Multi-channel SMO with pending and limited queue. Single channel SMO with waiting. Single channel system with unlimited queue

In practice, one-channel CMOs with a queue are quite common (a doctor serving patients; a pay phone with one booth; a computer that fulfills user orders). In theory queuing single-channel QS with a queue also occupy a special place (it is to such QS that most of the analytical formulas for non-Markov systems obtained so far belong). Therefore, we will pay special attention to a single-channel QS with a queue.

Let there be a single-channel QS with a queue that is not subject to any restrictions (neither in the length of the queue, nor in the waiting time). This QS receives a flow of applications with intensity X; the service flow has an intensity inverse to the average service time of a claim.It is required to find the final probabilities of the states of the QS, as well as the characteristics of its efficiency:

Average number of requests in the system,

Average time spent by a request in the system,

Average number of applications in the queue,

Average time spent by an application in the queue,

The probability that the channel is busy (the degree of channel loading).

As for the absolute throughput A and relative Q, there is no need to calculate them: due to the fact that the queue is unlimited, each request will sooner or later be served, therefore, for the same reason

Decision. The states of the system, as before, will be numbered according to the number of requests in the QS:

The channel is free

The channel is busy (serving a request), there is no queue,

The channel is busy, one application is in the queue,

The channel is busy, applications are in the queue,

Theoretically, the number of states is unlimited (infinite). The state graph has the form shown in Fig. 20.2. This is a scheme of death and reproduction, but with an infinite number of states. Along all arrows, the flow of claims with intensity A transfers the system from left to right, and from right to left - the flow of services with intensity

First of all, let us ask ourselves, are there any final probabilities in this case? After all, the number of states of the system is infinite, and, in principle, the queue can grow indefinitely! Yes, it is so: the final probabilities for such a QS do not always exist, but only when the system is not overloaded. It can be proved that if strictly less than one, then the final probabilities exist, and at, the queue at grows indefinitely. This fact seems to be especially "incomprehensible" when It would seem that the system does not have unrealizable requirements: during the service of one request, on average, one request arrives, and everything should be all right, but in reality it is not.

With the QS, it copes with the flow of requests only if this flow is regular, and the service time is also not random, equal to the interval between requests. In this “ideal” case, there will be no queue in the QS at all, the channel will be continuously busy and will regularly issue serviced requests. But as soon as the flow of requests or the flow of services become at least a little random, the queue will already grow indefinitely. In practice, this does not happen just because "an infinite number of applications in the queue" is an abstraction. These are the gross errors that can result from replacing random variables with their mathematical expectations!

But let us return to our single-channel CMO with no limited queue... Strictly speaking, the formulas for the final probabilities in the scheme of death and reproduction were derived by us only for the case of a finite number of states, but let us take liberty - we will use them for an infinite number of states. Let's calculate the final probabilities of states by formulas (19.8), (19.7). In our case, the number of terms in formula (19.8) will be infinite. Let's get an expression for

The series in formula (20.11) is a geometric progression. We know that the at series converges - it is an infinitely decreasing geometric progression with a denominator. For, the series diverges (which is an indirect, albeit not rigorous, proof that the final probabilities of states exist only for). Now we assume that this condition is satisfied, and Summing the progression in (20.11), we have

(20.12)

The probabilities are found by the formulas:

whence, taking into account (20.12), we finally find:

As you can see, the probabilities form a geometric progression with the denominator. Oddly enough, the maximum of them is the probability that the channel will be free at all. No matter how loaded the system with the queue, if only it can cope with the flow of requests at all, the most probable number of requests in the system will be 0.

Let's find the average number of applications in the QS. Here you have to tinker a little. Random variable Z - the number of applications in the system - has possible values \u200b\u200bwith probabilities

Its mathematical expectation is

(20.14)

(the sum is taken not from 0 to but from 1 to since the zero term is equal to zero).

Substitute into formula (20.14) the expression for

Now we take out the sign of the sum:

Here again we will apply a "little trick": there is nothing else but a derivative of pore from the expression means,

Reversing the operations of differentiation and summation, we get:

But the sum in formula (20.15) is nothing but the sum of an infinitely decreasing geometric progression with the first term and denominator; this sum is equal to and its derivative. Substituting this expression in (20.15), we get:

(20.16)

Well, now let's apply Little's formula (19.12) and take the average residence time of an application in the system:

Find the average number of customers in the queue Let us reason as follows: the number of customers in the queue is equal to the number of customers in the system minus the number of customers under service. This means (according to the rule of addition of mathematical expectations), the average number of requests in the queue is equal to the average number of requests in the system minus the average number of requests under service. The number of claims under service can be either zero (if the channel is free) or one (if it is busy). The mathematical expectation of such a random variable is equal to the probability that the channel is busy (we designated it). Obviously equal to one minus the probability that the channel is free;

Therefore, the average number of claims under service is

the operation or efficiency of the queuing system are as follows.

For CMO with rejections:

For CMO with unlimited waiting both the absolute and the relative throughput lose their meaning, since each arriving claim will sooner or later be served. For such a CMO important indicators are:

For Mixed type both groups of indicators are used: as relative and absolute bandwidthand the characteristics of the expectation.

Depending on the purpose of the queuing operation, any of the above indicators (or a set of indicators) can be selected as an efficiency criterion.

Analytical model A QS is a set of equations or formulas that allow one to determine the probabilities of the states of the system in the course of its operation and to calculate performance indicators based on the known characteristics of the incoming flow and service channels.

There is no general analytical model for an arbitrary QS... Analytical models are developed for a limited number of special cases of QS. Analytical models that more or less accurately represent real systems are usually complex and difficult to visualize.

Analytical modeling of the QS is greatly facilitated if the processes occurring in the QS are Markov (the flows of applications are the simplest, the service times are distributed exponentially). In this case, all processes in the QS can be described by ordinary differential equations, and in the limiting case, for stationary states - by linear algebraic equations and, having solved them, determine the selected performance indicators.

Let's consider examples of some QS.

2.5.1. Multichannel QS with failures

Example 2.5... Three traffic inspectors check the waybills of truck drivers. If at least one inspector is free, the passing truck is stopped. If all the inspectors are busy, the truck drives by without stopping. The flow of trucks is the simplest, the check time is random with an exponential distribution.

This situation can be simulated by a three-channel QS with failures (out of turn). The system is open-loop, with homogeneous applications, single-phase, with absolutely reliable channels.

Description of states:

All inspectors are free;

One inspector is busy;

Two inspectors are occupied;

Three inspectors are busy.

The system state graph is shown in Fig. 2.11.

Figure: 2.11.

On the graph: - the intensity of the flow of trucks; - the intensity of document checks by one car inspector.

Simulations are carried out to identify the part of vehicles that will not be tested.

Decision

The sought-after part of the probability is the probability that all three inspectors will be employed. Since the state graph represents a typical scheme of "death and reproduction", we will find it using dependencies (2.2).

The throughput of this post of traffic inspectors can be characterized by relative throughput:

Example 2.6... A group of three officers was assigned to receive and process reports from the reconnaissance group in the intelligence department of the association. The expected intensity of the flow of reports is 15 reports per hour. Average processing time of one report by one officer -. Each officer can receive reports from any reconnaissance group. The released officer processes the last of the reports received. Incoming reports must be processed with a probability of at least 95%.

Determine if the assigned team of three officers is sufficient to complete the assigned task.

Decision

The group of officers works as a refusal system consisting of three channels.

Stream of reports with intensity can be considered the simplest, since it is a total of several reconnaissance groups. Service intensity ... The distribution law is unknown, but this is unimportant, since it has been shown that for systems with failures, it can be arbitrary.

The description of the states and the state graph of the QS will be similar to those given in example 2.5.

Since the state graph is a "death and reproduction" scheme, there are ready-made expressions for the limiting state probabilities for it:

The attitude is called reduced intensity of the flow of applications... Its physical meaning is as follows: the value is the average number of claims arriving at the QS during the average service time of one claim.

In the example .

In the considered QS, a failure occurs when all three channels are busy, that is. Then:

As probability of failure in the processing of reports is more than 34% (), it is necessary to increase the personnel of the group. Let's double the composition of the group, that is, the CMO will now have six channels, and we will calculate:

Thus, only a group of six officers will be able to process incoming reports with a 95% probability.

2.5.2. Multichannel CMO with waiting

Example 2.7... There are 15 crossing means of the same type on the river crossing section. The flow of equipment arriving at the crossing is on average 1 unit / min, the average time for crossing one unit of equipment is 10 minutes (taking into account the return of the crossing means).

Assess the main characteristics of the crossing, including the probability of an immediate crossing immediately upon the arrival of a vehicle.

Decision

Absolute throughput , that is, everything that approaches the ferry is almost immediately ferried.

Average number of crossing facilities in operation:

Ferry utilization and downtime rates:

A program was also developed to solve the example. The time intervals for the arrival of equipment at the crossing, the time of the crossing are taken to be distributed according to the exponential law.

The utilization rates of the ferry after 50 runs are practically the same: .

The maximum length of the queue is 15 units, the average time spent in the queue is about 10 minutes.

Consider a single-channel waiting queuing system.

We will assume that the incoming flow of requests for service is the simplest flow with intensity λ.

The service flow rate is μ. Service duration is a random variable subject to an exponential distribution law. Service flow is the simplest Poisson flow of events.

A claim arriving at a time when the channel is busy enters the queue and awaits service. We will assume that the queue size is limited and cannot accommodate more thanm applications, i.e. application that found at the time of its arrival in the CMOm +1 orders (m waiting in line and one in service) leaves the QS.

The system of equations describing the process in this system has a solution:

(0‑1)

The denominator of the first expression is a geometric progression with the first term 1 and the denominator ρ, from which we obtain

For ρ \u003d 1 you can resort to direct calculation

(0‑8)

Average number of tickets in the system.

Since the average number of applications in the system

(0‑9)

where is the average number of customers being serviced, then knowing it remains to find... Because there is only one channel, then the number of serviced claims can be either 0 or 1 with probabilitiesP 0 and P 1 \u003d 1- P 0 respectively, whence

(0‑10)

and the average number of applications in the system is

(0‑11)

Average waiting time for a request in the queue.

(0‑12)

that is, the average waiting time for a claim in the queue is equal to the average number of claims in the queue divided by the intensity of the stream of claims.

Average time spent by a request in the system.

The time that a request remains in the system is the sum of the waiting time for the request in the queue and the service time. If the system load is 100%, then \u003d 1 / μ, otherwise \u003dq / μ. From here

(0‑13)

The content of the work.

Preparing the experiment toolkit .

It is carried out in the same way in accordance with the general rules.

Calculation on an analytical model.

1. IN microsoft application Excel prepare a table of the following form.

2. In the columns for the parameters of the CMO table, write down the initial data, which are determined according to the rule:

m \u003d 1,2,3

(maximum queue length).

For each valuem it is necessary to find the theoretical and experimental values \u200b\u200bof the QS indicators for the following pairs of values:

= <порядковый номер в списке группы>

3. Enter the appropriate formulas in the columns with the indicators of the analytical model.

Simulation model experiment.

1. Set the start mode with exponentially distributed service time by setting the value of the corresponding parameter to 1.

2. For every combinationm , and run the model.

3. Enter the launch results into the table.

4. Enter the formulas for calculating the average value of the indicator in the corresponding columns of the tableP open, q and A.

Analysis of results .

1. Analyze the results obtained by theoretical and experimental methods by comparing the results with each other.

2. For m \u003d 3, plot dependencies on one diagramP open based on theoretically and experimentally obtained data.

Optimization of CMO parameters .

Solve the problem of optimizing the size of the number of seats in the queuem for a device with an average service time \u003d from the point of view of obtaining maximum profit. Take as problem conditions:

- income from servicing one request equal to 80 USD / hour,

- the cost of maintaining one device is equal to $ 1 / hour.

1. For calculations, it is advisable to create a table:

The first column is filled with the values \u200b\u200bof natural numbers (1,2,3 ...).

All cells of the second and third columns are filled with values \u200b\u200band.

Formulas for the columns of the table of section 0 are transferred to the cells of columns from the fourth to the ninth.

Enter the values \u200b\u200bin the columns with the initial data of the Income, Expense, Profit sections (see above).

In the columns with the calculated values \u200b\u200bof the Income, Expense, and Profit sections, write down the calculation formulas:

- number of requests per unit of time

N r \u003d A

- total income per unit of time

I S \u003d I r * N r

- total consumption per unit of time

E S \u003d E s + E q * (n-1)

- profit per unit of time

P \u003d I S - E S

where

I r - income from one application,

E s - consumption for the operation of one device,

E q - the cost of operating one place in the queue.

Graphs for P open,

- table with data to find the bestm and value m opt,

- graph of the dependence of profit per unit of time fromm.

Control questions :

1) Give short description single-channel QS model with limited queue.

2) What indicators characterize the functioning of a single-channel QS with failures?

3) How the probability p is calculated0 ?

4) How probabilities p are calculatedi?

5) How to find the probability of a request refusal?

6) How do I find the relative bandwidth?

7) What is the absolute bandwidth?

8) How is the average number of applications calculated in the system?

9) Give examples of QS with a limited queue.

Tasks .

1) The port has one cargo berth for unloading ships. The flow rate is 0.5 calls per day. The average time for unloading one vessel is 2 days. If there are 3 vessels in the queue for unloading, then the arriving vessel is sent for unloading to another berth. Find indicators of the performance of the berth.

2) The railway station information desk receives telephone inquiries with an intensity of 80 applications per hour. The call center operator answers the incoming call on average 0.7 minutes. If the operator is busy, the client receives the message "Wait for an answer", the request is placed in a queue, the length of which does not exceed 4 requests. Give an assessment of the work of the help desk and the option of its reorganization

Theme. Queuing systems theory.

Each QS consists of a number of service units, which are calledservice channels (these are machines, transport carts, robots, communication lines, cashiers, salespeople, etc.). Any CMO is designed to serve someflow of applications (requirements) arriving at some random moments in time.

Classification of QS by the way of processing the input stream of applications.

Queuing systems

With rejections

(no queue)

With a queue

Unlimited queue

Limited queue

With priority

In order of arrival

Relative priority

Absolute priority

By service time

By the length of the queue

Classification by way of functioning:

open, i.e. the flow of applications does not depend on the internal state of the QS;

closed, i.e. the input flow depends on the state of the QS (one repair worker serves all the channels as they fail).

Multichannel CMO with waiting

System with limited queue length. Consider channel queuing system with waiting, which receives the flow of requests with the intensity ; service rate (for one channel) ; number of places in line

The states of the system are numbered according to the number of requests, linked by the system:

no queue:

- all channels are free;

- one channel is busy, the rest are free;

- busy -channels, the rest are not;

- everyone is busy - no free channels;

there is a queue:

- all n-channels are busy; one application is in the queue;

- all n-channels, r-claims in the queue are busy;

- all n-channels, r-claims in the queue are busy.

The GSP is shown in Fig. 9. Each arrow has the corresponding intensity of the streams of events. Along the arrows from left to right, the system is always transferred by the same flow of requests with intensity , along the arrows from right to left, the system transfers the service flow, the intensity of which is times the number of busy channels.

Figure: 9. Multichannel CMO with waiting

Probability of failure.

(29)

The relative throughput complements the failure probability to one:

Absolute throughput of the CMO:

(30)

Average number of busy channels.

The average number of requests in the queue can be calculated directly as the mathematical expectation of a discrete random variable:

(31)

where .

Here again (the expression in brackets) the derivative of the sum of a geometric progression occurs (see above (23), (24) - (26)), using the relation for it, we obtain:

Average number of applications in the system:

Average waiting time for a request in the queue.

(32)

Just as in the case of a single-channel queuing system with waiting, we note that this expression differs from the expression for the average queue length only by the factor , i.e.

.

Average residence time of a request in the system, the same as for a single-channel QS .

Systems with unlimited queue length. We considered channel QS with waiting, when no more than m-claims can be in the queue simultaneously.

As before, when analyzing systems without restrictions, it is necessary to consider the obtained relations for .

Probability of failure

We obtain the average number of requests in the queue for from (31):

,

and the average waiting time is from (32): .

Average number of applications .

Example 2. A petrol station with two dispensers (n \u003d 2) serves a traffic flow with an intensity \u003d 0.8 (cars per minute). Average service time for one machine:

There is no other gas station in this area, so the queue of cars in front of the gas station can grow almost indefinitely. Find the characteristics of the CMO.

QS with limited waiting time. Previously, we considered systems with waiting limited only by the queue length (the number of m-customers simultaneously in the queue). In such a queue, a request that has grown into a queue does not leave it until it waits for service. In practice, there are QSs of a different type, in which an application, after waiting for a while, can leave the queue (the so-called "impatient" applications).

Consider a QS of this type, assuming that the waiting time limit is a random variable.

Poisson "exit flow" with intensity:

If this flow is Poisson, then the process in the QS will be Markov. Let us find the probabilities of states for it. The numbering of system states is associated with the number of requests in the system, both serviced and queued:

no queue:

- all channels are free;

- one channel is busy;

- two channels are occupied;

- all n-channels are busy;

there is a queue:

- all n-channels are busy, one request is in the queue;

- all n-channels are busy, r-claims are in the queue, etc.

The graph of system states and transitions is shown in Fig. ten.

Figure: 10. CMO with limited waiting time

We mark this graph as before; all arrows leading from left to right will have the intensity of the flow of requests ... For out-of-order states, the arrows leading from them from right to left will, as before, have the total flow rate for servicing all busy channels. As for the states with a queue, the arrows leading from them from right to left will have the total intensity of the service flow of all n-channels plus the corresponding intensity of the flow of leaving the queue. If there are r-customers in the queue, then the total intensity of the exit flow will be equal to .

Average number of applications in the queue: (35)

Each of these applications is subject to a “flow of exits” with an intensity ... Hence, from the average - applications in the queue will, on average, leave without waiting for service, - applications per unit of time and only per unit of time on average will be served - applications. The relative throughput of the CMO will be:

Average busy channels we still get by dividing the absolute throughput A by Closed CMO

Until now, we have considered systems in which the incoming stream is not connected in any way with the outgoing one. Such systems are called open-loop. In some cases, after a delay, serviced requests are returned to the entrance. Such QS are called closed. The polyclinic serving this area, the team of workers assigned to the group of machines are examples of closed systems.

The same finite number of potential requirements circulates in a closed QS. Until a potential requirement has been realized as a service requirement, it is considered to be in a delay block. At the moment of implementation, it enters the system itself. For example, workers are serving a group of machines. Each machine is a potential requirement, becoming a reality at the moment of its breakdown. While the machine is working, it is in the delay block, and from the moment of breakdown until the end of the repair - in the system itself. Every worker is a service channel. = \u003d P 1 + 2 P 2 + ... + (n- 1 ) P n- 1 + n ( 1 -P The input of a three-channel QS with refusals receives a flow of applications with an intensity \u003d 4 requests per minute, service time for one channelt obsl\u003d 1 / μ \u003d 0.5 min. Is it profitable from the point of view of the QS throughput to force all three channels to serve requests at once, and the average service time is reduced by three times? How will this affect the average residence time of an application in the CMO?

Example 2 . / μ \u003d 2, ρ /n =2/3<1.

Objective 3:

Two workers serve a group of four machines. Stops of a working machine occur on average after 30 minutes. The average setup time is 15 minutes. Runtime and setup time are exponentially distributed.

Find the average share of free time for each worker and the average operating time of the machine.

Find the same specifications for a system in which:

a) two machines are assigned to each worker;

b) two workers always serve the machine together, and with double intensity;

c) the only faulty machine is served by both workers at once (with double intensity), and when at least one more faulty machine appears, they begin to work separately, each serving one machine (first, describe the system in terms of death and birth processes).

In commercial activities, CMOs with waiting (queue) are more common.

Consider a simple one-channel QS with a limited queue, in which the number of seats in the queue m is a fixed value. Consequently, a claim arriving at the moment when all the places in the queue are occupied is not accepted for service, does not enter the queue, and leaves the system.

The graph of this QS is shown in Fig. 3.4 and coincides with the graph in Fig. 2.1 describing the process of "birth - death", with the difference that in the presence of only one channel.

The labeled graph of the "birth - death" process of a service, all intensities of service flows are equal

The states of the QS can be represented as follows:

S0 - the service channel is free,

S, - the service channel is busy, but there is no queue,

S2 - the service channel is busy, there is one customer in the queue,

S3 - the service channel is busy, there are two requests in the queue,

Sm + 1 - the service channel is busy, all m places in the queue are occupied, any next customer is rejected.

To describe a random QS process, you can use the rules and formulas set forth earlier. Let's write expressions that determine the limiting probabilities of states:

In this case, the expression for p0 can be written more simply, using the fact that the denominator is a geometric progression with respect to p, then after the appropriate transformations we get:

c \u003d (1- s)

This formula is valid for all p different from 1, but if p \u003d 1, then p0 \u003d 1 / (m + 2), and all other probabilities are also equal to 1 / (m + 2).

If we assume m \u003d 0, then we move from considering a single-channel QS with waiting to the already considered single-channel QS with denial of service.

Indeed, the expression for the limiting probability p0 in the case m \u003d 0 has the form:

po \u003d m / (l + m)

And in the case of l \u003d m has the value p0 \u003d 1/2.

Let us determine the main characteristics of a single-channel queuing system with waiting: the relative and absolute throughput, the probability of failure, as well as the average queue length and the average waiting time for an application in the queue.

A request is rejected if it arrives at a time moment when the QS is already in the state Sm + 1 and, therefore, all the places in the queue yes are occupied and one channel serves

Therefore, the probability of failure is determined by the probability of occurrence

States Sm + 1:

Potk \u003d pm + 1 \u003d сm + 1 * p0

The relative throughput, or the fraction of serviced claims arriving per unit of time, is determined by the expression

Q \u003d 1- potk \u003d 1- сm + 1 * p0

the absolute bandwidth is:

The average number of requests L standing in the queue for service is determined by the mathematical expectation of a random variable k - the number of requests in the queue

the random variable k takes the following only integer values:

• 1 - there is one application in the queue,
• 2 - there are two applications in the queue,

t-in the queue all seats are occupied

The probabilities of these values \u200b\u200bare determined by the corresponding probabilities of the states, starting from the state S2. The distribution law of a discrete random variable k is depicted as follows:

Table 1. The law of distribution of a discrete random variable

The mathematical expectation of this random variable is:

Loch \u003d 1 * p2 + 2 * p3 + ... + m * pm + 1

In the general case, for p ≥ 1, this sum can be transformed using geometric progression models to a more convenient form:

Lp \u003d p2 * 1- pm * (m-m * p + 1)* p0

In the special case for p \u003d 1, when all the probabilities pk are equal, one can use the expression for the sum of the terms of the numerical series

1 + 2 + 3 + m \u003d m (m + 1)

Then we get the formula

L "och \u003d m (m + 1)* p0 \u003d m (m + 1)(p \u003d 1).

Applying similar reasoning and transformations, we can show that the average waiting time for servicing a claim on the queue is determined by the Little formulas

Pt \u003d Lp / A (at p? 1) and T1p \u003d L "pt / A (at p \u003d 1).

Such a result, when it turns out that Pt ~ 1 / l, may seem strange: with an increase in the intensity of the flow of applications, it is as if the queue length should increase and the average waiting time decreases. However, it should be borne in mind that, firstly, the value of Loc is a function of l and m and, secondly, the QS under consideration has a limited queue length of no more than m requests.

An application received by the QS at the time when all channels are busy receives a refusal, and, therefore, the time of its “waiting” in the QS is equal to zero. This leads in the general case (at p ≥ 1) to a decrease in Tochrostom l, since the share of such requests increases with increasing l.

If we abandon the limitation on the length of the queue, i.e. tend m -\u003e\u003e?, then the cases p< 1 и р?1 начинают существенно различаться. Записанные выше формулы для вероятностей состояний преобразуются в случае р < 1 к виду

When k is large enough, the probability pk tends to zero. Therefore, the relative throughput will be Q \u003d 1, and the absolute throughput will be equal to A - l Q - l, therefore, all incoming requests are served, and the average queue length will be equal to:

Lp \u003d p21-p

and the average waiting time according to Little's formula

Point \u003d Loch / A

In the limit p<< 1 получаем Точ = с / м т.е. среднее время ожидания быстро уменьшается с увеличением интенсивности потока обслуживания. В противном случае при р? 1 оказывается, что в СМО отсутствует установившийся режим. Обслуживание не успевает за потоком заявок, и очередь неограниченно растет со временем (при t > ?). Therefore, the limiting probabilities of states cannot be determined: for Q \u003d 1, they are equal to zero. In fact, the CMO does not fulfill its functions, since it is not able to service all incoming applications.

It is easy to determine that the share of serviced applications and the absolute throughput, respectively, are on average s and m, however, an unlimited increase in the queue, and, consequently, the waiting time in it, leads to the fact that after a while, applications begin to accumulate in the queue for an unlimited time.

As one of the characteristics of the QS, the average time Tcmo of the request stay in the QS is used, which includes the average time spent in the queue and the average service time. This value is calculated using Little's formulas: if the queue length is limited, the average number of applications in the queue is:

Lсmo \u003d m + 1;2

Tsmo \u003d Lсmo;for p? 1

And then the average time spent by a claim in the queuing system (both in the queue and under service) is:

Tsmo \u003d m + 1at p? 1 2m

 

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