The dynamics of financial flows shows that at any time a company can be responsible for its obligations. Markov SP, with discrete state

3. The dynamics of financial flows shows that at any time the Company can be responsible for its obligations.

4. Project results (the discount factor in the calculations is taken at the level of 8% per year):

results from project implementation (Figure 6.4.);

accumulated results from project implementation (Figure 6.5.);

From the last presented schedule, it can be seen that the start date for the return of funds is 2001 (the second year from the start of the project) and the payback period is 7 years (taking into account discounting - 9 years).

The accumulated discounted profit is $ 1,466,000.

7 RISK STRUCTURE AND PREVENTION MEASURES 7.1 Main risk factors

The main factors that give rise to the main risks of the project and create a real threat to the existence of the company are:

transition from state financing to joint financing of an object with commercial structures (change in the status and organization of work);

high rates of planned growth of services (setting up a fundamentally new business);

the market is occupied by other, currently more powerful competing organizations, extraordinary efforts are needed to conquer a market niche in six months or a year.

7.2 Structure and analysis of risks and measures to minimize them 7.2.1 Political risks

Associated with the instability of economic, tax, banking, land and other legislation in the Russian Federation, lack of support or opposition from the government, etc.

Risk mitigation measures:

development of internal tax policy;

formation of the business environment (partners, consortia, financial and industrial groups);

active participation of founders in interaction with power structures;

giving the institution a medical status.

7.2.2 Legal risks

They are associated with imperfect legislation, unclear documents, unclear legal measures in case of disagreement between the founders (for example, in a foreign court, etc.), delaying the deadline by the Contractor.

Risk mitigation measures:

clear and unambiguous wording of the relevant articles in the documents;

involvement of specialists with practical experience in this area for the preparation of documents;

allocation of the necessary financial resources to pay for highly qualified lawyers and translators.

7.2.3 Technical risks

They are associated with the complexity of the work and the lack of a technical project at the moment.

Possible incomplete use of equipment and delay in the commissioning of technical systems.

Risk mitigation measures:

accelerated development (or obtaining guarantees from suppliers) of the technical coordination of equipment and technical complexes;

conclusion of contracts on a turnkey basis with sanctions for discrepancies and missed deadlines;

insurance of technical risks.

7.2.4 Production risks

They are primarily associated with the possibility of delays in the commissioning of new technical means and insufficiently high quality of the services provided.

The potential for producing quality services in the future is high.

A significant risk may be the lack of highly qualified personnel (for the provision of hotel services).

Risk mitigation measures:

clear scheduling and project management;

accelerated development of the design concept, including quality criteria;

development and use of a well-thought-out service quality control system at all stages of its creation;

justification and allocation of sufficient financial resources for the purchase of high-quality equipment;

training of qualified personnel (including abroad).

7.2.5 Internal socio-psychological risk

When establishing this type of business, the following social and psychological risks may arise:

social tension in the team;

shortage, turnover of professional staff;

the presence of a destructive position.

Risk mitigation measures:

selection of professional staff (including testing), if necessary - training;

development of a mechanism to stimulate employees, including participation in the results of the Company's work;

a system of end-to-end multilevel awareness of the team and managers;

development of an effective approach to the formation and distribution of the wage fund.

7.2.6 Marketing Risks

They are associated with possible delays in entering the market, wrong (without taking into account market needs) choice of services, wrong choice of marketing strategy, mistakes in pricing policy, etc.

Delays in entering the market can be caused both by the production and technical reasons discussed above, and by the company's unwillingness to effectively realize and promote its technical, production, artistic and other potential to the market, which requires a marketing program that meets international standards and the service that implements it.

Since at the moment there is no full-scale marketing program, the assessment of the degree of solving marketing problems is low. At the same time, for a firm that aims to win a share of the market from competing firms, marketing tasks should be of the first priority.

Analysis of competitors shows that the competition will be tough, competitors have a number of advantages. In this regard, it is necessary to carefully realize your main advantages and focus on them the main efforts and resources.

Risk mitigation measures:

building a strong marketing service;

development of a marketing strategy;

development and implementation of product (assortment) policy and subordination to it of the activities of all divisions (for example, through the development and use of results-based management technology);

development and implementation of a marketing program;

carrying out a full range of marketing research, etc.

7.2.7 Financial risks

They are primarily associated with the provision of income, which depends primarily on advertising, as well as with the attraction of investments.

The working version of the financial plan (Appendix 1) assumes that the main financial receipts are provided through the use of numbers. Reducing the price or room occupancy of the hotel complex leads to serious difficulties in the implementation of the project.

Risk mitigation measures:

urgent research of the requirements of service consumers;

development and use of a well-thought-out quality control system for services at all stages of their creation;

justification and allocation of sufficient financial resources for the creation and purchase of high-quality equipment;

use of the approach to diversify sources of income, primarily through the bundle of "office numbers";

entering the stock market.

Another major financial risk factor is the need to receive large investments on time.

The presence of investments is a prerequisite for the start of the project: as long as they are delayed, the start of the project will be delayed.

Thus, investment is the toughest and most vital factor.

Risk mitigation measures:

variety of proposed project financing schemes;

development of an investment and financial strategy, the purpose of which is to enter the zone of profitable functioning;

carrying out a set of measures to find investment and credit resources.

Next steps for developers and project owners:

in-depth problem diagnostics of the project;

taking a set of measures to find investment and credit resources;

organization of collective work of top and middle-level management with consultants to develop a strategy and a specific program of events, primarily related to marketing, advertising and diversification and providing:

establishment of JSC;

high economic efficiency of the project;

minimization of risk;

formation and organizational design of teams for the implementation of the developed activities;

search for strategic foreign partners with experience in creating such institutions and capable of providing technical and investment support.

# FILE: Buisnes-Plan.INF
# THEME: Business Plan "CREATION OF A HOTEL COMPLEX"
#SECTION: Management
#PURPOSE: Business Plan
# FORMAT: WinWord
#

Table 3.2.

Qualitative characteristics of hotels in Moscow

№

Hotel name

Hotel address

Books

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Jobs and Processes

Every program running on Linux is called process... Linux as a multitasking system is characterized by the fact that many processes belonging to one or more users can run simultaneously. You can display a list of currently running processes with the command ps , for example, as follows:

/ home / larry # ps PID TT STAT TIME COMMAND 24 3 S 0:03 (bash) 161 3 R 0:00 ps / home / larry #

Note that by default the command ps lists only those processes that are owned by the user who started it. To see all the processes running in the system, you need to issue the command ps -a . Process numbers (process ID, or PID) listed in the first column are unique numbers that the system assigns to each running process. The last column, entitled COMMAND, indicates the name of the command that is running. In this case, the list contains processes that were started by the user larry. There are many other processes running in the system, their full list can be viewed with the command ps -aux ... However, among the commands run by user larry, there is only bash (the shell for user larry) and the command itself ps ... You can see that the bash shell is running concurrently with the command ps ... When the user entered the command ps , bash started executing it. After the command ps has finished its work (the process table is displayed), control is returned to the bash process. The bash shell then displays a prompt and waits for a new command.

The running process is also called task (job). Process and task are used interchangeably. However, the process is usually called a task when they mean job control (job control). Job control is a shell function that provides the user with the ability to switch between multiple jobs.

In most cases, users start only one task - this will be the command they entered last in the command shell. However, many shells (including bash and tcsh) have the functions job management (job control) that allow you to run multiple commands at the same time, or assignments (jobs) and switch between them as needed.

Job management can be useful if, for example, you are editing a large text file and want to temporarily interrupt the editing to do some other operation. With the job management features, you can temporarily leave the editor, return to the shell prompt, and perform other actions. When they are done, you can go back to working with the editor and find it in the same state in which it was left. Job management functions have many more useful uses.

Foreground and background

Assignments can either be on foreground (foreground), or background (background). There can be only one task in the foreground at any given time. The task in the foreground is the task with which you interact; it receives keyboard input and sends output to the screen (unless, of course, you have redirected input or output elsewhere). In contrast, background jobs do not receive input from the terminal; as a rule, such tasks do not need user interaction.

Some tasks take a very long time to complete, and nothing interesting happens during their work. Examples of such tasks are compiling programs, as well as compressing large files. There is no reason to look at the screen and wait for these tasks to complete. These tasks should be run in the background. During this time, you can work with other programs.

To control the execution of processes in Linux, a transfer mechanism is provided signals... Signal is the ability of processes to exchange standard short messages directly through the system. The signal message does not contain any information, except for the signal number (for convenience, instead of the number, you can use the name predefined by the system). In order to send a signal, a process just needs to use a system call kill () , and in order to receive a signal, nothing is needed. If a process needs to react in a special way to a signal, it can register handler, and if there is no handler, the system will respond for it. As a rule, this leads to the immediate termination of the process that received the signal. Signal handler starts asynchronously , immediately after receiving the signal, whatever the process is doing at that time.

Two signals - number 9 ( KILL ) and 19 ( STOP ) - always processed by the system. The first one is needed in order to kill the process for sure (hence the name). Signal STOP suspends process: in this state, the process is not removed from the process table, but it is not executed until it receives signal 18 ( CONT ) - after which it will continue to work. In Linux command shell, the signal STOP can be passed to an active process using an escape sequence Ctrl -Z .

Signal number 15 ( TERM ) serves to interrupt the job. When interruption (interrupt) the job, the process dies. Job interruption is usually done with an escape sequence Ctrl -C ... There is no way to recover an interrupted job. You should also be aware that some programs intercept the signal. TERM (using a handler), so the keystroke Ctrl -C (o) may not terminate the process immediately. This is so that the program can destroy traces of its work before it is terminated. In practice, some programs cannot be interrupted in this way at all.

Background mode and job destruction

Let's start with a simple example. Consider the yes command, which at first glance may seem useless. This command sends an endless stream of lines of y to standard output. Let's see how this command works:

/ home / larry # yes y y y y y

The sequence of such lines will continue indefinitely. You can kill this process by sending it an interrupt signal, that is, by pressing Ctrl -C ... Let us act differently now. To prevent this endless sequence from being displayed on the screen, redirect the standard output of the yes command to / dev / null. As you probably know, the / dev / null device acts like a black hole: all data sent to that device is lost. With the help of this device it is very convenient to get rid of too abundant output of some programs.

/ home / larry # yes\u003e / dev / null

Now nothing is displayed on the screen. However, the shell prompt is not returned either. This is because the yes command is still running and is sending its y messages to / dev / null. You can also destroy this job by sending it an interrupt signal.

Now, let's say you want the yes command to continue working, but the shell prompt should return to the screen so you can work with other programs. To do this, you can put the yes command in the background, and it will work there without communicating with you.

One way to put a process in the background is by appending the & character to the end of the command. Example:

/ home / larry # yes\u003e / dev / null & + 164 / home / larry #

The message is job number (job number) for process yes. The command shell assigns a job number to each running job. Since yes is the only executable job, it is assigned the number 1. The number 164 is the process identification number (PID), and this number is also given to the process by the system. As we will see later, the process can be referenced by specifying both of these numbers.

So now we have a yes command process running in the background and continuously sending a stream of y's to / dev / null. In order to find out the status of this process, you need to execute the command jobs which is an internal shell command.

/ home / larry # jobs + Running yes\u003e / dev / null & / home / larry #

We can see that this program really works. In order to find out the status of the task, you can also use the command ps as shown above.

In order to transmit a signal to the process (most often there is a need interrupt job work) the utility is used kill ... This command is given either the job number or the PID as an argument. An optional parameter is the number of the signal to send to the process. By default, a signal is sent TERM ... In the above case, the job number was 1, so the command kill% 1 will interrupt the job. When a job is accessed by its number (not PID), then this number must be preceded by a percent character ("%") on the command line.

Now let's enter the command jobs again to check the result of the previous step:

/ home / larry # jobs Terminated yes\u003e / dev / null

In fact, the job is destroyed, and when you enter the jobs command the next time, there will be no information about it on the screen.

You can also kill a job using the process identification number (PID). This number, along with the job ID, is indicated when the job starts. In our example, the PID value was 164, so the command kill 164 would be equivalent to the command kill% 1 ... When using PID as an argument to the kill command, you do not need to enter the "%" character.

Suspending and continuing jobs

First, start the process with the yes command in the foreground, as we did before:

/ home / larry # yes\u003e / dev / null

As before, because the process is in the foreground, no shell prompt is returned to the screen.

Now instead of interrupting the task with the keyboard shortcut Ctrl -C , the task can be suspend (suspend, literally - to suspend) by sending it a signal STOP ... To pause a task, press the appropriate key combination, usually Ctrl -Z .

/ home / larry # yes\u003e / dev / null Ctrl -Z + Stopped yes\u003e / dev / null / home / larry #

A suspended process is simply not executed. It does not consume the computing resources of the processor. A paused job can be started from the same point as if it had not been paused.

To resume the job in the foreground, you can use the command fg (from the word foreground - foreground).

/ home / larry # fg yes\u003e / dev / null

The command shell will once again display the name of the command, so the user will know which task he is currently running in the foreground. Suspend this task again by pressing the keys Ctrl -Z , but this time let's run it into the background with the command bg (from the word background - background). This will cause the process to behave as if it was started with a command with an & at the end (as you did in the previous section):

/ home / larry # bg + yes $\u003e $ / dev / null & / home / larry #

This returns the shell prompt. Team now jobs should show that the process yes really works at the moment; this process can be killed with the command kill as it was done before.

To suspend a task running in the background, you cannot use the keyboard shortcut Ctrl -Z ... Before pausing a task, it must be brought to the front with the command fg and only then pause. Thus, the command fg can be applied to either suspended jobs or a job running in the background.

There is a big difference between jobs in the background and suspended jobs. The paused task does not work - it does not consume the processing power of the processor. This task does nothing. A suspended task takes up a certain amount of the computer's RAM, after a while the kernel will pump this part of the memory to the hard disk " poste restante". On the contrary, a task in the background runs, uses memory, and performs some actions that you may need, but you can work with other programs at this time.

Jobs running in the background might try to display some text on the screen. This will interfere with working on other tasks.

/ home / larry # yes &

Here, the standard output has not been redirected to / dev / , so an endless stream of y characters will be printed to the screen. This thread cannot be stopped because the keyboard shortcut Ctrl -C does not affect tasks in the background. In order to stop this issue, use the command fg , which will bring the task to the foreground, and then destroy the task with the keyboard shortcut Ctrl -C .

Let's make one more remark. Usually the command fg and the team bg affect the jobs that were last suspended (these jobs will be marked with a + next to the job number if you enter the command jobs ). If one or more jobs are running at the same time, jobs can be brought to the foreground or in the background by specifying fg or commands bg their identification number (job ID). For example, the command fg% 2 brings job number 2 to the front and the command bg% 3 puts job number 3 in the background. Use PIDs as Command Arguments fg and bg it is impossible.

Moreover, to bring the task to the fore, you can simply specify its number. So the command %2 will be equivalent to the command fg% 2 .

It is important to remember that the job control function belongs to the shell. Commands fg , bg and jobs are internal shell commands. If, for some reason, you are using a command shell that does not support job control functions, then you will not find these (and similar) commands in it.

5.1. Random processes and their classification

A random process (SP) is a process or phenomenon whose behavior over time and the result cannot be predicted in advance. Examples of random processes: the dynamics of changes in the exchange rate or stocks, the revenue or profit of the organization over time, the volume of sales of goods, etc.
If a random process can change its state only at a strictly defined moment in time, then it is called a process with discrete time.
If a change in state is possible at an arbitrary moment in time, then this is an SP with continuous time.
If at any moment of time the SP is a discrete random variable (its value can be enumerated and two adjacent values \u200b\u200bcan be selected), then this is a process with a discrete state.
If at any moment of time the state can change continuously, smoothly and it is impossible to distinguish two neighboring states, then this is an SP with a continuous state.
Thus, 4 types of joint ventures are possible:
1) SP with continuous time and continuous state (example: the air temperature at some point in time, changes smoothly at any time).
2) SP with continuous time and discrete state (example: the number of visitors in a store, changes in multiples of one at any time).
3) JV with discrete time and continuous state (example: the dynamics of the exchange rate, changes smoothly at the time of currency trading).
4) SP with discrete time and discrete state (example: the number of passengers in transport changes in multiples of one and only at certain points in time, at stops).
Consider some system S, in which at a given time t about the joint venture is running. This process is called Markov process if for any moment of time t> t about, the behavior of the system in the future depends only on the state of the system at a given time at t= t about, and does not depend in any way on how, when and in what states it was in the past at t< t about . In other words, the “past” of the Markov process does not affect the “future” in any way (only through the “present”).

5.2. Streams of events.

The simplest type of SP is event streams. A stream of events is a sequence of events of the same type that occur at random times (for example, phone calls, shop visitors, cars passing through an intersection, etc.). They belong to the SP with a discrete state and continuous time. Mathematically, the flow of events can be depicted as random points on the time axis.

If the events in a stream occur one by one, and not in groups of several events, then such a stream is called ordinary. A stream of events is called a stream without consequences if for any non-overlapping time intervals style \u003d "color: red"\u003e the number of events in one interval does not in any way affect how many and how events will occur in another interval. An ordinary flow without consequence is called a Poisson flow. The most important characteristic of any stream of events is its intensity - the average number of events that occurred in the stream per unit of time.
Closely related to intensity is a quantity that has the meaning of the average time interval between two events. If the intervals between adjacent events are random variables that are independent of each other, then such a stream of events is called a Palm stream.
If the intensity of the flow of events does not depend on time, then such a flow is called stationary. If events occur in a stream at regular intervals, then it is called regular.
The stationary Poisson flow is called the simplest flow. In economic modeling, Poisson flows are mainly used, including the simplest ones. The following theorems are valid for them:
1) The number of events that occurred in the Poisson stream is a random variable, distributed according to Poisson's law. The probability that in a Poisson flow with intensity over the time interval ( t 1 ; t 2) it will happen exactly k events is equal to:
where .
If the flow is the simplest, then .
2) The interval between events or the waiting time for the next event T in the Poisson stream there is a random variable distributed according to the exponential law, that is, the probability that the next event will occur no earlier t, is equal to:
.
If the flow is simple, then
Example : The store is visited by an average of 20 customers per hour. Determine the probability that: a) there will be 2 buyers in 5 minutes; b) there will be at least 3 buyers in 10 minutes; c) there won't be a single buyer in 3 minutes.
Decision. Choosing 1 minute per unit of time, the intensity of the Poisson flow of store buyers (20 buyers per hour or 1/3 of the buyer per minute).
and) k=2, t 1 =0, t 2 =5,

b) k ≥3, t 1 =0, t 2 \u003d 10, we find the probability of a reverse event that will be less than 3 buyers;
.
c) by the second theorem t \u003d 3, .

5.3. Markov SP, with discrete state

In modeling probabilistic (stochastic) economic systems, the Markov SP is very often used. Consider a discrete state continuous-time SP. Then all his states can be listed: S 1 ,S 2 ,…, S n.
All possible transitions between states can be described using a state graph.
The state graph is an ordered graph, the vertices of which are possible states S i and between two states there is an edge - an arrow, if a direct transition between states is possible.
For example, a store can be in the following states:
S 1 - there are clients that are being served,
S 2 - no clients,
S 3 - goods are accepted,
S 4 - accounting of the goods, which sometimes occurs after its acceptance.
Then the work of the store can be described by the state graph

To calculate the main characteristics of the system, it is necessary to know the probabilistic indicators during the transition between states.
Consider 2 states S iand S j... The intensity of the transient flow is the average number of transitions from the state S i in state S j per unit of time that the system spends in the state S i... If the average time is known T ij, which the system spends in S i before going to S j, then you can write:.
Transient flux intensities are indicated on the state graph next to the corresponding arrows. The main task in such models is to determine the probabilities of states, which have the meaning of the average fraction of the time that the system spends in this state.
To find the probabilities of states, a system of equations is drawn up
(*)
This system can be composed according to the following rules:
1) The number of equations in the system is equal to the number of states.
2) Every state S j corresponds to the equation with number j.
3) On the left side of each equation is the sum of intensities (stand above the arrows) for all arrows entering the state S j multiplied by the probabilities of the states from which the arrows emerge;
4) On the right-hand side of the equations, the sum of the intensities emerging from S j shooter, this sum is multiplied by the probability P j.
However, the system of equations (*) is degenerate and to find a unique solution in this system, any one equation must be replaced by the normalization condition:
.
Example 1: The company's automated assembly line breaks down once a month on average and is repaired on average 3 days. In addition, on average, 2 times a month, it undergoes maintenance, which lasts an average of 1 day. On average, one in three cases of maintenance detects a problem and the line is repaired. Determine the average profit of the line per month, if for one day of uptime the profit is 15 thousand rubles. One day of technical processing costs 20 thousand rubles, and one day of repair - 30 thousand rubles.
Decision. Let us find the probabilities of states equal to the fractions of the operating time, repair and maintenance. Let be:
S 1 - the line is working,
S 2 - maintenance,
S 3 - renovation.

We compose a system of equations. In the state S 1 includes 2 arrows: from S 2 with an intensity of 20 and from S 3 with intensity 10, so the left side of the first equation is:. From the state S 1 there are two arrows with intensities 2 and 1, so the right side of the first equation of the system will take the form:. Similarly, based on the states S 2 and S 3 we compose the second and third equations. As a result, the system will look like:

However, this system is degenerate and for its solution it is necessary to replace any one (for example, the first) equation by the normalization condition:. As a result, we get the system:

We express from the 1st and 2nd equations R 1 and R 3 through R 2: , and substituting the result into the 3rd equation, we find :,,. We multiply the probabilities by 30 days of the month and find that on average a line operates 24.3 days per month, maintenance - 1.6 days, repairs - 4.1 days. It follows that the average profit will be 24.3 × 15-1.6 × 20-4.1 × 30 \u003d 209.5 thousand rubles.
Example 2: The travel agency has a salesperson and a manager. On average, 2 clients come to the agency per hour. If the seller is free, he serves the client, if he is busy, then the manager serves the client, if both are busy, the client leaves. Average service time by a seller is 20 minutes, by a manager - 30 minutes. Each client brings an average profit of 100 rubles.
Determine the agency's average profit for 1 hour, and the average number of lost clients per hour.
Decision. We determine the state of the system:
S 1 - the seller and manager are free,
S 2 - the seller is busy, the manager is free,
S 3 - the seller is free, the manager is busy,
S 4 - both are busy.
Building a state graph:

We compose a system of equations, replacing the 4th equation with the normalization condition:

Solving the system of equations, we find:
.
Therefore, the seller is engaged in the service P 2 + P 4 \u003d 0.25 + 0.15 \u003d 0.4, that is, 40% of the time. If he served 100% of the time, then he would serve 3 clients per hour, but in reality: 3 × 0.4 \u003d 1.2 and makes a profit for 1 hour 120 rubles. The manager works P 3 + P 4 \u003d 0.11 + 0.15 \u003d 0.26, i.e. 26% of the time and therefore will serve 2 × 0.26 \u003d 0.52 clients per hour and make a profit of 52 rubles per hour. The average profit for 1 hour will be 172 rubles. Clients are lost in state S 4. Since P 4 \u003d 0.15, 15% of 2 clients or 0.3 clients are lost per hour. Losses are 30 rubles per hour due to lost customers.

5.4. The processes of death and reproduction.

In many economic systems in which the joint venture operates, situations arise when from any (except for the first and last) state S i possible transition only to neighboring states S i +1 and S i -1 . such processes are called death and reproduction processes and are described by a state graph.

The intensities are called the intensities of reproduction, and m i - the intensity of death. To find the probability of each state, the following formulas are used:
, (+)
, , …, .
Example 5.1. There are 5 cars in the car service. Each of them breaks down on average 4 times a year and repairs last an average of 1 month. Determine what proportion of time all cars are in good order and the average number of good cars at an arbitrary point in time.
Decision. Enter the system states:
S 0 - all cars are broken,
S 1 - 1 vehicle is operational,
S 2 - 2 cars are operational,
S 3 - 3 cars are operational,
S 4 - 4 cars are operational,
S 5 - 5 cars are serviceable.
Let's build a graph of states and arrange the transition intensities.
For example, to go from S 1 in S 0 we have a situation: 1 car is operational and it breaks down, this happens 4 times a year, i.e. the intensity is 4. To go from S 2 in S 1: 2 cars are operational and each of them breaks down 4 times a year, i.e. the intensity is 8. The rest of the death rates are arranged by analogy.
To go from S 4 in S 5 we have a situation: 1 car is faulty and it is being repaired, this lasts 1 month or 12 times a year, i.e. the intensity is 12. To go from S 3 in S 4 we have a situation: 2 cars are faulty and each of them can be repaired with an intensity of 12, i.e. the total intensity is 24. The rest of the breeding intensities are arranged by analogy.

We calculate by the formulas (+) the probabilities of states, equal to the average fraction of the time the system is in these states.

, = 0,088, , ,
All cars are serviceable in condition S 5, the average proportion of time when cars are serviceable is 0.24. The average number of serviceable cars is found as the mathematical expectation:

Example 5.2... The organization accepts applications from the public for repair work. Applications are accepted by phone, via two lines and are served by two dispatchers. If one line is busy, the application is automatically switched to the second. If both lines are busy, the request is lost. The average number of servicing one request is 6 minutes. On average, one application brings a profit of 30 rubles. What is the profit per hour? Is it advisable to organize a third channel with a third dispatcher if its maintenance costs 150 rubles per hour?
Decision... Consider first a system with two channels.
Let's introduce the possible states:
S 0 - no orders (both phones are free),
S 1 - one request is being served (one phone is busy),
S 2 - two requests are served (both phones are busy).
The state graph will look like:

Find the probabilities of the states. According to the given formulas (+):

On average, 54% of applications are lost per hour, or 0.54 × 30 \u003d 16.2 applications. 13.8 requests per hour are served and the average profit is 13.8 × 30 \u003d 414 rubles.
Consider now the situation with three lines. In this case, three operators serve 3 telephone lines, and the incoming call comes to any free line. The following states are possible:
S 0 - no applications (three phones are free),
S 1 - one request is served (one phone is busy),
S 2 - two requests are served (two phones are busy),
S 3 - three requests are served (all phones are busy).

Using the formulas (+), we find the probabilities of states:
,
.
On average, 35% of applications are lost, or 10.4 applications per hour. Serving 19.6 applications. The average profit is 588 rubles per hour. The profit increased by 174. At a cost of 150 rubles per hour, it is expedient to introduce the third service channel.

Methods for the mathematical description of a Markov random process occurring in a system with discrete states depend on at which points in time - known in advance or random - transitions ("jumps") of the system from state to state can occur.

A random process is called a process with discrete time if the system transitions from state to state are possible only at strictly defined, pre-fixed times:. In the intervals between these moments, the system S retains its state.

A random process is called a process with continuous time if the transition of the system from state to state is possible at any, beforehand unknown, random moment

Let us first of all consider a Markov random process with discrete states and discrete time.

Let there be a physical system S, which can be in states:

moreover, transitions ("jumps") of the system from state to state are possible only at the moments:

We will call these moments “steps” or “stages” of the process and consider the random process occurring in the system S as a function of an integer argument: (step numbers).

The random process occurring in the system consists in the fact that at successive moments of time the system S finds itself in one or another state, behaving, for example, as follows:

In the general case, at moments, the system can not only change the state, but also remain in the same state, for example:

Let us agree to denote the event that after the steps the system is in the state For any k events

form a complete group and are inconsistent.

The process taking place in the system can be represented as a sequence (chain) of events, for example:

Such a random sequence of events is called a Markov chain if, for each step, the probability of transition from any state to any state does not depend on when and how the system entered the state

We will describe the Markov chain using the so-called state probabilities. Let at any time moment (after any step) the system S can be in one of the states:

that is, one of the complete group of incompatible events will take place:

Let's denote the probabilities of these events:

The probabilities after the first step

Probabilities after the second step; and generally after the step:

It is easy to see that for each step number k

since these are the probabilities of inconsistent events forming a complete group.

We will call the probabilities

probabilities of states; Let us set the task: find the probabilities of the states of the system for any k.

Let us represent the states of the system in the form of a graph (Fig. 4.6), where the arrows indicate the possible transitions of the system from state to state in one step.

A random process (Markov chain) can be imagined as if the point representing the system S randomly moves (wanders) along the state graph, jumping from state to state at moments and sometimes (in the general case) and delaying a certain number of steps in the same state. For example, the transition sequence

can be depicted on a state graph as a sequence of different positions of a point (see the dotted arrows depicting state-to-state transitions in Figure 4.7). The "delay" of the system in the state at the third step is depicted by an arrow leaving the state and returning to it.

For any step (moment of time or number, there are some probabilities of the system transition from any state to any other (some of them are equal to zero if a direct transition in one step is impossible), as well as the probability of the system delay in this state.

We will call these probabilities the transition probabilities of the Markov chain.

A Markov chain is called homogeneous if the transition probabilities do not depend on the step number. Otherwise, the Markov chain is called inhomogeneous.

Consider first a homogeneous Markov chain. Let the system S have possible states. Suppose that for each state we know the probability of transition to any other state in one step (including the probability of a delay in this state). Let us denote the probability of transition in one step from the state S, to the state there will be the probability of the system delay in the state.We write the transition probabilities in the form of a rectangular table (matrix):

Some of the transition probabilities can be zero: this means that in one step the transition of the system from state to state is impossible. Along the main diagonal of the matrix of transition probabilities are the probabilities that the system will not leave the state but will remain in it.

Using the events introduced above, the transition probabilities can be written as conditional probabilities:

From this it follows that the sum of the terms in each row of matrix (2.3) must be equal to one, since no matter what state the system is in before a step, the events are inconsistent and form a complete group.

When considering Markov chains, it is often convenient to use the state graph, on which the arrows have the corresponding transition probabilities (see Fig. 4.8). We will call such a graph a "labeled state graph".

Note that in Fig. 4.8, not all transition probabilities are indicated, but only those of them that are not equal to zero and change the state of the system, that is, with the "delay probability", it is unnecessary to put on the graph, since each of them complements the sum of the transition probabilities corresponding to all arrows coming out of this state. For example, for the graph in Fig. 4.8

If from state S; not a single arrow emanates (transition from it to any other state is impossible), the corresponding delay probability is equal to one.

Having at our disposal a labeled state graph (or, which is equivalent, a matrix of transition probabilities) and knowing the initial state of the system, we can find the probabilities of states

after any step.

Let's show how this is done.

Suppose that at the initial moment (before the first step) the system is in a certain state, for example, Then, for the initial moment (0) we will have:

that is, the probabilities of all states are equal to zero, except for the probability of the initial state, which is equal to one.

Let's find the probabilities of the states after the first step. We know that before the first step, the system is known to be in the state

Hence, in the first step, it will go to states with probabilities

written in the row of the matrix of transition probabilities. Thus, the probabilities of states after the first step will be:

Let's find the probabilities of the states after the second step:

We will calculate them using the formula of total probability, with hypotheses:

After the first step, the system was able to

The probabilities of hypotheses are known (see (2.4)); conditional probabilities of transition to a state for each hypothesis are also known and are written in the matrix of transition probabilities. Using the formula for total probability, we get:

or, much shorter,

In formula (2.6), the summation is formally extended to all states, in fact, only those of them for which the transition probabilities are nonzero, that is, those states from which a transition to a state (or a delay in it) can occur.

Thus, the probabilities of states after the second step are known. Obviously, after the third step, they are defined in the same way:

and generally after the step:

So, the probabilities of states after a step are determined by the recurrence formula (2.8) in terms of the probabilities of states after a step; those, in turn, through the probabilities of states after the step, and so on.

Example 1. A target is fired with four shots at times

Possible target (system) states:

The target is unharmed;

The target is slightly damaged;

The target received significant damage;

The target is completely hit (cannot function). The labeled system state graph is shown in Fig. 4.9.

At the initial moment, the target is in a state (not damaged). Determine the probabilities of target states after four shots. Solution. From the state graph we have;

The dynamics of financial flows shows that at any time a company can be responsible for its obligations. Markov SP, with discrete state

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