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On distribution of cycles in a Markovian queueing system
Microelectron. Reliab., Vol. 32, No. 10, pp. 1483-1487, 1992. Printed in Great Britain.
0026-2714/9255.00 + .00 © 1992 Pergamon Press Ltd
TECHNICAL NOTE ON DISTRIBUTION OF CYCLES IN A MARKOVIAN QUEUEING SYSTEM RAJIV BHANDARI
Department of Statistics, M.L.N. College, Yamuna Nagar 135001, Haryana, India
(Received for publication 30 May 1991) Abstract--This paper studies the probability distribution of the number of cycles completed in a single-server queueing system with Poisson arrivals, exponential service times and having a limited capacity. The expressions for the number of cycles completed in the time interval (0, t] and the distribution of the number of busy periods completed in time interval (0, t] are obtained explicitly. These distributions are contemplated through the joint probability distribution of the number of customers in the system at time t and the number of times the system reaches its extreme state in the interval (0, t].
INTRODUCTION
MODEL DESCRIPTION AND NOTATION
Though a good deal of work has already been done on a single-server limited capacity queueing system with Poisson arrivals and exponential service times, there are still many important aspects of the system which need to be exploited. With this view in mind, two important aspects of the system, namely the number of cycles completed in time interval (0, t] and the number of busy periods completed in time interval (0, t], are considered. A cycle is said to be completed when the queueing process, starting from state zero, re-reaches the state zero after covering all other possible states at least once; a busy period is one in which the queueing process starting with state zero re-reaches the zero state irrespective of the number of times that the other states are covered. To obtain the said distributions, the joint probabilities, Pro,n.j(t), of the number of customers in the system at time t, the number of times the system reaches state zero in the time interval (0, t] and the number of times the system reaches state N in the time interval (0, t] are obtained. The probability Bo,~(t) of completing exactly one busy period in the interval (0, t] is obtained as the marginal probability Po,2.(t). Then the probability Bo,,(t) of completing exactly r busy periods in the time interval (0, t], is given by the rth convolution of Po,2.(t). The probability B.,r(t) of completing r (not exactly) busy periods in the time interval (0, t] is obtained as the convolution of B0,,(t) and P,~,,(t) by taking a summation over m, m -- 1, 2 . . . . . N. The probability Co,,(t) of completing exactly r cycles in the time interval (0, t] is given by the rth convolution of Co, l(t). On the same lines, the probability C., ,(t) of completing r cycles in time interval (0, t] is obtained by the convolution of Co,,(t) and Pm, l(t).
The customers arrive, one by one, at a counter in accordance with a Poisson process and a parameter 2 > 0. The customers are served singly by a single server and the service times are independently, identically exponentially distributed with a parameter g > 0. This server is busy if there is at least one customer in the system. The capacity of the system is N. By capacity, we mean the space available for customers. Thus the customer arriving when the system has more than N customers is lost. The servicing process starts at time zero with zero customers in the system. Let the random variables X(t), Y(t) and Z ( t ) denote the number of customers in the system at time t, the number of times the system reaches state zero in a time interval (0, t], and the number of times the system reaches the state N, the capacity of the system, in a time interval (0, t], respectively. The joint probability of X(t), Y(t) and Z ( t ) is defined as
~R
s2/10-j
Pm, n,~(t) = prob[X(t) = m, Y(t) = n and Z(t) = j ] ,
O<~m<~N.
(1)
marginal probabilities, Pm, n,.(t), P .... j(t), Pro, .,j(t), P,, .... (t), P., n, .(t), and P .... j(t) may be
The
obtained by taking summations over ranges of the different specific suffixes. B0,,(t) = the probability of completing exactly r busy periods in a time interval (0, t], B., ,(t) = the probability of completing not exactly r busy periods in a time interval (0, t], Co,,(t)=the probability of completing exactly r cycles in a time interval (0, t], C., ,(t) = the probability of completing not exactly r cycles in a time interval (0, t], where r = 0, 1, 2 . . . . .
1483
1484
Technical Note where
The joint probability generating functions
F,.,.(z;t)= ~ P~,,.,y(t)zY=Fm,.,
[z[~
6ij
(2)
j=O
= V1 L0
if i = j if i e j,
and
G.,(y,z;t)= ~ F..,.(z;t)y"=G.,
F'(t) = d F(t).
ly[~
nffil N
H ( x , y , z ; t) = ~, G . ( y , z ; t)x" = H .
(4)
ratiO
A Laplace transform (LT) of the function F ( t ) is defined as
F*(s)=
f(t)e-~'dt,
Taking the Laplace transform of equation (7) and using the initial conditions P0,~,0(0) = 1,
sP*,o,/(s)=O,
O<~m<~N,
O<~j,
sP ~,.,j(s ) = - 2P ~.,j(s ) + (1 - 6t")laPl,.- l,j(s ) + foj, l<~n,
t >0
O<~j,
se~,..j(s) = --(2 + la)P*..j(s) + 2P*.,_ 1,.,j(s) and Re(s)>0.
+ laP* + ~,.,As),
(5)
l<.m<~N-1, EQUATIONS GOVERNING THE PROBABILISTIC ANALYSIS OF THE SYSTEM Simple probability reasoning provides the following equations for the joint distribution of X(t), Y(t) and Z(t), as defined by equation (1)
Pm, o , j ( t + d t ) = O ,
O<~m<~N
l<~n,
O<.j,
sPS, .,j(s) = - laPS, .,j(s) + 2P S_ 1,.,j- t(s), l~
O
By multiplying equation (8) by the appropriate powers of z, summing over j and using equation (2), we have (s + 2)F~,.(z; s) = la(1 -- fln)F~._ l(z; s) + fit.,
Po, 7,y(t + d t ) = 0, 0 < j
l~
Po,.,j(t + dt) = [1 - 3. dt]Po,.,j(t)
(s + 2 +la)F~,n(z; s) = 2F*_ ,,.(z; s) + laF*+ t,.(z; s),
+ [1 - Jt"]laPl,._ l,j(t) dt + 0(dt), l~
l<.m<~N-1,
0~
(s +la)F*,.(z;s)=2zF*=l..(z;s),
P.,.,j(t + dt) = [1 - (2 + la) dt]P,.,.,j(t)
+ laPin+~,.,j(t) dt + 0(dt), l <~n,
l<~n, 1 <~n.
(9)
Again multiplying equation (9) by the appropriate powers of y, summing over n and using equation (3). we have
+ 2P,._ l, .,j(t) dt
l <<.m <~N - 1 ,
(8)
(s + ).)G* (y, z; s) = layG'~'(y, z; s) + y,
O <<.j ,
(s + 2 + la)G*(Y, z; s) = 2G* l(Y, z; s) PN, .,j(t + dt) = [1 - la dt]PN..,j(t)
+ laG*+ l(y,z; s),
+ 2 P u _ l , . , j _ l ( t ) d t + 0(dt), l~
Ptc,.,o(t + dt) = 0,
l <~m < ~ N - 1 , 0
I ~
(s + l a ) G * ( y , z ; s ) = 2 z G S _ l ( y , z ; s ) . (6)
From equation (6) we have the following differencedifferential equations governing the system
P'.,,o,j(t)=O,
Further, multiplying equation (10) by the appropriate powers of x, summing over m and using equation (4), we have
O<<.m<<.N, O<~j
yx=
P'o,.,j( t ) = - 2Po,..j( t ) + (1 - 6t")laPt,.- l,j( t ),
P'.,,../(t) = --(2 + la)P,.,.,j(t)
[ -- .~ ( l - - z )
]
I - x ( 1 - y ) G*,
J'Z(1--X)] ~* xN+l -; "+g tI ~-1
lay(l-x) S + .~ (ii)
l<.n,
O<.j,
P'u,.,l(t) = --laPu,.4(O + )'PN- 7,.,j- 1(t),
0
1 <~n,
s+2
(s + 2 +la)x - 2 x 2 - l a
+ AP ..... j (t) + laP. + ~..,j(t),
P'u,..o(t) = O,
laL
n*=
1 <.n, O<.j
l<~m<~N--l,
(10)
where
(7)
a ~ = s ~Y+~ G ~, +
Y s+).
Technical Note and
a If
1485
By differentiating equation (15) n times with respect to y, puting y = 0 and dividing by n, we have Z n 2 (-'J. . . + , _ . / B , ( ) ~ .
2z +/at,_
"1" s
Now H* only exists, for all finite values of x, if the zeros of the denominator are also the zeros of the numerator. The zeros of the denominator are: (s + 2 +/z)-F x/(s + 2 + #)2-- 4)qz • ~(s) = 22 , i = 1, 2.
(12) Here, the value of the square root with a positive real part is taken, that is, ~(s) has Re(s) > 0. a~ has the positive sign before the radical. By simple application of Rouche's theorem, we can show that the denominator of H* has exactly one zero within the unit disc and there is no zero on the boundary. This zero must be #~(s) since 1~2(s)l < I"l(s)l. Now, H* is a polynomial in x, therefore, for H* to remain finite, the numerator of equation (11) must vanish for %(s), i = 1, 2. Hence
2,i \-I , _ 2[(1 - z )
n =
t~2111
NN+I
x,-#-l~,i
2
2
+s--~-~iy =0,
i=ls2.
~(~, -
= _
~)v*,(~)
2 .....
(16)
_
( - 1) ~+'
~,.-"
iffilkffi0
IO~ N - I l
--
O~N - l - l n + j - k 2 I
× " -~¢--~ ×
2
j
~=tk-0
~ ,+2
.-(),
\s + 2 ]
j-k
/k~--~--~.}L~-~-,
× L n'~-a~"~ J
'
~<----rj
_1
(13)
(17)
where [j
By eliminating G*_ min equation (13), we have
if n~ j .
r=
On simplifying the serial expansion of equation (17), we have
1 2 G~'= Y'~ (--l)k+lAk+l(Z) k-0 s+2
, (._y(.__?(
Y × o~_~ B, (z)y + B2(z)'
(14)
P-'"s's'=z-* () :ks+2J\s+lx/t~J/g/ nffil
where
xX
}.,.+, A2(z) = 2 l - Z s + l z
1,
'
By comparing the coefficientsof # in equation (16), we have
2z_q-:,) l
s+~
.- ,=+,- -<-- ,>:" ., ~<, tB,(:))
x,:.
×
ll[l,(1
+2
r
k n+j-2k
X X
i=O kfOaffiO
$+
x
m
E ( - l ) a+b+*
E bffiO
dffiO
d
X m -Ira- 1 - 2 i + n + j + 2 ( N - 2 l a + N N - 1~ + 2Ndl
s, (z) = ~
[A, (z)~l - A~(z)~l],
B 2 ( z ) ~-. - ~ [ A
I (z)r# 2 -
+(
\xl~)
A 2 ( z ) l ~ I ].
\s+~iks+2J
r--[ m--2 k n + j - 2 k - I
By comparing the coefficients of x = in equation (11), we have
× Z Z Z
2 ( , , - 0h)G* : ,~, (-- 1)'- '[y ~-~+A,~-("- ') + / l ( y s+ - ~ atT(m-'l -- ~Tm)G~] ,
2
kffi0 iffi0affi0
bffi0
×
k 7\
j- k
x
b
× R-lm-2-2i+n+j-
m----0,1 . . . . . N.
(15)
o~
Z ( - 1 ) a+b+k d=0 :\a:
d 1 + 2 ( N - 2 ) a + 2 ( N - l)b+ 2Nd]
(is)
1486
Technical Note Using equation (20) in (18), we have, after taking a LT of equation (18):
where ~1 =
/~R,
P.... At) =~ ~2 ~-- ~ ]/~-gR ~ -i tn - I
r
×E
Let
i=0
/'
1
Y/
1
k"
,
k
n + j - 2k
E E
ao
E
E ( - 1 ) °+'~
b=O
k=Oa=O
d=O
(19) x
then by taking the Laplace transform of equation (19), using Erdelyi tables [1] and tables of integrals, series and products [3], we have
x
j-k
(:)(. +J--)(. b
d
T y " ( t ' v ) = " g£=1O( \n - g ' j - - l + J - - 2 )
x Tj..(t, v , ) + # / + " - ' ( 4 )
[e-a'(# _ 2)._g+j_l 1
x
m-2r-I
k n+j-2k-I
EEE I=O
g
~o
1
t
× tE E .-m-~_.(-1) "~Oq=ol.g --l.
x
E
k=O a~O b=O
(.-,)(. k
"+j
oo
E ( - l ) "÷~÷k
d=O
j-k
x (2x/~--#)qt'+q g!
(v/-~t)'r(g
+q - l+v F(v + l)F(g + q - l + 1 + v)
X
~,.(t. v2), where
x ~&(v + ~,g + q - I + vj2v + 1, g+q-l+l+vj4w/~t)-v
(21)
v, = [m - 1 - 2i + j + 2 ( N - 2)a + 2 ( N - 1) 1 g!
x b + 2Nd],
e-t~_
v2 = [ m - 2 - 3i + n + j -
1 +2(N-2)a
+ 2
n-g+j-2n+j-r-2
×
X
y
Y
r=O
I=0
x (N - l)b + 2Nd],
Z ( - I ) - , -~-,-~ q=O
iv(z)=
1 (n - - g + j - - r -- 2)!
the modified Bessel function of the first kind, and
(n+j-r-2)( ×
t
2F2(a, b, c, d; z) = ~,
,,=o
)k~ - U
(w/~t)r(n+ j + q -- r -- l -- 1 + v) F(v + 1)F(n + j + q - r x 2F2(v +½, n + j + q - r - l -r
(a).(b).
g n
(c).(d). n!"
DISTRIBUTION OF THE NUMBER OF BUSY PERIODS COMPLETED IN TIME INTERVAL (0, tl
( 2 w / ~ - 2)q t" +j +q - r - 2 qt
2v + 1, n + j + q
kYok!(v + k + 1)'
1 + v) l + v;
Since
B~.l(s) = P~2(s) = (s +# 2)2 ×
~ - ~
j
-l+v;4,f~t)]. x ~ (oc~ - = ~ ) J ( = ~ + ' (20)
j=O
- a~ +')-y,
(22)
Technical Note and on simplifying the serial expansion of equation (22), we have
l)k+'\ k(J'/ \ ~(j+l) t
B~,(')=# ~ ~ ~ ~( x(\~/.~,/
T~,2(s, v),
1487
NUMBER OF CYCLES COMPLETED IN TIME INTERVAL (0, t] Since Co*,,(s)
(23)
= Po*,~(s) -
et,2,,(s)
fl
(~'I --
~f2) 2
(~ + ~)~ ( : f - : f ) ( : f +' - ~f+,) where
T~,2(s, v ) = ( s - ~ )2R -v.
(.V
(24)
By taking the Laplace transform of equation (24) using Erdelyi tables and tables of integrals, series and products, we have
(
I'
On simplifying the serial expansion of equation (28), we have
C~|(s)'=A(~/-~)i-'l~o~=oi~lo(-~=o l)S+l(21)
T°'2(t'v)=ve-Attq+l~p f O q =~O (-- 1)1 - P ( ~ ) x(i--l)( × (2,~
X
- ,)q q[
By taking the Laplace transform of equation (29), using Erdelyi tables and tables of integrals, series and products, we have / /~'V-1 2 ~ i-I C°"(t)=2~4~) t~o~o/~o~o(--l)/+t(2/)
( x / ~ t ) r ( q - p + 1 + v)
F(v + 1)F(q - - p + 2 + v)
x 2F2(v + l , q _ p 2v+l,
+ 1 +v;
q--p+2+v;
4 x / ~ t ). (25)
Using equation (25) in (23), we have, after taking a LT of equation (23):
,
Hence Co,,(t) is the rth convolution of Co,i (t), is given by
[ro,2(t, v3)-To,2(t, v4)], (26)
where
C.,,(t)--,~=~ jo Co.,(x)Pm.l(t - x)dx,
(31)
which yields the required probability of completing r cycles in a time interval (0, t].
v3 = [2(Nk + (N + 1)1) + j + r + I],
Acknowledgement--The author is grateful to Dr (Miss)
v4 -- [2(Nk + (N + 1)/) + 2N + j - r -4- 1].
Sharda Kumafi, Chairperson, Department of Statistics and Operational Research, Kurukshetra University, Kurukshetra, for her keen interest in this work.
After obtaining Bo,, (t), the r th convolution of B0, ~(t), we have
B. ,,(t) -
Bo,,(x)Pm, l(t - x) dx,
(27)
m-lJO
which gives the required probability of completing r (not exactly) busy periods in a time interval
(0, t].
where v5 -- [2(N(j + 1) + (N + l)k + l) + (i - 1)].
, =/z 1 k+, J J+ B° l(t) r-~Oj~Ok~OI~O(--) ( k ) t 1 1) x \~j
k _) 0,2(S, V). (29)
REFERENCES 1. A. Erdelyi, Tables of Integral Transforms, Vol. 1, McGraw-Hill, New York (1954). 2. T. L. Saaty, Elements of Queueing Theory with Applications, McGraw-Hill, New York (1961). 3. I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Academic Press, New York (1965).
0026-2714/9255.00 + .00 © 1992 Pergamon Press Ltd
TECHNICAL NOTE ON DISTRIBUTION OF CYCLES IN A MARKOVIAN QUEUEING SYSTEM RAJIV BHANDARI
Department of Statistics, M.L.N. College, Yamuna Nagar 135001, Haryana, India
(Received for publication 30 May 1991) Abstract--This paper studies the probability distribution of the number of cycles completed in a single-server queueing system with Poisson arrivals, exponential service times and having a limited capacity. The expressions for the number of cycles completed in the time interval (0, t] and the distribution of the number of busy periods completed in time interval (0, t] are obtained explicitly. These distributions are contemplated through the joint probability distribution of the number of customers in the system at time t and the number of times the system reaches its extreme state in the interval (0, t].
INTRODUCTION
MODEL DESCRIPTION AND NOTATION
Though a good deal of work has already been done on a single-server limited capacity queueing system with Poisson arrivals and exponential service times, there are still many important aspects of the system which need to be exploited. With this view in mind, two important aspects of the system, namely the number of cycles completed in time interval (0, t] and the number of busy periods completed in time interval (0, t], are considered. A cycle is said to be completed when the queueing process, starting from state zero, re-reaches the state zero after covering all other possible states at least once; a busy period is one in which the queueing process starting with state zero re-reaches the zero state irrespective of the number of times that the other states are covered. To obtain the said distributions, the joint probabilities, Pro,n.j(t), of the number of customers in the system at time t, the number of times the system reaches state zero in the time interval (0, t] and the number of times the system reaches state N in the time interval (0, t] are obtained. The probability Bo,~(t) of completing exactly one busy period in the interval (0, t] is obtained as the marginal probability Po,2.(t). Then the probability Bo,,(t) of completing exactly r busy periods in the time interval (0, t], is given by the rth convolution of Po,2.(t). The probability B.,r(t) of completing r (not exactly) busy periods in the time interval (0, t] is obtained as the convolution of B0,,(t) and P,~,,(t) by taking a summation over m, m -- 1, 2 . . . . . N. The probability Co,,(t) of completing exactly r cycles in the time interval (0, t] is given by the rth convolution of Co, l(t). On the same lines, the probability C., ,(t) of completing r cycles in time interval (0, t] is obtained by the convolution of Co,,(t) and Pm, l(t).
The customers arrive, one by one, at a counter in accordance with a Poisson process and a parameter 2 > 0. The customers are served singly by a single server and the service times are independently, identically exponentially distributed with a parameter g > 0. This server is busy if there is at least one customer in the system. The capacity of the system is N. By capacity, we mean the space available for customers. Thus the customer arriving when the system has more than N customers is lost. The servicing process starts at time zero with zero customers in the system. Let the random variables X(t), Y(t) and Z ( t ) denote the number of customers in the system at time t, the number of times the system reaches state zero in a time interval (0, t], and the number of times the system reaches the state N, the capacity of the system, in a time interval (0, t], respectively. The joint probability of X(t), Y(t) and Z ( t ) is defined as
~R
s2/10-j
Pm, n,~(t) = prob[X(t) = m, Y(t) = n and Z(t) = j ] ,
O<~m<~N.
(1)
marginal probabilities, Pm, n,.(t), P .... j(t), Pro, .,j(t), P,, .... (t), P., n, .(t), and P .... j(t) may be
The
obtained by taking summations over ranges of the different specific suffixes. B0,,(t) = the probability of completing exactly r busy periods in a time interval (0, t], B., ,(t) = the probability of completing not exactly r busy periods in a time interval (0, t], Co,,(t)=the probability of completing exactly r cycles in a time interval (0, t], C., ,(t) = the probability of completing not exactly r cycles in a time interval (0, t], where r = 0, 1, 2 . . . . .
1483
1484
Technical Note where
The joint probability generating functions
F,.,.(z;t)= ~ P~,,.,y(t)zY=Fm,.,
[z[~
6ij
(2)
j=O
= V1 L0
if i = j if i e j,
and
G.,(y,z;t)= ~ F..,.(z;t)y"=G.,
F'(t) = d F(t).
ly[~
nffil N
H ( x , y , z ; t) = ~, G . ( y , z ; t)x" = H .
(4)
ratiO
A Laplace transform (LT) of the function F ( t ) is defined as
F*(s)=
f(t)e-~'dt,
Taking the Laplace transform of equation (7) and using the initial conditions P0,~,0(0) = 1,
sP*,o,/(s)=O,
O<~m<~N,
O<~j,
sP ~,.,j(s ) = - 2P ~.,j(s ) + (1 - 6t")laPl,.- l,j(s ) + foj, l<~n,
t >0
O<~j,
se~,..j(s) = --(2 + la)P*..j(s) + 2P*.,_ 1,.,j(s) and Re(s)>0.
+ laP* + ~,.,As),
(5)
l<.m<~N-1, EQUATIONS GOVERNING THE PROBABILISTIC ANALYSIS OF THE SYSTEM Simple probability reasoning provides the following equations for the joint distribution of X(t), Y(t) and Z(t), as defined by equation (1)
Pm, o , j ( t + d t ) = O ,
O<~m<~N
l<~n,
O<.j,
sPS, .,j(s) = - laPS, .,j(s) + 2P S_ 1,.,j- t(s), l~
O
By multiplying equation (8) by the appropriate powers of z, summing over j and using equation (2), we have (s + 2)F~,.(z; s) = la(1 -- fln)F~._ l(z; s) + fit.,
Po, 7,y(t + d t ) = 0, 0 < j
l~
Po,.,j(t + dt) = [1 - 3. dt]Po,.,j(t)
(s + 2 +la)F~,n(z; s) = 2F*_ ,,.(z; s) + laF*+ t,.(z; s),
+ [1 - Jt"]laPl,._ l,j(t) dt + 0(dt), l~
l<.m<~N-1,
0~
(s +la)F*,.(z;s)=2zF*=l..(z;s),
P.,.,j(t + dt) = [1 - (2 + la) dt]P,.,.,j(t)
+ laPin+~,.,j(t) dt + 0(dt), l <~n,
l<~n, 1 <~n.
(9)
Again multiplying equation (9) by the appropriate powers of y, summing over n and using equation (3). we have
+ 2P,._ l, .,j(t) dt
l <<.m <~N - 1 ,
(8)
(s + ).)G* (y, z; s) = layG'~'(y, z; s) + y,
O <<.j ,
(s + 2 + la)G*(Y, z; s) = 2G* l(Y, z; s) PN, .,j(t + dt) = [1 - la dt]PN..,j(t)
+ laG*+ l(y,z; s),
+ 2 P u _ l , . , j _ l ( t ) d t + 0(dt), l~
Ptc,.,o(t + dt) = 0,
l <~m < ~ N - 1 , 0
I ~
(s + l a ) G * ( y , z ; s ) = 2 z G S _ l ( y , z ; s ) . (6)
From equation (6) we have the following differencedifferential equations governing the system
P'.,,o,j(t)=O,
Further, multiplying equation (10) by the appropriate powers of x, summing over m and using equation (4), we have
O<<.m<<.N, O<~j
yx=
P'o,.,j( t ) = - 2Po,..j( t ) + (1 - 6t")laPt,.- l,j( t ),
P'.,,../(t) = --(2 + la)P,.,.,j(t)
[ -- .~ ( l - - z )
]
I - x ( 1 - y ) G*,
J'Z(1--X)] ~* xN+l -; "+g tI ~-1
lay(l-x) S + .~ (ii)
l<.n,
O<.j,
P'u,.,l(t) = --laPu,.4(O + )'PN- 7,.,j- 1(t),
0
1 <~n,
s+2
(s + 2 +la)x - 2 x 2 - l a
+ AP ..... j (t) + laP. + ~..,j(t),
P'u,..o(t) = O,
laL
n*=
1 <.n, O<.j
l<~m<~N--l,
(10)
where
(7)
a ~ = s ~Y+~ G ~, +
Y s+).
Technical Note and
a If
1485
By differentiating equation (15) n times with respect to y, puting y = 0 and dividing by n, we have Z n 2 (-'J. . . + , _ . / B , ( ) ~ .
2z +/at,_
"1" s
Now H* only exists, for all finite values of x, if the zeros of the denominator are also the zeros of the numerator. The zeros of the denominator are: (s + 2 +/z)-F x/(s + 2 + #)2-- 4)qz • ~(s) = 22 , i = 1, 2.
(12) Here, the value of the square root with a positive real part is taken, that is, ~(s) has Re(s) > 0. a~ has the positive sign before the radical. By simple application of Rouche's theorem, we can show that the denominator of H* has exactly one zero within the unit disc and there is no zero on the boundary. This zero must be #~(s) since 1~2(s)l < I"l(s)l. Now, H* is a polynomial in x, therefore, for H* to remain finite, the numerator of equation (11) must vanish for %(s), i = 1, 2. Hence
2,i \-I , _ 2[(1 - z )
n =
t~2111
NN+I
x,-#-l~,i
2
2
+s--~-~iy =0,
i=ls2.
~(~, -
= _
~)v*,(~)
2 .....
(16)
_
( - 1) ~+'
~,.-"
iffilkffi0
IO~ N - I l
--
O~N - l - l n + j - k 2 I
× " -~¢--~ ×
2
j
~=tk-0
~ ,+2
.-(),
\s + 2 ]
j-k
/k~--~--~.}L~-~-,
× L n'~-a~"~ J
'
~<----rj
_1
(13)
(17)
where [j
By eliminating G*_ min equation (13), we have
if n~
r=
On simplifying the serial expansion of equation (17), we have
1 2 G~'= Y'~ (--l)k+lAk+l(Z) k-0 s+2
, (._y(.__?(
Y × o~_~ B, (z)y + B2(z)'
(14)
P-'"s's'=z-* () :ks+2J\s+lx/t~J/g/ nffil
where
xX
}.,.+, A2(z) = 2 l - Z s + l z
1,
'
By comparing the coefficientsof # in equation (16), we have
2z_q-:,) l
s+~
.- ,=+,- -<-- ,>:" ., ~<, tB,(:))
x,:.
×
ll[l,(1
+2
r
k n+j-2k
X X
i=O kfOaffiO
$+
x
m
E ( - l ) a+b+*
E bffiO
dffiO
d
X m -Ira- 1 - 2 i + n + j + 2 ( N - 2 l a + N N - 1~ + 2Ndl
s, (z) = ~
[A, (z)~l - A~(z)~l],
B 2 ( z ) ~-. - ~ [ A
I (z)r# 2 -
+(
\xl~)
A 2 ( z ) l ~ I ].
\s+~iks+2J
r--[ m--2 k n + j - 2 k - I
By comparing the coefficients of x = in equation (11), we have
× Z Z Z
2 ( , , - 0h)G* : ,~, (-- 1)'- '[y ~-~+A,~-("- ') + / l ( y s+ - ~ atT(m-'l -- ~Tm)G~] ,
2
kffi0 iffi0affi0
bffi0
×
k 7\
j- k
x
b
× R-lm-2-2i+n+j-
m----0,1 . . . . . N.
(15)
o~
Z ( - 1 ) a+b+k d=0 :\a:
d 1 + 2 ( N - 2 ) a + 2 ( N - l)b+ 2Nd]
(is)
1486
Technical Note Using equation (20) in (18), we have, after taking a LT of equation (18):
where ~1 =
/~R,
P.... At) =~ ~2 ~-- ~ ]/~-gR ~ -i tn - I
r
×E
Let
i=0
/'
1
Y/
1
k"
,
k
n + j - 2k
E E
ao
E
E ( - 1 ) °+'~
b=O
k=Oa=O
d=O
(19) x
then by taking the Laplace transform of equation (19), using Erdelyi tables [1] and tables of integrals, series and products [3], we have
x
j-k
(:)(. +J--)(. b
d
T y " ( t ' v ) = " g£=1O( \n - g ' j - - l + J - - 2 )
x Tj..(t, v , ) + # / + " - ' ( 4 )
[e-a'(# _ 2)._g+j_l 1
x
m-2r-I
k n+j-2k-I
EEE I=O
g
~o
1
t
× tE E .-m-~_.(-1) "~Oq=ol.g --l.
x
E
k=O a~O b=O
(.-,)(. k
"+j
oo
E ( - l ) "÷~÷k
d=O
j-k
x (2x/~--#)qt'+q g!
(v/-~t)'r(g
+q - l+v F(v + l)F(g + q - l + 1 + v)
X
~,.(t. v2), where
x ~&(v + ~,g + q - I + vj2v + 1, g+q-l+l+vj4w/~t)-v
(21)
v, = [m - 1 - 2i + j + 2 ( N - 2)a + 2 ( N - 1) 1 g!
x b + 2Nd],
e-t~_
v2 = [ m - 2 - 3i + n + j -
1 +2(N-2)a
+ 2
n-g+j-2n+j-r-2
×
X
y
Y
r=O
I=0
x (N - l)b + 2Nd],
Z ( - I ) - , -~-,-~ q=O
iv(z)=
1 (n - - g + j - - r -- 2)!
the modified Bessel function of the first kind, and
(n+j-r-2)( ×
t
2F2(a, b, c, d; z) = ~,
,,=o
)k~ - U
(w/~t)r(n+ j + q -- r -- l -- 1 + v) F(v + 1)F(n + j + q - r x 2F2(v +½, n + j + q - r - l -r
(a).(b).
g n
(c).(d). n!"
DISTRIBUTION OF THE NUMBER OF BUSY PERIODS COMPLETED IN TIME INTERVAL (0, tl
( 2 w / ~ - 2)q t" +j +q - r - 2 qt
2v + 1, n + j + q
kYok!(v + k + 1)'
1 + v) l + v;
Since
B~.l(s) = P~2(s) = (s +# 2)2 ×
~ - ~
j
-l+v;4,f~t)]. x ~ (oc~ - = ~ ) J ( = ~ + ' (20)
j=O
- a~ +')-y,
(22)
Technical Note and on simplifying the serial expansion of equation (22), we have
l)k+'\ k(J'/ \ ~(j+l) t
B~,(')=# ~ ~ ~ ~( x(\~/.~,/
T~,2(s, v),
1487
NUMBER OF CYCLES COMPLETED IN TIME INTERVAL (0, t] Since Co*,,(s)
(23)
= Po*,~(s) -
et,2,,(s)
fl
(~'I --
~f2) 2
(~ + ~)~ ( : f - : f ) ( : f +' - ~f+,) where
T~,2(s, v ) = ( s - ~ )2R -v.
(.V
(24)
By taking the Laplace transform of equation (24) using Erdelyi tables and tables of integrals, series and products, we have
(
I'
On simplifying the serial expansion of equation (28), we have
C~|(s)'=A(~/-~)i-'l~o~=oi~lo(-~=o l)S+l(21)
T°'2(t'v)=ve-Attq+l~p f O q =~O (-- 1)1 - P ( ~ ) x(i--l)( × (2,~
X
- ,)q q[
By taking the Laplace transform of equation (29), using Erdelyi tables and tables of integrals, series and products, we have / /~'V-1 2 ~ i-I C°"(t)=2~4~) t~o~o/~o~o(--l)/+t(2/)
( x / ~ t ) r ( q - p + 1 + v)
F(v + 1)F(q - - p + 2 + v)
x 2F2(v + l , q _ p 2v+l,
+ 1 +v;
q--p+2+v;
4 x / ~ t ). (25)
Using equation (25) in (23), we have, after taking a LT of equation (23):
,
Hence Co,,(t) is the rth convolution of Co,i (t), is given by
[ro,2(t, v3)-To,2(t, v4)], (26)
where
C.,,(t)--,~=~ jo Co.,(x)Pm.l(t - x)dx,
(31)
which yields the required probability of completing r cycles in a time interval (0, t].
v3 = [2(Nk + (N + 1)1) + j + r + I],
Acknowledgement--The author is grateful to Dr (Miss)
v4 -- [2(Nk + (N + 1)/) + 2N + j - r -4- 1].
Sharda Kumafi, Chairperson, Department of Statistics and Operational Research, Kurukshetra University, Kurukshetra, for her keen interest in this work.
After obtaining Bo,, (t), the r th convolution of B0, ~(t), we have
B. ,,(t) -
Bo,,(x)Pm, l(t - x) dx,
(27)
m-lJO
which gives the required probability of completing r (not exactly) busy periods in a time interval
(0, t].
where v5 -- [2(N(j + 1) + (N + l)k + l) + (i - 1)].
, =/z 1 k+, J J+ B° l(t) r-~Oj~Ok~OI~O(--) ( k ) t 1 1) x \~j
k _) 0,2(S, V). (29)
REFERENCES 1. A. Erdelyi, Tables of Integral Transforms, Vol. 1, McGraw-Hill, New York (1954). 2. T. L. Saaty, Elements of Queueing Theory with Applications, McGraw-Hill, New York (1961). 3. I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Academic Press, New York (1965).