CC BY
38
Single server queues with several services
1 A. Krishnamoorthy 1,a, 2 Vladimir Vishnevsky 2,b
3 Manjunath A. S. 3'c, 4 Dhanya Shajin 4-d •
1 Department of Mathematics Cochin University of Science and Technology Cochin-682022, India.
2 V. A. Trapeznikov Institute of Control Sciences of Russian Academy of Sciences, 65 Profsoyuznaya Street, Moscow 117997, Russia. 3Department of Mathematics Government Victoria College, Palakkad-678001, India.
4Department of Mathematics Amrita School of Arts and Sciences Amrita Vishwa Vidyapeetham, Brahmasthanam Kochi-682024, India.
Abstract
A service station offers several services of which one is the required and the others are harmful (or destructive) for each customer. At the time when selected for service customers enter in correct mode of service according to a Bernoulli process with parameter p which is the probability of being selected in correct mode. Arrival follows Markovian arrival process and service time is Phase type distributed in both undesired and required phases. An exponentially distributed threshold clock starts ticking if a customer enters to incorrect mode and the service is terminated if the clock realizes before the customer is transferred to correct service mode. The rate of loss of customers, rate of customers leaving with correct service starting with incorrect service are computed.
Keywords: desired/undesired service states, random threshold clock, Markovian arrival process
1 Introduction
In the analysis of classic queueing models it is assumed that the server is completely aware of the exact service requirement of a customer (see Gross and Harris [2]). Quite often only one type of service is offered by the system and so conflict does not occur. It is also true that the customer
1 Author for correspondence: email: achyuthacusat@gmail.com a Research is supported by the Department of Science and Technology, Government of India under grant INT/RUS/RSF/P-15 and Kerala State Council for Science, Technology & Environment (No.001/KESS/2013/CSTE).
2 email: vishn@inbox.ru b Research supported by the Russian Science Foundation and the Department of Science and Technology (India) via grant No. 16-49-02021 for the joint research project by the V. A. Trapeznikov Institute of control Science and the CMS College Kottayam.
3 email: manjunadem@gmail.com ° Research is supported by the University Grants Commission, Government of India, under Faculty Development Programme (Grant No. F.FIP/12£h Plan/KLMG003TF05) in Department of Mathematics, Cochin University of Science and Technology, Cochin-682022, India.
4 email: dhanya.shajin@gmail.com d Research is supported by Kerala State Council for Science, Technology & Environment (No.001/KESS/2013/CSTE).
knows the type of service he needs. Thus there is no conflict on the service provided to the customer. However, there are several real life situations where the server and/customer are (is) not knowledgeable about the exact service requirement. This is especially the case when several types of services are available at a service station. As concrete example we have vehicles for repair at service stations, patients consulting physicians for diagnosis and medication. If the right service required is not identified and instead the diagnosis turned out to be wrong the result could be disastrous. A wrong diagnosis and consequent service provided may sometimes turn out to be even fatal/ may result in the equipment getting service, rendered totally unusable. It is this type of problem that we analyze in this note.
In real life there are several service providing system offering a multitude of service. Neither the server nor the customer may be fully aware of the exact service requirement. This, very often, results in irreperable damage to the customer being served. We study such a system in this paper with each customer requiring a specific service. However, due to wrong diagnosis service does not start necessarily for the required one.
The article is organized as follows. In section 2, the mathematical model is described. This section also provides the steady state analysis and some performance measures. Several cases of the above model are considered in Sections 3 and 4. An illustration is given in Section 5. Numerical illustrative example is described in Section 6. In Section 7 we extend the above results to the case of arbitrarily distributed service time in the undesired and desired stages of service. However, we restrict n1 = n2 = 1. Further we relabel the undesired service as preliminary and desired as main services to suit certain context of application.
List of notations and abbreviations used:
CTMC: Continuous time Markov chain
LIQBD: Level independent quasi-birth and death process
MAP: Markovian arrival process
LST: Laplace Stieltjes Transform
e: Column vector of 1's with appropriate order
2 Model Description
The assumptions leading to the formulation of the mathematical model are
• An infinite capacity queueing system where a single server is providing both unwanted (incorrect) and required (correct) services.
• Arrival of customers to the system is according to the MAP (Markovian arrival process). In a MAP, the customers arrival is directed by an irreducible CTMC (continuous time Markov chain) [$t,t > 0} with the state space {1,2,...,m}. The transition intensities of the Markov chain [$t, t > 0} which are accompanied by arrival of k customers are described by the matrices Dk,k = 0,1. Vector q of the stationary distribution of the process {<pt,t > 0} is the unique solution to the system
q(D0 + D1) =qD = 0andqe = 1. (1)
Fundamental rate A of the MAP is given by A = yD^.
• A customer is selected for desired (required) service with probability p or to the incorrect service with probability q = 1 — p.
• ( ft 1,S1) of order n1 gives the PH-representation for the duration of the correct service time distribution when the service of a customer starts in correct service mode. Let 10 be such that S1e + S° =0. Let = ft 1(—S1)-1e be the mean of this PH-representation.
• (ft 2,S2) of order n2 gives the PH-representation for the duration of the incorrect service time distribution when the service of a customer starts in incorrect service mode. The rate
(vector) of loss is given by 0 and the rate (vector) of getting into correct service mode is given by S0 . Note that S2e + S0 + S0 = 0. Let p'2 = ft 2(-S2)-1e be the mean of this PH-representation.
• ( ft 3,S3) of order n3 gives the PH-representation for the duration of the correct service time distribution when the customer has gone through incorrect service initially. Let 3 be such that S3e + S30 = 0. Let n'3 = ft 3(S3)-1e be the mean of this PH-representation. [NOTE: One can take this to be same as ( ft 1,S1) but it looks more meaningful in most applications that the service time after going through incorrect one to be different from the use of directly getting into required service. Just something to keep in mind.]
• The service time of a customer can be modeled as a PH-distribution with representation ( ft ,S) of order n = n1 + n2 + n3, where
ft =(p ft^.q ho) (2)
5 =
'Si 0 0
0 S2 S0 ft :
0 0 S3
(3)
Let 0 be such that Se + S0 = 0 and S0 is given by 0 =[ 0
v
Let N(t) be the number of customers in the system, N*(t) the mode of service going on whether direct admission to required/ undesired or one that came from undesired service designated by 1,2 and 3 respectively, S(t) the phase of service and A(t) the phase of arrival at time t. With these the process {(N(t),N*(t),S(t),A(t)),t >0} is a continuous time Markov chain with state space H. = {0,1,2,...}, where
0 = {(0,r); 1 <r <m} (in the level zero we need consider only the phase of arrival) and
i = {(i,j, k,r); i > 1,1 < j < 3,1 < k < n}, 1<r <m}. Thus the infinitesimal generator of this CTMC is a LIQBD and is of the form
Q =
(^0 Aw
V
^01 Ai
a2
A0 Ai
An
\
)
where A0i= P® Di ,AW= S0 ® Im ,A0= In® Di ,Ai= S® D0 ,A2 = S°p ® Im .
2.1 Stability Condition
Consider A(= A0+ A1+ A2), the generator matrix of the Markov chain corresponding to the phase changes.
{(pi Pi+Si)®D q° 02® I,
m m
A = ( S + S0p )®D =
V0 Pi®',
p0 Pi®I,
12 H2
q0 P2® 1,
(q0 P2 + S2)®d 500 p3®i,
\
S3®D
Let n = ( n 1,n 2,n 3) be the steady-state probability vector of (S + S0 ft ). Then
n (S + S0 ft ) = 0and n e = l.
From the relation n (S + S0 ft ) = 0 we have
n 1(p0 P1+S1)+ n 2P 2 P1+ n 3P 0 01 = 0, n 1q 0 p2+ n 2(q 0 P2+S2)+ n 3q 0 02 = 0, n 2 S0 03+ n 3S3 = 0.
Multiplying equation (8) by e on right hand side we get
Putting this in equation (6) yields
^ 3 3 = n 2 S2 .
n i 0 = n 2^2^.
(4)
(5)
(6)
(7)
(8)
(9) (10)
0
Substituting relations (9) and (10) in equation (7) gives
2( 0 P2 + S2 H2 + ^2
This implies, for an arbitrary constant c
" 2 - ^P2(-J2
Substituting for n 2 in relation (10) we have
n 2 ( 2 P2+ S0 P2+S2)-0. (11)
n 2 - cP2(-S2)-1. (12)
n 1 = C-fP1(—S1)-1. (13)
Denote by 5 = P2(—S2)-1 S° the probability that a customer starting with incorrect service leaves the system after getting correct service. Then equation (9) gives
n 3 = cSp3(—S3)-1. (14)
From the normalizing condition n e = 1, the value of c is computed as
c ¿2 +Sv's] . (15)
Now from (1) and (5) we get the steady state probability vector of A as ft = n
Theorem 2.1 The system is stable if and only if
A<( n S0 p ® Im )e. (16)
Proof. The queueing system under study with the LIQBD type generator given in (2) is stable if and only if rate of left drift is less than the rate of right drift (see Neuts [6]), that is,
ft A0e < ft A2e. (17)
The left drift rate is ft ( In® D1 )e which when simplified reduces to A. Now, the right drift rate is ( n ®q)( S0 ft ® Im )e.
Let p =-£-. Then from (16), we have p < 1.
2.2 Steady-State probability vector
A brief outline for the computation of the stationary probability vector of the system is as follows. Let x denote the steady-state probability vector of the generator Q. Then
xQ = 0andxe = 1. (18)
Assuming that the stability condition (16) holds and partitioning x as x = (x0,x1,x2,...) we obtain
xn = x1Rn-1,n>1 (19)
where R is the minimal non negative solution to the matrix quadratic equation R2A2 + RA1 + A0 = 0. The two boundary equations involving x0 are
X0D0 + X1A10 = 0, (20)
XoAO1+X1[A1+RA2] = 0 (21)
These together with the normalizing condition in (18) gives
x1 = x0VwhereV = —A01[A1 + RA2]-1 (22)
Xo[I + V(I — R)-1]e = 1. (23)
To see how the system performs, it is instructive to define y = Xj. Then y = (^ y2 y3) where the yt, i = 1,2,3 indicates mode of service of the customer in service along with other system phases.
2.3 System Performance Measures
1. Probability that system is idle, Pidie = x0e = 1 — p.
2. Rate of loss of customers, Rioss = y2 S2° = Aq(1 — 5).
3. Probability that a customer is lost, Pioss = q(1 — 5).
4. Mean number of customers in the system, nNS = ixte.
= 1
5. Mean number of customers in the queue is given by p.NQ = Y,T=2 (i — l)xie.
yi S0 = Ap.
6. Probability that the server is serving in required mode, Pc = y1e + y3e = p — Aq^'2
7. Probability that the server is serving in unwanted mode, Pi = y2e = Aqp.'2.
8. Rate at which customers leave with required service starting in desired service mode, Rc =
Ji
9. Rate at which customers leave with correct service starting with unwanted service, R, = y3 S$ = AqS.
10. Expected waiting time in the system Ws = .
11. We define the system reliability at any time as the probability of customers in service is in desired mode of service Vreiiabuity = (Yi + ys)e.
3 Case of Poisson arrival and phase type service
In this section we consider the system with Poisson arrival process and service times are phase type distributed (see Section 2). Then {(N(t),N*(t),S(t)), t > 0} is a continuous time Markov chain with state space {0,1,2,...} where
i = {(i,j, k),1 < j < 3,1 < k < nrffori > 1.
/-A A ft \
Thus the infinitesimal generator is of the form Q' =
S0 S — AI AI
S0 p S —AI AI
\
Theorem 3.1 The system is stable if and only if p' <1 where
p' = A[p^i, + q(^2, + S^3,)].
(24)
Proof. From the relation (17) we have A < n S0 ft e where n = ( u i,K 2, ft 3) (with u i s as given in (12)-(14)) is the steady state probability vector of S + . The right drift n e =
I3=i $0 .
Multiplying (6) by e on right hand side we get
Y3=i n, S° =c--1 n, Sie = 1 ft) fti e (from(13)) = ^ where c is given in (15). Hence the condition for system stability is given by
A <—-i-— (25)
The generator matrix corresponding to the phase changes is S + S° ft and the stationary probability vector is n = (
ft i,U 2, ft 3).
Theorem 3.2 The steady-state probability vector x = (x0,xi,x2,---) of Q' is given by xo = 1 — p',xi = (1-p') ft R\i > 1,
where R is given by
AI — Ape — Si —Aqe 0
R=A-Ape fti AI — Aqe P2—S2 — S° &
-Ape —Aqe AI — S3
(26)
(27)
Proof. Let x be the steady-state probability vector of Q'. Then xQ' = 0 and xe = 1. The steady-state equations are given by
-Ax0 + xiS0 = 0, (28)
Xx0p + x1(S-Xl) + x2S0p=0, (29)
Axi-1 +Xi(S- AI) + xi+1S0p = 0,fori > 2. (30)
From (28) we have
x1S0 = Ax0. (31)
Multiplying equations (29) and (30) by the column vector e on the right hand side leads to
^+1S° = Axtefori > 1. (32)
Since Xi+1S0P = Ax{Bfori > 1 where B = e. 0, from (29) and (30) we obtain
x1 (AI — AB — S) = Ax0p (33)
and
x(AI — AB — S) = Axi-1,fori > 2. (34)
Denoting (AI — AB — S) by K, relation (33) takes the form x1 = Ax0pK-1, provided K is invertible. We now prove the nonsingularity of K.
Let the vector u be in the left kernal of K. Then
Au — uS — A(ue)P = 0. (35)
Suppose ue = 0. Then (35) reduces to u(AI — S) = 0. But (AI — S) is nonsingular and hence u = 0. If ue ± 0, normalize u by setting ue = 1. Post multiplying (35) by e gives
uS° = 0. (36)
Substituting for ue, (35) reduces to u = Afi(A! — S)-1. From (36) we have
A0(AI — S)-1S0 = 0. (37)
In (37) P(AI — S)-1S° is the Laplace-Stieltjes transform at s = A(> 0), of the probability distribution F(t) = 1 — fiexp(St)e for t > 0. Therefore (37) cannot hold and hence u = 0. Thus K is nonsingular.
The irreducibility of the representation (fi,S) leads to the irreducibility of K, so that the matrix R in (27) is positive.
We have sp(R) < 1, if p' < 1. Therefore the quantity x0 is given by the normalizing equation
x0 + x0pR(I — R)-1e = 1.
Substitution for R leads to
x0 - Ax0p(A'B + S)-ie = 1. (38)
The inverse of (AB + S) is calculated as
(AB + S)-i = S-i(I + ABS-i)-i = S-i Y,™=0 (-1)nAn(BS-i)n
in=0
= S-i[I - A[£™=0 (-1)nAn(BS-i)n]BS-i] = S-i[I -AZ™=0 p'nBS-i] = S-i[I - A(1 - p')-iBS-il
From (38) we have
-i 0 0
x0 - Ax0p(AB + S)-ie = x0 — Ax0p[S-i(I - A(1 - p')-iBS-i)]e
= x0 — Ax0pS-ie + A2x0(1 - p)-ipS-iBS-ie
= x0 + P'*0 + P'2 (1 — P')*0 = (1 — P')*0 = 1,
so that x0 = (1 — p').
Letting y = xt, it is obtained that y = p' n . In the sequel partition y = (y1,y2,y3), so that yt = p' n i,1 < i < 3.
4 Poisson arrival with exponential service
In this section we consider customers arrive according to the Poisson process with rate A and desired (correct) service time follows exponential distribution but the undesired (incorrect) service follows phase type distribution. Let N(t) be the number of customers in the system N*(t) the type of service and S(t) the phase of service at time t. Then {(N(t),N*(t),S(t)),t >0} is a continuous time Markov chain with state space {0,1,2,...} where
i = {(i, 1,0), (i, 3,r + 1)} U {(i, 2,j),1 <j< r}fori > 1. Thus the infinitesimal generator is of the form 0 12 3....
0
A bn
A1 A0 A2 A1 A0
where b0 = A(p, q0, 0), c0 = -A — n 0
§0
\,A0 = AI
A, =
S — AI
50
¿2
0
A — p.
\,A2 =
MP 0
P q §00 0
№ MP 0
c
0
0
with Se + S0 +S0 = 0
2
4.1 Stability condition
Consider A = A0+A1+ A2
'—p.q ^q0 0
p S0 S +q S00 S0 № MP —V
(39)
the generator matrix of the Markov chain corresponding to the phase changes. Let
n = (n0, ft, nr+1) be the steady state probability matrix of A. Solving the relations
nA = 0, ne = 1 (40)
we obtain
—^q n0 + pft S 0 + up nr+1 = 0 (41)
^qn00 + ft( S +q S 0 P)+y.q nr+10 = 0 (42)
ft S 0—nnr+1 = 0 (43)
Equation (43) gives
V.nr+1 = ft S 0 (44)
Putting this in equation(41),
Vq n0=p(ft S 0+ii S 2) (45)
Substituting these in equation(42) and simplifying we get
fi(S + q S 0 P) + pH ( S 0ft + S 0 P) + qH S 0 0 = 0 ^ ft ( § + § 0 P+ s 2 P) = 0
so that
ft = c 0(— S )-1 (46)
c being a constant and is computed from the normalizing condition. Let 5 be the probability that a customer getting correct service from incorrect services and q the probability of staying back in incorrect services. Then
8 = 0 (— S )-1 S 0 (47)
and
V = (p(— s )-1e)-1 (48)
Then the probability that a customer leaves the system without getting required service is
1 — 8 = 0 (— § )-1 § 0 (49)
and the mean time a customer stay back in incorrect services is
'~=(P(— S Y1*) (50)
The normalizing equation is n0+ft e + nr+1 = 1. Substituting for the components of n which are now computed as
pc ^ c cS .
n0 = —, it e =-, nr+1 = — (51)
w v M
we get - + - + — = 1 which shows
MVP
C= . (52)
Theorem 4.1 The system is stable if and only if A < ^ c.
Proof. The condition for the stability of the system is nA0e < nA2e. Simplification gives nA0e=A. Now A2e = 5° . Therefore nA2e = ^n0+ii( S 0+ S 0) Substituting for p.n0, right hand side becomes ift( S °+ S 2). Using equation(46) and the fact that ( S )-i( S ° + S 0) = e, the result follows. Hence the system is stable if p < 1 where p = A -.
4.2 Steady-State probability Vector
Let the steady state probability vector x of Q be x = (x*,x(1),x(2),...) be such that xQ = 0,xe = 1. Partitioning gives x(i) = (x0(i),-^ x(i),xr+i(i)). The relation xQ = 0 gives the following
system of equations.
—Ax* + x(l)c0 = 0 (53)
x*b0 + x(1)^1 + x(2)A2 = 0 (54)
Fori > 1, x(i — 1)A0 + x(i)A1 + x(i + 1)A2 = 0 (55)
From the matrix geometric structure we obtain
x(i) =x(1)Ri-1,i>1 (56)
where R is the minimal non negative solution to the matrix quadratic equation R2A2 + RA1 + A0 = 0. Equation (53) shows
1
x* = -x(1)c0. (57)
Equation (54) together with normalizing condition gives
x*b0 + x^)^ + RA2) = 0 (58)
subjecttox*e + x(1)(/ — R)-1e = 1. (59)
Substituting for x*,
x(1)(Ai+RA2+-xCobo) = 0 (60)
subjecttox(1) (jc0 + (I — R)-1e) = 1. (61)
But c0b0 = AA2 which implies
x(1)(Ai + RA2+A2) = 0 (62)
subjecttox(1) (jc0 + (I — R)-1e) = 1. (63)
4.2.1 Computation of R R can computed explicitly along the following lines. We have
A,
= lpS0
VP
HqP q S0P HqP
¥
p qp 0
(64)
<>o
so that
A2e =
= c0 (65)
Also from the relation RA2e = A0e, we obtain
RA2e = Ae. (66)
Now, R2A2 = R2 (si) (p qP 0) = R2A2e(p qP 0).
Substituting for RA2 from (66), we get
R2A2 = RAe(p qP 0) (67)
Therefore
ARe (p qP 0) + RA1 + AI = 0. (68)
This gives
(\i + Aq —AqP 0 \-1
R = A ( —Ape AI — Aqep — § — S2°) . (69)
\—Ap —MP A + ^ J
Lemma 4.2 x* = 1— p so that x(1)(I — R)-1e = p.
Proof. Multiplying by e on the right side of equation (54) and simplifying we get the
relation
f—A — ^\ m.
Ax* + x(1) ( S — AI+ S°) +x(2)(S10J = 0. (70)
Equation (53) gives
Ax* = x(1) (^f). (71)
Putting this in (70) the following relation is obtained.
x(2) (s°) = Ax(1)e. (72)
Multiplying equation(55) on right side by e and recursive use of the relation results in
x(i) (si) = Ax(i — 1)e for i > 3. (73)
Adding (71), (72) and (73)
Zr=1*(i)(si)=*. (74)
Adding the system of equations (55) with equation (54) and using the fact that x*b0 = x(1)A2 we get
Hr=1 *(Q A = 0. (75)
But the relation (40) says
x(i) = d n forsomeconstant c (76)
which in turn gives
2=1 x(i) =(1 — X*) n (77)
fi\
Multiplying on the right side by ( S°| and using the relation in (??)
ZT=1x(i) (s?) = (1— X*) ± (78)
KHJ P
The result follows from (74) and (78).
4.3 System Performance measures
5)
1. Probability that the system is idle = x*
2. Rate of loss = T,?=1 — x(i) S 0 = Aq(1
3. Probability of loss = q(l - S)
4. Mean number of customers in the system = tx(i)e = x(1)(I 5
R)-
Mean number of customers in the queue = (i - 1)x(i)e = x(1)(I - R) 2e - x(1)(I - R) 1e 6. Probability that the server is busy serving in correct mode
10) = x(1)(I - R)-1 (1) = p(n0 + nr
\ _
lr+1) = P~~
7. Probability that the server is busy serving in correct mode
0
IZiX(i)
= P
lq
— ne = —
v
5 An illustration
In this section we consider a queueing model consisting of two service stations - preliminary service and main service. Customers arrive to this system according to a MAP (Markovian Arrival Process) with representation (D0,D1) of order m. A customer, which taken for service is directly selected for main service with probability p or to the preliminary service with probability q(= 1 -p). A threshold clock starts ticking if a customer enters to preliminary service. When the duration of preliminary service exceeds the threshold clock, the customer move out of the system, else he goes to main service. The threshold clock follows exponential distribution with parameter Service times of the customers at these stations follow phase type distributions with representation (a,SP), (y,Sm) and of order a,b respectively. Write Sp + (e = -SPe and S^ = -SMe where e is a column vector of 1's of appropriate order. Hence service time of a customer can be modeled as a phase type distribution with representation ((, U) of order a + 2b such that Ue + U° = 0 where
f = (PY qa 0)
u =
0 0
0 Sp SpY ), yo =
0 0 \ ^m
Let N(t),N*(t),S(t),A(t) denote respectively the number of customers in the system, nature of service, phase of service and phase of arrival at time t.
e
!1 mainservice
2 preliminaryservice
3 onethatcomefrompreliminaryservice
The process H = {(N(T),N*(t),S(t),A(t)),t >0} is a continuous time Markov chain with state space {(n, i,j, k); i = 1,3,1 <j<b,1<k<m}U {(n, 2,j, k);1 < j < a,1 < k < m} for n> 1. Note that when N(t) = 0, the only other component in the state vector is A(t). Thus the infinitesimal generator of H is of the form
/Do Aoi \
Q* =
Aio Ai Ao
A2 Ai Ao
\ " " "J
(79)
where Aoi = f ® Di,Aw = U0 ® Im,Ao = Ia+2b ®D1,Ai = U® Do,A2 = U°$®Im.
The infinitesimal generator Q* given by (79) is of the same form as Q of the model described initially. Thus the analysis of the Markov chain with infinitesimal generator Q* can be done in the same way as for Q.
The significance of this model is as follows: customer arriving to a single server belong to two categories, though they join the same. Only while taken for service the category will be revealed. Call them category 1 and category 2, respectively. Category 1 are qualified for the main service without undergoing preliminary service. However, category 2 have to be given the preliminary service before admitted to mean service. However, if such customers do not get service in preliminary before realization of the timer (random clock), they get disqualified and so leave the system forever. On the other those among category 2, completing service successfully in preliminary are immediately admitted to main service. On completion of that service such customers leave the system.
Remark 5.1 In telecommunication it is this type of situation that is often encountered. Packages have to identify the server in idle state; then wait for a while. But in the mean time another message may get through, making the server busy. Then the customer (packet) under consideration has to go through a series of contention windows. These passages could be regarded as unwanted service. In case the process of going through contention windows exceeds a threshold time limit, the message will not get served.
Remark 5.2 The problem discussed in Madan [3 ] and Medhi [5] could be arrived at from our model as follows. Suppose that we reverse the order of preliminary and main service, that is, main service first and preliminary (hereafter we call it optional) service next. Then after completion of main service, the customer asks for an optional service with probability 1 — q (this optional service time has exponential distribution in Madan [3]). With probability q, the customer leaves the system immediately after main service completion. This model is also the same as a queue with instantaneous feedback after a service and immediate commencement of his service (feedback restricted to one). This feedback policy is referred to as queues with instantaneous feedback as head of the queue.
6 Numerical illustration
The following numerical illustration is based on the description in Section 2.
We fix parameters n1 = 2,n2 = 3,n3 = 4,pi = (0.4 0.6),p2 = (0.3 0.5 0.2),p3 (0.2 0.3 0.3 0.2),
S, =
* 6 8 *
r0 = =
with S1e + S0 = 0,
=
* 5 5
6 * 6
5 7 *
r0 = •¿2 =
c0 = • ¿2 =
with S_2 e + S_2A0 + S_2A0 = 0,
* 7 8 9 6
6 * 7 7 7
Ss = 6 6 * 6 r0 = 3 = 8
8 7 6 * 9
with S3e + SS = 0.
For the arrival process, we consider the following two sets of values for D0 and D1 as follows. The arrival processes labeled MNCA and MPCA respectively, have negative and positive correlation for two successive inter-arrival times with values -0.48891 and 0.48891. The standard deviation of the inter-arrival times of these two arrival processes are, respectively, 0.2819 and 0.2819.
1. MAP with negative correlation (MNCA):
Do =
'-5.0111 0 0
5.0111
-5.0111
0
0 0
-1128.75
,D1 =
0
0.05011 1117.4625
0
4.96099 112875 I
V p 1 l.nss Vns Pc P, Re R, ws
.4 0.2136 7.5229 0.5242 0.3921 2 1.9320 1.5046
.5 0.1780 4.9744 0.5483 0.3267 2.5 1.6100 0.9949
.6 0.1424 3.6690 0.5724 0.2614 3 1.2880 0.7338
.7 0.1068 2.8654 0.5965 0.1960 3.5 0.9660 0.5731
.8 0.0712 2.3138 0.6206 0.1307 4 0.6440 0.4628
.9 0.0356 1.9069 0.6447 0.0653 4.5 0.3220 0.3814
Table 1: Effect of p for MNCA
2. MAP with positive correlation (MPCA):
5.0111 5.0111 0 0 0 0
D = . 0 -5.0111 0 D 4.96099 0 0.05011
0 (0 0 -1128.75 r 1 \ 11.2875 0 1117.4625
V Ploss Vns Pc P, Rc R,
.4 0.2136 546.8179 0.5242 0.3921 2 1.9320 109.3646
.5 0.1780 349.9587 0.5483 0.3267 2.5 1.6100 69.9924
.6 0.1424 250.7699 0.5724 0.2614 3 1.2880 50.1545
.7 0.1068 191.0008 0.5965 0.1960 3.5 0.9660 38.2005
.8 0.0712 151.0402 0.6206 0.1307 4 0.6440 30.2083
.9 0.0356 122.4351 0.6446 0.0653 4.5 0.3220 24.4873
Table 2: Effect of p for MPCA
The output in Tables 1 and 2 are on expected lines. Note that Pioss decreases with increasing value of p. The value of Pc(Rc) steadily increases with p and values of Pi(Ri) and Ws decrease with increase in value of p, as expected.
The main comparison in Tables 1 and 2 is between values of in MNCA and MPCA. Both decrease with increase in value of p. However, MNCA has much smaller values compared to their MPCA counter parts. This indicates that positive correlation in the arrival process results in accumulation of large number of customers in the system.
7 M/G/l Model
In this section we consider an M/G/l system with two service stations - preliminary service and main service. Customers arrive to this system according to a Poisson process with rate X. A customer, when taken for service, is directly selected for main service with probability p or to the preliminary service with probability q (= l-p). A threshold clock starts ticking if a customer enters to preliminary service. When the duration of preliminary service exceeds the threshold clock, the customer moves out of the system, else he goes to main service. The threshold clock follows exponential distribution with parameter Ç. Here the service times, Vp,Vm of the preliminary and main services are independent having general distributions with distribution function G1 (. ), G2 (. ), LST G1 (. ), G* (. ) respectively.
The (total) service time V of a unit is
Vf withprobabilityq • P(G1(.) > exp(£)) Vp withprobabilityq • P(G1(.) < exp(£)) Vm withprobabilityp
V =
where Vf is the duration of threshold clock realization.
Thus
G(t) = P(V<t) = q [£ (e-^u(1 - Gi(u))du + f e-^uGi(u)dG2(t -u)]+p f dG2(u) and LST G*(s) of V is given by G*(s) = f™ e-stdG(t).
Remark 7.1 This modelling closely resembles the protocol IEEE 802.11. This is so because of a message generated has to wait before checking for idle server; if server is busy it has to go through a series of contention windows and then look for idle server. In case this process takes longer duration than the life of message (before its significance is lost), then the message does not serve any purpose. In the opposite case it is transmitted before its expiry time.
Remark 7.2 Assume the random clock to be of infinite duration (ie., its rate of realization goes to zero). Now interchange the roles of preliminary and main services (in this case, we call the preliminary service, which is the second one now, as optional service). Invariably main service is given for all customers. Thus the main service is followed by an optional service to which customers, on completion of main service, proceed with probability q. Then our model reduces to Madan [3] with exponentially distributed optional service and to Medhi [5] in the case of arbitrarily distributed optional service time.
v.
Transient solution
The supplementary variable technique (see Cox [1], Medhi [4]) could be used to get the transient solution. Denote by h(x) = dG(x), the hazard function of the service time distribution
■> v J 1-G(x)'
G(.) and the probability density function of V is given by
g(x) = h(x)exp[-N (x)}
where
N(x) = f* h(u)du (N(0) = Oand-^N(x) = h(x)). If V is the total service time, then h(x)dx = P(service will be completed in (x,x + dx) given that service time exceeds x) and E(V) = f xg(x)dx = -G*(1)(0).
The supplementary variable X(t) considered is defined below. Let N(t) = systemsizeattimet
X(t) = timealreadyspentinserviceuptotofaunitreceivingservice
pn(t) = P(N(t) = n)withp0(0) = 1
pn(t,x)dx = P(N(t) = n,x < X(t) < x + dx), n>1
Pn(t) = f" pn(t,x)dx, Q(t,z) = Y,n=o Pn(t)zn, Q(t,x,z) = Y%=i Pn(t,x)zn
Now we have
p0(t + St) = [1 - ASt + o(St)]p0(t) + J™ p1(t,x)h(x)dxSt.
AsSt^0, dP0(t) = -*P0(t) + Ç Pi(t,x)h(x)dx. (80)
ForSx > 0, p1(t + St,x + Sx) = [1 - ASt + o(5t)][1 - h(x)Sx + o(Sx)]p1(t,x). Subtracting and adding a term p1(t,x + Sx) to the LHS, then dividing by St(Sx) and taking as St ^ 0(5x ^ 0), we get
(dt + d)^,*) = -(¿ + h(x))P1(t,x). (81)
Forn > 0, (d + d)Pn(t,x) = -(A + h(x))pn(t,x)+Apn-1(t,x). (82)
We have the following boundary conditions:
and
P1(t, 0) = T P2(t,x)h(x)dx + AP0(t) (83)
Pn(t,0) = j0 pn+1(t,x)h(x)dx, n>2. (84)
Multiplying (82) by zn, n = 2,3,... and (81) by z, then adding all the terms we get
(dt + d)^ Vn(t,x)zn = -(A + h(x))Y%=1 pn(t,x)+AY^=2 Pn-1(t,x) (85)
(dt + d)Q(t,x,z) = -&-te + h(x))Q(t,x,z). (86)
Now multiplying (84) by zn, n = 2,3,... and (83) by z, then adding the terms we have
Q(t,0,z) = J" (Z™=1 pn+1(t,x)zn)h(x)dx + AzP0(t). (87)
Now
J0°° (Zn=1 pn+1(t,x)zn)h(x)dx = J0°° (1)Zn=1 pn+1(t,x)zn+1h(x)dx
= J0°° (1) ES=1 Vn(t,x)zn - P1(t,x)z]h(x)dx
= (1) J0°° [Q(t-X,z) - P1(t,x)z]h(x)dx
= (1) LT Q(t,x,z)h(x)dx - z(p'0(t) + AP0(t))]by(80)
Thus (87) reduces to
Q(t, 0,z) = (1) [J0°° Q(t,x,z)h(x)dx - z(p'0(t) + AP0(t))] + ^(t)
= (1) iJo" Q(t,x,z)h(x)dx - z(p'0(t) + AP0(t)) + Az2P0(t)]
zQ(t,0,z) = J™ Q(t,x,z)h(x)dx-zp'0(t)+Az(z-1)p0(t). (88)
The partial differential equation (86) can be solved using the boundary condition (88) and the normalizing condition £™=0 pn(t) = 1.
7.1 Steady state distribution
Let
limt^œpn(t) = pn, n> 0
and
limpn(t,x) = pn(x), x>0,n>1
= p0(x) = 0, x> 0. Then {pn,n > 0} gives the distribution of the general time system size.
Let
Q(x,z) =^Pn(x)zn = ^ [lim pn(t,x)jz"
= lim
t^m
^Pn(t, X)2
= lim Q(t,x,z)
t^m
and Then
Q(z) = J0° Q(x,z)dx. (80) * Ap0 = r pi(x)h(x)dx
(81)and(8182) * -^Pn(x) = -(A + h(x))pn(x) + Apn-1(x), n>1
(83) * pi(0) = J0° p2(x)h(x)dx + Ap0 (84) * pn(0) = f0 Pn+i(x)h(x)dx, n > 2.
The partial differential equation (86) and the boundary condition (88) reduces to
^Q(x.z) = -(A-Az + h(x))Q(x,z)
zQ(0,z) = f Q(x,z)h(x)dx + Az(z - 1)p0
and
From relation (93)
f
dQ(x,z) Q(x,z)
Po + Q(1) = 1.
= f — Az + h(x))dx
log(Q(x, z)) = logc(-A(1 - z)x - N(x)) Q(x, z) = cexp(-A(1 - z)x - N(x))
Q(0,z) = c
(89)
(90)
(91)
(92)
(93)
(94)
(95)
Q(x,z) = Q(0,z)exp(-A(1 - z)x - N(x))
Substituting (96) in (94) we get
zQ(0,z) = fm Q(0, z)e(-À(1-z)x-N(x))h(x)dx + Az(z - 1)p0
= Q(0,z) f" e-X(1-z)x[e-N(x)h(x)]dx + Az(z - 1)p0 = Q(0,z)G*(A(1 - z)) + Az(z - 1)p0.
Thus
Now from (96) we have
Q(0,z) =
Q(z) =f0mQ(x,z)dx
Xz(z-1)po z-G'(X-Xz)
f" Q(0,z)e(-A(1-Z)x-N(x'>'>dx Q(0,z)fm e(-X(1-z)xe-N(x)dx
Q(0,z) A(1-z)
[1 - f^ e-X(1-z)x(e-N(x)h(x))dx]
(96)
(97)
From (97) and (98) we get
Q(z)=^â))[1-G\A-Az)]
(98)
n=1
n=1
n
-n=1
Using L'Hospital rule, we get
Q(z) =
z[G\X-Xz)-1]p0 z-G'(X-Xz)
Q(1) =limQ(z)
Z^1
= Po'
1+AG,(1)(A-Az)
= Po
AE(V) 1-AE(V)
(99)
[G*(A-Az)-1]+zAG<1'>(A-Az)
From (95) we obtain Hence
Po = 1-AE(V).
Q(z) =
z[G*(A-Az)-1][1-AE(V)] z-G'(A-Az) '
(100)
7.2 Busy period
Let T be the length of a busy period (starting with a customer arrival to an idle server, until the becomes idle again). Define B(t) = P(T < t). Then B(t) satisfies the relation
B(t) = £ ££=o (-^fe-XuB'k(t - u)dG(u) (101)
The Laplace Stieltjes Transform (LST) of busy period B(t) be denoted by B*(s). That is, B*(s) = e-stdB(t) (forRe(s) > 0)
= C e-St £ X«=o - u)dG(u)dt
= 0 £ e-s(t-u)B*k(t -u)dtdG(u)
= 0 ZT=o (-^~e-Xue-su(B*(s))kdG(u)
= So e-(x+s-XB'(s))udG(u)
Therefore
B'(s) = G*(A + s-AB*(s)). (102)
From this the mean and higher moments of the number of customers in the system can be computed.
Conclusion:
We examined a queueing model offering n distinct services to which arrival is according to a MAP forming a single line. Service time has phase type distribution. A single server serves the customers. The service station provides two types of services - one is desirable and other is unwanted for each customer. If the service starts in an undesirable state then a clock also simultaneously starts ticking. In case this clock realizes before the exact requirement of the server is realized, then that customer leaves the system forever without being eligible for the service that he actually requires. On the other extreme, in case the correct identification of required service occurs before realization of clock, then the customer is served in that state and then leaves the system. In case right at the beginning of service the exact requirement of service is identified, then the customer starts getting that service right at the time when taken for service. Several system performance measures are evaluated. Applications of the model in hospital, telecommunication etc are indicated. Stochastic decomposition of the system state is analyzed. Some particular cases are indicated.
In a future work we propose to extend the model to multi server case.
References
[1] Cox, D. R. (1955): The analysis of non-Markovian stochastic processes by the inclusion of supplementary variables. Proc. Cambridge Phil. Soc. 51, 433-441.
[2] Gross, D. and Harris, C. M. (1988): Fundamentals of Queueing Theory, John Wiley and Sons, New York.
[3] Madan, K. C. (2000): An M/G/1 queue with second optional service. Queueing systems 34, 37-46.
[4] Medhi, J. (1994): Stochastic Processes. 2nd ed. Wiley, New York and Wiley Eastern, New Delhi.
[5] Medhi, J. (2002): A Single Server Poisson Input Queue with a Second Optional Channel. Queueing systems 42, 239-242.
[6] Neuts, M.F. (1981): Matrix-Geometric Solutions in Stochastic Models, Johns Hopkins University Press, Baltimore
