λi . (10)
j∈S i∈C(S)
Proposition 2. The equations (7) and (10) uniquely determine the values of
νj (S) by the following recursion for all S ⊆ {1, ..., J} and j ∈ S:
(j)
X νk
νj (S) = ν(S) 1+ (k)
. (11)
k∈S\{j} νj
The proof is by induction on size of S (see Proposition 1 in the paper [2]).
This result shows that the stationary probabilities (6) can be expressed ex-
plicitly, using (10) and (11).
Proposition 3. There exist assignment probabilities Pi,j (S), for all S ⊆ {1, ..., J},
j ∈ S, and i ∈ C(S), which satisfy the following relations:
X
νj (S) = λi Pi,j (S). (12)
i∈C(j)
The proof of this statement can be found in Propositions 2-6 in the [2]. It is
important to note that if we choose the assignment probabilities satisfying (12)
then the system will be reversible and have the stationary distribution (6) (see
Section 3 in [2]).
To obtain the probabilities in stability condition (1) we should sum up the
stationary probabilities (6) as follows:
X
Pk = π({j1 , ..., jJ−k }) for all k ∈ {1, ..., J}. (13)
j1 ,...,jJ−k ∈{1,...,J}
Substituting this expression in inequality (1), we obtain the stability condition
of the extended system in an explicit form.
4 An Example
In this Section, we consider an example of the new system to show in detail how
to find stationary and assignment probabilities.
Example 1.
Fig. 1. Example
Let J = 3, I = 2, C(1) = {1, 2}, C(2) = {1, 2} and C(3) = {1} (see Fig.1.).
By (6) we have:
µ1 µ2 µ3
π({1}) = π(∅) , π({2}) = π(∅) , π({3}) = π(∅) ,
ν1 ({1}) ν2 ({2}) ν3 ({3})
µ1 µ2
π({1, 2}) = π(∅) ,
ν1 ({1}) ν2 ({1, 2})
µ1 µ3
π({1, 3}) = π(∅) ,
ν1 ({1}) ν3 ({1, 3})
µ2 µ3
π({2, 3}) = π(∅) ,
ν2 ({2}) ν3 ({2, 3})
µ1 µ2 µ3
π({1, 2, 3}) = π(∅) .
ν1 ({1}) ν2 ({1, 2}) ν3 ({1, 2, 3})
It is easy to see that:
ν1 ({1}) = λ1 + λ2 , ν2 ({2}) = λ1 + λ2 , ν3 ({3}) = λ1 .
By relation (11) we have:
ν({1, 2}) = λ1 + λ2 , ν({2, 3}) = λ1 + λ2 , ν({1, 3}) = λ1 + λ2 ,
ν({1, 2, 3}) = λ1 + λ2 .
From formula (11) it follows that,
−1
ν2 ({2}) λ1 + λ2
ν1 ({1, 2}) = ν({1, 2}) 1 + = ,
ν1 ({1}) 2
−1
(λ1 + λ2 )2
ν3 ({3})
ν1 ({1, 3}) = ν({1, 3}) 1 + = .
ν1 ({1}) 2λ1 + λ2
Similarly, we obtain:
λ1 + λ2
ν2 ({1, 2}) = ,
2
(λ1 + λ2 )2
ν2 ({2, 3}) = ,
2λ1 + λ2
λ1 (λ1 + λ2 )
ν3 ({1, 3}) = ν3 ({2, 3}) = ,
2λ1 + λ2
−1
ν2 ({2, 3}) ν3 ({2, 3}) (λ1 + λ2 )(2λ1 + λ2 )
ν1 ({1, 2, 3}) = ν({1, 2, 3}) 1 + + = ,
ν1 ({1, 3}) ν1 ({1, 2}) 2(3λ1 + λ2 )
−1
ν1 ({1, 3}) ν3 ({1, 3}) (λ1 + λ2 )(2λ1 + λ2 )
ν2 ({1, 2, 3}) = ν({1, 2, 3}) 1 + + = ,
ν2 ({2, 3}) ν2 ({1, 2}) 2(3λ1 + λ2 )
−1
ν1 ({1, 2}) ν2 ({1, 2}) λ1 (λ1 + λ2 )
ν3 ({1, 2, 3}) = ν({1, 2, 3}) 1 + + = .
ν3 ({2, 3}) ν3 ({1, 3}) 3λ1 + λ2
Now we can write down the stationary probabilities π(S) by substituting the
obtained above rates νi (S) in expression (6):
µ1 µ2 µ3
π({1}) = π(∅) , π({2}) = π(∅) , π({3}) = π(∅) ,
λ1 + λ2 λ1 + λ2 λ1
2µ1 µ2
π({1, 2}) = π(∅) ,
(λ1 + λ2 )2
µ1 µ2 (2λ1 + λ2 )
π({1, 3}) = π(∅) ,
λ1 (λ1 + λ2 )2
µ2 µ3 (2λ1 + λ2 )
π({2, 3}) = π(∅) ,
λ1 (λ1 + λ2 )2
2µ1 µ2 µ3 (3λ1 + λ2 )
π({1, 2, 3}) = π(∅) , (14)
λ1 (λ1 + λ2 )3
where, recall, we use normalization condition to find π(∅). Now, using (13), it is
easy to write down the stationary probabilities Pi :
P0 = π({1, 2, 3}),
P1 = π({1, 2}) + π({2, 3}) + π({1, 3}),
P2 = π({1}) + π({2}) + π({3}),
P3 = π(∅). (15)
Finally, from (12), we can obtain the assignment probabilities as follows:
λ1 P1,1 ({1, 2}) + λ2 P2,1 ({1, 2}) = ν1 ({1, 2}),
λ1 P1,1 ({1, 3}) + λ2 = ν1 ({1, 3}),
λ1 P1,1 ({1, 2, 3}) + λ2 P2,1 ({1, 2, 3}) = ν1 ({1, 2, 3}),
λ1 P1,2 ({1, 2}) + λ2 P2,2 ({1, 2}) = ν2 ({1, 2}),
λ1 P1,2 ({2, 3}) + λ2 = ν2 ({2, 3}),
λ1 P1,2 ({1, 2, 3}) + λ2 P2,2 ({1, 2, 3}) = ν2 ({1, 2, 3}),
λ1 P1,3 ({1, 3}) = ν3 ({1, 3}),
λ1 P1,3 ({2, 3}) = ν3 ({2, 3}),
λ1 P1,3 ({1, 2, 3}) = ν3 ({1, 2, 3}),
P1,1 ({1, 2}) + P1,2 ({1, 2}) = 1,
P1,1 ({1, 3}) + P1,3 ({1, 3}) = 1,
P1,2 ({2, 3}) + P1,3 ({2, 3}) = 1,
P2,1 ({1, 2}) + P2,2 ({1, 2}) = 1,
P1,1 ({1, 2, 3}) + P1,2 ({1, 2, 3}) + P1,3 ({1, 2, 3}) = 1,
P2,1 ({1, 2, 3}) + P2,2 ({1, 2, 3}) = 1.
Solving this system we obtain:
λ1 + λ2
λ1 P1,1 ({1, 2}) + λ2 P2,1 ({1, 2}) = ,
2
λ1
P1,1 ({1, 3}) = ,
2λ1 + λ2
λ1 + λ2
P1,3 ({1, 3}) = ,
2λ1 + λ2
λ1
P1,2 ({2, 3}) = ,
2λ1 + λ2
λ1 + λ2
P1,3 ({2, 3}) = ,
2λ1 + λ2
P1,2 ({1, 2}) = 1 − P1,1 ({1, 2}),
P2,2 ({1, 2}) = 1 − P2,1 ({1, 2}),
(λ1 + λ2 )(2λ1 + λ2 )
λ1 P1,1 ({1, 2, 3}) + λ2 P2,1 ({1, 2, 3}) = ,
2(3λ1 + λ2 )
(λ1 + λ2 )
P1,3 ({1, 2, 3}) = ,
3λ1 + λ2
2λ1
P1,2 ({1, 2, 3}) = − P1,1 ({1, 2, 3}),
3λ1 + λ2
P2,2 ({1, 2, 3}) = 1 − P2,1 ({1, 2, 3}). (16)
This system has infinity number of solutions, because we can vary some prob-
abilities, for example P1,1 ({1, 2}) and P1,1 ({1, 2, 3}). We give some solutions of
this system in Section 5 below.
5 Simulation
To check theoretical results presented above, we conduct discrete-event simula-
tion of the system described in Example 1. In addition, we assume that
λ1 = λ2 = 10, µ1 = µ2 = µ3 = 10.
Now we can calculate theoretical stationary probabilities (14):
π(∅) = 1/6, π({1}) = 1/12,
π({2}) = 1/12, π({3}) = 1/6,
π({1, 2}) = 1/12, π({1, 3}) = 1/8,
π({2, 3}) = 1/8, π({1, 2, 3}) = 1/6. (17)
Also we can write down the stationary probabilities Pi from (14) and (15)
P0 = 1/6, P1 = 1/3, P2 = 1/3, P3 = 1/6. (18)
S = {1, 2} S = {1, 3} S = {2, 3} S = {1, 2, 3}
P1,1 P1,2 P2,1 P2,2 P1,1 P1,3 P1,2 P1,3 P1,1 P1,2 P2,1 P2,2 P1,3 δ(t)
1 1 1
0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 3 3
0.5 0.5 3
0.0322
1 1 2
0.5 0.5 0.5 0.5 0.0 1.0 0.0 1.0 6 6
0.5 0.5 3
0.0459
1 2 1 2 2 1 2 1 1 2 1 2 1
3 3 3 3 3 3 3 3 3 3 3 3 3
0.0870
0.0 1.0 0.0 1.0 1.0 0.0 1.0 0.0 0.0 1.0 0.0 1.0 0.0 0.1559
Table 1. The assignment probabilities violating equation system (16): row 1 is uniform
distribution; row 2 is uniform servers workload; row 3 is high workload on server 2; row
4 is maximum workload on server 2.
To compare these probabilities with simulation results, we use the Euclidean
distance between two probabilities distribution:
s X
δ(t) = (π(S) − π ∗ (S, t))2 ,
S⊆{1,...,J}
where
T (S, t)
π ∗ (S, t) = , t > 0,
t
S = {1, 2} S = {1, 3} S = {2, 3} S = {1, 2, 3}
P1,1 P1,2 P2,1 P2,2 P1,1 P1,3 P1,2 P1,3 P1,1 P1,2 P2,1 P2,2 P1,3 δ(t)
1 2 1 2
0.1 0.9 0.9 0.1 3 3 3 3
0.1 0.4 0.65 0.35 0.5 0.0016
1 2 1 2
0.1 0.9 0.9 0.1 3 3 3 3
0.3 0.2 0.45 0.55 0.5 0.0013
1 2 1 2
0.1 0.9 0.9 0.1 3 3 3 3
0.5 0.0 0.25 0.75 0.5 0.0010
1 2 1 2
0.5 0.5 0.5 0.5 3 3 3 3
0.1 0.4 0.65 0.35 0.5 0.0009
1 2 1 2
0.5 0.5 0.5 0.5 3 3 3 3
0.3 0.2 0.45 0.55 0.5 0.0012
1 2 1 2
0.5 0.5 0.5 0.5 3 3 3 3
0.5 0.0 0.25 0.75 0.5 0.0009
1 2 1 2
0.9 0.1 0.1 0.9 3 3 3 3
0.1 0.4 0.65 0.35 0.5 0.0013
1 2 1 2
0.9 0.1 0.1 0.9 3 3 3 3
0.3 0.2 0.45 0.55 0.5 0.0012
1 2 1 2
0.9 0.1 0.1 0.9 3 3 3 3
0.5 0.0 0.25 0.75 0.5 0.0015
Table 2. Cases of assignment probabilities that satisfy equation system (16): we can
vary only 2 marked probabilities; δ(t) is much less than in Table 1.
Fig. 2. Difference between theoretical and simulated stationary probabilities (Sample
mean δ(t)): case a) corresponds to row 3 in Table 1; case b) corresponds to row 1 in
Table 1; case c) corresponds to row 1 in Table 2.
and T (S, t) is the time, in interval [0, t), when the system is in the state S. Also
we introduce l(t) (loss rate), the number of customer losses per time unit in
interval [0, t).
Fig. 3. Number of losses per time unit in interval [0, t) (Sample mean l(t)): case a)
corresponds to row 4 in Table 1; case b) corresponds to row 3 in Table 1; case c)
corresponds to row 1 in Table 2; case d) corresponds to row 2 in Table 1.
Fig. 4. Sample mean queue size of class-2 customers for Pareto service times for case
row 1 in Table 2. Stability condition (1) is ρ2 < 1.50.
We obtain the sample mean based on 100 paths of δ(t) and present their
stationary values in Tables 1 and 2. We consider some different sets of assign-
ment probabilities: 4 cases violate equation system (16) (see Table 1) and 9 cases
satisfy it (see Table 2). To satisfy equation system (16) we can vary P1,1 ({1, 2})
and P1,1 ({1, 2, 3}) (free variables, marked columns) while other probabilities de-
pend on free variables or are uniquely determined. It is easy to see that δ(t) in
Table 1 is much larger than in Table 2, that confirms that stationary probabili-
ties (17) are found correctly (for cases in Table 2) and the process is reversible.
One can see on Fig.2 that δ(t) converges for both in the cases when the system
(16) is satisfied and in the cases when it is violated. We also show that number
of customer losses l(t) also converges but for the cases when the system (16)
is violated l(t) is not much larger than for the cases when the system (16) is
satisfied (see Fig.3). Moreover, case d) shows that we can obtain l(t) less than
for the cases when the system (16) is satisfied. This example shows that product
form distribution is not optimal to minimize losses.
Finally, we calculate the stability condition (1) and, provided this condition is
satisfied, check stability by simulation queue size of class-2 customers for Pareto
service times. In our example the number of servers J = 3, and from (18) and
(1) we obtain the upper bound for ρ2 as
ρ2 < 1.5. (19)
In the experiments, we construct the sample mean queue size Q2 (t) based on
300 paths queue size of class-2 customers for different values of ρ2 (see Fig.4). It
is easy to see that, if (19) does not hold, then Q2 (t) increases linearly to infinity
reflecting strong instability. On the other hand, when condition (19) is satisfied
then we see that all paths are stable.
6 Conclusion
We consider a modified Erlang loss system with multi-class multi-server first pri-
ority customers and a queue for second priority customers. In this work we show
that the stationary distribution describing this system has the product form and
is the same as in the system considered in the paper [2]. Furthermore, we use
these stationary probabilities to calculate the stability condition of the entire
system expressed as the upper bound for the traffic intensity generated by the
second priority customers. Simulation illustrates that the assignment probabili-
ties, satisfying conditions in the considered example, indeed imply the product
form of the stationary distribution. Moreover, we show that the stability condi-
tion holds for the example studied in Section 4. On the other hand, simulation
also shows that the product form distribution setting is not optimal to minimize
the losses in the system.
7 Acknowledgement
The authors thanks his adviser Prof. Evsey Morozov for help and useful com-
ments.
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