Aijrstem15 513

Page 1

American International Journal of Research in Science, Technology, Engineering & Mathematics

Available online at http://www.iasir.net

ISSN (Print): 2328-3491, ISSN (Online): 2328-3580, ISSN (CD-ROM): 2328-3629 AIJRSTEM is a refereed, indexed, peer-reviewed, multidisciplinary and open access journal published by International Association of Scientific Innovation and Research (IASIR), USA (An Association Unifying the Sciences, Engineering, and Applied Research)

A BULK ARRIVAL QUEUEING MODEL OF THREE STAGES OF SERVICE WITH DIFFERENT VACATION POLICIES, SERVICE INTERRUPTION AND DELAY TIME S. Maragathasundari Assistant Professor, Department of Mathematics, Velammal Institute of Technology, Chennai, Tamil Nadu, India Abstract: We study a M/G/1 queueing system with bulk arrival. Here the service is given in three stages of service. All the three stages of service are compulsory to all the arriving customers. After the completion of service the server takes a vacation. An added assumption Extended vacation is considered in this model. Moreover service interruption is considered. Repair process does not start immediately. There is a delay in repair. Service and all the other parameters follows general distribution. Breakdown is exponentially distributed. Using supplementary variable technique we derive the probability generating functions for the number of customers in the system. Some of the performance measures are calculated. Mathematics Subject Classification: 60K25, 60K30 Keywords: Batch arrival, Three stages of services, extended vacation time, delay time and Steady state distribution, Queue size. I. Introduction An overview on the conceptual aspects for the stage service in queueing model is provided in this model.The motivation of this model comes from numerous versatile applications. Many authors have studied queues with server vacations and breakdowns. Le et al [2] studied reliability analysis of an M/G/1 queue with breakdowns and vacations. Khalaf., Madan & Lucas.[4] studied about M[x]/G/1 Queue with Bernoulli schedule, General vacation times, random breakdowns, and general repair times. Borthakur and Choudhury [6] discussed about generalized vacation in his batch arrival Possion queue. Chodhury[8] in his batch arrival queueing system considered the aspect of additional service channel and derived the performance measures of the queueing system Badamchi Zadeh and Shahkar. G.H.[10] investigated a two phase queue system with Bernoulli feedback and Bernoulli schedule server vacation. Maragathasundari and Srinivasan, .[12]made a analysis in M/G/1 feedback queue with three stage multiple sever vacation. In this paper, we consider a batch arrival queueing system with three stages of service. After the completion of three stages of service completion the server goes for vacation . The server after completion of completion, he might take the optional extended vacation with probability đ?‘&#x; or may return to the system to serve the customers with probability 1 − đ?‘&#x;. Breakdown arrives at random. Due to various reasons , repair process does not follow immediately. There is a delay in repair Service time, vacation time, extended vacation & Repair time follows general distribution. This paper is organized as follows. Model assumptions are given in section2. Steady state condition is given in section 3. Queue size distribution at a random epoch is given in section 4. The average queue size and the average waiting time are computed in section 5. Conclusion is given in section 6. II. Model Assumptions Customers arrive at the system in batches of variable sizes in a compound Poisson process, and one by one service on a ‘first come-first served’ basis is implemented. Let đ?œ†đ?‘?đ?‘– đ?‘‘đ?‘Ą(đ?‘– = 1,2,3 ‌ . ) be the first order probability that a batch of đ?‘– customers arrives at the system during a short duration of time (đ?‘Ą, đ?‘Ą + đ?‘‘đ?‘Ą), where 0 ≤ đ?‘?đ?‘– ≤ 1 and ∑∞ đ?‘–=1 đ?‘?đ?‘– = 1 and đ?œ† > 0 is the mean arrival rate of the batches. There is a single server and the service time follows general(arbitrary) distribution with distribution function đ?‘€đ?‘– (đ?‘Ł)and density function đ?‘šđ?‘– (đ?‘Ł). Let Âľđ?‘– (đ?‘Ľ)đ?‘‘đ?‘Ľ be the conditional probability density of service completion of the đ?‘– đ?‘Ąâ„Ž stage of service during the interval (đ?‘Ľ, đ?‘Ľ + đ?‘‘đ?‘Ľ), given that the elapsed time is đ?‘Ľ, so that đ?‘šđ?‘– (đ?‘Ľ) Âľđ?‘– (đ?‘Ľ) = đ?‘– = 1,2 ,3 (1) 1 − đ?‘€đ?‘– (đ?‘Ľ) đ?‘Ł and đ?‘šđ?‘– (đ?‘Ł) = Âľđ?‘– (đ?‘Ł)đ?‘’ − âˆŤ0 Âľđ?‘– (đ?‘Ľ)đ?‘‘đ?‘Ľ (2) As soon as a service is completed, the server may take a vacation. The server’s vacation time follows a general(arbitrary) distribution, with the distribution function đ??ľ1 (đ?‘ ) and density function đ?‘?1 (đ?‘ ). Let đ?›˝(đ?‘Ľ)đ?‘‘đ?‘Ľ be

AIJRSTEM 15-513; Š 2015, AIJRSTEM All Rights Reserved

Page 52

S. Maragathasundari, American International Journal of Research in Science, Technology, Engineering & Mathematics, 11(1), JuneAugust, 2015, pp. 52-56

the conditional (đ?‘Ľ, đ?‘Ľ + đ?‘‘đ?‘Ľ), so that

probability

of

đ?›˝(đ?‘Ľ) =

a

completion

đ?‘?1 (đ?‘Ľ) 1 − đ??ľ1 (đ?‘Ľ)

of

a

vacation

during

the

interval (3)

đ?‘Ą

and đ?‘?1 (đ?‘ ) = đ?›˝(đ?‘ )đ?‘’ − âˆŤ0 đ?›˝(đ?‘Ľ)đ?‘‘đ?‘Ľ (4) After the completion of vacation, the server has the option of taking extended vacation with probability r. The server’s Extended vacation time follows a general(arbitrary) distribution, with the distribution function đ??ľ2 (đ?‘ ) and density function đ?‘?2 (đ?‘ ). Let đ?œƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ be the conditional probability of a completion of a Extended vacation during the interval (đ?‘Ľ, đ?‘Ľ + đ?‘‘đ?‘Ľ), so that đ?‘?2 (đ?‘Ľ) đ?œƒ(đ?‘Ľ) = (3đ?‘Ž) 1 − đ??ľ2 (đ?‘Ľ) đ?‘Ą

đ?‘?2 (đ?‘ ) = đ?œƒ(đ?‘ )đ?‘’ − âˆŤ0 đ?œƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ (4đ?‘Ž) The server may breakdown at random, and breakdowns are presumed to arise according to the Poisson stream, with the mean breakdown rate đ?›ź > 0. Further, we assume that once the system breaks down, the customer whose service is interrupted goes back to the head of the queue. The server’s delay time follows a general(arbitrary) distribution, with the distribution function đ??ľ3 (đ?‘ ) and density function đ?‘?3 (đ?‘ ). Let đ?›ż(đ?‘Ľ)đ?‘‘đ?‘Ľ be the conditional probability of a completion of a delay during the interval(đ?‘Ľ, đ?‘Ľ + đ?‘‘đ?‘Ľ), so đ?‘? (đ?‘Ľ) that đ?›ż(đ?‘Ľ) = 3 (5đ?‘Ž) 1−đ??ľ3 (đ?‘Ľ) đ?‘Ą

đ?‘?3 (đ?‘ ) = đ?›ż(đ?‘ )đ?‘’ − âˆŤ0 đ?›ż(đ?‘Ľ)đ?‘‘đ?‘Ľ (5đ?‘?) The server’s repair time follows a general(arbitrary) distribution, with the distribution function đ??ľ4 (đ?‘ ) and density function đ?‘?4 (đ?‘ ). Let đ?›ž(đ?‘Ľ)đ?‘‘đ?‘Ľ be the conditional probability of a completion of a repair during the interval(đ?‘Ľ, đ?‘Ľ + đ?‘‘đ?‘Ľ), so that đ?‘?4 (đ?‘Ľ) đ?›ž(đ?‘Ľ) = (6 1 − đ??ľ4 (đ?‘Ľ) đ?‘Ą

đ?‘?4 (đ?‘ ) = đ?›ž(đ?‘ )đ?‘’ − âˆŤ0 đ?›ž(đ?‘Ľ)đ?‘‘đ?‘Ľ III. Steady State Condition Then, connecting the states of the system at time đ?‘Ą + đ?‘‘đ?‘Ą with those at time đ?‘Ą and then taking the limit as đ?‘Ą → ∞,we obtain the following set of steady state equations governing the system đ?œ• (1) (1) (1) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡1 (đ?‘Ľ) + đ?›ź)đ?‘ƒđ?‘› (đ?‘Ľ) = đ?œ† ∑đ?‘›âˆ’1 đ?‘–=1 đ??śđ?‘– đ?‘ƒđ?‘›âˆ’1 (đ?‘Ľ) đ?œ•đ?‘Ľ đ?‘› đ?œ• (1) (1) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡1 (đ?‘Ľ) + đ?›ź)đ?‘ƒ0 (đ?‘Ľ) = 0 đ?œ•đ?‘Ľ 0 đ?œ• (2) (2) (2) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡2 (đ?‘Ľ) + đ?›ź)đ?‘ƒđ?‘› (đ?‘Ľ) = đ?œ† ∑đ?‘›âˆ’1 đ?‘–=1 đ??śđ?‘– đ?‘ƒđ?‘›âˆ’1 (đ?‘Ľ) đ?œ•đ?‘Ľ đ?‘› đ?œ• (2) (2) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡2 (đ?‘Ľ) + đ?›ź)đ?‘ƒ0 (đ?‘Ľ) = 0 đ?œ•đ?‘Ľ 0 đ?œ• (3) (3) (3) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡3 (đ?‘Ľ) + đ?›ź)đ?‘ƒđ?‘› (đ?‘Ľ) = đ?œ† ∑đ?‘›âˆ’1 đ?‘–=1 đ??śđ?‘– đ?‘ƒđ?‘›âˆ’1 (đ?‘Ľ) đ?œ•đ?‘Ľ đ?‘› đ?œ• (3) (3) đ?‘ƒ (đ?‘Ľ) + (đ?œ† + đ?œ‡3 (đ?‘Ľ) + đ?›ź)đ?‘ƒ0 (đ?‘Ľ) = 0 đ?œ•đ?‘Ľ 0 đ?œ• đ?‘‰ (đ?‘Ľ) + (đ?œ† + đ?›˝(đ?‘Ľ))đ?‘‰đ?‘› (đ?‘Ľ) = ∑đ?‘›âˆ’1 đ?‘–=1 đ?œ†đ??śđ?‘– đ?‘‰đ?‘›âˆ’đ?‘– (đ?‘Ľ) đ?œ•đ?‘Ľ đ?‘› đ?œ• đ?‘‰ (đ?‘Ľ) + (đ?œ† + đ?›˝(đ?‘Ľ))đ?‘‰0 (đ?‘Ľ) = 0 đ?œ•đ?‘Ľ 0 đ?œ• đ??¸ (đ?‘Ľ) + (đ?œ† + đ?œƒ(đ?‘Ľ))đ??¸đ?‘› (đ?‘Ľ) = ∑đ?‘›âˆ’1 đ?‘–=1 đ?œ†đ??śđ?‘– đ??¸đ?‘›âˆ’đ?‘– (đ?‘Ľ) đ?œ•đ?‘Ľ đ?‘› đ?œ• đ??¸ (đ?‘Ľ) + (đ?œ† + đ?œƒ(đ?‘Ľ))đ??¸0 (đ?‘Ľ) = 0 đ?œ•đ?‘Ľ 0 đ?œ• đ?œ•đ?‘Ľ đ?œ• đ?œ•đ?‘Ľ đ?œ• đ?œ•đ?‘Ľ đ?œ• đ?œ•đ?‘Ľ

(7) (8) (9) (10) (11) (12) (13) (14) (15) (16)

đ??ˇđ?‘› (đ?‘Ľ) + (đ?œ† + đ?›ż(đ?‘Ľ))đ??ˇđ?‘› (đ?‘Ľ) = ∑đ?‘›âˆ’1 đ?‘–=1 đ?œ†đ??śđ?‘– đ??ˇđ?‘›âˆ’đ?‘– (đ?‘Ľ)

(17)

đ??ˇ0 (đ?‘Ľ) = 0

(18)

đ?‘…đ?‘› (đ?‘Ľ) + (đ?œ† + đ?›ž(đ?‘Ľ))đ?‘…đ?‘› (đ?‘Ľ) =

∑đ?‘›âˆ’1 đ?‘–=1 đ?œ†đ??śđ?‘–

đ?‘…đ?‘›âˆ’đ?‘– (đ?‘Ľ)

đ?‘…0 (đ?‘Ľ) + (đ?œ† + đ?›ž(đ?‘Ľ))đ?‘…0 (đ?‘Ľ) = 0

đ?œ†đ?‘„

∞ ∞ (3) = âˆŤ0 đ?‘…0 (đ?‘Ľ)đ?›ž(đ?‘Ľ)đ?‘‘đ?‘Ľ + âˆŤ0 đ?‘ƒ0 (đ?‘Ľ)đ?œ‡3 (đ?‘Ľ)đ?‘‘đ?‘Ľ ∞ ∞ + âˆŤ0 đ?‘‰0 (đ?‘Ľ)đ?›˝(đ?‘Ľ)đ?‘‘đ?‘Ľ + âˆŤ0 đ??¸0 (đ?‘Ľ)đ?œƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ

(19) (20) (21)

The above equations are to be solved subject to the boundary conditions ∞ (3) ∞ (1) đ?‘ƒđ?‘› (0) = âˆŤ0 đ?‘ƒđ?‘›+1 (đ?‘Ľ)đ?œ‡3 (đ?‘Ľ)đ?‘‘đ?‘Ľ + (1 − đ?‘&#x;) âˆŤ0 đ?‘‰đ?‘›+1 (đ?‘Ľ)đ?›˝(đ?‘Ľ)đ?‘‘đ?‘Ľ

AIJRSTEM 15-513; Š 2015, AIJRSTEM All Rights Reserved

Page 53

S. Maragathasundari, American International Journal of Research in Science, Technology, Engineering & Mathematics, 11(1), JuneAugust, 2015, pp. 52-56 ∞

∞

+ ∫0 𝐸𝑛+1 (𝑥)𝜃(𝑥)𝑑𝑥 + ∫0 𝑅𝑛+1 (𝑥)𝛾(𝑥)𝑑𝑥 + 𝜆𝐶𝑛+1 𝑄

∞ (1) = ∫0 𝑃𝑛 (𝑥)𝜇1 (𝑥)𝑑𝑥 ∞ (2) = ∫0 𝑃𝑛 (𝑥)𝜇2 (𝑥)𝑑𝑥 ∞ (3) 𝑉𝑛 (0) = ∫0 𝑃𝑛 (𝑥)𝜇3 (𝑥)𝑑𝑥 ∞ 𝐸𝑛 (0) = 𝑟 ∫0 𝑉𝑛 (𝑥)𝛽(𝑥)𝑑𝑥 ∞ (𝑖) 𝐷𝑛 (0) = 𝛼 ∑3𝑖=1 ∫0 𝑃𝑛−1 (𝑥)𝑑𝑥 ∞ 𝑅𝑛 (0) = ∫0 𝐷𝑛 (𝑥)𝛿(𝑥)𝑑𝑥 (2) 𝑃𝑛 (0) (3) 𝑃𝑛 (0)

(22) (23) (24) (25) (26)

(𝑖) =𝛼 ∑𝑛𝑖=1 𝑃𝑛−1

(27) (28) (29) (30)

𝐷0 (0) = 0 𝑅0 (0) = 0

IV. Queue Size distribution at a random epoch Now multiply equation (7) by 𝑍 𝑛 & summing over 𝑛 from 1 𝑡𝑜 ∞, adding the result to equation (8), and using generating function we get 𝜕 (1) (1) 𝑃 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝜇1 (𝑥) + 𝛼)𝑃𝑞 (𝑥, 𝑧) = 0 (31) 𝜕𝑥 𝑞 Similarly, 𝜕 (2) (2) 𝑃𝑞 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝜇2 (𝑥) + 𝛼)𝑃𝑞 (𝑥, 𝑧) = 0 (32) 𝜕𝑥 𝜕

(3) (3) 𝑃 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝜇3 (𝑥) + 𝛼)𝑃𝑞 (𝑥, 𝑧) 𝜕𝑥 𝑞 𝜕 𝑉 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝛽(𝑥))𝑉𝑞 (𝑥, 𝑧) = 0 𝜕𝑥 𝑞 𝜕 𝐷 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝛿(𝑥))𝐷𝑞 (𝑥, 𝑧) = 0 𝜕𝑥 𝑞 𝜕 𝑅 (𝑥, 𝑧) + (𝜆 − 𝜆𝑐(𝑧) + 𝛾(𝑥))𝑅𝑞 (𝑥, 𝑧) = 0 𝜕𝑥 𝑞

=0

Applying the same process for the boundary conditions, we get ∞ (3) ∞ (1) 𝑧𝑃𝑞 (0, 𝑧) = ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇3 (𝑥)𝑑𝑥 + (1 − 𝑟) ∫0 𝑉𝑞 (𝑥, 𝑧) 𝛽(𝑥)𝑑𝑥 ∞ ∞ + ∫0 𝐸𝑞 (𝑥, 𝑧)𝜃(𝑥)𝑑𝑥 + ∫0 𝑅𝑞 (𝑥, 𝑧)𝛾(𝑥)𝑑𝑥 + 𝜆(𝐶(𝑧) − 1) 𝑄 Similarly we have, ∞ (1) (2) 𝑃𝑞 (0, 𝑧) = ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇1 (𝑥)𝑑𝑥 ∞ (3) = ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇3 (𝑥)𝑑𝑥 ∞ (3) 𝑉𝑞 (0, 𝑧) = ∫0 𝑃𝑞 (𝑥, 𝑧)𝜇3 (𝑥)𝑑𝑥 (𝑖) 𝐷𝑞 (0, 𝑧) = 𝛼𝑧 ∑3𝑖=1 𝑃𝑞 (𝑧) ∞ 𝑅𝑞 (0, 𝑧) = ∫0 𝐷𝑞 (𝑥, 𝑧)𝛿(𝑥)𝑑𝑥 (3) 𝑃𝑞 (0, 𝑧)

(33) (34) (35) (36)

(37) (38) (39) (40) (41) (42)

Integrating equation (31) from 0 𝑡𝑜 𝑥, we get ∞ (1) (1) 𝑃𝑞 (𝑥, 𝑧) = 𝑃𝑞 (0, 𝑧)𝑒 −(𝜆−𝜆𝐶(𝑧)+𝛼)𝑥−∫0 𝜇1(𝑡)𝑑𝑡 Again integrating Equation (43) by parts with respect to 𝑥 yields, ̅1 (𝑠) 1−𝑀 (1) (1) 𝑃𝑞 (𝑧) = 𝑃𝑞 (0, 𝑧) [ ] , 𝑠 = 𝜆 − 𝜆𝐶(𝑧) + 𝛼 𝑠 ∞ −(𝜆−𝜆𝐶(𝑧)+𝛼)𝑥 Where ̅​̅​̅​̅ 𝑀1 (𝑠) = ∫ 𝑒 𝑑 ̅​̅​̅​̅ 𝑀1 (𝑥)the Laplace Stieltjes is transform of first stage of service time

(43) (44)

0

𝑀1 (𝑥). Now multiply both sides of the Equation (43) by 𝜇1 (𝑥) & integrating ∞ (1) (1) ̅1 (𝑠) ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇1 (𝑥)𝑑𝑥 = 𝑃𝑞 (0, 𝑧)𝑀 Similar process to be carried out for equations (26)-(31), Hence we have ̅1 (𝑠)(1−𝑀 ̅2 (𝑠)) 1−𝑀

(2) (2) 𝑃𝑞 (𝑧) = 𝑃𝑞 (0, 𝑧) [

]

𝜃 ∞ (2) (1) ̅1 (𝑠)𝑀 ̅2 (𝑠) And ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇1 (𝑥)𝑑𝑥 = 𝑃𝑞 (0, 𝑧)𝑀 ̅ (𝑠)𝑀 ̅2 (𝑠)(1−𝑀 ̅3 (𝑠)) 𝑀 (3) (1) 𝑃𝑞 (𝑧) = 𝑃𝑞 (0, 𝑧) [ 1 ] 𝑠 ∞ (3) (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠) And ∫0 𝑃𝑞 (𝑥, 𝑧) 𝜇3 (𝑥)𝑑𝑥 = 𝑃𝑞 (0, 𝑧)𝑀 (1−𝐵̅1 (ℎ)) 𝑉𝑞 (𝑧) = 𝑉𝑞 (0, 𝑧) [ ] , ℎ = 𝜆 − 𝜆𝐶(𝑧) ℎ ̅ (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠) (1−𝐵1(ℎ)) = 𝑃𝑞 (0, 𝑧)𝑀 ℎ ∞ (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)𝐵̅1 (ℎ) ∫0 𝑉𝑞 (𝑥, 𝑧) 𝛽(𝑥)𝑑𝑥 = 𝑃𝑞 (0, 𝑧)𝑀 1−𝐵̅2 (ℎ) 𝐸𝑞 (𝑧) = 𝐸𝑞 (0, 𝑧) [ ] ℎ

AIJRSTEM 15-513; © 2015, AIJRSTEM All Rights Reserved

(45) (46) (47) (48) (49)

(50) (51)

Page 54

S. Maragathasundari, American International Journal of Research in Science, Technology, Engineering & Mathematics, 11(1), JuneAugust, 2015, pp. 52-56 ̅

(1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)𝐵̅1 (ℎ) (1−𝐵2(ℎ)) = 𝑟𝑃𝑞 (0, 𝑧)𝑀

(52)

ℎ ∞ (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)𝐵̅1 (ℎ)𝐵̅2 (ℎ) ∫0 𝐸𝑞 (𝑥, 𝑧) 𝜃(𝑥)𝑑𝑥 = 𝑟𝑃𝑞 (0, 𝑧)𝑀 1−𝐵̅ (ℎ) 𝐷𝑞 (𝑧) = 𝐷𝑞 (0, 𝑧) [ 3 ] ℎ ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠) 1−𝐵̅3 (ℎ) 1−𝑀 (1) = 𝛼𝑧𝑃𝑞 (0, 𝑧) [ ][ ] 𝑠 ℎ ∞

∫0 𝐷𝑞 (𝑥, 𝑧) 𝛿(𝑥)𝑑𝑥 =

= ∞

(54)

(1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)]𝐵̅3 (ℎ) 𝛼𝑧𝑃𝑞 (0,𝑧)[1−𝑀

1−𝐵̅4 (ℎ)

𝑅𝑞 (𝑧) = 𝑅𝑞 (0, 𝑧) [

(53)

(55)

𝑠

]

ℎ (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)]𝐵̅3 (ℎ) 1−𝐵̅ (ℎ) 𝛼𝑧𝑃𝑞 (0,𝑧)[1−𝑀 4

[

]

(56)

𝑠 ℎ (1) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)]𝐵̅3 (ℎ)𝐵̅4 (ℎ) 𝛼𝑧𝑃𝑞 (0,𝑧)[1−𝑀

∫0 𝑅𝑞 (𝑥, 𝑧) 𝛾(𝑥)𝑑𝑥 = 𝑠 Therefore from equation 37,we have 𝑠𝜆𝑄(𝐶(𝑧)−1) (1) 𝑃𝑞 (0, 𝑧) = 3 ̅ ̅ ̅ ̅

(57) (58)

̅ ̅ ̅ 𝑠[𝑧−∑𝑖=1 𝑀𝑖 (𝑠)[1+𝐵1 (ℎ)+𝑟𝐵1 (ℎ)𝐵2 (ℎ)]−𝛼𝑧[1−∑3 𝑖=1 𝑀𝑖 (𝑠)]𝐵3 (ℎ)𝐵4 (ℎ)]

Let 𝑆𝑞 (𝑧) be the generating function of the queue size (1) (2) (3) 𝑆𝑞 (𝑧) = 𝑃𝑞 (0, 𝑧) + 𝑃𝑞 (0, 𝑧) + 𝑃𝑞 (0, 𝑧) + 𝑉𝑞 (𝑧) + 𝐸𝑞 (𝑧) + 𝑅𝑞 (𝑧) + 𝐷𝑞 (𝑧) ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠))(𝑠 + 𝛼ℎ𝑧 − 𝛼ℎ𝑧𝐵̅3 (ℎ)𝐵̅4 (ℎ)) (1 − 𝑀 ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)𝐵̅1 (ℎ) −𝑄 [ ] +𝛼ℎ𝑧 + 𝑟𝑠𝑀 (1 − 𝐵̅2 (ℎ)) = ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)(1 − 𝐵̅1 (ℎ) + 𝑟𝐵̅1 (ℎ)𝐵̅2 (ℎ))] 𝑠[𝑧 − 𝑀 ̅ ̅ ̅1 (𝑠)𝑀 ̅2 (𝑠)𝑀 ̅3 (𝑠)] −𝛼𝑧𝐵3 (ℎ)𝐵4 (ℎ)[1 − 𝑀 In order to find 𝑄, we use the normalization condition𝑆𝑞 (1) + 𝑄 = 1 And hence the utilization factor 𝜌 is determined. 𝜌 < 1 is the stability condition under which the steady state exists.

(59)

(60)

V. The Average Queue Size and Average Waiting Time Let 𝐿𝑞 denote the mean number of customers in the queue under the steady state.Then 𝑑

𝐿𝑞 = 𝑆𝑞 (𝑧)│𝑧=1 (61) 𝑑𝑧 Since the formula gives0⁄0 for m, then we write 𝑆𝑞 (𝑧) given in (59) as 𝑆𝑞 (𝑧) = 𝑁(𝑧)⁄𝐷(𝑧) Where 𝑁(𝑧) and 𝐷(𝑧) are the numerator and denominator of the right hand side of (59) respectively. Then we use 𝐷′ (1)𝑁 ′′ (1) − 𝑁 ′ (1)𝐷′′ (1) 𝐿𝑞 = (62) 2(𝐷′ (1))2 ̅​̅​̅​̅2 (𝛼)𝑀′ ̅​̅​̅​̅3 (𝛼) − 𝛼𝜆𝐸(𝐼) − 𝑟𝑠𝜆𝐸(𝑠)𝐸(𝐼)] ̅​̅​̅​̅′1 (𝛼)𝑀′ 𝑁 ′ (1) = −𝑄[𝛼(𝜆𝐸(𝐼))𝑀 (63) ′′ (1) ′ ′ ̅​̅​̅​̅2 (𝛼)𝑀′ ̅​̅​̅​̅3 (𝛼)] + (𝜆𝐸(𝐼) (𝑀 ̅​̅​̅​̅2 (𝛼)𝑀′ ̅​̅​̅​̅3 (𝛼))) − 𝛼 − 𝛼𝜆𝐸(𝐼) ̅​̅​̅​̅1 (𝛼)𝑀′ ̅​̅​̅​̅1 (𝛼)𝑀′ 𝑁 = −𝑄(𝜆𝐸(𝐼))[𝑀 ̅​̅​̅​̅′1 (𝛼)𝐸(𝑠) + 𝛼𝑀 ̅​̅​̅​̅′ 2 (𝛼)𝐸(𝑠) + 𝛼𝑀 ̅​̅​̅​̅′ 3 (𝛼)𝐸(𝑠) + 𝛼𝐸(𝑣)𝐸(𝑆))] + 𝑟[(𝜆𝐸(𝐼)2 (𝐸(𝑠) + 𝛼𝑀 2 + ((𝜆𝐸(𝐼) 𝐸(𝑠) − 𝜆𝐸(𝐼(𝐼 − 1))𝛼𝐸(𝑠) ̅​̅​̅​̅′1 (𝛼) + 𝛼𝐸(𝑠)𝑀 ̅​̅​̅​̅′ 2 (𝛼) + 𝛼𝐸(𝑠)𝑀 ̅​̅​̅​̅′ 3 (𝛼) − 𝛼𝐸(𝑠)𝐸(𝑣) + (𝜆𝐸(𝐼)2 (𝛼𝐸(𝑠)𝑀 2 )) + 𝛼𝐸(𝑠 (64) ′ (1) 𝐷 = (−𝜆𝐸(𝐼))[1 − 𝑟] ̅​̅​̅​̅ ̅2 (𝛼)𝑀 ̅3 (𝛼) 𝑀′1 (𝛼)𝑀 1 + 𝑟 + 𝜆𝐸(𝐼) ( ) ̅​̅​̅1 (𝛼)𝑀 ̅​̅​̅​̅′ 2 (𝛼)𝑀 ̅3 (𝛼) + 𝑀 ̅​̅​̅1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅​̅​̅​̅′ 3 (𝛼) ] +𝛼 [ (65) +𝑀 +(𝜆𝐸(𝐼))[𝐸(𝑣) + 𝑟𝐸(𝑣) + 𝑟𝐸(𝑠)] ̅​̅​̅​̅′1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅3 (𝛼) + 𝑀 ̅​̅​̅​̅′ 2 (𝛼)𝑀 ̅3 (𝛼) + 𝐷′′ (1) = −𝜆𝐸(𝐼(𝐼 − 1)[1 − 𝑟] + (−𝜆𝐸(𝐼))[1 − (−𝜆𝐸(𝐼))𝑟(𝑀 ′ ′ ′ ̅2 (𝛼)𝑀 ̅​̅​̅​̅3 (𝛼))] + [(−𝜆𝐸(𝐼))(1 − 𝛼 − 𝜆𝐸(𝐼)𝑟)(𝑀 ̅​̅​̅​̅1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅3 (𝛼) + 𝑀 ̅​̅​̅1 (𝛼)𝑀 ̅​̅​̅​̅2 (𝛼)𝑀 ̅3 (𝛼) + 𝑀 ̅​̅​̅1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅​̅​̅​̅′ 3 (𝛼))] + (𝜆𝐸(𝐼))2 [𝐸(𝑣) + 𝑟𝐸(𝑣) + 𝑟𝐸(𝑠)] − 𝛼[−𝜆(𝐸(𝐼(𝐼 − 1))𝑀 ̅​̅​̅​̅′1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅3 (𝛼)𝑟 + 𝑀 2 2 ̅​̅​̅​̅′1 (𝛼)𝑀 ̅​̅​̅​̅′ 2 (𝛼)𝑀 ̅3 (𝛼)𝑟 + 2(𝜆𝐸(𝐼)) 𝑀 ̅​̅​̅​̅′1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅​̅​̅​̅′ 3 (𝛼)𝑟 + (−𝜆𝐸(𝐼))[𝑀 ̅​̅​̅​̅′1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅3 (𝛼) + +2(𝜆𝐸(𝐼)) 𝑀 ′ ′ ̅​̅​̅​̅ ̅ ̅​̅​̅​̅ ̅​̅​̅ ̅ 𝑀 2 (𝛼 )𝑀3 (𝛼) + 𝑀 3 (𝛼)𝑀 1 (𝛼)𝑀′2 (𝛼)][(−𝜆𝐸(𝐼)) + 𝑟(−𝜆𝐸(𝐼))[𝐸(𝑣) + 𝐸(𝑠)] − 𝜆𝐸(𝐼(𝐼 − ′′ (𝛼) − 𝜆𝐸(𝐼(𝐼 − 1)[𝑀 ̅​̅​̅​̅ (𝛼)𝑀 ̅​̅​̅1 (𝛼)𝑀 ̅​̅​̅​̅′ 2 (𝛼)𝑀 ̅3 (𝛼)𝑟 + (−𝜆𝐸(𝐼))[𝑀 ̅2 (𝛼)𝑀 ̅​̅​̅1 (𝛼)𝑀 ̅​̅​̅​̅ ̅2 (𝛼)𝑀 ̅​̅​̅​̅′ 3 (𝛼)] + 1)𝑀 3 1 2 ̅​̅​̅​̅′1 (𝛼)𝑀 ̅2 (𝛼)𝑀 ̅3 (𝛼) + ̅​̅​̅ ̅​̅​̅​̅′ 2 (𝛼)𝑀 ̅3 (𝛼) + ̅​̅​̅ ̅2 (𝛼)𝑀 ̅​̅​̅​̅′ 3 (𝛼))]] + 𝛼 [(𝜆𝐸(𝐼)) (𝐸(𝐷) + 𝐸(𝑅))[(𝑀 𝑀 1 (𝛼)𝑀 𝑀 1 (𝛼)𝑀

AIJRSTEM 15-513; © 2015, AIJRSTEM All Rights Reserved

Page 55

S. Maragathasundari, American International Journal of Research in Science, Technology, Engineering & Mathematics, 11(1), JuneAugust, 2015, pp. 52-56

̅̅̅̅′1 (đ?›ź)đ?‘€ Ě…2 (đ?›ź)đ?‘€ Ě…3 (đ?›ź) + đ?‘€ Ě…Ě…Ě…1 (đ?›ź)đ?‘€ ̅̅̅̅′ 2 (đ?›ź)đ?‘€ ̅̅̅̅′ 3 (đ?›ź) + Ě…Ě…Ě… Ě…2 (đ?›ź)đ?‘€ ̅̅̅̅′ 3 (đ?›ź)][1 − đ?œ†đ??¸(đ??ź)đ??¸(đ??ˇ) − đ?œ†đ??¸(đ??ź)[đ?‘€ đ?‘€ 1 (đ?›ź)đ?‘€ ′ ′ Ě…Ě…Ě…Ě…Ě…Ě…Ě… Ě…2 (đ?›ź)đ?‘€ Ě…Ě…Ě…3 (đ?›ź) + đ?‘€ Ě…Ě…Ě…1 (đ?›ź)đ?‘€ Ě…Ě…Ě…Ě…2 (đ?›ź)đ?‘€ Ě…3 (đ?›ź) + đ?‘€ Ě…Ě…Ě…1 (đ?›ź)đ?‘€ Ě…Ě…Ě…2 (đ?›ź)đ?‘€ ̅̅̅̅′ 3 (đ?›ź)] đ?œ†đ??¸(đ??ź)đ??¸(đ?‘…)] + đ?œ†đ??¸(đ??ź(đ??ź − 1)][đ?‘€ ′ 1 (đ?›ź)đ?‘€ (66) Where primes and double primes in (62) denote the first and second derivatives at đ?‘§ = 1, respectively.Carrying out the derivatives at đ?‘§ = 1 and if we substitute the values of đ?‘ ′ (1), đ?‘ ′′ (1), đ??ˇâ€˛ (1) đ?‘Žđ?‘›đ?‘‘đ??ˇâ€˛â€˛ (1) into (62), we obtain đ??żđ?‘ž in closed form. Further, the mean waiting time of a customer could be found using đ?‘Šđ?‘ž = đ??żđ?‘ž â „đ?œ† and other performance measures can be determined using Littles formula. VI. Conclusions In this paper we have investigated a batch arrival queueing model of three stage heterogeneous service with different types of vacation policies .All the three stages of service are compulsory to all the arriving customers. The server has the option of undergoing a extended vacation. The Steady state results and some performance measures of the queueing system are derived. As a future work, repair process in terms of stages and stand by server could also be considered as additional aspects in this model. References [1]. [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

K.C Madan, An M/G/1 queue with second optional service, Queueing Systems, 34,(2000),37-46 Cramer. M (1989), “Stationary Distributions in a queueing system with vacation times and limited service�, Queueing Systems Theory apply, Vol.4, No.1, pp.57-68. Takagi, H. (1990), “Time dependent analysis of M/G/1 vacation models with exhaustive service�, Queueing Systems, Vol.6, No.1, pp. 369-390. Khalaf. R.F, Madan. K. C & Lucas. C.A (2011b), “An M[x]/G/1 Queue with Bernoulli schedule, General vacation times, random breakdowns, general and general repair times�, Applied mathematical sciences,Vol.5,No.1,pp.35-51. Choi, Doo and Kim, Tae-Sung, Analysis of two phase queueing system with vacation and Bernoulli feedback, Stochastic Analysis and Applications, 21,(2003),1009-1019. Borthakur. A and Choudhury. G (1997),“On a batch arrival Possion queue with generalized vacation, Sankhya Ser.B, Vol.59, pp.369-383. A. Federgruen and K. C. So, Optimal maintenance policies for single server queueing systems subject to breakdowns, Operation Research, 38 (1990), 330-343. Chodhury.G (2003), “A batch arrival queueing system with an additional service channel�, International Journal of Information and Management Sciences, Vol. 14, pp.17–30. Choudhury and Madan, (2005), A two stage batch arrival queueing system with a modified Bernoulli schedule vacation under Npolicy, Mathematical and computer Modelling, Vol.42, pp.71-85. Badamchi Zadeh.A and Shahkar. G.H (2008),“A two phase queue system with Bernoulli feedback and Bernoulli schedule server vacation�, Information and Management sciences, Vol.19, pp.329-338. ] K.C. Madan and A.Z. Abu-Dayyeh, On a single server queue with optional phase type server vacations based on exhaustive deterministic service and a single vacation policy, Applied Mathematics and Computation, 149 (2004), 723-734. Maragathasundari and S. Srinivasan, Analysis of M/G/1 feedback queue with three stage multiple sever vacation, Applied mathematical sciences, Vol.6, (2012), pp.6221-6240. K.C. Chae, H.W. Lee and C.W. Ahn, An arrival time approach to M/G/1- type queues with generalized vacations, Queueing Systems, 38 (2001), 91-100.

AIJRSTEM 15-513; Š 2015, AIJRSTEM All Rights Reserved

Page 56

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.