a single server retrial queue with impatient customers

DOI: 10.15415/mjis.2012.11006

Mathematical Journal of Interdisciplinary Sciences

Vol. 1, No. 1, July 2012 pp. 67–82

©2012 by Chitkara University. All Rights

Reserved.

A Single Server Retrial Queue with Impatient Customers

P. C. Garg and Sanjeev Kumar

Department of Statistics, Punjabi University, Patiala-147002. E-mail: [email protected]

Abstract

In the present paper, a single server retrial queue with impatient customers is studied. The primary arrivals and repeating calls follow Poisson distribution. The service time is exponentially distributed. Explicit time dependent probabilities of exact number of arrivals and departures from the orbit are obtained by solving the differential-difference equations recursively. Steady state solution of the number of busy servers is obtained. The numerical results are graphically displayed to illustrate the effect of arrival rate, retrial rate and service rate on different probabilities against time. Some special cases of interest are also deduced.

Keywords: Retrial, arrivals, departures, impatient.

1. INTRODUCTION

When a incoming call finds the server free at the time of its arrival, this call begins to be served immediately and leaves the system after service completion. However, if the server is busy at the time of a incoming

call then with probability 1-a1 the call leaves the system without service and with

probability a1 > 0 joins the secondary queue, named as orbit and retries from

there with intensity θ to get service. If an incoming repeated call finds the server free, it is served and leaves the system after service. Otherwise, if the server is busy at the time of a repeated call arrival with probability 1-a

2 the customer leaves

the system without service and with probability a2 retries for service again.

Falin and Templeton (1997) studied a single server retrial queueing system and obtained the steady state probabilities for the number of units in the system when the server is free and busy. Garg et al. (2008) obtained the time dependent probabilities for the exact number of arrivals & departures for the secondary queue of a single channel retrial queueing system. In this paper, the concept of impatience is applied to obtain the time dependent probabilities for the exact number of arrivals and departures from the orbit for a single server retrial queueing system by a given time recursively in different cases i.e. when server is busy and idle. Numerical results along with graphs are also found.

The queueing system investigated in this paper is described by the following assumptions:

1. The primary calls arrive in a Poisson distribution with parameter λ . 2. The service time follows an exponential distribution with parameter µ .

Garg, P. C. Kumar, S.

68

Mathematical Journal of Interdisciplinary Sciences, Volume 1, Number 1, July 2012

3. The repeated calls follow a Poisson distribution with parameter θ . 4. The stochastic processes involved viz. (a) arrival of calls (b) service times (c) retrial times are statistically independent.

2. DEFINITIONS

p ti j, , ( )0 =Probability that there are exactly i arrivals in the orbit and j departures from the orbit by time t when server is idle.p ti j, , ( )1 =Probability that there are exactly i arrivals in the orbit and j departures

from the orbit by time t when server is busy.Initially

p p i j p ii j i0 0 0 1 0 10 1 0 0 0 0, , , , , ,( ) ; ( ) , ; ( ) , .= = < = ≥t

The difference-differential equations governing the system are

d

dtp p t p t0 0 0 0 0 0 0 0 10, , , , , ,( ) ( ) ( )=− +λ µ (2.1)

d

dtp t i j p t p t j i j ii j i j i j, , , , , ,( ) , , ,0 0 1 0=− + −( )( ) ( )+ ( ) ≤ ≥λ θ µ >> 0 (2.2)

d

dtp t a i j a p t p t ai j i j i j, , , , , ,1 1 2 1 0 11( )=− + + −( ) −( )( ) ( )+ ( )+λ µ θ λ λ pp t

i j p t i j a p t

i j

i j i j

−

− −

( )

+ − +( ) ( )+ − +( ) −( ) ( )

1 1

1 0 2 1 11 1 1

, ,

, , , ,θ θ ,,

.0 2 3≤ ≤ ( )j i

Using the Laplace transformation f s of f t( ) ( ) given by

f s e f t dt Re sst( )= ( ) ( )>−∞

∫ ,0

0

in the equations (2.1)-(2.3) along with the initial conditions. We have

s p s p s+( ) ( )= ( )+λ µ0 0 0 0 0 1 1, , , , (2.4)

s i j p s p s j i j ii j i j+ + −( )( ) ( )= ( ) ≤ ≥ >λ θ µ, , , , , , , (0 1 0 0 22 5. )

s a i j a p s

p s a p s

i j

i j i j

+ + + −( ) −( )( ) ( )= ( )+ ( )+−

λ µ θ

λ λ

1 2 1

0 1 1 1

1 , ,

, , , , ii j p s

i j a p s j i

i j

i j

− +( ) ( )+ − +( ) −( ) ( ) ≤ ≤

−

−

1

1 1 0

1 0

2 1 1

θ

θ, ,

, , ,

(2.6)

A Single Server Retrial Queue

with Impatient Customers

69


3. MAThEMATICAl ANAlySIS

p ss a

s a s0 0 01

1, , ( )=

+ ++ +( ) +( )−

λ µλ µ λ λµ

(3.1)

p ss a s0 0 0

1, , ( )= + +( ) +( )−

λλ µ λ λµ

(3.2)

p s a i z z

b

b

x

i

bb i

x

0 0 0 11 1

0 102

, , ( )( )=

+ + +

=

∞

==

∞

∑ ∏∑ λ µ λµ11

0

00

0

1

1

2 2

0

=

∞

=

∞

− −

=

+

∑∑

∏ + × + +( ) >

z

m im

i

s s i im i( ), ,, ,η λ θξ

(3.3)

p s a i z

b

b

x

i

bb i

x

0 0 0 11

0 100 21

, , ( )( )=

+ +

=

∞

==

∞

=∑ ∏∑ λ µ λµ

∞∞

=

∞

−

=

+

∑∑

∏ + >

z

m im

i

s im i

0

01

2 20 0( ), ,, ,η ξ (3.4)

p ss i j

p s j ii j i j, , , , ,0 1 1( )=+ + −( )( )

( ) ≤ ≤µ

λ θ (3.5)

p si jzx

i j

z o

z

j

j

j

, ,000

1

1

1

1

1

1

1

1111

( )==

∞

==

+

=

+

=

− +

=

∞

∑∑∑∑∑−

α

α

α

α

α∑∑ ∑∑∑∑

− +− +

=

∞

=

∞

( )

w

w w

w

w w

zx

jj

s

jj

a

α

α α

α

α α

α α α θ α

2

2

1

2 1

1

11

00

1 2 1 λλ

λµ λ µ

α

β

µ

αα

α α

α

1

000

1

1

221

1 + +

=

∞

=

∞

=

∞− +

∑∑∑∑+

( )

z

bbbw

w w

ze y

z

j

j j

e ae++

=

+ + +∏ −( ) ( )

×

+(

δ

θ

δλµ α1 0 1 2

1

121, ,

,

xe xe

jb b b

k q

a

s r

))

+( )−

=

− +

=

−+

∏∏u

kq

j

m j b

k qQ Q m j

s,

, ,

, ,1

2 4 4

1

1

1

α α ξ

ηα

bb

m

b j

j iα

α1

1

1

2 3 2

1=

+ +

∏

≤ ≤ (3.6)

Where

= =+ ++

+

− =

∞

=

∞

=

∞−

∑∑∑ , ,δαααα

α

α

α

x ywwww

w v

v vvvv

v 1

01 11

1

000

++

∑=≠

1 x y

x y

ϒk q

q q q q q

q q

k k

a k, =

+ + −( ) − + ≤ ≤ − +

+ + − + −

+ +

+

λ α θ α α α α

λ µ α α

1 2 3 2 4 4

3 3

1 1

1 1(( ) −( ) ≤ ≤ − +

+θ α α α1 1 2 22 1k q q


70


uw k

w kk q

k q q

k q q q

Q

Q Q

, =≤ ≤ − +

− + ≤ ≤ −

+ − +

− + + ++ −

α

α α

α α

α α α1 1

3 1 1

1 2 2

2 3 21 3

44 4αq +

η

λ α θ

λ µ α θ

λ α

αm j b

m

m j

m

m j

a a j m j

, ,1

1 1

1 1 1 21 2

2

=

+ −( ) ≤ ≤

+ + −( ) −( ) + ≤ ≤

+−

− jj j m j

a m j a j m b j

m b j b

θ

λ µ θ

λ θ

α

α

2 1 3

3 1 3 1 3

3

1 2 1

1

+ ≤ ≤

+ + −( ) −( ) + ≤ ≤ +

+ − −( ) αα α

α

α

λ

λ µ

1 1

1

1

3 1 2 3

2 3 1

2 3 21

+ + ≤ ≤ +

= + +

+ = + +

j m b j

m b j

a m b j

ξ

δ

α

δ

m j b

m

m j

x

m j

Z m j

Z j m j

j m j

b

xm j

m Zj

, ,

,

,

1

0

1

1 2

2 1 31

3=

≤ ≤

+ ≤ ≤

+ ≤ ≤

− +

−

−

−

++ + ≤ ≤ +

+ + ≤ ≤ +

+ = + +

− −

1 3 1 3

3 1 2 3

1 2 3 1

1

1 1 1

1

3

j m b j

b b j m b j

Z m b j

Z

m b j

α

α α

α

α

++ = + +

+ + ++

1 2 3 21

1

m b j

w w wv v v

α

α α α

++

+

− +

+ − + + +

=

+ + + + − + = = −1

1

1

1 1 1 12 2

α α

α α α α αα α β αv v

v v v v v

y

w w w i jq q j ,

Following Bateman (1954), we note that the Laplace inverse transform of

Q p

p p

t e

m l l

d

dp

Q p pm t

kl

m

k

nl

l

k l kk( )

( )=

−( ) −( )( ) −−

==

−

−∑∑α α

! !111

1

1

kkm

p

i k

k

k

p p

( )( )

≠ ≠=α

α α, for i k

Where p p p p p Q pm m

n

mn( )= −( ) −( ) −( ) ( )α α α1 21 2

& is a polynomial

of degree < + + −m m mn1 2 1 , we obtain inverse Laplace transform of



71


s r sk qkQ

j uk q

m j b

Q Q

+( )

+

=

− +

=

−+

∏∏ ,

,

, ,1

2 4 4

1

1

1

α α

ηα(( )

+

−

=

+ +

= ++ +

∏ξ

αα

η

m j b

m

b j

m ji j s

, ,

&11

1

2 3 2

3 12 3 2Π mm j i

m j i

, ,

, ,( )−ξ are

fu u u u u u u

j1 1 2 2 2 2 4 1 4 1 1 2 2 2 2 3 4 2 4 2 2 1 4, , , , , , , , , ,, α α α α α α− + − + + − jj j j b b j j b

j j j i

t

gj i

+ + +

+ +

4 1 1 2 13 2 1

3 1 3 2

, , , , , , ,

, ,

( )

&, , ,

ξ α ξ α α

ξ ξ

ξξ2 3 2i j j i t+ + , , ( )

Denoting the convolution as usual by* and the inverse transform of equations (3.1)-(3.6) are

p t e cosha a

t ea

ta

0 0 0

1

2

212 2

11 11 2 1

2, , ( )=+( )+ + +( )

+−

+( )+−

+λ µ λλ µ λµ 11

2

1

212

1

212 2

11

1 2 1

1 2

( )

−( )++( )+ +( )

+( )+ + +

µ

λ µ

λ λµ

λ µ λµ

t

a

a asinh

a a 11

2

( )t (3.7)

p ta a

e

sinha

at

0 0 02

12 2

1

1

2

212

2

1 2 1

1

1

, , ( )=+( )+ + +( )

+

−+( )+λ

λ µ λµ

λ

λ µ

(( )+ + +( )µ λµ212 1

23 8

at ( . )

p t aii z z

b

b

x

i

bb i

x

, ,0 0 11 1

0 102

( )= ( )

+ + +

=

∞

==

∞

∑ ∏∑ λ µ λµ11

1 0 2 0 2 2 0

00

0

=

∞

=

∞

− +( )

∑∑

∗ ( )( ) >+

z

i te g t ii i i i

λ θξ ξ ξ, , , , , , ,, (3.99)

p t aii z

b

b

x

i

bb i

x

, ,0 1 11

0 100 21

( )= ( )

+ +

=

∞

==

∞

=∑ ∏∑ λ µ λµ

∞∞

=

∞

∑∑

+( ) >

z

g t ii i i i

0

1 0 2 0 2 2 00ξ ξ ξ, , , , , , ,, (3.10)

P t e P t j ii j

i j t

i j, , , , , ( . )0 1 1 3 11( )= ∗ ( )( ) ≤ ≤− + −( )( )µ λ θ

P ti j

i j

z w

w w

w

w

j

j

j

, ,11

1

1

1

0

1

1 2

3 2

1

2

( )=− =

+

=

− +

=

∞ − +

∑∑∑ ∑

α

α

α

α

α

α

α

α −− +

=

∞

=

∞

=

∞

==

+

∑∑∑∑∑∑

( )

w

zxzx

jj

jj

α

α

α

αα α α θ α

1

111

2

1

1

0000

1

1

1

1 2 1 λλ µ

λµ λ µ

α

β

αα

α α

α α

1

000

1

1

121

1

+ +

=

∞

=

∞

=

∞− +

∑∑∑∑+

( )

z z

bbbw

w w

ze y

j

j j

e e++ + + +

=

−( ) ( )

∏ δ δ

λµ α1 0 1 2 1

1 1 2

1 21

, ,

, ,

xe xea

f

b b b

e

j

u u

,, , , , , , , ,, , ,1 2 2 4 1 4 1 1 2 2 2 2 3 4 2 2 1 4 4 1 u u u u uj j j j bα α α α α α ξ− + − + − + αα ξ α ξ α α1 2 1 2 1

3 21

1 3 12

, , , , ,

( . )

j b b j j bt

j i

+ +( )

≤ ≤


72


P ti j

i j

z w

w w

w

w

j

j

j

, ,11

1

1

1

0

1

1 2

3 2

1

2

( )=− =

+

=

− +

=

∞ − +

∑∑∑ ∑

α

α

α

α

α

α

α

α −− +

=

∞

=

∞

=

∞

==

+

∑∑∑∑∑∑

( )

w

zxzx

jj

jj

α

α

α

αα α α θ α

1

111

2

1

1

0000

1

1

1

1 2 1 λλ µ

λµ λ µ

α

β

αα

α α

α α

1

000

1

1

121

1

+ +

=

∞

=

∞

=

∞− +

∑∑∑∑+

( )

z z

bbbw

w w

ze y

j

j j

e e++ + + +

=

−( ) ( )

∏ δ δ

λµ α1 0 1 2 1

1 1 2

1 21

, ,

, ,

xe xea

f

b b b

e

j

u u

,, , , , , , , ,, , ,1 2 2 4 1 4 1 1 2 2 2 2 3 4 2 2 1 4 4 1 u u u u uj j j j bα α α α α α ξ− + − + − + αα ξ α ξ α α1 2 1 2 1

3 21

1 3 12

, , , , ,

( . )

j b b j j bt

j i

+ +( )

≤ ≤

4. STEADy STATE SOlUTION OF ThE NUMBER OF BUSy SERVERS

DefineP s P s k sP s Pn k j n j kj n n, , , , ,, , ;( )= ( ) = ( )→ →+=

∞∑ 0 10 0 0 as s 0 and substituting

in equations (2.4) to (2.6), we get

λ θ µ+( ) =n P Pn n, ,0 1 (4.1)

λ µ θ

λ λ θ θ

a n a P

P a P n P n

n

n n n

1 2 1

0 1 1 1 1 0

1

1 1 1

+ + −( )( )= + + +( ) + +( ) −− +

,

, , , aa Pn2 1 1 4 2( ) + , ( . )

Define z z

z z z (4.3)

P P m

P P P

mn

n n m( )= =

( )+ ′( ) = ( )=

∞∑ 0

0 0 1

0 1, , ,

λ θ µ

λ µ θ

λ λ θ θ

a P a P

P a P P a

1 1 2 1

0 1 1 0

1

1

+( ) ( )+ −( )) ′( )= ( )+ ( )+ ′ ( )+ −

z z z

z z z z 22 1 4 4( ) ′ ( )P z ( . )

To get the steady state distribution of the number of busy serversP Pm = (Number of busy servers = m), m = 0,1 put z=1 in above equations, we have λ θ µP N P0 0 1+ = (4.5)

Where N0 is the number of repeated calls when no server is busy. Denote the

ratio θNP

r0

00= which equals the rate of flow of repeated calls given that the

number of busy servers is zero. Equation (4.5) can be written as

Pr

P10

0=+λµ

(4.6)

And from the normalizing condition P P0 1 1+ = , we have

Pr0

0

=+ +µ

µ λ (4.7)



73


Thus the steady state distribution of the number of busy servers in this model is identical to the steady distribution of the number of busy servers in the Erlang loss Model.

5. NUMERICAl EXAMPlE

The numerical results are generated by using BASIC program following the work of Bunday (1986). Assume that the mean service time is 1.In table 1(a) for retrial rate θ= = =. , . & . ,09 03 041 2a a values of probabilities of no arrival and no departures are found for two different cases i.e. when the server is free

& busy for two different values of ρλµ

=

= 0 1 0 2. , . .

Table 1(a)

tP0,0,0(t) P0,0,1(t)

ρλµ

=

= 0 1 0 2. , . . = .1 ρ

λµ

=

= 0 1 0 2. , . . = .2 ρ

λµ

=

= 0 1 0 2. , . . = .1 ρ

λµ

=

= 0 1 0 2. , . . = .2

0 1 1 0 01 0.939322 0.883419 6.06E-02 0.1161642 0.919014 0.847899 0.080661 0.1508623 0.912107 0.836674 0.087314 0.1611454 0.90965 0.83273 0.089506 0.1641125 0.908668 0.830968 0.090218 0.1648856 0.908178 0.829863 0.090438 0.1650047 0.907849 0.828953 9.05E-02 0.1649248 0.90757 0.828113 9.05E-02 0.1647759 0.907357 0.827178 9.05E-02 0.164627

Table 1(b)

t P3,0,0(t) P3,0,1(t) P2,1,0(t) P2,1,1(t) P3,1,0(t) P3,1,1(t) P3,2,0(t) P3,2,1(t)

0 0 0 0 0 0 0 0 0

1 4.95E-11 1.11E-12 2.14E-10 1.17E-09 1.96E-13 1.11E-12 1.13E-15 5.07E-15

2 3.70E-10 3.15E-11 6.84E-09 1.75E-08 1.06E-11 3.15E-11 1.85E-13 5.97E-13

3 9.35E-10 1.77E-10 4.28E-08 6.89E-08 8.79E-11 1.77E-10 2.72E-12 8.07E-12

4 1.57E-09 5.08E-10 1.39E-07 1.61E-07 3.54E-10 5.08E-10 2.03E-11 3.87E-11

5 2.17E-09 1.04E-09 3.18E-07 2.88E-07 9.48E-10 1.04E-09 8.72E-11 1.15E-10

6 2.74E-09 1.78E-09 5.95E-07 4.37E-07 1.92E-09 1.78E-09 2.39E-10 2.90E-10

7 3.20E-09 2.60E-09 9.67E-07 6.06E-07 3.43E-09 2.60E-09 5.96E-10 5.24E-10

8 3.68E-09 3.69E-09 1.44E-06 7.82E-07 5.23E-09 3.69E-09 1.11E-09 9.92E-10

9 3.99E-09 4.55E-09 1.98E-06 9.72E-07 7.81E-09 4.55E-09 2.13E-09 1.35E-09


74


In table 1(b) for retrial rate θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a various probabilities of number of arrivals and departures are calculated when the server is free and busy.

Table 1(c)

tP3,2,0(t)

θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.1 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.2 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.3 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.4

0 0 0 0 0

1 2.35E-09 8.82E-09 1.86E-08 5.57E-08

2 1.84E-07 6.46E-07 1.28E-06 3.65E-06

3 1.95E-06 6.35E-06 1.17E-05 3.23E-05

4 9.24E-06 2.79E-05 4.79E-05 1.28E-04

5 2.85E-05 7.97E-05 1.28E-04 3.34E-04

6 6.76E-05 1.75E-04 2.62E-04 6.76E-04

7 1.34E-04 3.23E-04 4.55E-04 1.16E-03

8 2.37E-04 5.30E-04 7.12E-04 1.77E-03

9 3.80E-04 8.01E-04 9.38E-04 2.65E-03

Table 1(d)

tP3,2,1(t)

θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.1 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.2 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.3 θ ρ= = = =. , . , . & . ,09 0 1 03 041 2a a.4

0 0 0 0 0

1 1.63E-08 6.14E-08 1.31E-07 3.94E-07

2 6.29E-07 2.22E-06 4.43E-06 1.29E-05

3 4.42E-06 1.45E-05 2.71E-05 7.69E-05

4 1.57E-05 4.81E-05 8.41E-05 2.35E-04

5 3.92E-05 1.11E-04 1.83E-04 5.06E-04

6 7.87E-05 2.08E-04 3.22E-04 8.86E-04

7 1.37E-04 3.38E-04 4.93E-04 1.36E-03

8 2.17E-04 5.01E-04 7.27E-04 1.95E-03

9 3.18E-04 7.13E-04 6.66E-04 2.20E-03



75


Table 1(e)

tP3,2,0(t)

ρλµ

=

= 0 1 0 2. , . . = .4 ρ

λµ

=

= 0 1 0 2. , . . = .5 ρ

λµ

=

= 0 1 0 2. , . . = .6 ρ

λµ

=

= 0 1 0 2. , . . = .7

0 0 0 0 0

1 2.64E-08 1.85E-07 6.11E-08 4.30E-07

2 1.84E-06 6.45E-06 4.07E-06 1.45E-05

3 1.75E-05 4.12E-05 3.72E-05 9.01E-05

4 7.45E-05 1.34E-04 1.54E-04 2.88E-04

5 2.08E-04 3.08E-04 4.19E-04 6.52E-04

6 4.49E-04 5.71E-04 8.85E-04 1.20E-03

7 8.16E-04 9.24E-04 1.58E-03 1.93E-03

8 1.33E-03 1.39E-03 2.57E-03 2.95E-03

9 1.85E-03 1.71E-03 3.17E-03 3.08E-03

In tables 1(c) & 1(d) for ρ= = =0 3 4 51 2. , . & . ,a a probabilities of 3 arrivals and 2 departures is calculated for two different cases when the server is free & busy at different retrial rates θ= 0 1 0 2 0 3 0 4. , . , . , . . In tables 1(e) & 1(f) for θ= = =0 2 4 51 2. , . & . ,a a probabilities of 3 arrivals and 2 departures is calculated for two different cases when the server is free & busy at different values of ρ= = =0 3 4 51 2. , . & . ,a a 0.4,0.5,0.6,0.7.

Table 1(f)

tP3,2,1(t)

ρλµ

=

= 0 1 0 2. , . . = .4 ρ

λµ

=

= 0 1 0 2. , . . = .5 ρ

λµ

=

= 0 1 0 2. , . . = .6 ρ

λµ

=

= 0 1 0 2. , . . = .7

0 0 0 0 0

1 1.20E-07 8.49E-07 2.10E-07 1.50E-06

2 7.63E-06 2.76E-05 1.28E-05 4.70E-05

3 6.71E-05 1.67E-04 1.08E-04 2.77E-04

4 2.69E-04 5.24E-04 4.21E-04 8.49E-04

5 7.15E-04 1.17E-03 1.09E-03 1.86E-03

6 1.48E-03 2.11E-03 2.19E-03 3.30E-03

7 2.57E-03 3.31E-03 3.75E-03 5.17E-03

8 4.11E-03 5.54E-03 5.96E-03 8.23E-03

9 6.74E-03 1.20E-02 8.11E-03 2.77E-02


76


1.2

1

0.8

0.6

0.4

0.2

00 2 4

t

6 8 10

PSeries1

Series2

Series3

Series4

1.2

1

0.8

0.6

0.4

0.2

00 2 4

t

6 8 10

Series1

Series2

Series3

Series4

Figure 2(a)

In figure 2(a) for retrial rate θ= = =. , . & . ,09 03 041 2a a values of probabilities of no arrival and no departures are plotted versus time for two different cases

i.e. when the server is free & busy for two different values of ρλµ

=

= 0 1 0 2. , . .

When there is no arrival, no departure & server is free, probabilities decrease with time. Also when there is no arrival, no departure & server is busy, probabilities increase with time in starting moments and then decrease with time.

0

0 108642

E-09

E-09

E-09

E-09

E-09

E-09

5

4

3

2

1

-1

Series1

Series2

P

t

0

00 108642

E-09

E-09

E-09

E-09

E-09

E-09

5

4

3

2

1

-1

Series1

Series2

P

t

Figure 2.1(b)



77


0 8

t

10642

P

Series1

Series2

0.000002

0.000002

0.000001

0.000001

0.000000

0

-5E-0700 8

t

10642

P

Series1

Series2

0.000002

0.000002

0.000001

0.000001

0.000000

0

-5E-07

Figure 2.2(b)

In figure 2.1(b)-2.4(b) for retrial rateθ ρ= = = =. , . , . & . ,09 0 1 03 041 2 a a various probabilities of number of arrivals and departures are plotted versus time when the server is free and busy. In the initial moments, probabilities of exact number of arrivals and departures, when the server is busy, are greater than those probabilities of exact number of arrivals and departures, when the server is free. With the passage of time, probabilities of exact number of arrivals and departures, when the server is free, become greater than those probabilities of exact number of arrivals and departures, when the server is busy.

9E-09

E-09E-09E-09E-09

E-09E-09

E-09

E-09

E-09

0

0 2

21

14 6 8 10

-

8

567

43

t

Series1

Series2

P

9E-09

E-09E-09E-09E-09

E-09E-09

E-09

E-09

E-09

0

0 2

21

14 6 8 10

-

8

567

43

t

Series1

Series2

P

Figure 2.3(b)


78


2.5

1.5

E-09

E-09

E-09

E-09

E-09

E-10

2

0

1

5

5-

P

t

Series1

Series2

8 106420

2.5

1.5

E-09

E-09

E-09

E-09

E-09

E-10

2

0

1

5

5-

P

t

Series1

Series2

8 1064200

Figure 2.4(b)

In figures 2(c) & 2(d) for ρ= = =0 3 4 51 2. , . & . ,a a probabilities of 3 arrivals and 2 departures versus time for two different cases when the server is free & busy at different retrial rates θ= 0.1,0.2,0.3,0.4 are shown and these probabilities increase with the increasing value of θ=.

In figures 2(e) & 2(f) for θ= = 0.2, a1

= .4 & a2 = .5, probabilities of 3

arrivals and 2 departures versus time for two different cases when the server is free & busy at different retrial rates ρ = 0.4,0.5,0.6,0.7 are shown and these probabilities increase with increase in ρ

6

0.0025

0.003

108420

0.002

0.001

0

0.0005

0.0005

0.0015

-

Series1

Series2

Series3

Series4

P

6

0.0025

0.003

108420

0.002

0.001

0

0.0005

0.0005

0.0015

-

Series1

Series2

Series3

Series4

P

Figure 2(c)



79


0.0025

0.002

0.0015

0.001

-0.0005

0.0005

0

0

P

t

108642

Series1

Series2

Series3

Series4

0.0025

0.002

0.0015

0.001

-0.0005

0.0005

0

0

P

t

108642

Series1

Series2

Series3

Series4

Figure 2(d)

0 42 6 8 10

t

P

0

0.0005

-0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

Series4

Series3

Series2

Series1

00 42 6 8 10

t

P

0

0.0005

-0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

Series4

Series3

Series2

Series1

Figure 2(e)

0.03

0.025

-0.005

0.005

0.01

0.015

0.02

0

0 2 4 6 8 10

Series1

Series2

Series3

Series4

t

P

0.03

0.025

-0.005

0.005

0.01

0.015

0.02

0

00 2 4 6 8 10

Series1

Series2

Series3

Series4

t

P

Figure 2(f)


80


6. SPECIAl CASES

1. When the customers are served singly according to a Poisson distribution with parameter µ , the Laplace transform of the P ti j, , ( )0 and P ti j, , ( )1 probabilities and can be obtained by putting a a1 2 1= = in equations (3.1)-(3.6), we get

P s

s

s s0 0 0, , ( )( )( )

=+ +

+ + + −λ µ

λ µ λ λµ (6.1)

P ss s0 0 1, , ( )

( )( )=+ + + −

λλ µ λ λµ

(6.2)

P ss i j

P s j i i ji j i j, , , ,( )( )

( ), , ,0 1 0 0=+ + −

≤ > ≥µ

λ θ (6.3)

P ss s

i

i z zxi b

z z i b b

x

, ,0 1

11

1 1 1 2( )=

( )+( ) + +( )

+ +=

+ + + + + +

λ µ λµ

λ λ µ

Π++

=

∞

=

∞

=

∞

=

∞

+ +( )

+ +( ) + +( )

∑∑∑∑ b bbbbz

b

i

i s

s s i

λ θ

λ θ λ θ

1

21

2

0000

2

,

bbi

i≥1 6 4 ( . )

P si i

i

iyyyzz ii

, ,1 1 200000 212

( )=( ) ( )=

∞

=

∞

=

∞

=

∞

=

∞

∑∑∑∑∑µλθ α α α

zzyz

y z z y z

i

i

11

2

11 0001

1

1

1

1

1

1

=

∞

=

∞

=

∞

=

+

=

+

=

+ + +

∑∑∑∑∑∑−

( )

α

α

α

α

α

λ µ λµ xx x ei

e ey z y z i y z y

x

i

s s

s

+ − + +( ) − + + + + +( )( )

=

+( ) + +( )

+ +

=∏ λ λ µ

λ

1 1 2

1

1Σ

αα θ λ α θr

z

r

y

r

ir rs−( )( ) + +( )

− − +( )

=∏ 1

1

1

(6.5)

P si j

i j

jyyyzz jj

, ,1

1 20000 211

( )=( ) ( )

( )=

∞

=

∞

=

∞

=

∞

∑∑∑∑

λ µθ

α α α

==

∞

=

∞

=

∞

=

∞

=

+

=

+

=

− +

+

∑∑∑∑∑∑∑−− 00001

1

1

1

1

1

1

11

2

1 rrz

i j

z

i jj

j

j α

α

α

α

α

λ µµ λµ λµ λ

λ µ

z x x c ui j

uz y

x

jr

c

i jz r

s

s

( ) ( ) +( )

+ +( )

+

= =

−− + +( )∏ ∏ −

−

1 1

1 1Σ

−− + + + + + +( )( )

−

−−

=

+ + −( )( ) + +(

z i j r z y

r

z

r

ui j

u ej

e e

rs s

1 1 1

1

Σ Σ

λ α θ λ α θ)) ≤ <− +( )

=∏ y

r

jr j i

1

1

1 6 6, ( . )



81


The results (6.1) – (6.6) coincide with the results of Garg, Bansal & Srivastava (2009).2. After getting the above case (i.e. when the customers are served singly

according to a Poisson distribution with parameter µ by putting a1 = a2= 1), when the concept of retrial is ended further by fixing z ,z ,z , z0 1 2 1 1 i j j jx x y y− , , , , , ,

to zero and letting P s P s P s P s P si j i j, , , , , , , ,,1 0 0 1 0 0 0 0 0( )= ( ) ( )= ( )= ( )

11

1 11 2 3

12s s s s

jj

j

j

+ +=( )+ +( )

=+

+ +

=+λ µ

α α α α λθ

λ µ λ λ α θ,

11

3s

s

s

+ +

= =++ +

λ α θ

λλ µ

in equations (3.1)-(3.4), we obtain

P ss0 0

1, ( )= +λ (6.7)

P ss s

ii

i0 0 1, ,( )=+( ) + +( )

≥λ

λ λ µ (6.8)

P ss s

i k i k

k ii j

i j

,

!

!( )=

+ +

+

−( ) + −( )λλ µ

µλ

1

!!

( . )

k

j k

k

s

s

j i

j

=

−

∑+( )+ +( )

≤ ≤

0

1

1 6 9

λ

λ µ

Which coincide with equation (5) of Pegden and Rosenshine (1982).

REFERENCES

[1] Bateman, H. (1954). Tables of integral transforms, Vol.1, Mcgraw-Hill Book Company, New York.[2] Bunday, B.D. (1986). Basic Queueing Theory, Edward Arnold (Publishers) Ltd., London.[3] Falin, G.I. and Templeton, J.G.C. (1997). Retrial queues, Chapman and Hall, London. http://dx.doi.org/10.1007/978-1-4899-2977-8[4] Garg, P. C., Srivastava, S.K. & Bansal, S.K. (2009). Explicit time dependent solution of a two

state retrial queueing system, Pure and Applied Mathematika Sciences, LXIX (1-2), 33–50.[5] Pegden, C. D. & Rosenshine, M. (1982). Some new results for the M/M/1 queues, Management

Science, 28(7), 821–828. http://dx.doi.org/10.1287/mnsc.28.7.821

P. C. Garg is Professor, Department of Statistics, Punjabi University, Patiala.

Sanjeev Kumar, is Professor, Department of Statistics, Punjabi University, Patiala.

a single server retrial queue with impatient customers

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