Master’s Thesis: INSURANCE RISK AND QUEUEING …alexandria.tue.nl/extra1/afstversl/wsk-i/siani2011.pdf · INSURANCE RISK AND QUEUEING MODELS WITH THRESHOLD DEPENDENCE By: Jean-Claude

Departement of Mathematicsand Computer Science.

Eindhoven University of Technology.

Masters Thesis:INSURANCE RISK AND QUEUEING

MODELS WITH THRESHOLDDEPENDENCE

By:Jean-Claude Siani Emaga.

Supervisors:Prof.dr.ir. O.J. Boxma.

Dr. A. Lopker.

January 3, 2011

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To My Lord Jesus-Christ

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Acknowledgements

This Master thesis is the result of nine months of intensive work, during which Ireceived support and orientation from some people to whom I really want to expressmy gratitude. To prof. Onno Boxma who proposed to me this project because Iwas interested in something related to finance or insurance. He then asked me ifI was interested in the risk model connected to queueing theory. This sounded tome as an opportunity to work in a concrete field of mathematics. This also offeredthe opportunity to work in the friendly environment of EURANDOM with PHDstudents and post-docs. Onno gave me constructive arguments which significantlystrengthened the quality of this thesis. He guided me in the right way to fulfill thisjob. Many thanks to Dr Andreas Lopker for his guidance and fruitful discussionsabout some proofs. Further I am grateful to Dr Marion Blonk who gave to methe opportunity to follow this master program by supporting the tuition fees. Iwould also like to thank the administrative staff of EURANDOM for organizing theworkshop that I attended during my stay in this institute. Lastly, I thank my wifeand my son for their support, my family in Holland; namely Remco Lagarde and hiswife, Dino Casarotto and his wife and M.L. van Langen-Hopmans for their supportand their prayers.

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Abstract

In this master thesis we consider a dependent setting of the classical queueing model.We assume that the distribution of the service time depends on the inter-arrival timebetween customers according to some threshold. An exact solution for the workloaddistribution (virtual waiting time) is obtained in the queueing system where thethreshold is deterministic and the service time is set to zero, if the inter-arrival timeis less than the threshold. The result is compared with the stationary ruin proba-bility under a threshold of an insurance risk model. For the case of an exponentialthreshold, the Laplace-Stieltjes transform of the sojourn time of a customer is de-rived. The effect of the threshold dependence on the mean performance measures isillustrated by several numerical examples.

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Keywords: Ruin Probability, Threshold, Dependence, Laplace-Stieltjes transform,Risk theory, Queueing model with dependencies, Duality.

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Contents

1 Introduction 2

2 Ruin Theory 7

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Beekmans convolution formula . . . . . . . . . . . . . . . . . . . . . 9

2.3 Explicit expressions for the ruin probabilities . . . . . . . . . . . . . . 11

3 Ruin probability in an insurance risk model with threshold 13

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.2 Exponential claim size distribution . . . . . . . . . . . . . . . . . . . 15

3.3 The case where no claim is accepted before some deterministic thresh-old of the inter-arrival time . . . . . . . . . . . . . . . . . . . . . . . 17

3.4 Duality D1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.5 Duality D2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.6 Sample path argument for duality D1 . . . . . . . . . . . . . . . . . . 25

3.7 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 A queueing model in which the service time depends on the inter-arrival time 28

4.1 Laplace-Stieltjes transform of the sojourn time of a customer . . . . 29

4.2 Laplace-Stieltjes transform of the waiting time . . . . . . . . . . . . . 33

4.3 Performance measures . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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4.3.1 Mean sojourn time . . . . . . . . . . . . . . . . . . . . . . . . 35

4.3.2 Mean number of customers in the system . . . . . . . . . . . . 35

4.3.3 Numerical illustrations . . . . . . . . . . . . . . . . . . . . . . 37

4.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5 Conclusions and Recommendations 44

5.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5.2 Recommendations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

J.-C. Siani Emaga January 3, 2011 1 of 46

Chapter 1

Introduction

Nowadays Queueing theory and Risk theory are well developed branches of appliedprobability theory. A large body of literature, which is still growing rapidly, existson these subjects. It is Erlang (Cf. Brockmeyer, Halstrom, Jensen [1948]), a Danishmathematician, who may be considered as the founder of Queueing theory. His stud-ies, in the period 1909-1920, are now classical queueing theory. His classical queueingmodels rely on the assumption of independence between service time and inter-arrivaltimes of the customers. This independence assumption is also made for the classicalCramer-Lundberg model to describe the surplus process of an insurance portfolio. Ageneral connection between ruin models and queueing models is the equality betweenthe survival probability and the stationary workload distribution of the queueingmodel, under independence assumption ([10] pages 161-162). In contrast to the clas-sical assumption, we suppose that the claim sizes and the inter-occurrence times aredependent. Recently various results have been obtained concerning the asymptoticbehavior of the probability of ruin with dependent claims size. Hansjorg Albrecherand Onno J. Boxma (2004) investigated the following dependence structure: Thetime between two claim occurrences depends on the previous claim size. They de-rived an exact solution for the probability of survival by means of Laplace-Stieltjestransforms. Albrecher and Boxma (2005) studied the Gerber-Shiu function in themodel with dependence between claim size and inter-arrival time. Isaac Kwan andHailiang Yang (2007) studied the reverse dependence structure of the model devel-oped by Albrecher and Boxma (2004); a closed form for the ruin probability wasfound for the special case where the claim size distribution is exponential. IsaacKwan and Hailiang Yang considered also a model similar to that in Albrecher andBoxma (2004), but with a fixed threshold. By solving a system of delay differentialequations they found that when the claim size follows an exponential distribution,an explicit solution for the ruin probability can be obtained.


Motivated by a related model in risk theory (Albrecher and Boxma 2004, Kwan andYang 2007), we consider in this master thesis a generalization of the classical M/G/1and G/M/1 queue with some dependence structure.

The following notations are used to describe the various queueing and risk modelsin this thesis. In general a/b/1 denotes a queueing model with arrival distributiona and service distribution b. The symbols a and b take values from the set {M,G}denoting as usual either an exponential distribution or a general distribution. Thisis the well known Kendall notation, where e.g. M/G/1 denotes a Markovian queuewith exponential inter-arrival times and general service times. All studied queueswork under the FIFO regime, i.e. customers are served according to their arrivaltime. For the risk models we use a similar notation, namely a/b for a risk modelwith inter-arrival time distribution a and claim size distribution b.

To describe the various dependencies, we use arrow symbols. For instance a/b /1

denotes a a/b/1 queue, where the service time of the nth customer depends on theinter-arrival time between the (n 1)th and the nth customer or a /b symbolizesa risk model where the inter-arrival time of the nth claim and the (n + 1)th claimdepends on the next claim size.

Symbol Interpretation for queueing model

a/b /1 Service times depend on the previous inter-arrival times

a/b /1 Service times depend on the next inter-arrival times

a /b/1 Inter-arrival times depend on the next service timesa /b/1 Inter-arrival times depend on the previous service times

For the risk models a similar convention holds.

Symbol Interpretation for risk model

a/b Claim sizes depend on the previous inter-arrival times

a/b Claim sizes depend on the next inter-arrival times

a /b Inter-arrival times depend on the next claim sizesa /b Inter-arrival times depend on the previous claim sizes

Note that dependence does not refer to dependence in the probabilistic sense,but should be interpreted as a dependency of the distribution of one quantity on the

outcome of the other. For example in the a/b /1 queue, the service time distribution


of the nth customer is completely determined by the previous inter-arrival time, whilein the a /b/1 model the inter-arrival distribution between customers n and (n + 1)is determined by the service time of the (n+ 1)th customer.

In case of a dependency on a threshold, the type of threshold is denoted by a sub-script. Basically two types of threshold dependence appear in this thesis, namelydependence on a deterministic threshold (with symbol ad) and dependence on an

exponential threshold (with symbol aM). For example the symbol M/Gd/1 denotes

a queue, where the service time distribution of the nth customer depends on hisinter-arrival time in a way such that it is equal to some distribution F1 wheneverthe inter-arrival time is below a deterministic threshold, and is equal to a differentdistribution F2 otherwise.

We will also explore two different types of dualities between queueing models andrisk models in this thesis. Both dualities, named D1 and D2, map a specific queueingmodel to a specific risk model, thereby acting in a characteristic way on the presentdistributions and dependencies.

Figure 1.1: The duality D1 between a risk model (below) and a queueing model(above)

Duality one is defined as follows. Starting with a queueing model a/b/1 with inter-arrival times A1, A2, . . . (A1 denoting the time between the first and the secondcustomer) and service times B1, B2, . . ., a new risk model a/b is obtained by thefollowing principle: we fix some N {1, 2, 3, . . .} and let the inter-arrival times ofthe claims be AN , AN1, AN2, . . .. The claim sizes of the dual risk model are given by


BN , BN1, . . .. It turns out that this duality maps a/b/1 models to a/b models, butflips the dependencies, since the process (A1, B1), (A2, B2), . . . is mapped to its timereversed version (AN , BN), (AN1, BN1), . . .. To indicate these duality relations wewrite

a/b /1

D1 a/b , (1.1)

a/b /1

D1 a/b ,

a /b/1 D1 a /b,a /b/1 D1 a /b.

Figure 1.2: The duality D2 between a risk model (below) and a queueing model(above)

The second duality is defined in a simple way by letting the inter-arrival times andthe claim sizes of the dual risk model be given by the service and inter-arrival timesof the original queueing model, that is, the queueing model a/b/1 is mapped to therisk model b/a. Here the dependencies do not change their direction, so that we can


write

a /b/1 D2 b/a . (1.2)a /b/1 D2 b/a ,

a/b /1

D2b /a,

a/b /1

D2b /a,

The studied queueing model in this thesis mainly originates from the risk model

M/Gd, so we will actually apply only the dualities (1.1) and (1.2).

The thesis is organized as follows.

In the introductory chapter we describe the classical Cramer-Lundberg modelof risk theory and show Beekmans convolution formula. We also show howformulas for the ruin probabilities can be found.

Chapter 3 deals with several queueing and risk models related to the M/Gd

model.

In Section 3.1 we introduce the risk model M/Gd described by Kwan and

Yang (2007), which is of great importance for this thesis.

Section 3.2 deals with the special case M/Md.

In Section 3.3 we investigate another special case of the M/Gd model,

namely the case where the claim sizes are either zero, if the inter-occurrencetime is smaller than a threshold a, or exponential, if not.

In Section 3.4 the duality D1 is applied to the M/Md ruin model. It turns

out that the resulting M/Md/1 queuing model is equivalent to a standard

G/M/1 queue.

We apply duality D2 to the model M/Md in Section 3.5 and obtain a

Md/M/1 queuing model. This queue is in fact equivalent to a standardM/G/1 queue.

In Section 3.6 we generalize the duality argument of Section 3.4 to the

M/Gd ruin model and the dual model M/

Gd/1.

Chapter 4 is entirely devoted to an M/GM/1, numerical illustrations are given

and the effect of dependence is investigated.


Chapter 2

Ruin Theory

2.1 Introduction

In this chapter we introduce ruin theory. We consider the development in time ofthe capital U(t) of an insurer. When the capital becomes negative, we say that ruinoccurs. Let (u) denote the probability that this happens, provided that the annualpremium and the claims process remain unchanged; here u = U(0) is the initialcapital. Our interest throughout this chapter is to compute this probability for twotypes of claim distributions:

1. Exponential distributions and sums, mixtures and combinations.

2. Distribution with only a finite number of values.

In addition we wish to derive an exponential bound: (u) eRu, for some R > 0.

We define the surplus process or risk process as follows:

U(t) = u+ ct S(t), t 0, (2.1)

where

U(t)= the insurers capital at time t;

u = U(0) = the initial capital;

c = the (constant) premium income per unit of time;

S(t)=X1 +X2 + . . .+XN(t) , the aggregate claim amount

with

N(t)= the number of claims up to time t, assumed to follow a Poisson process with rate , and

Xi = the size of the ith claim, assumed to be non-negative.


Let T denote the point in time at which the ruin occurs. So,

T =

{min{t|t 0} if U(t) < 0 for some t,

if U(t) 0 for all t.The probability that T is finite is called the ruin probability. It is written as follows:

(u) = P [T

We assume that (k1)(u) eRu. By (2.4) we get:

k(u)

0

0

exp{R(u+ ct x)}dP (x)etdt

= eRu

0

exp{t(+Rc)}dt

0

eRxdP (x)

= eRu

+ cRmx(R)

= eRu,

(2.5)

where the last equality follows from (2.3). Now if we take the limit when k weget the desired inequality.

Theorem 2.1.2. The ruin probability for u 0 satisfies

(u) =eRu

E[eRU(T )|T u occurs if and only if a finite point in time t exists such thatU(t) 0. This means that L > u and T < are equivalent, as a consequence wehave (u) = 1FL(u) where FL(u) is the distribution function of L. Let the randomvariables Lj, j = 1, 2, . . . denote the amounts by which the j

th record low is less thanthe (j 1)th one then we have L = L1 + L2 + . . . + LM , where M is a randomnumber of new records. From the memoryless property of the Poisson process, theprobability that a particular record low is the last one is the same every time. Thisimplies that M follows a geometric distribution. The parameter of M is (1 (0)),that is the probability of non-ruin starting with a capital of u = 0. The amountsof improvements L1, L2, . . . are independent and identically distributed. So L has acompound geometric distribution.


Theorem 2.2.1 (Distribution of the capital at time of ruin). If the initial capital uequals 0, then for all y > 0 we have

P [U(T ) (y dy,y), T y|T

Recursive formula for ruin probability The ruin probability in u can be ex-pressed in term of the ruin probabilities at smaller initial capitals as follows:

(u) =

c

u0

(1 P (y))(u y)dy + c

u

(1 P (y))dy (2.11)

Proof. We have

(u) = P [T 0],

(2.12)

but

P [M > 0] = 1 P [M = 0] = 1 p = 11 +

(2.13)

and the density of FL1 is given by

fL1(y) =1 P (y)

1, y > 0, (2.14)

so

(u) =1

1 +

0

P [T

Example 2.3.1. Ruin probability for exponentially distributed claim size.Let the claims distribution be exponential(1) then

0

eru((u))du = 11 +

( 11r 1)

(1 + (1 + )r 11r )

=

(1 + )[ (1 + )r]

=

r,

where = 1+

= 1 . We see that except for the constant this is a mgf of anexponential distribution with parameter .

Remark 2.3.2. Kaas et al.([1] page 91) present a discrete-time model counterpartof the continuous-time model. In fact they consider a more general risk model, nowonly on the discrete time points 0, 1, 2, . . ., and let Un, n = 0, 1, 2, . . . denote thesurplus at time n. Letting Gn denote the profit between time point n 1 and n, wehave

Un = u+G1 +G2 + . . .+Gn, n = 0, 1, 2, . . .

The authors define a discrete-time version of the ruin probability (u) and the ruin

time T as follows: T = min{n : Un < 0}; (u) = P [T

Chapter 3

Ruin probability in an insurancerisk model with threshold

3.1 Introduction

In this chapter we study the ruin probability in an insurance risk model of type

(M/Gd) where the claim size is dependent on the previous inter-arrival claim time.

A similar model with dependence was first introduced by Albrecher and Boxma(2004). But in their model, the inter-arrival time depends on the previous claim sizethrough a threshold. It was possible to get the ruin probability as the inversion ofa Laplace transform, but it was not possible to obtain a closed form solution. Inthis chapter we consider a model developed by Isaac K. M. Kwan and Hailiang Yang(2007). We also study the case where all claims are rejected if the inter-arrival timebetween two successive claims is less than some deterministic threshold a.

Let Ti be the time between the (i 1)th claim and the ith claim. We assume thatthe number of claims follows a Poisson process with parameter , thus the Ti areexponentially distributed with parameter . As in the previous chapter we defineTu = inft0 {t|U(t) 0} to be the time of ruin. We also define (u) = 1 (u) asthe survival probability. In the following, we assume that if Ti a then Bi followsa distribution with cumulative distribution function (cdf) F (x) and pdf f(x) and ifTi > a, Bi follows a distribution with cdf G(x) and pdf g(x). We define also F (x)(resp. G(x)) to be the tail probability of F (x) (resp. of G(x)). We assume thefollowing net profit condition is satisfied:They should be a > 0 such that

ct = (1 + )E[St], > 0,

so by Walds Identity we have ct = (1 + )t((1 ea)F + eaG).


where is the safety load, F and G are respectively the mean of F and G . Let usstart with the computation of (u). By conditioning on the time of occurrence T1 ofthe first claim, we obtain

(u) = P [T =|u(0) = u]

=

a0

P [T =|u(0) = u, T1 = t]etdt+ a

P [T =|u(0) = u, T1 = t]etdt

=

a0

u+ct0

P [T =|X = x, u(0) = u, T1 = t]f(x)etdxdt+ a

u+ct0

P [T =|X = x, u(0) = u, T1 = t]g(x)etdxdt

=

a0

u+ct0

(u+ ct x)f(x)etdxdt+ a

u+ct0

(u+ ct x)g(x)etdxdt.

We substitute t = vuc

. Then

c(u) =

u+cau

v0

(v x)f(x)evuc dxdv +

u+ca

v0

(v x)g(x)evuc dxdv.

We differentiate the left and right hand side with respect to u:The first part gives

d

du

u+cau

v0

(v x)f(x)evuc dxdv

=

u+ca0

(u+ ca x)eaf(x)dx u

0

(u x)f(x)dx

+

c

u+ca0

v0

(v x)evuc f(x)dxdv

and the second part gives

d

du

u+ca

v0

(v x)g(x)evuc dxdv

= u+ca

0

(u+ ca x)eag(x)dx+ c

u+ca

v0

(v x)evuc g(x)dxdv


We finally obtain

cd(u)

du=

u+ca0

(u+ ca x)eaf(x)dx u

0

(u x)f(x)dx+

c

u+ca0

v0

(v x)evuc f(x)dxdv

u+ca

0

(u+ ca x)eag(x)dx+

c

u+ca

v0

(v x)evuc g(x)dxdv

= (

c

u+ca0

v0

(v x)evuc f(x)dxdv +

c

u+ca

v0

(v x)evuc g(x)dxdv

)+

ea u+ca

0

(u+ ca x)(f(x) g(x))dx u

0

(u x)f(x)dx.

We observe that the first part of this last equation is just (u). It follows that wehave proved: (see also [4])

Theorem 3.1.1 (Survival Probability). The survival probability satisfies the integro-differential equation

cd(u)

du(u) = ea

u+ca0

(u+ cax)(f(x) g(x))dx u

0

(ux)f(x)dx.

(3.1)

3.2 Exponential claim size distribution

Let F (x) = 1 e1x and G(x) = 1 e2x be the cumulative distributions of theclaim sizes, i.e. we consider the model M/

Md. Then the ruin probability satisfies

the third order differential equation [4]:

cd3(u)

du3+(c1+c2)

d2(u)

du2+2(c1)

d(u)

duea(12)

d(u+ ca)

du= 0.

(3.2)

Proof. By (3.1) we get

cd(u)

du(u) = ea

u+ca0

(u+cax)(1e1x2e2x)dx u

0

(ux)1e1xdx.

(3.3)


We differentiate both sides of (3.3). To simplify the differentiation we set v=u+ca-x;it follows that

cd2(u)

du2 d(u)

du= ea

[(u+ ca)(1 2) 21

u+ca0

(u+ ca x)e1xdx]

+ea22

u+ca0

(u+ ca x)e2xdx

1(u) + 11 u

0

(u x)e1xdx

= ea(u+ ca)(1 2) + ea22 u+ca

0

(u+ ca x)e2xdx

1[cd(u)

du (u) + ea2

u+ca0

(u+ ca x)e2xdx] 1(u)

= ea(u+ ca)(1 2) c1d(u)

du

+2ea(2 1)

u+ca0

(u+ ca x)e2xdx.

(3.4)

Rearranging the terms,

cd2(u)

du2+ (c1 )

d(u)

du ea(1 2)(u+ ca)

= ea2(2 1) u+ca

0

(u+ ca x)e2xdx.

Now we differentiate this last equation with respect to u. Let

t(u) = ea2(2 1) u+ca

0

(v)e2(u+cav)dv.

Then,dt(u)

du+ 2t(u) = e

a2(2 1)(u+ ca). (3.5)

Now we substitute

t(u) = cd2(u)

du2+ (c1 )

d(u)

du ea(1 2)(u+ ca)

in Equation (3.5) and obtain

cd3(u)

du3+((1 +2)c)

d2(u)

du2+2(1c)

d(u)

du(12)ea

d(u+ ca)

du= 0.

By using the relation (u) = 1 (u) we obtain (3.2) and this completes the proof.


3.3 The case where no claim is accepted before

some deterministic threshold of the inter-arrival

time

Let the claim size B equal zero if the inter-occurrence time between claims is lessthan some threshold a and let B have a pdf g(x) if the claim inter-occurrence timesis more than a. This means that F (x) = 1 for all x 0. By an integration of (3.1)from 0 to u and using the fact that (u) 1 as u to find (0), we obtain thefollowing integral equation:

c(u) = ea[

u+ca

G(x)dx+

u+ca0

(u+ ca x)G(x)dx]. (3.6)

Let G(x) = ex, then (3.6) becomes

c(u) = ea[

u+ca

exdx+

u+ca0

(u+ ca x)exdx]

(3.7)

=

eae(u+ca) + ea

u+ca0

(u+ ca x)exdx. (3.8)

Now we differentiate (3.8) with respect to u.We have

cd(u)

du=

eae(u+ca) + ea[

d

du

u+ca0

(v)e(u+cav)dv]

= eae(u+ca) + ea[(u+ ca)

u+ca0

(v)e(u+cav)dv]

= c(u) + ea(u+ ca).

Bringing all terms to the left side we obtain

cd(u)

du+ c(u) ea(u+ ca) = 0. (3.9)

This is in accordance with (3.2) because if we divide the left and the right side of(3.2) by 1 and let 1 tend to we obtain (3.9). By Taylors Theorem we have

(u+ ca) =k=0

(ca)k(k)(u)

k!.


Thus we have just proved that for this simple case the ruin probability satisfies thefollowing differential equation:

cd(u)

du+ c(u) ea

k=0

(ca)k(k)(u)

k!= 0. (3.10)

Suppose that equation (3.10) has as solution

(u) = AeRu. (3.11)

By substituting this solution into (3.10) we obtain the following characteristic equa-tion

Rc eaeRca + c = 0. (3.12)The non-linearity of this equation makes it difficult to solve. So even for this simplecase where the threshold is deterministic, the ruin probability is still difficult toobtain.

Is it possible to derive, as in the classical model when the claim size is exponentiallydistributed, an explicit expression for the ruin probability? This question can beanswered if we can find a negative solution for the above equation. In fact equation(3.12) has exactly one negative solution R, when a > 0, > 0, > 0, c > 0 and > ea/c.

Proof. We write = ea/c. Note that by the net profit condition > > 0. Toprove that (3.12) has exactly one negative solution, we first state Rouches Theoremwhich will be useful (See [5] page 652).

Theorem 3.3.1 (Rouches Theorem). If f(z) and g(z) are analytic inside and on aclosed contour D and |f(z)| < |g(z)| on D then g(z) and f(z) + g(z) have the samenumber of zeros inside D.

The functions g(z) = z+ and f(z) = ezca are analytic in the contour D delimitedby the left-half ball with center at the origin and with radius r because they can beexpressed as power series there. On the boundaries we have |f(z)| < |g(z)| . Indeedfor z = iy with y real, |f(z)| = |ecaiy| = < a] =P [a < T t]P [T > a]

=ea et

ea= 1 e(ta), t > a.

(3.18)Consequently the Laplace-Stieltjes transform of the inter-arrival time of customers


for which Ti a is given by

Xs(s) = E[es(

Ni=1X

is)] =

k=0

E[es(Ni=1X

is)|N = k]P [N = k]

=k=0

(E[esXs ])k(1 ea)kea

=k=0

( a0

exssexs

1 eadxs

)k(1 ea)kea

=(s+ )ea

s+ e(s+)a.

(3.19)

The Laplace-Stieltjes transform of Xl is given by

s+ esa. (3.20)

We know that the Laplace-Stieltjes transform of a convolution of two distributionsis equal to the product of the Laplace-Stieltjes transforms. So the Laplace-Stieltjestransform of the inter-arrival time of those customers which receive service is :

T (s) = Xs(s) Xl(s) =

+ se(s+)a. (3.21)

We know that (see [6] page 279) if is the service intensity and A the mean of theinter-arrival time of a G/M/1 queue, then in the steady state: (a) The distributionof the waiting time W is a mixture of an atom at 0 and an exponential distributionwith intensity on (0,) with weights 1 , resp. .

(b) The distribution of the workload V is a mixture of an atom at 0 and an exponentialdistribution with intensity on (0,) with weights 1 , resp. . Here =(A)

1 < 1 and = 1

where is the positive solution of the equation 1 =

0exA(dx), where A(x) is the cdf of the inter-arrival times. Thus

P [V > t] = et, (3.22)

andP [W > t] = et. (3.23)


In the above G/M/1 we have = ea

and =

. It follows that

P [W > u] =

eu

=+R

eRu

=

eae(a+u)R

= (u), (3.24)

as required. The line before the last follows from (3.12).

Now we verify that the previous result of the workload is in compliance with Formula(3.14).

By ([10] page 153 (1.4)), the ruin probability for the stationary case can be expressedin terms of the one for the zero-delayed case by integrating the following:

s(u) = G(u+ s) +

u+s0

(u+ s y)G(dy) w.r.t. Aeq, (3.25)

where s(u) is the ruin probability for the delayed case with T1 = s. Let now

Aeq(x) = x0 (1A(u))du

Abe the equilibrium distribution and eq be the Laplace-Stieltjes

transform of the equilibrium distribution, then

eq(s) =

0

esx(1 A(x))

Adx. (3.26)

Hence we obtain

eq(s) =1 (s)As

=

sease(+s)a

+ se(+s)a= (s)esa. (3.27)

This is the Laplace-Stieltjes transform of the difference X a, so that Aeq(x) =A(x + a), i.e. for this particular distribution, the equilibrium distribution is thedistribution of a shift by a.

We now derive a relation between the stationary ruin probability and the zero delayedone. We have

(0)(u) =

0

(G(u+ x) +

u+x0

(u+ x y)dG(y))dAeq(x)

=

0

(e(u+x) +

u+x0

(u+ x y)eydy)dAeq(x)

=

a

(e(ua+x) +

ua+x0

(u a+ x y)eydy)dA(x)

=

0

(e(ua+x) +

ua+x0

(u a+ x y)eydy)dA(x).

(3.28)


The last step is due to the fact that dA(x) = 0 for x < a (the inter-arrival time isalways larger than a). This can be recognized as (u a), so (0)(u) = (u a) =ea

eRu. By setting R = , we obtain the same result as for the workload. So we

conclude that this result is in compliance with Equation (3.14).

3.5 Duality D2

In this section we construct another type of duality between the risk model thatwas introduced in Section 3.3 and the standard M/G/1 queue. Here we reverse thedistribution of both models as follows: the claim size become the inter-arrival timeof customers in the M/G/1 and the service time is just the claim inter-occurrencetime of the risk model described in Section 3.3.

Let us consider an initially empty single server queueing system, where we have twotypes of customers: Exceptional customers and normal customers. Each customerbrings an exponential amount of work with parameter . Normal customers havea service time less than some threshold a and the exceptional customers have aservice time larger than the threshold. If the service time is less than a then theinter-arrival time is set to zero, otherwise the inter-arrival time is exponential withrate . Exceptional customers enter the system immediately upon arrival and beginthe service if the server is empty; otherwise they wait in the queue in front of theserver. In other words, customers are served in groups composed of n customerswith successive service times less than the threshold plus one customer with largeservice time (the exceptional customer). So the number of customers in a group isa geometric number and the service time is a geometric sum of exponential randomvariables with parameter .

Let Ai be a sequence of i.i.d. exponential random variables having rate , represent-ing the inter-arrival time between the ith and (i + 1)th customer, where Ai is setto zero if the service time of the customer i is less than the threshold; let Bi be asequence of i.i.d. exponential random variables for the service time with rate . Wehave seen above that this dependent M/M/1 queue is equivalent to a normal G/M/1queue. Because in this duality we take as inter-arrival time for a customer the size ofthe claims and for service time the inter-occurrence time between claims, the abovedescribed model is just an M/G/1.

Let be the inter-arrival intensity, B the mean service time of a M/G/1, and = B < 1 the occupation rate. Then in steady-state the distribution of theworkload V and the waiting time W are identical (By PASTA).


The Laplace-Stieltjes transform of the service time is given by see (3.21):

B(s) =

+ se(s+)a.(3.29)

For a M/G/1 the Laplace-Stieltjes transform for the waiting time is given by ([7]page 66)

W (s) =(1 )s

B(s) + s

=(1 )s

( +se(s+)a

) + s

= (1 ) + ea(1 )

ea(esa ) + s. (3.30)

Theorem 3.5.1. The stationary waiting time is zero with probability 1 and isequal to W+ with probability , where W+ is a random variable with density

(1 )e(x+a)g(x+ a), x 0. (3.31)

Here g is the density of a distribution G with Laplace-Stieltjes transform

G(s) =

+ se(+s)a

and is a solution of = ea().

For more properties of G see [8] pages 141-163.

Proof. According to formula (3.30) the Laplace-Stieltjes transform of the stationarywaiting time is given by

W (s) = (1 ) + e(s)a (1 )ea + (s )e(+(s))a

.

We multiply the numerator and the denominator of the fraction on the right handside by

ea and obtain

W (s) = (1 ) + (1 )esa

+ (s )ea(+s)

= (1 ) + (1 )esa

+ (s )ea(+s)

= (1 ) + esaG(s )G()

.

(3.32)


The line before the last one follows from the relation = ea() and the last

line from the fact that G() = (1 )1. So if G(x) is a distribution functionwith density g(x) and Laplace-Stieltjes transform G(s) then

G(s)G()

is the Laplace

transform of the density (1 )exg(x). It follows that esa G(s)G()

is the Laplace-

Stieltjes transform of the random variable with the density given in (3.31).

3.6 Sample path argument for duality D1

Let Wn be the waiting time of the nth customer in a G/G/1-type queue with apossible dependency between Un, the service time of the n customer, and An, theinter-arrival time between the nth customer and the (n + 1)th customer. ThenW1 = 0 and Wn fulfills Lindleys equation, i.e.

Wn+1 = (Wn + Zn)+,

where Zn = Un An (see Figure 3.2). Thus {Wn} evolves as a random walk withincrements Z1, Z2, . . . as long as the random walk only takes non-negative values,and it is reset to 0 once it hits 0.

Before we explain the relation between the G/G/1 queue and a dual risk process, westate the following well known result.

Lemma 3.6.1. (see [10] page 49) Let (Zi)i=1,2,... be a sequence of identically dis-tributed random variables (not necessarily independent) and let Sn = Z1 +Z2 + +Zn1 with S1 = 0. Define the Lindley process generated by Z1, Z2, . . . , by W1 = 0and Wn+1 = (Wn + Zn)

+. Then for every N 1

WN = SN minn=1,...,N

Sn. (3.33)

and WNd maxn=1,...,N Sn where

d denote equality in distribution.

Proof. We are going to show that QN = SN minn=1,...,N Sn fulfills Lindleys equa-tion. We have

QN+1 = SN+1 minn=1,...,N+1

Sn

=

{0 if SN+1 < minn=1,...,N Sn

SN+1 minn=1,...,N Sn else

=

{0 if SN+1 minn=1,...,N Sn < 0QN + ZN else

= (QN + ZN)+.


Since the distributions of (SNS1, SNS2, . . . , SNSN1, 0) and (SN , SN1, . . . , S1)coincide, we have that

WN = max{SN S1, SN S2, . . . , SN SN1, 0}d max{SN , SN1, . . . , S1}.

Now consider the following risk process. We fix some N and let Yk = ANk UNk,where Ak denotes the time between the (k 1)th and the kth claim and where Ukis the size of the kth claim. Then, assuming that R0 = u, the surplus after the kthclaim is Rk := u+ Y1 + . . .+ Yk and u = inf{n : Rn < 0} is equal to the number ofclaims that leads to ruin.

Theorem 3.6.2 (see [10] page 49). The two events {u N} and {WN > u}coincide.

Proof. Let S1 = 0 and Sn = Z1 +Z2 + +Zn1 = Y1Y2 Yn1 = uRn1.By the previous lemma we have

WN > u SN minn=1,...,N

Sn > u SN minn=1,...,N1

Sn > u

SN Sm > u, for some m {1, 2, . . . , N 1} Zm + Zm+1 + . . .+ ZN1 > u, for some m {1, 2, . . . , N 1} YNm YNm1 . . . Y1 > u, for some m {1, 2, . . . , N 1} u+ YNm + YNm1 + . . .+ Y1 < 0, for some m {1, 2, . . . , N 1} RNm < 0, for some m {1, 2, . . . , N 1} u < N.

If P [WN > u] tends to a limit P [W > u] as n tends to infinity (i.e. Wn W indistribution), then we have

P [u u]. (3.34)

In this case we can apply the results from Theorem 3.1.1. Recall that for the thresholdmodel with Poisson inter-arrival times and claim size distributions F and G, thesurvival probability fulfills (3.1). We then conclude from (3.34) that the same is truefor the limiting distribution of the waiting times for a modified M/G/1 queue. Thus,letting H(x) = P [W x], we have the equation

cd

duH(u) H(u) = ea

u+ca0

H(u+ ca x)(f(x) g(x))dx

u

0

H(u x)f(x)dx. (3.35)


See the figure below for the sample path giving the ruin time and the waiting time.

Figure 3.2: Queueing process

Figure 3.3: Associated risk process with N = 5. Here (u) = 2.

3.7 Conclusion

In this chapter, we studied the special case of the model developed by Kwan andYang (2007). where they found a closed formula for the ruin probability in the case ofexponential claim size distribution. In our case we assume that no claim is acceptedif the claim inter-arrival time is below a certain deterministic threshold. We havestudied duality between this risk model and two queueing models. We found thesame duality result as in the classical setting.


Chapter 4

A queueing model in which theservice time depends on theinter-arrival time

The motivation for this model comes from the work of Yang and Kwan (2007) inan insurance risk model. They assume that the claim size of an insurance businessdepends on the time between claims. By the duality principe between the stationaryruin probability and the workload process in the classical risk model, we investigatethe corresponding queueing model where the service time of the current customerdepends on the inter-arrival time between the previous customer and the current cus-

tomer. We consider a M/GM/1 queue where customers numbered n = 0, 1, 2, . . . ,

arrive according to a Poisson process. The inter-arrival times An, n = 1, 2 . . . , be-tween the nth customer and (n+ 1)th customer are assumed i.i.d. and exponentiallydistributed. We let Bn, n = 1, 2, . . . , be a sequence of i.i.d. random variables, Bnrepresenting the service time of the nth customer. We assume further that if theinter-arrival time Ai is less than a threshold Ti, then the service distribution is G(x)otherwise the service distribution is F (x). The Ti are assumed to be i.i.d. randomvariables (independent from the inter-arrival and service times) with exponentialdistribution function T having parameter . We are interested in the relevant per-formance measures of this queueing model; in particular, in the mean performancemeasures such as the mean sojourn time, the mean number of customers and themean waiting time. This model is applicable in supermarkets where customers whopick only a small number of items move directly to the server and receive service;in this case customers with short inter-arrival time receive a small amount of servicetime. We begin by investigating the sojourn time of a given customer by computinghis Laplace-Stieltjes transform.


4.1 Laplace-Stieltjes transform of the sojourn time

of a customer

We now turn to the computation of how long a customer spends in the system, whichis known as sojourn time. The sojourn time Sn+1 of the (n+ 1)th arriving customeris evidently the service time of the (n+ 1)th customer if and only if he arrives afterthe departure of the nth arriving customer. Hence

Sn+1 =

{Bn+1 if Sn An 0,

Sn An +Bn+1 if Sn An > 0.

For n the distribution function of Sn has a limit. Let S be this limit randomvariable, then we have S

d d=(S A)+ + B. In the sequel we call BF the randomvariable of the service time with distribution F and BG the random variable of theservice time with distribution G.

What is exactly the stability condition of this queue? We know that the system willbe stable if E[A] > E[B], that is

1

>

+ E[BG] +

+ E[BF ].

Theorem 4.1.1. The Laplace-Stieltjes transform of the sojourn time of this M/GM/1

type queue is given by:

E[eS] = C1(+ )(+ )E[eBF ] ( )(E[eBF ] E[eBG ])C2

(+ )[( E[eBF ] )(+ ) + ( )

(E[eBF ] E[eBG ]

)] ,(4.1)

where C1 = E[eS], C2 = E[e

(+)S].

Proof. By conditioning successively on the inter-arrival time, the threshold and theservice time of a customer we have

E[eS] = E[e((SA)++B)]

= E[eB1SA] + E[e(SA+B)1S>A].


We compute the two expectations of the right hand side separately.

E[eB1SA] = E[eB1SA1AT ] + E[e

B1SA1A>T ]

= E[eBG ]

0

E[1SA1Au]eudu

+E[eBF ]

0

E[1SA1A>u]eudu

= E[eBG ]

0

u0

E[1St]etdteudu

+E[eBF ]

0

u

E[1St]etdteudu. (4.2)

The double integrals in (4.2) can be reduced to a single integral by changing theorder of integration. We obtain the following:

E[eB1SA] = E[eBG ]

0

E[1St]e(+)tdt

+E[eBF ]

0

E[1St](et e(+)t)dt

= E[eBG ]E[

S

e(+)tdt]

+E[eBF ]E[

S

(et e(+)t)dt]

= E[eBF ]E[eS] +

+ E[e(+)S](E[eBG ] E[eBF ]).

(4.3)

We also have

E[e(SA+B)1S>A] = E[e(SA+B)1S>A1AT ] + E[e

(SA+B)1S>A1A>T ]


Now expression E[e(SA+B)1S>A1AT ] becomes

E[e(SA+B)1S>A1AT ] = E[eBG ]

0

E[e(SA)1S>A1Au]eudu

= E[eBG ]

0

u0

E[e(St)1S>t]etdteudu

= E[eBG ]

0

( t

eudu)E[e(St)1S>t]e

tdt

= E[eBG ]

0

e(+)E[e(St)1S>t]dt

= E[eBG ]E[

0

e(+)te(St)1S>tdt]

= E[eBG ]E[ S

0

e(+)teSdt]

= E[eBG ]

+ E[eS e(+)S

]=

+ E[eBG ]

(E[eS] E[e(+)S]

). (4.4)


In a similar way we compute the expression E[e(SA+B)1S>A1A>T ]

E[e(SA+B)1S>A1A>T ] = E[eBF ]

0

E[e(SA)1S>A1A>u]eudu

= E[eBF ]

0

u

E[e(St)1S>t]etdteudu

= E[eBF ]

0

( t0

eudu)E[e(St)1S>t]e

tdt

= E[eBF ]

0

(1 et)E[e(St)1S>t]etdt

= E[eBF ]

0

(et e(+)t)E[e(St)1S>t]dt

= E[eBF ]

0

etE[e(St)1S>t

]dt

E[eBF ]

0

e(+)tE[e(St)1S>t

]dt

= E[eBF ][E(

S0

e(S+()t)dt)

+ E(

S0

e(S+(+)t)dt)]

= E[eBF ]

E(eS(1 e()S)

)E[eBF ]

+ E(eS(1 e(+)S)

)=

E[eBF ]

(E[eS] E[eS]

) +

E[eBF ](E[eS] E[e(+)S]

). (4.5)

So we finally get

E[eS] = E[eBF ](E[eS] +

E[e(+)S]) +

+ E[eBG ]E[e(+)S]

+E[eBF ]

(E[eS] E[eS])

(+ )

E[eBF ](E[eS] E[e(+)S])

+E[eBG ]

+ (E[eS] E[e(+)S]).

We collect all terms with E[eS] and obtain


E[eS] =

E[e

BF ]E[eS] + (+)(+)(E[e

BF ] E[eBG ])E[e(+)S](1

+ (

E[eBF ] + E[eBG ]))

= C1(+ )(+ )E[eBF ] ( )(E[eBF ] E[eBG ])C2

(+ )[( E[eBF ] )(+ ) + ( )

(E[eBF ] E[eBG ]

)] .This completes the proof of Theorem 4.1.1.

This Laplace-Stieltjes transform has two unknown parameters, namely C1 and C2.Let h() = E[eS]. To determine C1 we use the fact that

lim01E[eBF ]

= E[BF ], lim0

1E[eBG ]

= E[BG] and take the limit as 0of h(); this leads to:

C1 = 1E[BF ] + E[BG]

+ = 1 ,

where E[BF ] and E[BG] are respectively the expectation of the random variable BFand BG and =

(+)

(E[BF ] + E[BG]).

We realize that

E[eS]=

0exfs(x)dx=

0P [A > x]fs(x)dx=P [A > S]. It follows that E[e

S]is the probability that the system is empty on arrival. By the stability conditionfound above, C1 > 0.

4.2 Laplace-Stieltjes transform of the waiting time

In the preceding section we found the Laplace-Stieltjes transform for the sojourntime, which is useful for this section. The sample path of the process allows us tofind another relation between the sojourn time and the waiting time; this is givenby:

Wd= (S A)+. (4.6)


By conditioning on the sojourn time we get:

E[eW ] = E[e(SA)+

]

= P [A > S] + E[e(SA)1S>A]

= E[eS] +

0

E[e(SA)1S>A|S = u]dS(u)

= C1 + E[

A

eAeu]dS(u)

= C1 +

0

t

eteudS(u)etdt

= C1 +

0

((e()u 1)eudS(u))

= C1 +

(C1 E[eS])

=1

(C1 E[eS]). (4.7)

Example 4.2.1. (M/M/1) Suppose that the inter-arrival time is exponentially dis-tributed with parameter and the service time is exponentially distributed with mean1

. Then C1=(1 ) = 1 .

Thus

E[eW ] =1

((1 ) (1 )

(1 ) + )

= (1 ) + (1 )(1 ) +

. (4.8)

By the inversion of the Laplace-Stieltjes transform we obtain, FW (t) = P (W t) =(1)+(1e(1t), t 0. Hence, with probability (1) the waiting time is zero(i.e., the system is empty on arrival) and, given that the waiting time is positive (i.e.the system is not empty on arrival), the waiting time is exponentially distributed withparameter (1 ).

The Laplace-Stieltjes transforms found in the previous sections allow us to computethe mean performance measures of the model.

4.3 Performance measures

We denote by G, (2)G resp. F ,

(2)F the first and second moment of BG and BF

respectively.


4.3.1 Mean sojourn time

Using the Taylor series expansion of E[eBG ] and E[eBF ] in Theorem 4.1.1 weobtain

h() =f()

g(),

where f() and g() are given by

f() = C1[(+ ) F (+ ) +1

2(+ )

(2)F

2 + F2 +O(3)]

+[(G F ) +1

2(

(2)F

(2)G )

2 (G F )2 +O(3)]C2,

(4.9)

and

g() = [(+ )(F 1)(+ )

2

(2)F (F 1) +

2

(2)F

2 +O(3)]

+(G F )2 +1

22(

(2)F

(2)G ) (G F )

1

2(

(2)F

(2)G )

2 +O(3).

We know that E[S] = h(0); it follows that

E[S] =(g(0)g(0)

f(0)

f(0)

)h(0)

=(+ )[(

(2)G +

(2)F ) 2(1 G)] + 2[C1(+ )(1 + F (+ )) + 2(G F )C2]

2C1(+ )2.

4.3.2 Mean number of customers in the system

Let Nd denote the number of customers which arrive during the sojourn time of acustomer, Na the number of customers in the system just before an arrival and Nthe number of customers in the system in the steady state.

By the well-known step argument we get Nad= Nd and by PASTA we also have

Nad= N .

The generating function of Nd is given by:

E[zNd ] =

0

nd=0

zndet(t)nd

(nd)!dS(t)

=

0

e(1z)tdS(t)

= E[e(1z)S]. (4.10)


This implies that

E[Nd] =d(E[zNd ])

dz

z=1

= E[S]. (4.11)

Thus we have E[Nd] = E[S] = E[N ] by Littles Law.

In the sequel we suppose that BGd exp(1/G) and BF

d exp(1/F ). Our Laplace-Stieltjes transform is now

E[eS] = C1(+ )(+ )(1 + G) ( )(G F )C2(+ )g()

=f()

g(), (4.12)

where g() is given by

g() = GF3 + (G + F 2GF GF )2

+(1 2G + GF (2 + ) F (G + F ))(+ F 2G).

Lemma 4.3.1. Under the stability condition, the denominator g() of the aboveLaplace-Stieltjes transform (4.12) has exactly one positive real zero and two negativereal zeros.

Proof. Since GF > 0 is the leading coefficient of g() it follows that g() as and g() as . Furthermore we have g(0) = C1 < 0; thisimplies that there is at least one positive zero of g().

Next we show that we have two negative zeros. We suppose first that G < F .Then

g( 1F

) = (F G) > 0.

On the other hand if G > F then

g( 1G

) = (G F )1 + GG

> 0.

If G = F then

g( 1F

) = g( 1F

) = 0.

So in any of these cases g has two negative zeros. It follows that g() has exactlyone positive zero since g() is a polynomial of degree 3.


Since h() is an analytic function in the right half-plane, the positive zero of thedenominator of h() is also a zero of the numerator of h(); this allows us to computethe constant C2. Let be the positive zero of g() then by the argument stated abovewe have

C1(+ )( + )E[eBF ] ( )(E[eBF ] E[eBG ])C2 = 0.

This implies that

C2 =C1( + )( + )E[eBF ]( )

(E[eBF ] E[eBG ]

)=

(+ (F + G))( + )E[eBF ]( )

(E[eBF ] E[eBG ]

) ,where the last equality follows by replacing C1 by its value found above.

Example 4.3.2. For the special case where BF = BGd exp() we obtain C1 = 1 ,

and E[eS] = (1%)(1%)+ which is the well-known formula for the Laplace-Stieltjes

transform for the sojourn time of the standard M/M/1 queue.

4.3.3 Numerical illustrations

For the numerical result we compare our model with the standard M/G/1 where theservice distribution is a mixture of two exponentials distributions. So the servicedistribution is

B =

+ BG +

+ BF . (4.13)

In the first example we take very small, so that the threshold becomes very large

in the mean. This means that BGd exp( 1

G) is more probable to occur and we

approximately have an M/M/1 queue.

Example 4.3.3. Let Td exp(0.01), A d exp(1), F (x) = 1 e100x, G(x) =

1 e 108 x.Then the constants are given by: C1 =

209910100

and C2 = 0.206211 and the expectedsojourn time is E[S] = 3.8647.

In Figure 4.1 below, the Laplace-Stieltje transform of the sojourn time of the ap-proximated M/M/1 queue and the one with a mixture of service time distributionsare shown.In the second example we take large and this means that BF

d exp( 1F

) is more

probable to occur and again we approximately have an M/M/1 queue.


0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

indep

dep

LST sojourn time

Figure 4.1: Laplace-Stieltjes transform of sojourn time.

Example 4.3.4. Let Td exp(10), A d exp(2), F (x) = 1 e 264100x, G(x) = 1 ex.

Then the constants are given by: C1 = 0.0353535 and C2 = 0.00715273 and theexpected sojourn time is E[S] = 19.0154.

Figure 4.2 below is the Laplace-Stieltje transform of the approximated M/M/1 queueand the one with a mixture of service time distribution.

Example 4.3.5. Let now Td exp(4), A d exp(2), F (x) = 1e10x, G(x) = 1ex.

Then the constants are given by: C1 =15

and C2 = 0.12058 and the expected sojourntime is E[S] = 4.56029.

Here below are the Laplace-Stieltje transforms of this M/M/1 type queue and theone with mixture of service time distributions.


0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

indep

dep

LST sojourn time


0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

indep

dep

LST sojourn time



In the following numerical illustration, values of the mean sojourn time are obtainedfor both the dependence and independence case. Their values are compared andthe following measures are shown in the tables below: ESdep= dependent meansojourn time, ESind= independent mean sojourn time, EWdep= dependent meanwaiting time, ENdep=dependent mean number of customers , ENind=independentmean number of customers, ABS(difference)=absolute difference; Reld=absolutedifference/ESdep. For the numerical illustration, we have chosen: = 6, 1 = 1,2 = 1/4 in the first table, = 4, 1 = 1, = 2 in the second table and = 2,1 = 1/4, 2 = 5/8 in the third table.

1.0 1.2 1.4 1.6 1.8 2.0 2.2

0.2

0.4

0.6

0.8

1.0

Figure 4.4: Difference between dependent sojourn time and independent sojourntime as function of .


Table 4.1: Effect of the dependence when , 1, 2 are fixed .

ESdep ESind EWdep ENdep ENind ABS ABS/ESdep1 0.704218 0.662698 0.347075 0.704218 0.662698 0.04152 0.0581.5 1.49867 1.3375 1.09867 2.24801 2.00625 0.16117 0.10751.75 2.55983 2.22239 2.14047 4.4797 3.88917 0.33744 0.13182 6.1422 5.1875 5.7047 12.2844 10.375 0.9547 0.15542.01 6.47013 5.45847 6.03193 13.005 10.9715 1.01166 0.15632.02 6.83222 5.75761 6.39331 13.8011 11.6304 1.07461 0.15722.04 7.68264 6.46009 7.24234 15.6726 13.1786 1.22255 0.15912.15 22.1397 18.3941 21.6919 47.6004 39.5474 3.7456 0.16912.17 32.8646 27.2452 71.3161 59.122 32.4153 5.6194 0.17092.19 62.9018 52.0332 137.755 113.953 62.4512 10.8686 0.17272.2 114.808 94.8679 114.357 252.578 208.709 19.9401 0.17362.21 635.515 524.567 635.064 1404.49 1159.29 110.948 0.17452.211 1159.22 956.74 1158.77 2563.04 2115.35 202.48 0.1746

Table 4.2: Effect of the dependence when , , 1 are fixed.

2 ESdep ESind EWdep ENdep ENind ABS ABS/ESdep1/10 4.56029 3.8 4.16029 9.12058 7.6 0.70629 0.16671/9 4.9246 4.0963 4.5172 9.84921 8.19259 0.8283 0.16811/8 5.47185 4.54167 5.05519 10.9437 9.08333 0.93018 0.16991/7 6.38526 5.28571 5.95669 12.7705 10.5714 1.09955 0.17221/6 8.21465 6.77778 7.77021 16.4293 13.5556 1.43687 0.17491/5 13.7098 11.2667 13.2431 27.4196 22.5333 2.4431 0.17820.21 17.146 14.0758 16.6726 34.2919 28.1517 3.0702 0.17900.22 22.8738 18.76 22.3938 45.7477 37.52 4.1138 0.17980.23 34.3309 28.1317 33.8443 68.6619 56.2633 6.1992 0.18050.24 68.7048 56.2533 68.2114 137.41 112.507 12.4515 0.18120.245 137.454 112.502 136.958 274.908 225.003 24.952 0.18150.248 343.704 281.251 343.205 687.408 562.501 62.453 0.18170.249 687.454 5562.5 686.954 1374.91 1125 124.954 0.18170.2499 6874.95 5625 6874.45 13749.9 11250 1249.95 0.1818


Table 4.3: Effect of the dependence when , 2, 1 are fixed.

ESdep ESind EWdep ENdep ENind ABS ABS/ESdep1 1.544 1.75 1.1697 3.089 3.5 0.206 0.13342 3.450 4.625 3.0127 6.900 8.125 0.6125 0.17753 9.1757 10.85 8.7007 18.3515 21.7 1.6743 0.18243.5 20.664 24.3636 20.1754 41.328 48.7273 3.6996 0.179003.6 26.4118 31.1161 25.9207 528235 62.2321 4.7043 0.17813.7 35.9929 42.3684 35.4995 71.9858 84.7368 6.3755 0.17713.8 55.1574 64.8707 54.6617 110.315 129.741 9.7133 0.17613.9 112.655 132.373 112.157 225.311 264.746 19.718 0.17503.97 380.987 447.374 380.488 761.974 894.749 66.387 0.17423.98 572.654 672.375 572.154 1145.31 1344.75 99.721 0.17413.99 1147.65 1347.37 1147.15 2295.31 2694.75 199.72 0.17403.999 11497.7 13497.4 11497.2 22995.3 26994.7 1999.7 0.17393.9999 114998 134997 114997 229995 269995 19999 0.1739


4.4 Conclusion

The computation of the Laplace-Stieltjes transform of this M/GM/1 allows us to find

the mean performance measures of this queueing model for the case where the servicetime distribution is exponential. It would be nice to extend this to other servicetime distributions such as the Erlang, Phase-type or Hyper-exponential distribution.The result of these mean performance measures was compared with a mixture ofexponential service time distributions. In heavy traffic both mean sojourn timesdiffer up to a constant multiplicative factor.


Chapter 5

Conclusions and Recommendations

5.1 Conclusions

Risk and Queueing models are two classes of storage models which are closely con-nected. In this master thesis a dependent risk model was considered and the queue-ing model counterpart was investigated. According to the classical Cramer-Lundbergrisk model, a queueing model with the FIFO discipline has a straightforward interpre-tation in a risk model: The zero delayed ruin probability and steady-state waitingtime distribution coincide and the stationary ruin probability is also equal to thesteady-state distribution of the workload process. We saw in chapter 3 that for aspecial case of dependence this result remains valid. In chapter 4 the Laplace-Stieltjestransform of the sojourn time distribution for the M/G/1 queue with dependencewas found. According to the numerical results it seems that compared to a naturalM/G/1 queue, in heavy traffic their mean queue lengths (and mean sojourn times)differ by a constant multiplicative factor.

5.2 Recommendations

Queueing Systems with dependence remains a large area of research. In our analysis,we concentrated on dependence of claim sizes or service times on inter-arrival times.It would be interesting to study the case where the distribution of the inter-arrivaltime between two customers depends on the next service time in general and to com-pare the result with the work of H. Albrecher and O. Boxma (2004) who derive exactsolutions for the probability of ruin by means of Laplace-Stieltjes transforms. Also

the study of the queueing model M/Gd remains open. In Chapter 4, we were not able


to prove the existence of a positive root of the denominator of the Laplace-Stieltjestransform for the sojourn time when the service time has a general distribution. Werecommend to investigate the proof by means of Rouches theorem and to investigatethe sojourn time distribution of this dependence queue. It would be also interestingto study the heavy traffic behavior of this queueing with dependent structure.


Bibliography

[1] R. Kaas, M. Goovaerts, J. Dhaene, M. Denuit. 2008. Modern Actuarial RiskTheory using R. 2nd Edition, Springer.

[2] H. Albrecher, O.J. Boxma. 2004. A ruin model with dependence between claimsizes and claim intervals. Insurance: Mathematics and Economics, 35, 245-254.

[3] H. Albrecher, O.J. Boxma 2005. On the discounted penalty function in aMarkov-dependent risk model. Insurance: Mathematics and Economics, 37, 650-672.

[4] I.K.M. Kwan, H. Yang. 2007. Ruin probability in a threshold insurance riskmodel. Belgian Actuarial Bulletin, Vol 7, No.1.

[5] J.W. Cohen. 1982. The Single Server Queue (North-Holland series in appliedmathematics and mechanics; v.8). North-Holland Publishing Company, Ams-terdam.

[6] S. Asmussen. 2003. Applied Probability And Queues. Second edition, Springer.

[7] I. Adan and J. Resing. 2002. Queueing Theory. Lecture notes. Departement ofMathematics and Computer Science, Eindhoven University of Technology.

[8] J.S.H. van Leeuwaarden, A.H. Lopker and A.J.E. Janssen. 2010. Connecting Re-newal Age Processes with M/D/ Queues through Stick Breaking, StochasticModels, page 26.

[9] F. Dufresne, H.U. Gerber. 1989. Three methods to calculate the probability ofruin. ASTIN Bulletin, 19(1), 71-90.

[10] Soren Asmussen, Hansjorg Albrecher. 2010. Ruin Probabilities, Vol 14. SecondEdition, World Scientific.


AcknowledgementsAbstractContents1. Introduction2. Ruin theory3. Ruin probability in an insurance risk model with threshold4. A queueing model in which the service time depends on the inter-arrival time5. Conclusions and recommendationsBibliography

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Master’s Thesis: INSURANCE RISK AND QUEUEING …alexandria.tue.nl/extra1/afstversl/wsk-i/siani2011.pdf · INSURANCE RISK AND QUEUEING MODELS WITH THRESHOLD DEPENDENCE By: Jean-Claude