Scientists: our universe can die at any moment. Markov SP, with discrete state

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

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

results from the implementation of the project (Fig. 6.4.);

accumulated results from the implementation of the project (Fig. 6.5.);

From the last presented graph, 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 generate the main risks of the project implementation 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 the planned growth of services (setting up a fundamentally new business);

the market is occupied by other, currently stronger competing organizations, extraordinary efforts are needed to conquer a market niche in six months - 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, the lack of support or opposition from the government, etc.

Risk Mitigation Measures:

development of domestic tax policy;

formation of business external environment (partners, consortiums, financial and industrial groups);

active participation of founders in interaction with power structures;

giving the institution the status of a medical institution.

7.2.2 Legal risks

They are connected with the imperfection of the legislation, unclearly executed documents, the ambiguity of judicial measures in case of disagreements between the founders (for example, in a foreign court, etc.), delays by the Contractor.

Risk Mitigation Measures:

clear and unambiguous wording of relevant articles in documents;

attracting specialists with practical experience in this field to draw up documents;

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

7.2.3 Technical risks

Associated with the complexity of the work and the lack of a technical project at the moment.

There may be incomplete utilization of equipment and a delay in the introduction of technical systems.

Risk Mitigation Measures:

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

conclusion of contracts on a turnkey basis with sanctions for inconsistencies and failure to meet deadlines;

technical risk insurance.

7.2.4. Production risks

They are connected primarily with the possibility of delays in the commissioning of new technical facilities and the 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 implementation management;

accelerated development of the design concept, including quality criteria;

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

substantiation and allocation of sufficient financial resources for the acquisition 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 socio-psychological risks may arise:

social tension in the team;

shortage, turnover of professional staff;

having a destructive attitude.

Risk Mitigation Measures:

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

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

a system of end-to-end multi-level awareness of the staff and managers;

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

7.2.6 Marketing risks

Associated with possible delays in entering the market, incorrect (without taking into account market needs) choice of services, erroneous choice of marketing strategy, errors 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 on the market, which requires a marketing program that meets international standards and a service that implements it.

Since at the moment there is no full-scale program of marketing activities, the assessment of the degree of solution of marketing problems is low. While for a firm that aims to win market share from competing firms, marketing tasks should be a top 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 understand their main advantages and focus on them the main efforts and resources.

Risk Mitigation Measures:

creation of a strong marketing service;

development of a marketing strategy;

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

development and implementation of a program of marketing activities;

conducting a full range of marketing research, etc.

7.2.7 Financial risks

They are connected primarily 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 income is provided through the use of numbers. Reducing the price or occupancy of the rooms of the hotel complex leads to serious difficulties in the implementation of the project.

Risk Mitigation Measures:

urgent research into the requirements of consumers of services;

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

substantiation and allocation of sufficient financial resources for the creation and acquisition of high-quality equipment;

using the approach of diversifying sources of income, primarily through the “office-number” link;

entry to the stock market.

Another major factor of financial risk is the need to receive large investments in a timely manner.

The presence of investments is a necessary condition 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 operation;

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

Next steps for developers and project owners:

conducting in-depth problematic diagnostics of the project;

carrying out a set of measures to search for 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 activities, primarily related to marketing, advertising and diversification and providing:

establishment of JSC;

high economic efficiency of the project;

risk minimization;

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 able to provide 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

Task number 7

A random continuous variable has a probability density f(x) = 32*t*e

Required:

1.)Find its distribution function F(x).

2.) Construct graphs of the distribution function F(x) and the probability density f(x).

3.) Calculate the probability of hitting a random variable in (0.5; 2)

Solution.

1.)F(x) = 32*t*e dt = -e + 1

2.) Charts are shown below

3.) The probability of falling into a random interval is found as:

Р(0.5< < 2) = F(0.5) – F(2) = 0.0001

4.)

Task 8. The probability density f(x) of the random variable  is given. The random variable  is related to the random variable  by the functional dependence
. Find:

Mathematical expectation and variance of a random variable  using the probability density of a random variable ;

Probability density of a random variable  and build its graph;

The mathematical expectation and variance of the random variable  using the found probability density of the random variable .

Solution.

1. Mathematical expectation:

2. Probability density of a random variable :

3. Mathematical expectation:

Variance of a random variable :

The numerical characteristics calculated by different methods are the same.

Task 9. A system of two random variables (,) is given, the distribution law of which is given by Table 1. Find:

Laws of distribution of random variables  and ;

Mathematical expectations and variances of random variables  and ;

Solution.

distribution of a random variable :

(2)=0.18+0.15+0.08=0.51

(3)=0.04+0.12+0.12=0.28

(5)=0.06+0.05+0.10=0.21

distribution of a random variable :

(-1)=0.18+0.04+0.06=0.28

(0)=0.15+0.12+0.05=0.32

(1)=0.08+0.12+0.10=0.30

(2)=0.10

Dispersion of a random variable :

Mathematical expectation of a random variable :

Variance of a random variable :

correlation moment:

Correlation coefficient:

(2/0)=
;

(3/0)=

(5/0)=

Conditional distributions

Conditional mathematical expectations:

Task 10. The system of continuous random variables (,) is distributed uniformly in the area D, bounded by the lines x=1, y=0,
x>0; find:

Solution.

1. Since the distribution is uniform, then f(x;y)=const. The joint probability density is found from the normalization condition:

2. Probability densities of random variables  and :

.
; x;

; y[-2;0];

Mathematical expectations and variances of random variables  and :

;

;

Task 11. Find the mathematical expectation and variance of a random variable, =а+b+с, where (,) is the system of random variables from Problem 10. a=2; b=-3; c=3.

Solution.

Finding the mathematical expectation:

Dispersion:

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 organization's revenue or profit over time, sales volumes of goods, etc.
If a random process can change its state only at a strictly defined point in time, then it is called a process with discrete time.
If a change of state is possible at an arbitrary moment of time, then this is a continuous-time SP.
If at any moment in time the SP is a discrete random variable (its value can be enumerated and two neighboring values can be distinguished), then this is a process with a discrete state.
If, at any moment of time, the state can change continuously, smoothly, and two neighboring states cannot be distinguished, then this is a SP with a continuous state.
Thus, 4 types of joint ventures are possible:
1) SP with continuous time and continuous state (example: air temperature at some point in time, changes smoothly at any point in time).
2) SP with continuous time and discrete state (example: the number of visitors in the store, changes by a multiple of one at any time).
3) SP with discrete time and a continuous state (example: the dynamics of the exchange rate, 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 times, at stops).
Consider some system S, in which at a given time t about SP is running. This process is called Markov if for any moment of time t> t about, the behavior of the system in the future depends only on the state in which the system was at a given time when t= t about, and does not depend on how, when and in what states it was in the past when t< t O . In other words, the "past" of the Markov process does not affect the "future" in any way (only through the "present").

5.2. Event streams.

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

If the events in the stream occur singly, and not in groups of several events, then such a stream is called an ordinary one. A stream of events is called a stream without consequences if, for any non-overlapping time intervals style="color:red">, the number of events in one interval has no effect on how many and how events will occur in another interval. Ordinary flow without consequence is called Poisson flow. The most important characteristic of any stream of events is its intensity - the average number of events that occurred in the stream in one unit of time.
The intensity is closely related to the quantity , which 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 hold for them:
1) The number of events that occurred in the Poisson flow is a random variable distributed according to the Poisson law. The probability that in a Poisson flow with intensity over a time interval ( t 1 ; t 2) will happen exactly k events is equal to:
, where .
If the flow is simple, then .
2) Interval between events or waiting time for the next event T in the Poisson flow there is a random variable distributed according to the exponential law, i.e. the probability that the next event will occur no earlier than 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) in 3 minutes there will be no buyer.
Solution. By choosing 1 minute per unit of time, the intensity of the Poisson flow of shoppers is (20 shoppers per hour or 1/3 shoppers per minute).
a) k=2, t 1 =0, t 2 =5,

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

5.3. Markov SP, with discrete state

In modeling probabilistic (stochastic) economic systems, Markov SP is often used. Consider an SP with a discrete state and continuous time. Then all its states can be enumerated: S 1 ,S 2 ,…, S n.
You can describe all possible transitions between states using a state graph.
The state graph is an ordered graph whose vertices are the possible states Si 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 who are served,
S 2 - no clients,
S 3 - goods are received,
S 4 - accounting for goods, which sometimes occurs after its acceptance.
Then the operation 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 Si and S j. The intensity of the transition flow is the average number of transitions from the state Si into a state S j per unit of time that the system spends in the state Si. If the average time is known T ij, which the system spends in Si before moving to S j, then we can write: .
The intensities of transient flows 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 time that the system spends in this state.
To find the probabilities of states, a system of equations is compiled
(*)
This system can be built 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 the number j.
3) On the left side of each equation is the sum of the intensities (they stand above the arrows) for all the arrows entering the state S j multiplied by the probabilities of the states from which the arrows exit;
4) On the right side of the equations is the sum of the intensities coming from S j shooter, this sum is multiplied by the probability Pj.
However, the system of equations (*) is degenerate, and in order to find a unique solution in this system, any one equation must be replaced by the normalization condition:
.
Example 1: The automated assembly line of the enterprise fails on average once a month 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 out of three cases during maintenance, a problem is discovered and the line is repaired. Determine what average profit the line brings 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.
Solution. Let us find the probabilities of states equal to the shares of the time of work, repair and maintenance. Let:
S 1 - the line is working,
S 2 - maintenance,
S 3 - repair.

We compose a system of equations. In state S 1 includes 2 arrows: from S 2 with intensity 20 and out S 3 with intensity 10, so the left side of the first equation is: . Out of 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 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 with 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, the 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=209.5 thousand rubles.
Example 2: A salesperson and a manager work in a travel agency. 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. The average time of service by the seller is 20 minutes, by the manager - 30 minutes. Each client brings an average profit of 100 rubles.
Determine the average profit of the agency for 1 hour, and the average number of lost customers per hour.
Solution. We determine the state of the system:
S 1 - the seller and the 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.
We build 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 servicing 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 = 1.2 and makes a profit of 120 rubles in 1 hour. The manager works P 3 + P 4 =0.11+0.15=0.26, i.e. 26% of the time and therefore serves 2 × 0.26=0.52 clients per hour and makes 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, then 15% of 2 possible clients or 0.3 clients are lost per hour. Losses amount to 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 the first and last) state Si transition to neighboring states is possible Si+1 and Si-one . such processes are called death and reproduction processes and they are described by a state graph.

The intensities are called multiplication intensities, and m i- intensity of death. The following formulas are used to find the probability of each state:
, (+)
, , …, .
Example 5.1. The fleet has 5 cars. Each of them breaks down on average 4 times a year and repairs last an average of 1 month. Determine what fraction of the time all cars are serviceable and the average number of serviceable cars at an arbitrary point in time.
Solution. Enter system states:
S 0 - all cars are broken,
S 1 - 1 car is serviceable,
S 2 - 2 cars are serviceable,
S 3 - 3 vehicles are serviceable,
S 4 - 4 vehicles are serviceable,
S 5 - 5 vehicles are serviceable.
We construct a state graph and arrange the transition intensities.
For example, to go from S 1 in S 0 we have a situation: 1 car is serviceable and it breaks down, this happens 4 times a year, i.e. intensity is 4. For the transition from S 2 in S 1: 2 cars are serviceable and each of them breaks down 4 times a year, i.e. the intensity is equal to 8. The other death intensities are arranged by analogy.
To move from S 4 in S 5 we have a situation: 1 car is out of order and it is being repaired, this lasts 1 month or 12 times a year, i.e. the intensity is 12. For the transition from S 3 in S 4 we have a situation: 2 cars are out of order and each of them can be repaired with a rate of 12, i.e. the total intensity is 24. The remaining reproduction intensities are arranged by analogy.

Using the formulas (+), we calculate 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 state S 5 , the average share 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 population for repair work. Applications are accepted by phone, on two lines and they are served by two dispatchers. If one line is busy, the application will automatically switch 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?
Solution. Consider first a system with two channels.
Let's introduce possible states:
S 0 - no applications (both phones are free),
S 1 – one request is served (one phone is busy),
S 2 - two requests are served (both phones are busy).
The state graph will look like:

We find the probabilities of states. According to the above formulas (+):

On average, 54% of requests are lost per hour, or 0.54 × 30 = 16.2 requests. 13.8 applications are served per hour and the average profit is 13.8 × 30 = 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 applications are served (all phones are busy).

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

The methods of mathematical description of a Markov random process occurring in a system with discrete states depend on at what 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 transitions of the system from state to state are possible only at strictly defined, pre-fixed times: . In the time 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 random moment unknown in advance

Consider first of all a Markov random process with discrete states and discrete time.

Let there be a physical system S that can be in the 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 a random process occurring in the system S as a function of an integer argument: (step number).

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

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

We agree to denote an event consisting in the fact that after the steps the system is in the state For any k, the event

form a complete group and are incompatible.

The process occurring 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 came to the state

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

i.e., one of the complete group of incompatible events will occur:

Let us denote the probabilities of these events:

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 incompatible events forming a complete group.

We will call the probabilities

state probabilities; Let us pose the problem: to find the probabilities of the states of the system for any k.

Let's depict 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 graph of states, 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 point positions (see dashed arrows depicting 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 (point 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 delaying 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 delay in this state). Let us denote the probability of transition in one step from the state S, the state will be the probability of the system delay in the state. Write the transition probabilities in the form of a rectangular table (matrix):

Some of the transition probabilities may be equal to zero: this means that the transition of the system from state to state is impossible in one step. 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:

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

When considering Markov chains, it is often convenient to use a state graph on which the corresponding transition probabilities are marked on the arrows (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 listed, but only those that are not equal to zero and change the state of the system, i.e., with a “delay probability”, it is unnecessary to put on the graph, since each of them complements to unity the sum of the transition probabilities corresponding to all arrows emanating from this state. For example, for the graph in Fig. 4.8

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

Having at our disposal a labeled graph of states (or, equivalently, 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 it's 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:

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

Let us 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

This means that in the first step it will go into states with probabilities

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

Find the probabilities of states after the second step:

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

After the first step, the system was in a state

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

or, much shorter,

In formula (2.6), the summation formally extends to all states; in fact, it is necessary to take into account only those for which the transition probabilities are different from zero, 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 similarly:

and generally after the step:

Thus, the probabilities of states after a step are determined by the recurrent formula (2.8) in terms of the probabilities of states after a step; those, in turn, through the probabilities of states after the step, etc.

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

Possible states of the target (system):

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 the state (intact). Determine the probabilities of target states after four shots Solution. From the state graph we have;

MOSCOW, July 30 - RIA Novosti. Physicists of the IKBFU I. Kant considered one of the possible mathematical models of dark energy, and found out that the future of our Universe can be much more unpredictable and catastrophic than previously thought. The results of the study were published in the highly rated scientific journal "The European Physical Journal C" .

"Taking into account a new class of singularities (states in which one or another parameter becomes infinite) makes the future of our Universe unpredictable and dangerous. In this work, we have shown that some singularities can arise quite suddenly, almost at any time. Neither stars, nor even galaxies will not survive such a catastrophe,” said Artem Yurov, one of the authors of the study, professor at the Immanuel Kant Baltic Federal University.

At the end of the 20th and the beginning of the 21st century, a number of important discoveries were made in cosmology: indirect evidence of the inflationary expansion of the Universe, dark matter and energy, and gravitational waves were discovered. In 1998, scientists discovered that our universe is not just expanding, but expanding at an ever-increasing rate.

The reason for this acceleration, scientists believe the so-called "dark sector" of the universe. According to observational data, the total content of our Universe consists of only 4.9% of the baryonic matter familiar to us, the remaining 95.1% falls on the "dark sector", which consists of mysterious dark matter (26.8%) and even more mysterious dark matter. energy (68.3%).

There are three main hypotheses about what dark energy is. According to the first one, dark energy is a cosmological constant - a constant energy density that evenly fills the space of the Universe. The second hypothesis defines dark energy as a kind of quintessence - a dynamic field, the energy density of which can change in space and time. According to the third, dark energy is a manifestation of modified gravity at distances of the order of the size of the visible part of the Universe.

"The future of our Universe depends on which of these models is correct. If the second hypothesis is correct and dark energy is indeed the quintessence, then the future may turn out to be full of amazing and unpleasant surprises. In particular, singularities may appear right during accelerated expansion! For example, the average the pressure of the quintessence can suddenly "explode," said Professor Yurov.

The fact that such a catastrophe is possible was calculated in 2004 by a professor at the University of Cambridge, John Barrow. A more complete mathematical study of this issue allowed the physicists Sergey Odintsov, Shinichi Nojiri and Shinji Tsudzikawa to classify such possible catastrophic singularities of the future.

A group of physicists from the IKBFU Kant, under the guidance of Professor Artem Yurov, suggested and showed mathematically that there could be a whole class of singularities not covered by the Odintsov-Nojiri-Tsudzikawa classification. This means that our Universe may die suddenly. Foreign colleagues became interested in the study of Russian physicists, which was carried out with the support of Project 5-100. In particular, John Barrow addressed the authors with a letter.

"The model we are talking about is one of hundreds of models of the birth and death of our Universe. The authors from the Immanuel Kant Baltic Federal University correctly considered the model with a specific potential of the scalar field and showed that the scale factor can dramatically change its behavior. For specialists, this The work is of interest. It should be kept in mind for the future, since it apparently does not contradict modern observational data," stressed cosmologist, Professor of NRNU MEPhI Sergey Rubin.