Time Dependent Analysis of an 𝑴/𝑴/𝟐/𝑵 Queue With Catastrophes Текст научной статьи по специальности «Математика»

Time Dependent Analysis of an M/M/2/N Queue With

Catastrophes

Gulab Singh Bura and Shilpi Gupta

Department of Mathematics & Statistics Banasthali Vidyapith, Rajasthan (India).

E-mail: burastats@gmail.com

Abstract

We consider a Markovian queueing system with two identical servers subjected to catastrophes. When the system is not empty, catastrophes may occur and destroy all present customers in the system. Simultaneously the system is ready for new arrivals. The time dependent and the steady state solution are obtained explicitly. Further we have obtained some important performance measures of the studied queueing model.

Keywords: Markovian queueing system, catastrophes, limited capacity, Time Dependent Solution.

1 Introduction

During the last 40 years the attention of the queueing models has been focused on the effect of catastrophes, in particulars, birth and death models. The catastrophes arrive as negative customers to the system and their characteristic is to remove some or all of the regular customers in the system. The catastrophes may come either from outside the system or from another service station. For example, in computer networks, if a job infected with a virus, it transmits the virus to other processors and inactivities them [8]. Other interesting articles in this area include ([2],[6],[7]). In real life it is not necessary that a queueing system should have only one server. Practically they may have more than one server identical or non identical in their functioning .Krishna kumar et. al.[7] obtained the time dependent solution of two identical servers Markovian queueing system with catastrophes.Dharmaraja and kumar[3] consider a multi-server Markovian queueing system with heterogeneous servers and catastrophes.Jain and Bura [5]obtained the transient solution of an M/M/2/N queuing system with varying catastrophic intensity and restoration. We in this paper confine ourselves to a Markovian queueing system with two identical servers subjected to catastrophes.

Rest of the paper is organized as follows:In section 3, we describe the mathematical form of the model and obtained the time dependent solution of the model. In section 4, we obtain the time dependent performance measures of the system. Section 5 provides the steady state probabilities. In section 6, we obtain the expression for steady state mean and variance. Finally, the conclusion have been given in section 6.

2 Model description and analysis

We consider an M/M/2/N queueing system with first come first out discipline that is subjected to catastrophes at the service station. Customers arrive in the system according to a Poisson stream with parameter d.The service time distribution is independently identically exponential with parameter p. When the system is not empty, catstrophes occur according to a Poisson process of rate ^. Let X(t) denote the number of customers in the system at time t.

Define Pn(t) = P(X(t) = n);n = 0,1,2,...,N be the transient state probability that there are n customers in the system at time t, and P(z, t) = £n=o Pn(t)zn be the probability generating function.

From the above assumption, the probability satisfies the following system of the

differential- difference equations:

P'o(t) = -Apo(t) + PPi(t) + HlLi Pn(t)] ;n = 0 (2.1)

P'i(t) = -(* + P + OPi(t) + Apo(t) + 2pp2(t) ;n = 1 (2.2)

Pk(t) = ~(A + 2p + f)pn(t) + Apn-i(t) + 2ppn+i(t) ;n = 2,3, ...,(N - 1)

(2.3)

P,N(t) = -(2P + OPn^ + ApN-i(t) (2.4)

It is assumed that initially the system is empty i.e.

Po(0) = 1 Pn(0) = 0 ,n = 1,2.....N (2.5)

After Multiplying equations (2.1) to (2.4) by zn for all n> 0, then summed on n from n = 0 to N and adding, we have

ln=o Pk(t)zn = [Az + ?f-(A + 2p + f)]P(z, t)

+2P(1 - i)Po(t) + AzN(l - z)pN(t) + ppi(t)(z -V) + t (2.6)

It is easily seen that the probability generating function P(z,t) satisfies the following differential equation:

d [P(z, t)] = [Az + 2-f-(A + 2p + f)]P(z, t)

+2P(1 - i)Po(t) + AzN(1 - z)pN(t) + ppi(t)(z -1) + % (2.7)

with the initial condition

P(Z,0) = 1 (2.8)

The equation (2.7) can be considered as a first order differential equation in P(z, t) and by finding the integrating factor and using the initial condition (2.8),the solution of the equation (2.7) is obtained as

P(z, t) = 2p(1 -1) f Po(t - u)e(Az+Z)ue-(X+2^+^)udu

+AzN(1 - Z) f PN(t - u)e(Az+Z^ue-(A+2^+^)udu

+p(Z - 1) f Pi(t - u)e(Az+Z^ue-(A+2^+^)udu

+% f e(Az++T)ue-(X+2^+^)udu + e(Xz+T)te-(^+2^+^)t (2.9)

Using the Bessel function identity, if a = 2^A2p and

then,

exp(Az + 2-)t = Y,n=-™ In(at)(Pz)n

where ln(.) is the moddified Bessel function of order n. Substituting this equation in (2.9) and compairing the coefficient of zn on either side, we have, for n = 0,1, ...,N Pn(t) = 2pfin Jg Po(t - u)e-(X+2^+^>u[In(au) - filn+i(au)]du

+Wnf0 PN(t-u)e (я+2^+Ом[£ NIN-n(au) — /3 (N+1)I(N+1)-n(au}]du +^Pn f P1(t — u)e-(X+2v-+^u[fi-1In-1(au) — In(au)]du +%pn f e-(X+2^)uIn(au)du + pne-(X+2^+^)tIn(at) (2.10)

where we have used I-n(.) = In(.)

Here, we have obtained Pn(t) for n = 1,...,N — 1. However, this expression depends upon P0(t) and Pjq(t). In order to determine, P0(t) and PN(t) we introduce the Laplace transform. In the sequel, for any function f(.), let f*(s) denote its Laplace transform i.e. , f*(s) = f™ e-stf(t)dt

Substitute n = 0, in equation (2.10) we get

P0(t) = 2^ f P0(t — u)e-(X+2v-+^u[I0(au) — pi1(au)]du

+A f PN(t — u)e-(X+2v-+^u[p-NIN(au) — fi-(N+11(N+1)(au)]du +P f P1(^ — u)e-(X+2,v-+^u[fi-1I1(au) — I0(au)]du +% f e-(X+2v+^ul0(au)du + e-(X+2^+^^)tl0(at) (2.11)

Taking Laplace transform on both sides of equation (2.11) and solving for, P0 (s) we obtain,

ш+Чш2-а2

2

1—

2^

ш-Чш2-а

P*(s)=AP*N(s),M-^-a‘

2Л

N

+PPi(s)

2Л

ш-Чш2-а2 2Л

—1

+ -+1

where w = s + A + 2^ + %. After some algebra, the above equation can be expressed as

,N+1

1—

ш-ЧшА-а2

1

2Л

ш-Чш2-а2 2Л

рн*)=±рт1ш-^

2Л

pPi(s)

ш-Чш2-а2

ш-ЧшА-а2

2Л

1—

+а+1)

2Л J J \2ц.

ш-Чш2-а2

cs / V 2Л

By solving equation (2.12), we get,

©

(2.12)

р^=г^)М^

N+1

Pl(s) j ш-Чш2-а2

2Л

-Чш2

+ (i+l\(1:\\2L — ±(:

\s J \2ц) |s+f s+%\ 2Л

On inversion, this equation yields an expression for P0(t) which depends upon PN(t)

(2.13)

N-l

Po(t) = e-^1 + (f) 2 f0 PN(t — U)e-(x+2^u (-+1)lN+1(au)du

l

(y)2 fo p1(t — u)e-(X+2^+^^u (-£) l1(au)du

S

2

+^\e-it — ft e-ty+2p+$)ue-$(t-u)l1Y^ldu

—J^fo e-(X+2v+f)ue-f(t-u) '-^^du (2.14)

Substituting n=1 in equation (2.10), we get

P1(t) = 2y.fi f P0(t — u)e-(X+2^+^^>u[I1(au) — fil2(au)]du

+Afi f PN(t — u)e-(X+2^+^)u[fi-N IN-1(au) — fi-(N+1^IN(au)]du

G. Singh Bura, S. Gupta RT&A, No 1 (52)

TIME DEPENDENT ANALYSIS OF AN M/M/2/N QUEUE WITH Volume 14, March 2019

CATASTROPHES___________________________________________________________________

+^.@ f P1(t — u)e-(X+2v-+^u[p-1I0(au) — I1(au)]du +%P f e-(X+2^+^)uI1(au)du + ^e-(X+2^+^)tI1(at) (2.15)

Taking Laplace transform on both sides of equation (2.15) and solving for, P1 (s) we obtain,

Pi*(s)

Vm2

a

2

ш-V ш2-а2 4^

-Vm2

4^

+Л

ш-V ш2-а2 2Л

Po(s)

N-1 |

1

[i—MF-b^+^+i)

-Vm2

4^

(2.16)

Substituting n=N in equation (2.10), we get

PN(t) = 2^pN f P0(t — u)e-(A+2v-+^u[IN(au) — ^IN+1(au)]du +A f PN(t — u)e-(A+2v-+^u[I0(au) — @-1I1(au)]du +^PN f P1(t — u)e-(A+2v-+^u[fi-1IN-1(au) — IN(au)]du +%PN fo e-(X+2^+^ulN(au)du + pNe-(X+2^+^^)tIN(at)

By taking Laplace transform and solving for Pn(s), we obtain from equation (2.17),

fш+Vш2-а2 n* ,_ч 0 1 „ (ш-V ш2-а2^ (ш-V ш2 -a2'''

(2.17)

2

— ^(^[i — ^f^)}

4^

л / ш-V ш2-а2 sN+1

2^ ( 2Л )

ш-Vш2-а2 \ + ■1 2^

—( 2Л )

P№

After some algebra, equation (2.18) can be expressed as

[i — r(s)ms) = g*(s)

where

f"(s) = [l—

ш-Vш2-a2 4^

ш-V ш2-а2 4^

N+1

+

ш-V ш2-а2 2Л

N+1

ш-Vш2-а2 2^

(2.18)

(2.19)

(2.20)

2

cu

a

2

cu

a

(, , 1 ja-VGt2-a*\N+l j$

g(S) = l( ) (s

f ш-Vш2-a2

(i+i)

1 I ^1 ^ 2^ Л /ш-дш^-а2^)

+ [1 — ( 41 )} [I+? — T+f ( 2Л )}

+ Е1Ш + л

И-1ш

л

-Vш2

4^ N+1

4^

-Vш2

4^

—1

Pi(s)

-V’ш2

2Л

1—

-Vo)2

4pi

-Vш2

4^

N+1

Р№

equation (2.21) can be written as

g*(s) = ±(t+l)h*(s)

where

,N+1

h*(s) =

ш-Vш2-а2 4^

, L _ (ш-Уш2-а2^ ( 2у. Л ( 2у. р

[ ( 4^. )} {5+^ s+^\^+V~^2^a2)j

1

а

а

а

а

а

(2.21)

(2.22)

+ Е1Ш

+ A

-VuA

4^

N+1

P№

ш-Vш2-а2

-1

-1}-

ш-Чш^-а2

1 -

ш-V ш2-a2

, . .. (2.23)

4^ ) I \ 2A У ( \ 4^ 1' v '

On inversion, the equation (2.20), (2.23) and (2.22) yield an expression for f(t),h(t) and g(t) given by

X c-(A+2u+Ot h(at) | e-(A+2^+f)t(2N +

\2ц

±e-(A+2^+i;)t(2N + 3)1ш+з(аУ 12Ц

(N+1) (N+1)

f(t) = ^е-(Х+2^)г + e-(X+2^),:(2N + 2) -

-J^e-(X+2^+^)l(2N + 3)-

(N+1)

h(t) = d) 2 e-№+&(N + 1)^+(£)

t

(2.24)

[/o e-(X+2v+f)ue-f(t-u) {2g(N + 1) lN+1(au) - a(N + 2) ,N+2^au)} du

(N+1)

2

+ (£) 2 A / e-(A+2^+?)ue-?(t-u)(N + 3)

+ л{г)2 /о e-(X+2^+^uNImIEEIpi(t - u)du

IN+s(au)

du

k2^J

N+1

--(—) 2 /* e-(*+2V+t)u

I 2c-(A+2u+Otl2(at) t

(N + i),N+1(au)p1(t - U)du

9(t)=1(f + 1)h(t)

Since 0 < f*(s) < 1 so equation (2.19) can be written as

P*N(s) = g*(s)l?=o [f*(s)Y

On inversion, this equation yields an expression for PN(t) given by

PN(t) = g(t)*T1?=o [f(t)Tr

where [f(t)]*r is the r-fold convolution of f(t) with itself. We note that

[f(t)Y° = 1

(2.25)

(2.26)

(2.27)

(2.28)

a

и

3 Performance measures

Mean

we know that

m(t) = E[X(t)] = Jj^=i nPn(t) m(0) = £%=i nPn(0) = 0 m'(t) = Y£=i nPn(t)

From equation (3.2), (3.3) and (3.4),

m'(t) = (A + 2g + %) 2£=i nPn(t) + ANPN(t) + A Zn=i nPn-i(t)

+2g'Ln-i nPn+i(t) + gPi(t)

After some algebra, the above equation can be expressed as

m'(t) = -%m(t) + (A - 2g) + 2gP0(t) - APN(t) + gP1(t) (3.1)

The above equation can be considered as a first order linear differential equation in m(t). By finding the integrating factor and using the initial condition m(0) = 0, the solution of the above equation is obtained as follows:

m(t) = (^-2£L(1 - e-?t) - 2 / PN(u)e-^(t-u)du

+2g/ P0(u)e-^(t-u')du + g / P1(u)e-^(t-v-)du (3.2)

Variance

We know that

Var[X(t)]=E[X2(t)] — [E{X(t)}]2 Var[X(t)] = k(t) — [m(t)]2 (3.3)

where k(t) = E[X2(t)] = £%=1 n2Pn(t)

Also, k(0) = £%=1 n2Pn(0) = 0

and k!(t) = l£=1 n2pn(t)

From equation (3.2), (3.3) and (3.4),

k'(t) = — (A + 2p + OEn=i n2Pn(t) + AN2PN(t) + 2£n=i n2Pn-1(t) + 2мЕп-1 n2Pn+i(t) + pPi(t)

After some algebra, the above equation can be expressed as

k'(t) = —fk(t) + (A + 2p) — 2pP0(t) — A(2N + 1)PN(t)

+2(A — 2p)m(t) + pP1(t) (3.4)

The above equation can be considered as a first order linear differential equation in k(t). By finding the integrating factor and using the initial condition k(0) = 0, the solution of the above equation is obtained as follows:

k(t) = (1 — e-^) — A(2N + 1) f PN(u)e-S(t-u)du

—2p f P0(u)e-^(t-u^du + 2(A — p) f m(u)e-^(t-u^du +p f P1(u)e-^(t-u')du + C (3.5)

Substituting the above equation in equation (3.3), we get

Var[X(t)] = (1 — e-^) — A(2N + 1) f PN(u)e-^(t-u)du

—2p f P0(u)e-^(t-u^du + 2(A — p) f m(u)e-^(t-u^ du +p f P1(u)e-^(t-u')du — {m(t)}2

4 Steady state probabilities

In this section, we shall discuss the structure of the steady state probabilities.

Theorem-

For (f > 0, the steady state distribution {Pn:n > 0} of the M/M/2/N queue with catastrophe corresponds to

where

p

Po = PP1PN + (1—p)— fp1

Pn = 2oppn+1(l — p)p1Pn + aAp1l-n(1 — P1)PN + (1 — p)pn

+pa(1 — p)pn P1

p _ \{t+2K1-P)2}+v{(P~1-1)-Pi(1-P)}Pi]PN+1 N A\1-p-p^+1 pN+1(1-p)]

P=-

P1 =

(A+2p+^)-j (A+2p+^)2-8Ap 4p

(A+2p+^)-j (A+2p+^)2-8Ap 2A

(4.1)

(4.2)

(4.3)

(4.4)

(4.5)

_ 1

4 (Л+2ц+^)2—8Лц

Proof-

(4.6)

We have from equation (3.13),

N+1

ш —Ч ш2—а2\ P*(s)

2Л

Л (ш—Чш2—а2

K.2ftJ (^5+^ 5+^ у 2Я

+(!+l)(2^)l^?-^?

(4.7)

Multiplying equation (4.7) by s on both sides and taking limit as s ^ 0, we get

HmsP0* (s) = ±-P?*lp« - (1) l™0spi(s)

s^0

Л I ш—ЧшЧ—х2

ш—Чш^—а2

}

2A

+ (1)lims(-+l)[^^-—

\2цЛ s^0 \s ) (s+f s+l; V 2A

Using the property

limsPo* (s) = Po

s^0

After some algebra, the above expression becomes Po = ppiPN + (1- p)-fpi

By taking Laplace transform of the equation (3.10), for n = 1,2, ...,N- 1, we get,

1 -

(4.8)

Р„Ч5) = 2РР0(5)У—=)

,m*p^/' 1 (ш—Чш2—а2

+AP**(s) Ьр—f)

^ I.

P

Чш2—а2

+(!+1

W ш2—а2,> ш—Чш2—а2

4ц

2Л

п—1

1-

Чш2~ a2\n

4ц )

N—n

M

f ш—Чш2—а2

ш—Чш2—а2

4ц

4ц

ш—Чш2—а2 ~Л

P*(S)

-а2/ \ 4ц

Multiplying the above equation by s on both sides and taking limit as s ^ 0, we get

'пгл -^D N—n(1-pi)PN

(4.9)

limsPn*(s) = 2oppn(1 — p)P0 + oAp*

s^0

+p apn l(1 — p)Pl +о%рг

(4.10)

Substituting equation (4.8) in the above equation, and solving, we get Pn = 2appn+1pliv (1 — P)PN + olpl—n(1 — pi)PN

n = 1,2,..., N — 1 (4.11)

+pa(1 — p)p

nl 2^^]Pl + (1 — p)pn

Multiplying the equation (3.21) by s on both sides and taking limit as s ^ 0, after some algebra, we get

limsg*(s) = l[^ + 2p(1 — p)2]pN+1 + jPN+l[(p—l — 1) — Pi(1 — p)Wi (4.12)

s^0

Now taking limit as s ^ 0 in the equation (3.20), we get lim.f*(s) = P[1 + pl[+1pN(1 — p)]

s^0

Multiplying the equation (3.19) by s on both sides and taking limit as s ^ 0, we get lim.sP*^(s) =

s^0 s^0 1—j (s)

Substituting equation (4.12) and (4.13) in the above equation

Рл

_ \{^+2ц(1—р)2}+ц{(р 1 — l)—Pi(1—P)}Pi1p^

N л\1—р—рЦ_+1р!Я+1(1—р)\

Multiplying the equation (3.16) by s on both sides and taking limit as s ^ 0, we get

y2 — p{1

(4.13)

(4.14)

(4.15)

lim.sP^(s)

s^0

Ч P2

a2

о—Чш2—а2

4ц

2plims]1 — s^0 i

-Чш2

4ft

2

cu

a

ш-^а>2-а2

4^

№)

+Alims

s^0 \

a

s^0 Vs

/ ш-^ш2-а2

2Л

N-1

j-^ш2-а2 2Л

}Pi^(5)

+lims (- + l)

s^0 VS /

ш-^ш2-а2 4^

After some algebra, the above expression becomes

Pi - P(1 - P)} = 2P(1 — P)pPo + *(Pi-1 - P^Pn + KP

(4.16)

5 Steady state mean and variance

1

The corresponding values of the steady state mean and variance of the system length are

obtained by taking limit as t ^ ю in equation (4.2) and (4.3). These values are given by

1

m = E(X) = -[(A — 2^.) + 2^P0 — APN + ^PJ

Var(X) = -[(A + 2/л) + 2(A — /л)m — 2/лР0 — A(2N + 1)PN + ^P1] — m2 6 Conclusion

In the present paper, we have discussed the M/M/2/N queueing system subject to catastrophes. The transient as well as the steady state probabilities of the models have been determined analytically. Further, we have also obtained the performance measures of the system.

References

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