Научная статья на тему 'M/G/1 queue with system disasters and impatient customers when system is down'

M/G/1 queue with system disasters and impatient customers when system is down Текст научной статьи по специальности «Математика»

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СИСТЕМА МАССОВОГО ОБСЛУЖИВАНИЯ / QUEUEING SYSTEMS / НЕНАДЕЖНАЯ ЛИНИЯ ОБСЛУЖИВАНИЯ / UNRELIABLE SYSTEM / НЕТЕРПЕЛИВЫЕ ЗАЯВКИ / IMPATIENT CUSTOMERS / СТАЦИОНАРНЫЙ РЕЖИМ РАБОТЫ СМО / ВЕРОЯТНОСТНЫЕ ХАРАКТЕРИСТИКИ СМО / EQUILIBRIUM PROBABILITIES / RATE OF LOST CUSTOMERS / QUEUEING SYSTEMS WITH BREAKDOWNS

Аннотация научной статьи по математике, автор научной работы — Kovalenko A.I., Smolich V.P.

A queueing system of type M/G/1 with customers impatience is considered. The system as a whole suffers occasionally a disastrous breakdown, upon which all present customers are cleared from the system and lost. A repair process then starts immediately. The equilibrium probabilities of the system, expected number of customers in the system, proportion of customers served, rate of lost customers due to disasters and rate of abandonments due to impatience are obtained in the article.

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Текст научной работы на тему «M/G/1 queue with system disasters and impatient customers when system is down»

UDC: 519.872.1 MSC2010: 60K25

M/G/1 QUEUE WITH SYSTEM DISASTERS AND IMPATIENT CUSTOMERS WHEN SYSTEM IS DOWN © Kovalenko A. I., Smolich V. P.

V. I. Vernadsky Crimean Federal University

Taurida Academy Department of mathematics and informatics 4 Vernadskogo Avenue, Simferopol, Republic of Crimea, 295007, Russian Federation

e-mail: svp54@mail.ru

M/G/1 QUEUE WITH SYSTEM DISASTERS AND IMPATIENT CUSTOMERS WHEN SYSTEM IS DOWN.

Kovalenko, A. I. and Smolich, V. P.

Abstract. A queueing system of type M/G/1 with customers impatience is considered. The system as a whole suffers occasionally a disastrous breakdown, upon which all present customers are cleared from the system and lost. A repair process then starts immediately. The equilibrium probabilities of the system, expected number of customers in the system, proportion of customers served, rate of lost customers due to disasters and rate of abandonments due to impatience are obtained in the article.

Keywords: queueing systems, unreliable system, impatient customers, equilibrium probabilities, rate of lost customers, queueing systems with breakdowns.

Introduction

Consider a system operating as an M/G/1 queue. The system as a whole suffers random disastrous failures (catastrophes) such that, when a failure occurs, all present customers are flushed out of the system and lost. The system then goes through a repair process whose duration is random. Meanwhile, while the system is down and inoperative, the stream of arrivals continues. However, the new arrivals become impatient: each customer, upon arrival, activates his own timer, with random duration T, such that if the system is still down when the timer expires, the customer abandons the system never to return. Our goal is to analyze this system and calculate quality of service measures: proportion of customers served; expected number of customers in the system; rate of customers cleared and lost due to disasters; and rate of abandonments due to impatience when the system is down.

There is a considerable literature on queues with system's breakdowns. Models with customers impatience in queues also have been studied by various authors in the past.

For references on the subject see [1, 2]. Inspired by the work of Yechiali [1], we propose, here, a new methodology for investigating the system. In the current study we extend the analysis to deal with systems where service times and repair times are arbitrary (absolutely continuous) random variables.

1. The model

Customers arrive to an M/G/ 1-type queue according to a Poisson process with rate A. The service times are independent and identically distributed with common cumulative distribution function F^x) := Pj^i ^ x}, corresponding density function /(x), reliability function $1 (x) := 1 — F1(x). The system suffers disastrous breakdowns, occurring when the server is at its functioning phase, at a Poisson rate n. That is, the system life-time is exponentially distributed with mean 1/n. When the system fails, all customers present are rejected and lost. Upon failure, a repair process starts immediately. The repair times follow a general distribution F2(y), density function /2(y), reliability function $2(y). Customers arriving while the system is down become impatient: each customer activates an independent "impatience timer" T, exponentially distributed with mean 1/a, such that, if the repair process has not been completed by the time T expires, the customer abandons the system never to return. We suppose that inter-arrival periods, service times, "impatience times", server lifetimes and repair times are mutually independent.

Laplace Transform of an arbitrary function U(x) will be denoted by U*(s):

U *(s)

U(x)e dx.

We use as well the conditional completion rates (at time x) for the service and repair times, respectively:

^(x) = —

$j(x)

$i(x)'

Y(y) = —

Take notice of the relationships:

$1 (x) := P{^1 > x}

x < 0

exp (— /0 ^(t) dt) ' x ^ 0

fi(x) = ju(x)$i(x)' / $fc(x) dx = E(wfc)' k =1, 2

$1(s)

1 — /i (s)

1

s

j = l:

2. Integrq-differential equations

Л /х(х) Л /i(x) A /i(x) A

д(х) A

j = 0: L:

т(у)

X a A 2a А За A

Fig. 1. Transition-rate diagram

Let £(t) be the random process, depicting our system. Let j indicate the system's phase: j = 1 denotes that the system is functioning and serving customers, while j = 0 indicates that the system is down, undergoing a repair process. Let k denote the number of customers in the system.

Obviously, the moments of system's "crashes" form the regeneration points. The random variable w, time between successive regeneration points, is the sum of repair time w2 and exponentially distributed system life-time, so w has absolutely continuous distribution with finite expectation E(w) = E(w2) + -. Therefore, we can apply

n

fundamental theorem of renewal theory and derive the ergodicity of the process £(t).

Define the functions:

Pjk (t) : Qik(t,x) : Qok(t,y) :

= P{£(i) = (j,fc)}, j = 0, 1; k = 0,1, 2, = P{f(t) = (1, k), wi < x} = P{^(t) = (0,k), < y}; k = 0,1,2,.

qik(t,x) :

dQik (t,x) 5x

qok (t,x) :=

dQok (t,y) дУ

Observe that:

Pik(t) = Qik(t, = / qik(t,x) dx; k = 1, 2, 3,

Pok(t) = Qok(t, = / qok(t,y) dy; k = 0,1, 2,

By the method of the supplementary variable, we obtain the next integro-differential equations and boundary conditions:

X X

Pio(t) + (A + n)Pi0(t) = J qii(t,x)Mx) dx + y qoo(t,y)7(y) dy (6)

0 0

-dt--'--dx--+ (^(x) + A + n)qii(t,x) = 0 (7)

dqik(t,x) dqik(t,x) .,,>>.>.>>/.>> > /, ^ -dt--'--dx--+ (^(x) + A + n)qik (t, x) = Aqi,k-i (t, x), (8)

k = 2, 3,...

X X

qii(t, 0) = APio(t) + J qi2(t,x)^(x) dx + J qoi(t,yb(y) dy (9)

0 o

qik(t, 0) = J qi,k+i(t, x)^(x) dx + J qok(t,y)Y(y) dy, k = 2, 3,... (10)

oo

dqoo(t,y) . dqoo(t,y) . M . , „ s s

—dt— + —dy— +(A + 7 (y))qoo(t,y) = aqoi(t,y) (11)

dqok(t,y) . dqok(t,y) . /-> . / \ | i \ v

—dt— + —dy— +(A + 7(y) + ka)qok(t, y) =

= (k + 1)aqo,k+i(t, y) + Aqo,k-i(t,y), k = 1, 2, 3,... (12)

X

qoo(t, 0) = nPu(t), where Pu(t) := ^ Pik(t) (13)

k=0

qok(t, 0) = 0, k =1, 2, 3,... (14) Applying the ergodicity of process £(t), find the limits in (6) - (14) as t ^ x>. Let

Pjk := lim Pjk(t), gik(x) := lim qik(t,x), gok(y) := lim qok(t,y) (15)

t^X t^X t^X

denote the system's steady-state probabilities and corresponding limiting functions. Then (4) and (5) imply

X X

Pik = J gik(x) dx; k = 1, 2, 3,... Pok = J gok(y) dy; k = 0,1, 2,... (16)

oo

Given t ^ x>, the equations and boundary conditions (6) - (14) transform to:

X X

(A + n)pio = J gn(x)Mx)dx + J goo(yb (y) dy (17) 0 0

g1 i(x) + (^(x) + A + n)gii(x) = 0 (18)

g1 k(x) + (Mx) + A + n)gik(x) = Agi,k-i(x), k = 2, 3,... (19)

X X

gii(0) = APio + y g^(x)y(x) dx + J goi(y)Y(y) dy (20)

oo

gik(0) = J gi,k+i(x)^(x) dx ^y gofc(y)Y(y) dy, k = 2, 3,... (21)

oo

goo(y) +(A + y (y))goo(y) = agoi(y) (22)

go k (y) +(A + y (y) + ka)gofc (y) =

(k + 1)ago,k+i(y) + Ago,k-i(y), k = 1, 2, 3,... (23)

X

goo(0) = nPi^, where Pu := ^ Pik (24)

k=o

gok (0) = 0, k =1, 2, 3,... (25)

3. Generating functions

Let

X X

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Go(y, z) =: ^ gok(y)zk, Gi(x, z) =: ^ gik(x)zk (26)

k=0 k=1 define the Probability Generating Function (PGF) of the phase j, j = 0,1. (17)-(25) yields:

5G1(x, z)

dx

+ (y(x) + A(1 - z) + n)Gi(x,z) = 0 (27)

X X

Gi (0, z ) = —(A(1 - z) + n)Pi0 + J Go (y, z)y (y) dy + 1 J Gi(x, z)y(x) dx (28)

0 0

dGo(y,z) /-, ^dGo(y,z) . / \ n roo^ -dy--a(1 — z)-—-+ (A(1 — z) + y(y))Go(y, z) = 0 (29)

Go(0,z ) = nPi^ (30)

Applying standard methods for the solution of the first-order linear partial differential equation (29) with boundary condition (30), we have:

Go(y, z) = nPu ■ exp | — J 7(t) dt — a(1 — z)(1 — e-ay)

= nPi^ ■ $2(y) ■ exp j — a(1 — z)(1 — e-ay) ^ (31)

The solution of the equation (27) is given by:

ix

— j ^(t) dt — (A(1 — z) + n)x = Gi(0, z) ■ $i(x) ■ exp (—(A(1 — z) + n)x) (32)

Calculate integrals in (28):

X X

J Go(y,z)Y(y) dy = nPi^ -J /2(y)exp j — -A (1 — z)(1 — e-ay)j dy oo

X

A(z) := J /2(y)ex^| — a(1 — z)(1 — e-ay)} dy (33)

o

X

J Go(y,z)Y(y) dy = nPu ■ A(z) (34)

o

Denote

Then,

Observe that Next,

A(1) = 1 and A'(1) = - (1 — /2(a)) = A$2(a) (35)

a 2 2

j Gi(x, z)^(x) dx = Gi(0, z) J /i(x)e-(A(i-z)+n)xdx = Gi(0, z)/*(A(1 — z) + n) (36) oo Combining (34), (36) and (28), we get

G (0 z) = —(A(1 — z) + n)zPio + nzPi^ ■ A(z) Gi(0,z) = z — /*(A(1 — z) + ni

and therefore,

Gi (x z) = —(A(1 — z)+ n)zPi0 + nzPi^ ■ A(z) (x)e-(A(i-z)+n)x (37)

Gi(x,z)= z — /* (A(1 — z) + n) ^i(x)e ( )

To conclude the calculation of G0(y,z) and Gi(x,z) we need P10 and P^. To accomplish the purpose calculate the integral

x _ x _

/X r, X

Go(y, 1) dy = ^ / gok(y) dy = P0k =: p0^ 0 k=00 k=0

From (31) we find G0(y, 1) = qPu ■ $2(y). Hence

X X

I Go(y, 1) dy = nPu J $2(y) dy = nPu ■ EM, 00 where E(u2) is expected repair time. Therefore,

Po^ = vPu ■ EM, (38)

and since P0• + Pu = 1, then

= nE M = 1 ( )

^ 1 + nEM, 1 1 + nEM ( )

Investigate the denominator of (37) h(z) = z — f1(A(1 — z) + n). Observe that h(0) = — f(A + n) < 0 and h(1) = 1 — f(n) > 0. Thus 3zo G (0,1), such that h(zo) = 0. This root is unique in (0,1) since h"(z) = —A2f1*"(A(1 — z) + n) < 0 Vz G (0; 1). Hence the numerator of (37) must vanish at z0:

— (A(1 — zo) + n)zoPio + nzoPi • ■ A(zo) = 0,

implying that

nA(zo)Pu A(1 - zo) + n

Pic = . 4 ^ " . (40)

4. Quality of service measures

4.1. Expected number of customers in the system. Let L0 be the number of customers, given the system is down, and L1 be the number of customers in the operative system. Calculate the expectations E(L0),E(L 1). So far

Gj(y, z) dy = £ ( / j(У) dy \ zk = £ Pjkzk, J = 0,1

0 k=0 \0 / k=0

then

E(Lj) = lim d I Gj(y,z) dy

Using (31), we have:

X

E(Lo) = lim /nP* ■ $2(y) ■ exp ( — A(1 — z)(1 — e-ay)1 A(1 — e-ay)dy = z^-i J [a J a

o

X

AA = nP* ■ ~ $2(y)(1 — e-ay)dy = nP* ■ " (EM — $2(a))

aa o

E(Lo) — Po* ■ 1 — 7^ (41)

or (see (38))

^ ■ (1—fan

a \ E (^2)/ Substituting (40) in (37), we obtain:

Gi<x-z) = z — z) + n) $i(x)e-(A(i-z)+n)x (A(z) — .4(*o)^L^)

Thus

oo

Gi(x,z) dx =-J^' , , $i(A(1 — z) + n) fA(z) — A(zo) A(1 z) + n

o

z — /**(A(1 — z) + n) ^ ' "V A(1 — zo) + n

nzPl. 1 — /*(A(1 — z) + n) ( A(z) A(zo)

z — /*(A(1 — z) + n) VA(1 — z) + n A(1 — zo) + n Some computations give:

X

E(L i) = lim d [ G i(x,z) dx = -^^(Pio — Pi•) + APi1 + $2(«)1 (42) i dzJ 1 — /*1(n) V n /

o

Remark 1. In the partial case of exponentially distributed service (wi) and repair (w2) times

-(x) = - = const, 7 (x) = y = const, and

/i*(n) = -+-, $2(«) = , E M = 1,

- + n Y + a y

(41) with (42) imply

E (Lo) = ^, E (L i ) = - (Pio — Pi*) + APi i+ 1

Y + a n \n Y + a

which coincides with results of [1].

4.2. Proportion of customers served. The system suffers from two types of losses: (i) rejected customers due to system's disastrous failures and (ii) abandonments of impatient customers during the repair phase. When the system is in state (j,n),n ^ 1, the rate of failure is n and then n customers are lost. Thus, the unit-time rate of lost customers,

Riost, is given by

X

Riost = ^ nnPin = nE(Li) (43)

n=i

Similarly, the rate of abandonment due to impatience is

X

Raband = anPon = aE(Lo) (44)

n=i

Finally, the expected number of customers served per unit of time is

Rserved = A — Rlost — Raband (45)

Taking into account (41),(42), we get:

/ *(n)

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Rserved = A — aE(Lo) — » Ji (n) ,(Pio — Pi.) — An$2(a)Pi* — APi*

1 — Zi(n)

So far as A — aE(Lo) — An$2(a)Pi* — APi* = 0 (see (38),(41)),

/ *(n) / *(n)

Rserved = n 1 —i/*(n) (Pi* — Pio) = ^*(n)(Pi* — Pi0),

implying that the proportion of customers served is

/ *(n)

Pserved = -KnÍ)<P1• " <46)

Again, in the case of exponentially distributed service time,

Pserved = A (Pi* — Pi0).

Conclusion

We analyze the model of M/G/1 que with breakdowns and impatient customers and derive various quality of service measures: expected number of customers in the system, proportion of customers served, rate of lost customers due to disasters, rate of abandonments due to impatience.

References

1. YECHIALI, U. (2007) Queues with system disasters and impatient customers when system is down. Queueing Syst.. 56. p. 195-202.

2. ALTMAN, E. & YECHIALI, U. (2006) Analysis of customers' impatience in queues with server vacations. Queueing Syst.. 52. p. 261-279.

3. COOPER, R. B. (1981) Introduction to Queueing Theory. 2nd ed.. North-Holland, Amsterdam.

4. Коваленко, А. И., Марянин, Б. Д., Смолич, В. П. Система массового обслуживания с ненадежной линией и нетерпеливыми заявками // ТВИМ. — ТНУ, 2013. — №1. — C. 53-60. KOVALENKO, A. I., MARYANIN, B.D. & SMOLICH, V. P (2013) Queueing system with unreliable line and impatient customers. TVIM. №1. p. 53-60.

5. Смолич В. П. Метод дополнительной переменной в задачах ТМО и теории надежности / А. И. Коваленко, В. П. Смолич. — Lambert Academic Publishing (Германия), 2014. — 232 c. KOVALENKO, A. & SMOLICH, V. (2014) The supplementary variable method applying to the problems in queueing systems and reliability. Lambert Academic Publishing. Germany.

Статья поступила в редакцию 02.12.2015

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