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CHAPTER 2
ANALYSIS OF A BATCH ARRIVAL GENERAL BULK
SERVICE QUEUEING SYSTEM WITH MULTIPLE
VACATIONS, SETUP TIME AND SERVER’S CHOICE
OF ADMITTING RE-SERVICE
2.1 INTRODUCTION
In many practical situations one can observe that the leaving batch
of customers may request for re-service. Avi - Itzhak and Naor (1963) have
analyzed an M/G/1 queueing model with repair of service station on request
by a leaving customer. Rosenberg and Uri Yechiali (1993) analyzed a bulk
arrival general service single server queue with single and multiple vacations
under LIFO service regime. Huan et al (1995) discussed a computational
analysis of M(n)/G/1/N queues with setup time. The interdeparture time
distribution for each class in the /1i
/Gi
MX queue with setup times and
repeated server vacations was studied by Frans (1999). Choudhury (2000)
analyzed single server Poisson bulk arrival general service queue with a setup
period and a vacation period.
Madan and Baklizi (2002) considered an M/G/1 queueing model, in
which the server performs first essential service to all arriving customers. As
soon as the first service is over, they may leave the system with the
probability (1- ) and second optional service is provided with probability .
Arumuganathan and Ramaswami (2003) analyzed a non-Markovian bulk
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arrival general bulk service queueing systems with instantaneous Bernoulli
feedback and multiple vacations. Madan et al (2004) analyzed a single server
bulk arrival queue, in which the leaving batch of customers might opt for
re-service. They obtained various performance measures too. Hur et al (2005)
studied single server bulk arrival queueing system with vacations and server
setup. Arumuganathan and Judeth Malliga (2006) analyzed a bulk queue with
re-service of service station and set up time. Al-khedhairi and Lotfi Tadj
(2007) discussed a bulk service queue with a choice of service and re-service
under Bernoulli schedule. Lotfi Tadj and Ke (2008) studied a hysteretic bulk
quorum queue with a choice of service and optional re-service. Madhu Jain
et al (2010) discussed an optimal repairable MX/G/1 queue with multi -
optional services and Bernoulli vacation.
In the literature, queueing models with re-service, the server
accepts all requests for re-service, irrespective of other constraints like
number of customers in the queue, the cost of power and so on. But in reality,
one can observe that, the server may reject the request for re-service with
some probability. In many systems, before the commencement of service, the
server will do some preparatory work such as warming up the machine or
booting the computer, etc; in queueing terminology, such term is referred to
as setup time. Server vacation models are useful for the system in which the
server wants to utilize the idle time for different purposes. Application of
vacation models can be found in production systems, designing local area
networks and data communications systems. Addressing this, the proposed
model /1ba,/GXM queueing system with multiple vacations, setup time
and server’s choice of admitting re-service is developed.
In this chapter, a bulk queueing system with server’s choice of
admitting re-service, multiple vacations and setup time is considered. At a
service completion, the leaving batch may request for re-service with
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SERVICE
RE-SERVICE
MULTIPLE
VACATIONS
aQaQ
aQaQ
aQ
aQ
aQ
1
1
aQ
SETUP TIME
Bulk
arrival
aQ
probability and it is not mandatory to accept it; the server admits this
request with a probability . After the re-service or service completion
without request for re-service, if the queue length is less than a, the server
leaves for a secondary job (vacation) of random length. After this vacation, if
the queue length is still less than ‘a’, the server leaves for another vacation
and so on, until he finally finds at least ‘a’ customers waiting for service. At a
vacation completion epoch, if the server finds at least ‘a’ customers waiting
for service, he requires a setup time to start the service. After a setup time or
on service completion or on re-service completion, if the server finds at least
‘a’ customers waiting for service say , he serves a batch of min ( ,b )
customers, where b a . Analytical treatment of this model is obtained by the
supplementary variable technique. The model under study is schematically
represented in Figure 2.1.
Figure 2.1 Schematic Representation of the Queueing Model
(Q – Queue Length)
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The motivation of the model comes from a real life situation
observed in the Environmental Sensor Networks (ESN). ESN system can
potentially provide a new data for environmental science (eg. climate models)
as well as vital hazard warnings (eg. flood alerts) etc. This is particularly
important in remote or dangerous environments where many fundamental
processes have rarely been studied due to their inaccessibility. A sensor
network is designed to transmit the data from an array of sensors to a data
repository on a server. Monitoring the behavior of ice caps and glaciers is an
important part of our understanding of the Earth’s climate. The environmental
sensor nodes gather data such as glaciers, movement of stones and sediment
under the ice, temperature, pressure, vibration etc. autonomously and pass the
data to the cluster heads. The cluster heads pass the gathered messages
(customers) to the base station. After obtaining the required information, the
base station operator generates various reports and transmits (service) the
report to the application nodes. After getting the reports, the application node
may request some more reports with the same data (re-service). The request
may be accepted or rejected by the base station based on the importance of the
current report generation at that instant. If the number of messages received
by the base station to produce the report is inadequate, then the base station
will do some other associated work such as antivirus running, backup process,
etc., At the completion epoch of the associated work (vacations), if the
number of messages is inadequate, then, the base station will repeat the
associated works until the required number of information reaches to process
it. When the operator returns from the associated work and finds the required
number of messages available in the queue, then the operator begins some
preparatory work, such as checking the application nodes, type of reports
required, etc, for which some amount of time is required, called as setup time.
The above situation can be modeled as /1ba,/GXM queueing system with
multiple vacations, setup time and server’s choice of admitting re-service.
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For the proposed model, the probability generating function (PGF)
of the steady state queue size distribution at an arbitrary time epoch is
obtained using supplementary variable technique. Particular cases and some
special cases are discussed. Various performance measures are derived. A
cost model for the queueing system is developed. Numerical solution for
particular values of parameters is presented.
2.2 MATHEMATICAL MODEL
Let X be the group size random variable of the arrival, be the
Poisson arrival rate. gk
be the probability that ‘k’ customers arrive in a batch
and X (z) be its probability generating function (PGF). Let ‘ ’ be the
probability that a leaving batch request for re-service and ‘ ’ be the
probability that the server accepts a re-service. Let S(x) (s(x)) {~S( ) }[ S
0(x)]
be the cumulative distribution function (probability density function)
{Laplace-Stieltjes transform}[remaining service time] of service. Let V(x)
(v(x)) {~V( ) }[ V
0(x)] be the cumulative distribution function (probability
density function) {Laplace-Stieltjes transform}[ remaining vacation time] of
vacation. Let U(x) (u(x)) { U( )}[ U0(x)] be the cumulative distribution
function (probability density function) {Laplace-Stieltjes transform}
[remaining set up time] of set up. Let R(x) (r(x)) {~R( )}[ R
0(x)] be the
cumulative distribution function (probability density function) {Laplace-
Stieltjes transform} [remaining re-service time] of re-service. Nq(t) denotes
the number of customers waiting for service at time t, Ns(t) denotes the
number of customers under service at time t. The different states of the server
at time ‘t’ are defined as follows:
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0, if theserver is busy with service
1, if theserver is busy with re serviceC(t)
2, if theserver ison vacation
3, if theserver ison setup work
thZ(t) j if theserver ison j vacation startingfrom theidleperiod
To obtain the system equations, the following state probabilities are
define
0P (x, t)dt = Pr N (t) = i, N (t) = j, x S (t) x + dt,C(t) = 0 , a i b, j 0s qi, j
0R (x, t)dt Pr N (t) n, x R (t) x dt,C(t) 1 , n 0n q
0Q (x, t)dt = Pr N (t) = n, x V (t) x + dt,C(t) = 2, Z(t) = j , j 1, n 0qj,n
and
0U (x, t)dt Pr N (t) n, x U (t) x dt,C(t) 3 , n an q
Now, the following system equations are obtained for the queueing
system, using supplementary variable technique:
bP (x t, t t) = P (x, t) 1 t + (1- ) P (0, t)s(x) ti,0 i,0 m,im=a
b+ (1- ) P (0, t)s(x) t R (0, t)s(x) t U (0, t)s(x) t
m,i i im=a
a i b
jP (x - t, t + t) = P (x, t) 1 t + P (x, t) t, a i b - 1, j 1
i, j i, j i, j-k kk=1
bP (x t, t t) = P (x, t) 1 t + (1- ) P (0, t) s(x) tb, j b,j m,b+ jm=a
jb+ (1 - ) P (0, t) s(x) t + P (x, t) g t
m,b+ j b, j-k km=a k=1
+R (0, t) s(x) t + U (0, t) s(x) t; j 1b+ j b+ j
30
bQ (x - t, t + t) = Q (x, t) 1 t + (1 - ) P (0, t) v(x) t
1,0 1,0 m,0m=a
b(1 - ) P (0, t) v(x) t R (0, t) v(x ) t
m,0 0m=a
bQ (x - t, t + t) = Q (x, t) 1 t + (1 - ) P (0, t) v(x) tm,n1,n 1,n m=a
b n+ (1 - ) P (0, t) v(x) t + Q (x, t) tm,n 1,n-k km=a k=1
+R (0, t) v(x) t, 1 n a -1n
nQ (x - t, t + t) = Q (x, t) 1 t + Q (x, t) t, n a
1,n 1,n 1,n-k kk=1
Q (x - t, t + t) = Q (x, t) 1 t + Q (0, t) v(x) t, j 2j,0 j,0 j-1,0
Q (x - t, t + t) = Q (x, t) 1 t + Q (0, t) v(x) tj,n j,n j-1,n
n+ Q (x, t) t, 1 n a 1, j 2
j,n-k kk=1
nQ (x - t, t + t) = Q (x, t) 1 t Q (x, t) t, n a, j 2
j,n j,n j,n-k kk=1
bR (x - t, t + t) = R (x, t) 1 t + P (0, t) r(x) t
0 0 m, 0m = a
b nR (x - t, t + t) = R (x, t) 1- t + P (0, t) r(x) t + R (x, t) tn n m,n n-k km=a k=1
n 1
n-aU (x - t, t + t) = U (x, t) 1- t + U (x, t) t + Q (0, t) u(x) t, n an n n-k kk=1 =1 ,nl l
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2.3 STEADY STATE QUEUE SIZE DISTRIBUTION
From the above equations, the steady state queue size equations are
obtained as follows:
b bd- P (x) = - P (x) + (1- ) P (0) s(x) + (1- ) P (0) s(x)
i,0 i,0 m,i m,im=a m=adx
R (0)s(x) U (0)s(x); a i bi i
(2.1)
jd- P (x) = - P (x) + P (x)g , a i b - 1, j 1
i, j i, j i, j-k kdx k=1 (2.2)
b bd- P (x) = - P (x) + (1- ) P (0) s(x) + (1- ) P (0) s(x)
b, j b, j m,b+ j m,b+ jm=a m=adx
j+ P (x) g + R (0) s(x) + U (0) s(x); j 1
b, j-k k b+ j b+ jk=1
(2.3)
bd- Q (x) = - Q (x) + (1- ) P (0) v(x)
1,0 1,0 m,0m=adx
b+ (1 - ) P (0) v(x) + R (0) v(x)
m,0 0m=a
(2.4)
b bd- Q (x) = - Q (x) + (1- ) P (0) v(x) + (1- ) P (0) v(x)m,n m,n1,n 1,n m=a m=adx
n+ Q (x)g + R (0) v(x), 1 n a -1n1,n-k kk=1
(2.5)
nd- Q (x) = - Q (x) + Q (x)g , n a
1,n 1,n 1,n -k kdx k =1 (2.6)
d- Q (x) = - Q (x) + Q (0) v(x) j 2
j,0 j,0 j-1,0dx (2.7)
nd- Q (x) = - Q (x)+Q (0)v(x)+ Q (x)g , 1 n a-1, j 2
j,n j,n j-1,n j,n-k kdx k=1 (2.8)
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nd- Q (x) = - Q (x) + Q (x)g , n a, j 2
j,n j,n j,n-k kdx k=1 (2.9)
bd- R (x) = - R (x) + P (0) r(x)
0 0 m,0m=adx (2.10)
b nd- R (x) = - R (x) + P (0)r(x) + R (x) g , n 1n n m,n n-k km=adx k=1
(2.11)
n-ad- U (x) = - U (x) + U (x) g + Q (0) u(x), n an n n-k kdx k=1 =1 ,nl l
(2.12)
The Laplace –Stieltjes transforms (LST) of P (x)j,n
, Q (x)j,n
, U (x)n and
R (x)n are defined as
x - xP ( ) = e P (x)dx, Q ( ) = e Q (x)dxj,n j,n j,n j,n
0 0,
xU ( ) = e U (x)dxn n0
and xR ( ) = e R (x)dxn n0
Taking LST on both sides of the Equations (2.1) through (2.12), we have
b
am
b
am)(S
~)0(
i,mP)1()(S
~)0(
i,mP)1()(
0,iP~
)0(0,i
P)(0,i
P~
R (0)S( ) U (0)S( ), a i bi i
(2.13)
j
1k1j,1bia,
kg)(
kj,iP~
)(j,i
P~
)0(j,i
P)(j,i
P~
(2.14)
b
am)(S
~)0(
jb,mP)1()(
j,bP~
)0(j,b
P)(j,b
P~
b
am)(S
~)0(
jb,mP)1(
j
1k kg)(
kj,bP~
R (0)S( ) U (0)S( )b j b j
j 1 (2.15)
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bQ ( ) Q (0) Q ( ) (1 ) P (0)V( )
1,0 1,0 1,0 m,0m a
b(1 ) P (0)V( ) R (0) V( )
m,0 0m a
(2.16)
b bQ ( ) Q (0) Q ( ) (1 ) P (0)V( ) (1 ) P (0)V( )m,n m,n1,n 1,n 1,n m a m a
n- Q ( )g - R (0)V( ), 1 n a -1n1,n-k kk=1
(2.17)
nQ ( ) Q (0) Q ( ) Q ( ) g , n a
1,n 1,n 1,n 1,n k kk 1 (2.18)
Q ( ) Q (0) Q ( ) Q (0) V( ) j 2j,0 j,0 j,0 j 1,0
(2.19)
nQ ( ) Q (0) Q ( ) Q ( ) g Q (0) V( ),
j,n j,n j,n j,n k k j 1,nk 1
1 n a 1; j 2
(2.20)
nQ ( ) Q (0) Q ( ) Q ( )g , n a; j 2
j,n j,n j,n j,n k kk 1 (2.21)
bR ( ) R (0) R ( ) P (0)R( )
0 0 0 m,0m a (2.22)
b nR ( ) - R (0) = R ( ) - P (0)R( ) - R ( g ; n 1n n n m,n n-k km=a k=1
(2.23)
n-aU ( ) - U (0) = U ( ) - U ( g - Q (0)U( ); n an n n n-k k ,nk=1 =1 ll
(2.24)
2.3.1 Probability Generating Function
Lee (1991) developed a new technique to find the steady state
probability generating function (PGF) of the number of customers in the
queue at an arbitrary time epoch. To apply the technique, first the following
probability generating functions are defined.
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jP (z, ) = P ( )zi i,jj=0
andj
P (z,0) = P (0)z a i bi i,jj=0
nQ (z, ) = Q ( )zj j,nn=0
and nQ (z,0) = Q (0)z j 1j j,nn=0
nR(z, ) = R ( )znn=0
and nR(z,0) = R (0)znn=0
nU(z, ) = U ( )znn=a and nU(z,0) = U (0)znn=a
(2.25)
whereb
P (z, )ii=a
is the PGF of joint transform of the number of customers in
the queue during busy period. At = 0 , one can get the PGF of the number
of customers in the queue during busy period at an arbitrary time. Similarly,
the other joint transforms Q (z, )j
, R(z, ) and U(z, ) are defined.
The probability generating function P(z) of the number of
customers in the queue at an arbitrary time of the proposed model can be
obtained using the following equation.
b-1P(z) = P (z,0) + P (z,0) + Q (z,0) + R(z,0) + U(z,0)
bi ji=a j=1 (2.26)
In order to find P (z,0)i
, P (z,0)b
,Q (z,0)j
, R(z,0) and U(z,0) , the
following sequence of operations are done.
Multiplying (2.16) by z0, (2.17) by z
n (1 n a-1), (2.18) by z
n
(n a), summing up from n = 0 to and using (2.25), we get
a 1 b n n( X(z))Q (z, ) Q (z,0) V( ) (1 ) P (0)z R (0)zm,n n1 1 m an 0
(2.27)
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Multiplying (2.19) by z0, (2.20) by z
n (1 n a-1), (2.21) by z
n
(n a) , summing up from n = 0 to and using (2.25), we get
a-1 n( X(z))Q (z, ) = Q (z,0) - V( ) Q (0)z , j 2
j j j-1,nn=0 (2.28)
Multiplying (2.22) by z0, (2.23) by z
n ( n 1), summing up from
n = 0 to and using (2.25), we get
b( X(z))R(z, ) R(z,0) R( ) P (z,0)mm a
(2.29)
Multiplying (2.24) by zn ( n a) , summing up from n = a to and
using (2.25), we get
a 1 n( X(z))U(z, ) U(z,0) U( ) Q (z,0) Q (0)z
nn 01 l l,l (2.30)
Multiplying (2.13) by z0, (2.14) by z
j ( j 1) , summing up from
j = 0 to and using (2.25), we get
b( X(z))P (z, ) = P (z,0) -S( ){(1- ) P (0)
i i m,im=a
b+ (1- ) P (0) +R (0) U (0)}
m,i i im=a, a i b-1 (2.31)
Multiplying (2.13) by z0 with i = b, (2.15) by z
j ( j 1) , summing
up from j = 0 to and using (2.25), we get
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b bX(z))z P (z, ) = z P (z,0)b b
b b-1 j(1- ) (P (z,0) - P (0)z )m m,jm=a j=0
b b-1 j-S( ) + (1- ) (P (z,0) - P (0)z )m m,jm=a j=0
b-1 b-1n n+(R(z,0) - R (0)z ) +(U(z,0) - U (0)z )n nn=an=0
(2.32)
By substituting = – X(z) in the Equations (2.27) through
(2.32), we get
a-1 b n nQ (z,0) = V( X(z)) (1- ) P (0)z + R (0)zm,n n1 m=an=0
(2.33)
a-1 nQ (z,0) = V( X(z)) Q (0)z j 2
j j-1,nn=0 (2.34)
bR(z,0) = R( X(z)) P (z,0)mm=a
(2.35)
a-1 nU(z,0) = U( X(z)) Q (z,0) - Q (0)z
,nn=0=1 l ll (2.36)
b bP (z,0) = S( X(z)) (1- ) P (0) + (1- ) P (0) + R (0) +U (0)i m,i m,i i im=a m=a
a i b-1 (2.37)
and
b b-1 j(1- ) P (z,0) - P (0)zm m,jm=a j=0
b b-1 jbz P (z,0) = S( X(z)) (1- ) P (z,0) - P (0)zmb m,jm=a j=0
b-1 b-1n n+ R(z,0) - R (0)z + U(z,0) - U (0)zn nn=an=0
(2.38)
37
Letb
p = P (0), q = Q (0), u = U (0), r = R (0)i m,i i i i i i i
m = a =1l,
l
and
k 1 p q ri i i i
, c 1 p u ri i i i
From the Equation (2.38), we get
S( X(z)) f(z)P (z,0) =b b
z - (1- )S( X(z)) - S( X(z))R( X(z)) (2.39)
where f(z)=b-1 m(1- ) + R( X(z)) S( X(z)) - z cmm=a
a-1 m+ U( X(z)) V( X(z))k - q - (1- )p + r zm m m m
m=0
From the Equations (2.27) and (2.33), we get
a-11 nQ (z, ) = V( X(z)) - V( ) (1- )p + r zn n1 X(z) n=0 (2.40)
From the Equations (2.28) and (2.34), we get
a-11 nQ (z, ) = V( X(z)) - V( ) Q (0)z , j 2j j-1,nX(z) n=0
(2.41)
From the Equations (2.29) and (2.35), we get
b-11R(z, ) = R( X(z)) - R( P (z,0)mm=aX(z)
(2.42)
From the Equations (2.30) and (2.36), we get
a-11 nU(z, ) = U( X(z)) - U( ) Q (z,0) - q znX(z) n=0=1 ll (2.43)
38
From the Equations (2.31) and (2.37), we get
1P (z, ) = S( X(z)) -S( ) c , a i b 1i iX(z)
(2.44)
From the Equations (2.32) and (2.39), we get
S( X(z)) -S( ) f(z)P (z, ) =b b
X(z) z - (1- )S( X(z)) - S( X(z))R( X(z))(2.45)
Substituting P (z,0),i
P (z,0)b
, Q (z,0)j
, R(z,0) and U(z,0) from the
Equations (2.40) – (2.45) in the Equation (2.26), the probability generating
function of the queue size P(z) at an arbitrary time epoch is obtained as
b 1S( X(z)) 1 c
f (z) S( X(z)) 1ii aP(z)X(z) h(z)( X(z))
a 1 b 1iV( X(z)) 1 k z R( X(z)) 1 S( X(z))ci mi 0 m a
X(z) ( X(z))
R( X(z)) 1 S( X(z))f (z)
h(z)( X(z))
a 1 a 1n n[U( X(z)) 1] V( X(z)) k z q zn nn 0 n 0
( X(z))
39
On further simplification, we get
m m
m m
b
(S( X(z))-1) b-1 b m(z -z )cmR( X(z))-1 (S( X(z)) m=a
1-S( X(z)) a-1m+ (1- )p +r z
R( X(z))-1 S( X(z)) m=0
a-1b m+ (z -1) U( X(z)) V( X(z))k -q z
m=0
z -S( X(-
P(z)=
z)) a-1 m(k -q )zm mR( X(z))-1 S( X(z)) m=0
h(z)(- X(z)) (2.46)
where bh(z) z (1 )S( X(z)) S( X(z))R( X(z))
The probability generating function P(z) has to satisfy the condition
P(1) = 1. In order to satisfy the condition, applying L’Hospital’s rule in and
evaluating P(z)1z
lim and equating the expression to 1, we have to satisfy
b 1 2 E(X) E(S) + E(R) (b m)cm
m a
a 1 a 1(X,S, R) (1 )p r (X,S, R) q km m m m
m 0 m 0
a 12b E(X)E(V)k 2 E(X) b - E(X) E(S) + E(R)m
m 0
where
2 2 2 2(X,S, R) 2 mE(S)E(X) E (X)E(S ) E(S)E(X )
2 2 2 2 2 22 mE(R)E(X) 2 E (X)E(R) E(R)E(X ) E(R )E (X)
and
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2(X,S,R) 2m b E(X)E(S) E(X)E(R) b(b 1) E(X )E(S)
2 2 2 2 2 2 2 2E (X)E(S ) E (X)E(R ) 2 E (X)E(S)E(R)
2E(X )E(R) m(m 1) (m b)(m b 1) 2b E(X)E(G)
Since p , q , r and ui i i i
are the probabilities of ‘i’ customers being
in the queue at service completion epoch, vacation completion epoch, re-
service completion epoch and set up time completion epoch, respectively, it
follows that the left hand side of the above expression must be positive. Thus
P(1) = 1 is satisfied if and only if b - E(X) E(S) + E(R) 0 .
Define ‘ ’ asE(X) E(S) + E(R)
b. Thus < 1is the condition
to be satisfied for the existence of steady state for the model under
consideration.
2.3.2 Computational Aspects of Unknown Probabilities
Equation (2.46) gives the probability generating function P (z) of
the number of customers in the queue at an arbitrary time epoch, which
involves ‘b + 2a’ unknown probabilities namely, p ,p ,p ,.......p ,0 1 2 a 1
r ,r , r ,.......r ,0 1 2 a 1
q ,q ,q ,.......q ,0 1 2 a 1
c ,c ,.....,ca a 1 b 1. Using Lemma (2.1)
and theorems (2.1) and (2.2), q and ri i
; i = 0 to a-1 are expressed in terms of
pi; i = 0 to a-1.
Now, the Equation (2.46) contains only ‘b’
unknowns p ,p ,p ,.......p , c ,c ,.....,ca0 1 2 a 1 a 1 b 1. By Rouche’s theorem, the
expression bz (1 )S( X(z)) S( X(z))R( X(z)) has b-1
zeros inside and one on the unit circle 1z . Since P (z) is analytic within
41
and on the unit circle, the numerator of (2.46) must vanish at these points,
which gives ‘b’ equations and ‘b’ unknowns. These equations can be solved
by suitable numerical techniques.
Lemma 2.1
Let i be the probability that ‘i’ customers arrive during a vacation.
The probability generating function of i is given by
iz V X(z)i
i=0
Proof:
Conditioning on the actual vacation length, number of arrivals and
the group size, we get
m (m)ti e t g
i d V(t)i m=0 m !0
where(m)
gi
is the m – fold convolution of gi with itself (i.e., total of m
arrivals make ‘i’ customers)
Multiplying the above equation by z i
and taking the summation
from i = 0 to , we get
mti - t (m) i
z = e g z dV(t)i im!i=0 0 m=0 i=m
42
m mt) [X(z)]t
= e d V(t)m !0 m=0
= V X(z) (2.47)
Hence theLemma
Theorem: 2.1
The constants rn involved in P(z) are expressed in terms of pn as,
nr = pn i n - ii=0
, n=0,1,2,…a-1, whereiis the probability that ‘i’
customers arrive during a re-service period.
Proof:
From the Equation (2.35), we have
bR(z,0) P (z,0)R( X(z))mm a
b jnR (0)z = R( X(z)) P (0)zn m,jm=an=0 j=0
jnr z = R( X(z)) p zn jn=0 j=0
which, using Lemma 2.1, gives
jn nr z z p zn n jn 0 n 0 j 0
Equating the coefficients of zn
, n=0,1,2,…a-1, we get
43
nr = pn i n-ii=0
; n=0,1,2,…a-1 (2.48)
Hence the theorem
Theorem: 2.2
The constants qn involved in P(z) are expressed in terms of pn as,
n
0i ip
inbnq ; n = 0,1,2,3,……….a-1, where
(1 )0 0 0b
0 10
and
n n(1 ) bn n i i i n ii 0 i 1bn
10
; n = 1,2,3,…..…a-1 and i is
the probability that ‘i’ customers arrive during a vacation period.
Proof:
From the Equations (2.33) and (2.34), we have
1j)0,z(
jQ
a 1 a 1n nV( X(z)) (1 )p r z q zn n n
n 0 n 0
a 1n nq z V ( X(z)) (1 )p r q zn n n nn 0 n 0
which, using Lemma 2.1, gives
a 1n n nq z z (1 )p r q zn n n n n
n 0 n 0 n 0
a 1 n a 1n n(1 )p r q z (1 )p r q z
n i i i i n i i i in an 0 i 0 i 0
44
Equating the coefficients of zn on both sides of the above equation
for n = 0,1,2,3,…a-1, we getn
q (1 )p r qn n i i i ii 0
Substituting ri and solving it for qn, we get
n n n n 1(1 ) p p q
n i i n k i i k n i ii 0 i 0k 0 i 0qn1
0
Coefficient of pn in qn is(1 )
0 0 0 b01
0
Coefficient of pn-1 in qn is [ 1 + 1 Coefficient of pn-1 in qn-1]/1- 0
(1 ) b1 1 0 0 1 1 0
b11
0
Coefficient of pn-2 in qn is [ 2 + 1 Coefficient of pn-2 in qn-1
+ 2 Coefficient of pn-2 in qn- 2]/1- 0
(1 ) ( b b )2 2 0 1 1 0 2 1 1 2 0
b21
0
Proceeding like this, we get
Coefficient of p0 in qn is
n n(1 ) bn n i i i n ii 0 i 1 bn
10
Therefore,n
0i ip
inbnq ; n = 0,1,2,3,……….a-1 (2.49)
Hence the theorem
45
2.4 PERFORMANCE MEASURES
In this section, some useful performance measures of the proposed
model like, expected number of customers in the queue E(Q), expected
length of idle period E(I), expected length of busy period E(B) are derived
which are useful to find the total average cost of the system. Also, probability
that the server is on setup work P(U), probability that the server is on vacation
P(V) and probability that the server is busy P(B) are derived.
2.4.1 Expected Queue Length
The expected queue length E (Q) (i.e. mean number of customers
waiting in the queue) at an arbitrary time epoch, is obtained by differentiating
P(z) at z = 1and is given by
)Q(E)z(P1z
lim
b-1 a-1f X,S, R c + f X,S, R,U, V km m1 2m=a1 m=0
E(Q) =2 2 2 a-124 E (X)T
f (X,S, R) - f (X,S,R, U) (1- )p + r )1 m m3 4m=0
(2.50)
where S1 = E(X)E(S) , 2 2 2S2 = X (1)E(S) + E (X)E(S ) ,
2 2S3 = E (X)E(S)E(R) ,S4 = X (1)E(S) , 2 2S5 = X (1)E(X)E(S ) ,
3 3 3S6 = E (X)E(S ) , S7 = X (1)E(S) , 2 2S8 = X (1)E(S ) + X (1)E(R ) ,
2S9 = E(S) + E(R) X (1) , R1 = E(X)E(R) ,
2 2 2R2 = X (1)E(R) + E (X)E(R ) , 2 2R3 = E (X)E(X)E(S)E(R)
R4 = X (1)E(R) , 2 2R5 = X (1)E(X)E(R ) , 3 3 3R6 = E (X)E(R ) ,
R7 = X (1)E(R) , 2 2R8 = E(R )E(S) + E(R)E(S ) ,
46
V1 = E(X)E(V) , 2 2 2V2 = X (1)E(V) + E (X)E(V ) ,
U1 = E(X)E(U) , 2 2 2U2 = X (1)E(U) + E (X)E(U ) , 2 2U3 = E (X)E(U)E(V) ,
T1 = b -S1- R1 , T2 = S1+ R1 , T3 = S2 + R2 , T6 = S6 + R6 ,
T7 = S6 + S7 + 3S5 , 3 3T8 = 3R5 + 6R3 + R4 + R6 + 3 E (X)R8 + 6mS3 ,
f1 = 6 E(X) b(b -1) -S2 - R2 - 2 S3+ 6 X (1)T1 , f2 = 2m T2 + T3+ 2 S3 ,
2 3 3f3 = 4 b(b -1)(b - 2)E(X) -12 E(X) 3R3 + E (X)R8 - 4 E(X)T6 -8 X (1)T2
2 3 2+4 X (1)b -6 S9 + 6 b(b -1)X (1) -18 E (X)S8 ,
2 2E1 = (b - b + m - m )T2 + (b - m)(T3+ 2 S3) , E2 = 2(b - m) f1 T2 ,
2E3 = 3(m - m)T2 + 3mT3+ T7 + T8 ,
E4 = 3m(m -1)T1+ 3m(b(b -1) - S2 - ( R2 + 2S3)) ,
3 3E5 = b(b -1)(b - 2) + S6 -S4 -3S5 + (R6 -3 E (X)R8- R4 -6R3-3R5) ,
E6 = (m + b)(m + b -1)(m + b - 2) - m(m -1)(m - 2) ,
E7 = 2mT1+ b(b -1) -S2 - (R2 + 2S3) ,
E8 = (m + b)(m + b -1) - m(m -1) , 2E9 = 2(2mb + b - b)V1+ +3b(V2 + 2U3)
f (X,S, R) = 12 E(X)T1E1-E21
; f (X,S, R,V, U) = 4 E(X)T1E9-f1(2b)V12
f (X,S,R) = f1f2 + 3f3T2-4 E(X)T1E33
and
2f (X,S, R, U) = 4 E(X)T1 E4 + E5 - E6 - U1 3(2mb + b - b) - 3b U24
-f1 E7 - 2b U1- E8 - 3f3 T1- b
47
2.4.2 Expected Length of Idle Period
Let I be the idle period random variable, then the expected length of
the idle period is given by,
E(I) = E(I1) + E(U),
where I1 is the random variable denoting ‘ idle period due to multiple vacation
process’ and E(U) is the expected length of setup time.
To find E(I1), another random variable U1 is defined as,
0, if the server findsat least ‘a’ customers after the first vacationU =
1 1, if the server finds less than‘a’ customers after the first vacation
Now, the expected length of idle period due to multiple vacations
E(I1) is given by
E(I1) = E(I1 / U1 = 0)P(U1 = 0) + E(I1 / U1= 1)P(U1 = 1)
= E(V)P(U1 = 0) + (E(V) + E(I1))P(U1 = 1)
where E(V) is the expected vacation time.
Solving for E(I1), we have
E(V)E I
1P U 0
1
(2.51)
To find P (U1 = 0), we do some algebra using the Equation (2.33),
then
a 1n nQ (z,0) Q (0)z V( X(z)) (1 )p r zn n1 1,nn 0 n 0
a 1n nz (1 )p r zn n n
n 0 n 0
48
Equating the coefficients of zn (n=0,1,2,…..a-1) on both sides, we
get
nQ (0) (1 )p r
1,n i n i n ii 0
Therefore,
a 1P(U 0) 1 Q (0)
1 1,nn 0
a 1 n1 (1 )p r
i n i n in 0 i 0
(2.52)
Using (2.51) and (2.52), the expected length of idle period E(I) is obtained as
E(V)E(I) = + E(U)
a-1 n1- (1- )p + r
i n-i n-in=0 i=0
(2.53)
where E(V) is the expected vacation time
2.4.3 Expected Length of Busy Period
Let B be the busy period random variable. Let ‘T’ be the residence
time that the server is rendering service or under re-service. Therefore, T=S
with probability (1 ) and T=S+R with probability .
Another random variable J is defined as,
0, if the server finds less than‘a’ customers after a residence timeJ=
1, if the server findsat least ‘a’ customers after a residence time
49
Now, the expected length of busy period E(B) is given by
E(B) = E(B / J = 0)P(J = 0) + E(B / J = 1)P(J = 1)
Solving for E(B), we get
E(B)E(S)
E(R)P(J 0)
Thus, the expected length of busy period is obtained as
E(S)E(B) = + E(R)
a-1(1- )p + rn n
n=0
(2.54)
where E(R) is the expected re-service time and E(S) is the expected service
time.
2.4.4 Probability that the Server is on Setup Work
Let U be the setup period random variable and P (U) be the
probability that the server is on setup work.
From the Equation (2.43), we have
a-11 nU(z,0) = U( X(z)) -1 V( X(z)) (k - q )zn nX(z) n=0
Now, the probability that the server is on setup work is given by
P (U) = lim U(z,0)z 1
=
a-1 nU( X(z)) -1 V( X(z)) (k - q )zn nn=0
limz 1 X(z)
50
P (U) =a-1
E(U) (1- )p + rn nn=0
(2.55)
where E(U) is the expected setup period.
2.4.5 Probability that the Server is on Vacation
Let V be the random variable for multiple vacations and P (V) be
the probability that the server is on multiple vacations at time t.
From the Equations (2.40) and (2.41), we have
a-1 a-1V( X(z)) -1 n nQ (z,0) = (1- )p + r z + Q (0)zn n j-1,nj X(z) n=0 n=0j=2j = 1
a-1 a-1V( X(z)) -1 n n= (1- )p + r z + q zn n nX(z) n=0 n=0
a-1V( X(z)) -1 n= k znX(z) n=0
, where k 1 p q rn n n n
Now, the probability that the server is on vacation is given by
P(V) = lim Q (z,0)jz 1 j = 1
=a-1V( X(z)) -1 nlim k zn
z 1 X(z) n=0
P(V) =a-1
E(V) knn=0
(2.56)
where E(V) is the expected vacation period.
51
2.4.6 Probability that the Server is Busy
Let B be the busy period random variable and P (B) be the
probability that the server is busy at time t.
From the Equations (2.42), (2.44) and (2.45), we have
P(B) =b
lim P (z,0) R(z,0)iz 1 i=a
=b-1
lim P (z,0) + P (z,0) + R(z,0)i bz 1 i=a
Now, the probability that the server is busy is given by
P(B) =b-1 b-1f (1) E(R)
E(S) c + + f (1) + (T1)ci iT1 T1i=a i=a
(2.57)
where
m m
b-1 a-1f (1) = E(X)E(S) + E(R)E(X) - m c + (1- )p + r E(X)E(U)m
m=a m=0
and T1 = b - E(X) E(S) + E(R)
2.5 PARTICULAR CASES
In this section, some of the existing models are deduced as a
particular case of the proposed model.
Case (i): If no request for re-service (i.e. 0 ) and if there is no setup time
i.e., U( ( )) 1X z , then the Equation (2.46) reduces to
1a
0m
mzm
k)1bz(]1))z(X(V~
[
1b
amm
k)m
zb
z(]1))z(X(S~
[
))z(X)(z(h
1)z(P
52
where ))z(X(S~bz)z(h and mqmpmk , which exactly coincides
with the result XM / G(a,b) /1 and multiple vacations without setup time and
N-policy of Krishna Reddy et al (1998).
Case (ii): Considering single service (i.e. a = b = 1), no request for re-service
(i.e. 0 ) and if there is no setup time (i.e. U( ( )) 1X z ), then (2.46)
becomes
V( X(z)) -1 z -1 k0P(z) =
z -S( X(z)) X(z) - , where k = p + q
0 0 0
which coincides with the result XM / G /1 queueing system and multiple
vacations without N-Policy of Lee et al (1994).
2.5.1 Special Cases
The model so developed is general in nature as the service time,
set-up time and vacation time are arbitrary. But for practical purposes,
service time, set-up time and vacation time with particular distribution is
required. In this section, some special cases of the proposed model by
specifying vacation time random variable as exponential distribution, setup
time random variable as exponential distribution and bulk service time
random variable as hyper exponential distribution are discussed.
Case (i): /1ba,/GXM queueing system with setup time, server’s choice of
admitting re-service and exponential vacation time
If the vacation time is assumed as exponential with probability
density functionx
v(x) e , where is the parameter, then,
V( X(z))(1 X(z))
. Substituting this expression for V( X(z)) in
53
(2.46) and after some algebra, the PGF of the queue size distribution of this
special case of the queueing model is obtained as,
(S( X(z)) 1) b 1 b m(z z )cmm aR( X(z)) 1 (S( X(z))
1 S( X(z)) a 1 m(1 )p r zm m
R( X(z)) 1 S( X(z)) m 0
a 1 b m(z 1) U( X(z)) k q zm m(1 X(z))m 0
bz
P(z)
a 1S( X(z)) m(k q )zm mm 0R( X(z)) 1 S( X(z))
h(z)( X(z))
where bh(z) z (1 )S( X(z)) S( X(z))R( X(z))
Case (ii): /1ba,/GXM queueing system with multiple vacations, server’s
choice of admitting re-service and exponential setup time
In case of exponential setup time random variable with probability
density function uxu(x) u e , where ‘u’ is the parameter, then,
uU( X(z))
u (1 X(z)). Substituting this expression for U( X(z)) in
(2.46) and after some algebra, the PGF of the queue size distribution of this
special case of the queueing model is obtained as,
54
(S( X(z)) 1) b 1 b m(z z )cmm aR( X(z)) 1 (S( X(z))
1 S( X(z)) a 1 m(1 )p r zm m
R( X(z)) 1 S( X(z)) m 0
a 1 ub m(z 1) V( X(z))k q zm m
u (1 X(z))m 0
bz S( X(z
P(z)
a 1)) m(k q )zm m
m 0R( X(z)) 1 S( X(z))
h(z)( X(z))
and bh(z) z (1 )S( X(z)) S( X(z))R( X(z))
Case (iii): /1ba,/GXM queueing system with hyper exponential bulk
service time, setup time and server’s choice of admitting re-service
If the service time is assumed as hyper- exponential service time
with the probability density functionyx wxs(x) cye (1 c)we , where y and
w are the parameters, then,yc w(1 c)
S( X(z))y (1 X(z)) w (1 X(z))
.
Substituting this expression for S( X(z)) in (2.46) and after some algebra,
the PGF of the queue size distribution of this special case of the queueing
model is obtained as,
b 1 b m(z z )cmm a
P(z)
yc w(1 c)1
y (1 X(z)) w (1 X(z))
yc w(1 c)R( X(z)) 1
y (1 X(z)) w (1 X(z))
yc w(1 c)1
y (1 X(z)) w (1 X(z)) a 1 mz
m 0
a 1 b m(z 1) U( X(z)) V( X(z))k q zm m
m 0
(1 )p rm myc w(1 c)
R( X(z)) 1y (1 X(z)) w (1 X(z))
yc w(1 c)bz
y (1 X(z)) w (1 X(z)) a 1 m(k q )zm m
m 0
h(z)( X(z))
yc w(1 c)R( X(z)) 1
y (1 X(z)) w (1 X(z))
55
where
yc w(1- c)bh(z) = z - (1- ) +
y + (1- X(z)) w + (1- X(z))
yc w(1- c)- + R( X(z))
y + (1- X(z)) w + (1- X(z))
2.6 COST MODEL
Cost analysis is the most important phenomenon in any practical
situation at every stage. Cost involves startup cost, operating cost, holding
cost, setup cost, re-service cost and reward cost. It is quite natural that the
management of the system desires to minimize the total average cost and to
optimize the cost. Addressing this, in this section, the cost model for the
proposed queueing system is developed and the total average cost is obtained
with the following assumptions:
Cs : Startup cost per cycle
Ch
: Holding cost per customer per unit time
Co : Operating cost per unit time
Cr : Reward cost per cycle due to vacation
Crs : Re-service cost per unit time
Cu : Setup cost per cycle
Since the length of the cycle is the sum of the idle period and busy
period, from the Equations (2.53) and (2.54), the expected length of cycle,
cE(T ) is obtained as
56
E (Tc) = E (length of the Idle Period) +E (length of the Busy Period)
E (Tc) =E(V)
+ E(U)a 1 n
1 (1 )p ri n i n in 0 i 0
+E(S)
E(R)a 1
(1 )p rn nn 0
Now, the total average cost (TAC) per unit time is obtained as
Total Average Cost = Start-up cost per cycle + holding cost of number of
customers in the queue per unit time + Operating cost
per unit time * + re-service cost per unit time +
Setup cost per cycle – reward due to vacation per
cycle.
TAC =E(V) 1
C - C + C E(U) + C E(R) + C E(Q) + C us r rs ohP(U = 0) E(T )c
(2.58)
It is difficult to have a direct analytical result for the optimal value
a* (minimum batch size in M
X/G(a,b)/1 queueing system) to minimize the
total average cost. The simple direct search method to find optimal policy for
a threshold value a* to minimize the total average cost, is defined.
Step 1: Fix the value of maximum batch size ‘b’
Step 2: Select the value of ‘a’ which will satisfy the following relation
*TAC a TAC(a), 1 a b
Step 3: The value a* is optimum, since it gives minimum total average cost.
57
Using the above procedure, the optimal value of ‘a’ can be
obtained, which minimizes the total average cost function. Some numerical
example to illustrate the above procedure is presented in the next section.
2.7 NUMERICAL ILLUSTRATION
In this section, the consistency of the theoretical results obtained in
the sections 2.3 – 2.4 are justified numerically with the following assumptions
and notations:
Service time distribution is 2 – Erlang with parameter
Batch size distribution of the arrival is geometric with mean 2
Probability of requesting re-service
Probability of admitting re-service
Re-service time is exponential with parameter
Vacation time is exponential with parameter
Setup time is exponential with parameter u
Minimum service capacity a
Maximum service capacity b
2.7.1 Effects of Various Parameters on the Performance Measures
The effects of various parameters such as arrival rates, expected
queue length, expected idle period, expected busy period, probability that the
server is on vacation, probability that the server is busy, probability of request
for re-service, probability of server’s choice of admitting customers for re-
service and threshold value ‘a’ are analyzed numerically and presented in
Tables 2.1 – 2.3 and represented in Figure 2.2. All numerical results are
obtained using Mat Lab software.
58
The effects of different probabilities of server’s choice of admitting
re-service ‘ ’ for a fixed ‘a’ and ‘b’ on various performance measures are
obtained numerically. These results are tabulated in Table 2.1. From the
table, the following observations are made:
If the probability of server’s choice of admitting re-service
increases, then
the probability that the server is on vacation decreases
the probability that the server is busy increases
the expected queue length increases
the expected idle period decreases
the expected busy period increases
The effects of different arrival rates for a fixed ‘a’ and ‘b’ are
compared with various performance measures and are presented in Table 2.2.
From the table, it is clear that, if the arrival rate increases, then
the probability that the server is on vacation decreases
the probability that the server is busy increases
the expected queue length increases
the expected idle period decreases
the expected busy period increases
In Table 2.3 and Figure 2.2, for various threshold values ‘a’, the
expected queue length, the expected idle period and the expected busy period
are compared with different probabilities of server’s choice of admitting re-
service. From the table and the figure, the following points are observed:
59
When the probability of admitting re-service increases, the
mean queue size and the mean busy period increase
When the threshold value increases, the mean queue size and
mean busy period increase whereas the mean idle period
decreases
2.7.2 Optimal Cost
In this section, a numerical example is analyzed to illustrate how
the management of an Environmental Sensor Networks (ESN) can effectively
use the results obtained in the sections 2.3, 2.4 and 2.6 to make the decision
regarding the threshold value to minimize the total average cost.
It is assumed that, the maximum capacity of the ESN is 10 units
(i.e. b = 10 messages). If the ESN operator starts the process even for a single
message (i.e. a = 1) without waiting for further arrival, clearly, the operating
cost will increase. On the other hand, if they start the process until all 10
messages arrive, the holding cost may increase; hence, there must be some
value between 1 and 10 that will optimize the cost. An optimal policy
regarding the threshold value ‘a’ which will minimize the total average cost is
wished to be obtained.
The total average costs are obtained numerically with the following
assumptions:
Startup cost : ` 3.00
Operating cost per unit time : ` 5.00
Holding cost per customer : ` 0.25
Re-service cost per unit time : ` 2.00
Reward cost per unit time due to vacation : ` 1.00
Setup cost per unit time : ` 1.00
60
The effects of threshold value ‘a’ on the total average cost with b = 10
are reported in Tables 2.3 and 2.4 and represented in Figures 2.3 and 2.4.
From the Table 2.3 and the Figure 2.3, it is clear that, for an
environmental sensor networks center with the capacity of 10 messages at a
time, the management has to fix the threshold value a = 5 to minimize the
total average cost for the probability of requesting re-service 0.6
(i.e. 0.6 ) and the probability of server’s choice of admitting re-service 0.2
(i.e. 0.2 ). Similarly, the management has to fix the threshold value
a = 4 to minimize the total average cost for the probability of requesting
re-service 0.6 (i.e. 0.6 ) and the probability of server’s choice of admitting
re-service 0.4 (i.e. 0.4 ).
Similarly, the management has to fix the threshold value ‘a’ to
minimize the total average cost for various probabilities of requesting re-
service and different probabilities of server’s choice of admitting re-service.
These values are presented in Table 2.4 and represented in Figure 2.4.
2.8 CONCLUSION
In this chapter, a “ /1ba,/GXM queueing system with server’s
choice of admitting re-service, multiple vacations and setup time” is analyzed.
The probability generating function for the queue size at an arbitrary time
epoch is derived. Various performance measures are also obtained. Some
particular cases and special cases are also discussed. The theoretical
development of the model is justified with numerical results. The results so
obtained in this chapter can be used for managerial decision to optimize the
overall cost and search for the best operating policy in a waiting line system.
61
Table 2.1 Probability of Server’s Choice of Admitting Customers (Vs)
Performance Measures
(For 2.5 , a =3, b = 4, 2.0 , 0.6 , 6 , 5 , u = 4)
P(V) P(B) E(Q) E(I) E(B)
0.0 0.3062 0.5995 4.2853 0.5319 1.3258
0.2 0.2864 0.6269 4.7467 0.5228 1.4617
0.4 0.2651 0.6528 5.2338 0.5175 1.5626
0.6 0.2445 0.6796 5.8252 0.5108 1.7060
0.8 0.2239 0.7065 6.5247 0.5043 1.8749
1.0 0.2031 0.7334 7.3580 0.4982 2.0689
P (V) - Probability that the server is on multiple vacations; P (B) - Probability that the
server is busy; E(Q) – Expected queue length; E(I) – Expected idle period; E(B) – Expected
busy period
Table 2.2 Arrival Rate (Vs) Performance Measures
(For 2.0 , a =3, b = 4, 0.4 , 0.6 , 6 , 5 , u = 4)
P(V) P(B) E(Q) E(I) E(B)
1.5 0.5295 0.3659 2.6919 0.5625 1.2346
2.0 0.3916 0.5081 3.5863 0.5467 1.2866
2.5 0.2651 0.6528 5.2338 0.5175 1.5626
3.0 0.1488 0.7978 9.9946 0.4868 2.3810
3.5 0.0414 0.9418 44.3704 0.4597 7.5073
P (V) - Probability that the server is on multiple vacations; P (B) - Probability that the
server is busy; E(Q) – Expected queue length; E(I) – Expected idle period; E(B) – Expected
busy period
62
Table 2.3 Threshold Value (Vs) Performance Measures and Total
Average Cost
(For 0.5 , 2.0 , b = 10, 3 , 8 , u = 7)
a0.2; 0.6 0.4; 0.6
E(Q) E(I) E(B) TAC E(Q) E(I) E(B) TAC
1 0.2829 0.3260 1.5222 1.9854 0.2931 0.3242 1.5950 1.9897
2 0.5855 0.3115 1.9065 1.7934 0.5924 0.3108 1.9673 1.8142
3 0.9931 0.3024 2.3051 1.6931 0.9969 0.3020 2.3607 1.7199
4 1.4492 0.2961 2.7196 1.6502 1.4524 0.2960 2.7694 1.6806
5 1.9356 0.2917 3.1419 1.6492 1.9409 0.2916 3.1836 1.6828
6 2.4456 0.2884 3.5650 1.6800 2.4524 0.2884 3.6025 1.7146
7 2.9823 0.2860 3.9757 1.7382 2.9899 0.2860 4.0125 1.7726
8 3.5578 0.2842 4.3552 1.8237 3.5682 0.2842 4.3873 1.8588
9 4.1997 0.2829 4.6750 1.9417 4.2139 0.2830 4.7014 1.9780
10 4.9706 0.2822 4.8792 2.1101 4.9867 0.2823 4.9078 2.1462
E(Q) – Expected queue length; E(I) – Expected idle period; E(B) – Expected busy period;
TAC – Total average cost
Table 2.4 Threshold Value (Vs) Total Average Cost for various and
(For 0.5 , 2.0 , b = 10, 3 , 8 , u = 7)
a Total Average Cost
0.7; 1.0 1.0; 1.0 1.0; 0.5 1.0; 0.8
1 2.0087 2.0221 2.0000 2.0128
2 1.8872 1.9336 1.8557 1.9016
3 1.8229 1.8871 1.7795 1.8441
4 1.7998 1.8738 1.7498 1.8248
5 1.8091 1.8877 1.7553 1.8360
6 1.8456 1.9272 1.7898 1.8731
7 1.9063 1.9903 1.8497 1.9343
8 1.9935 2.0790 1.9357 2.0216
9 2.1130 2.2000 2.0545 2.1419
10 2.2833 2.3719 2.2238 2.3128
63
Figure 2.2 Threshold Value (Vs) Expected Queue Length
(For various probability of admitting re-service ‘ ’)
Figure 2.3 Threshold Value (Vs) Total Average Cost
(For various probability of admitting re-service ‘ ’)
64
Figure 2.4 Threshold Value (Vs) Total Average Cost for = 0.7 , = 1.0
(For 0.5 , 2.0 , b = 10, 3 , 8 , u = 7)
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