Journal of Applied Mathematics and Stochastic Analysis

VolumeΒ 2009, Article IDΒ 308025, 14 pages

http://dx.doi.org/10.1155/2009/308025

## Interloss Time in Loss System

Occupational Medicine Department (DML), National Institute for Occupational Safety and Prevention (ISPESL), Via Alessandria 220/E, 00198 Rome, Italy

Received 29 March 2009; Accepted 24 May 2009

Academic Editor: HoΒ Lee

Copyright Β© 2009 Pierpaolo Ferrante. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We consider the interloss times in the Erlang Loss System. Here we present the explicit form of the probability density function of the time spent between two consecutive losses in the model. This density function solves a Cauchy problem for the second-order differential equations, which was used to evaluate the corresponding laplace transform. Finally the connection between the Erlang's loss rate and the evaluated probability density function is showed.

#### 1. Introduction

In this paper, we treat the random variable representing the time spent between the th and the th lost unit or th interloss time, in the loss system.

The model is characterized by the Markov property of entering and exiting processes, by one service channel and by the system capacity to accommodate one customer at a time (for an overview see Medhi [1, page. 77]).

Our work has been inspired by the location problem of emergency vehicles (ambulances). Each vehicle can be regarded as an (or ) system, because its clients cannot wait in the queue.

In Emergency Medical Systems (EMSs), the nearest ambulance to the accident place is called βdistrict unitβ, and it assures the best performance to the system. If an emergency call arrives at EMS while its βdistrict unitβ is busy, the nearest ambulance among those available is dispatched (see Larson [2] or Larson [3]). The length of interloss times affects the performance of the system and provides an informative support on efficiency of EMS.

For the exponential loss model it was conducted, in Ferrante [4], a detailed description of the process of losses where represent the random number of losses in the time interval . Let be the conditional probability to lose clients in with customers in the system at time : the main results found in Ferrante [4] are the explicit values of the conditional probabilities of no losses in : and the iterative procedure to determine the distribution of the total number of losses in All is obtained by solving the inhomogeneous differential equations with and the initial conditions depending on the () customers in the system at . Furthermore, the generating probability functions of were evaluated and their explicit values are the following:

The aim of this work is to identify the type of the process of interloss times for the loss model and to find the differential equation which governs it, in order to determine the probability density functions with and the related properties.

Our results show unexpected connections among very different branches of probability such as random motion on hyperbolic space and queueing systems. In effect, the probabilities appearing below have a structure quite similar to the Hyperbolic distances of moving particles envisaged in Cammarota and Orsingher [5].

In Section 2, we establish that is a renewal process for the loss system and that the density functions (1.7) solve the second-order linear homogeneous differential equations

Let be the number of customers in the system at the moment , and let be the moment of the th loss with . The initial conditions for (1.8) depend on , and the renewal process has the following property:

for .

In Section 2, we also present the derivation of (1.8) and its solution conditionally by .

Let be the conditional density function of the 1th interloss time with (= 0,1) customers in the system at time :

For the model, the explicit values obtained for (1.10) are the following:

In Section 3, we compute the laplace transforms of (1.10): for , using (1.8).

The explicit values obtained for (1.12) are the following:

Finally, let be the conditional means of the 1th interloss time
with ,1; it has been checked that their values are
where *r* is the Erlang loss rate, and is the interarrival mean time.

#### 2. First Interloss Time

In the model, let be the number of customers in the system at the moment , let be the th interarrival time, let be the moment when the th client enters the system, let be the service time of the th served customer, let be the th interloss time. Furthermore, let be the arrival order of th loss happened in , starting from the th loss, and let be the moment when the th loss happen, with and .

If we consider that the system is busy at time , the event *βThe* 1th *interloss time is **β* is represented by Figure 1.

The random variable can be expressed as follows: where represents the arrival order of the 1th loss happened in , starting from zero.

Now, let be the conditional probability that the arrival order of 1th loss happened in with is equal to and it can be expressed as follows and can be computed conditionally by the moments when the served customers have arrived at the system where

Lemma 2.1. *The functions
**
with do not depend on t but on the time interval :
*

*Proof. * We proceed by showing that (2.7) is true for :

Then, we suppose that it is true for , and we obtain that

where

Finally, by the Markov property of the exponential distribution, the (2.7) appears

The conditional density function
can be evaluated as mean of convolution of exponential probability density functions, and thus we have that
At first, we state the following result concerning the evaluation of the integrals . Lemma 2.2. *The functions
**
satisfy the difference-differential equations
**
where , .**Proof. *We first note that
and therefore

In view of Lemma 2.2 we can prove also the following.

Theorem 2.3. * The function satisfies the second-order linear homogeneous differential equation
**
with the initial conditions
**
The explicit value of is
*

*Proof. *From (2.13), it follows that
and thus, in view of (2.17) and by letting , we have that
While the first condition is straightforward to verify, the second one needs some explanations: if we write
and observe that
by substituting (2.24) in (2.23) the second condition emerges.

The general solution to (2.22) has the form

By imposing the initial conditions (2.19) to (2.18) we obtain (2.20).

*Remark 2.4. *By (2.7) derive that the functions do not depend on *t*, but on the time interval , in fact if , we have that
for

Furthermore, by the Markov properties of the system, the random variables are independent, and is a renewal process with

*Remark 2.5. * If we get , because the 1th interloss time is greater than , , when nobody enters.

If , we have that because without exits and with the system busy at , the 1th interloss time has the same distribution of the interarrival time.

*Remark 2.6. *The probability density function can be expressed by the following hyperbolic functions:

Now, if we assume that the system is free at the starting point, the event *βThe* 1th *interloss time is **β* is represented by Figure 2.

Let be the conditional probability that the *n*th entered customer is the 1th lost at time *t*, when ,
it can be computed conditionally by the moments when the served customers have arrived at the system
where
The function can be computed as mean of convolution of exponential probability density functions. So we have that

Lemma 2.7. *The functions
**
satisfy the difference-differential equations
**
where .**Proof. *See proof of Lemma 2.2.

In view of Lemma 2.7 we can prove also the following.

Theorem 2.8. *The function satisfies the second-order linear homogeneous differential equation
**
with the initial conditions
**
The explicit value of is
*

*Proof. *By substituting in (2.21) and (2.22) with , (2.35) emerges. While the first condition is straightforward to verify, the second one needs some explanations: if we write
and observe that
by substituting (2.39) in (2.38), the second condition emerges.

By imposing the initial conditions (2.36) to (2.35), we obtain (2.37).

*Remark 2.9. *If , we get because the 1th interloss time is greater than , , when nobody enters.

If , we have that because without exits and with the system free at , the 1th interloss time is equal to the sum of 1th and 2th interarrival times.

*Remark 2.10. *The function can be expressed by the following hyperbolic function:

*Remark 2.11. *The function can also be computed as the convolution of the exponential density with rate and :

In fact, if we observe that by (2.7), (2.32), and (2.42) we obtain that

#### 3. Interloss Mean Time

In this section we compute the laplace transforms of the density functions (2.20) and (2.37), and we evaluate the averages of the 1th interloss time, conditionally by .

Theorem 3.1. *The laplace transform of **
satisfies the equation
**
and its explicit value is
*

*Proof. *By (2.18), (2.19), and the property
the results (3.2) and (3.3) emerge.

*Remark 3.2. *The 1th conditional interloss mean time
can be found by evaluating the derivative of (3.3) with respect to in , as follows:
In the model, the interloss mean time (3.5) is equal to the inverse of the Erlang loss rate.

Theorem 3.3. *The laplace transform of **
satisfies the equation
**
and its explicit value is
*

*Proof. *By (2.35), (2.36), and (3.4), the results (3.8) and (3.9) emerge.

*Remark 3.4. *The conditional interloss mean time
can be found by evaluating the derivative of (3.9) with respect to in , as follows:

*Remark 3.5. *By (2.41), the result (3.9) can be obtained using (3.3), as follows:

Now, let be the time spent between and the th loss, the conditional density functions with , can be evaluated by convolutions as follows: if , we have that while if , we have that By the independence between interloss times, the laplace transforms of (3.14) and (3.15) can be expressed as power of (3.3) as follows:

(i)if , (ii) if ,*Remark 3.6. *The conditional averages
with can be permuted by evaluating the derivatives of (3.16) and (3.17) with respect to in , thus obtaining(i)if , we have that
(ii)if , we have that

#### Acknowledgments

The author wants to acknowledge his PhD tutor Professor Enzo Orsingher for the theoretical support and Professor Francesca Guerriero and Dr. Patrizia Scano for their contributions.

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