Rule of minimum risk author. The problem of decision making under uncertainty. Operations under uncertainty

and

If in the initial conditions K1 > K2, then

and

For the minimum risk method at K0=2.29 we get the following

For the method of the minimum number of erroneous decisions with K0 =2,57:

For the maximum likelihood method at K 0 =2,37:

We summarize the results of the calculations in the final table.

Tasks for task number 2.

The task option is selected by the last two digits of the record book number. In all tasks, it is required to determine the boundary value K 0 , dividing objects into two classes: serviceable and faulty. The results of the decisions are plotted on a graph (Fig. 9.1), which is built on graph paper and pasted into the work.

So, the technical diagnostics of the object is carried out according to the parameter K. For a serviceable object, the average value of the parameter is given K 1 and standard deviation σ 1 . For faulty respectively K2 and σ 2 . In the source data, the ratio of prices is also given for each option C12/C21. Distribution K accepted as normal. In all options P1=0,9; P2=0,1.

Task options are given in table. 2.1-2.10.

Initial data for options 00÷09 (Table 2.1):

An object- gas turbine engine.

Parameter- vibration velocity (mm/s).

Faulty state- violation of the normal operating conditions of the engine rotor supports.

Table 2.1

Initial data for options 10÷19 (Table 2.2):

An object- gas turbine engine.

Parameter Cu ) in oil (g/t).

Faulty state- increased concentration Cu

Table 2.2

Initial data for options 20÷29 (Table 2.3):

An object- pumped up fuel pump of the fuel system.

Parameter- fuel pressure at the outlet (kg / cm 2).

Faulty state- deformation of the impeller.

Table 2.3

Initial data for options 30÷39 (Table 2.4):

An object- gas turbine engine.

Parameter- level of vibration overloads ( g ).

Faulty state- rolling out the outer race of the bearings.

Table 2.4

Initial data for options 40÷49 (Table 2.5):

An object- intershaft bearing of a gas turbine engine.

Parameter- readings of the vibroacoustic device for monitoring the condition of the bearing (µa).

Faulty state- the appearance of traces of chipping on the bearing raceways.

Table 2.5

Initial data for options 50÷59 (Table 2.6)

An object- gas turbine engine.

Parameter- iron content ( Fe ) in oil (g/t).

Faulty state- increased concentration Fe in oil due to accelerated wear of gear connections in the gearbox.

Table 2.6

Initial data for options 60÷69 (Table 2.7):

An object- oil for lubricating a gas turbine engine.

Parameter- optical density of oil, %.

Faulty state- reduced performance properties of the oil having optical density.

Table 2.7

Initial data for options 70÷79 (Table 2.8):

An object- fuel filter elements.

Parameter- copper impurity concentration ( Cu ) in oil (g/t).

Faulty state- increased concentration Cu in oil due to the intensification of wear processes of copper-plated spline connections of drive shafts.

Table 2.8

Initial data for options 80÷89 (Table 2.9)

An object- axial piston pump.

Parameter- the value of the pump performance, expressed by volumetric

Efficiency (in fractions of 1.0).

Faulty state- low volumetric efficiency associated with pump failure.

Table 2.9

Initial data for options 90÷99 (Table 2.10)

An object- aircraft control system, consisting of rigid rods.

Parameter- total axial play of the joints, microns.

Faulty state- increased total axial play due to wear of mating pairs.

Table 2.10

Laboratory work 2 "Operation and diagnostics of contact network supports"

Objective: get acquainted with the methods for determining the corrosion state of a reinforced concrete support of a contact network

Work order:

1) Study and draw up a brief report on the operation of the ADO-3 device.

2) Study and solve the problem using the minimum risk method (according to the options (by number in the journal)

3) Consider the special question of how to diagnose the condition of the supports (except for the angle of inclination).

P.p. 1 and 3 are performed by a team of 5 people.

Item 2 is carried out individually by each student.

As a result, it is necessary to make an individual electronic report and attach it to the blackboard.

Minimum risk method

In the presence of decision uncertainty, special methods are used that take into account the probabilistic nature of events. They allow you to assign the border of the tolerance field of the parameter to make a decision on diagnosing.

Let the condition of the reinforced concrete support be diagnosed by the vibration method.

The vibration method (Fig. 2.1) is based on the dependence of the decrement of damped vibrations of the support on the degree of reinforcement corrosion. The support is set in oscillatory motion, for example, by means of a guy cable and a drop device. The ejection device is calibrated to a predetermined force. An oscillation sensor, such as an accelerometer, is installed on the support. The decrement of damped oscillations is defined as the logarithm of the ratio of oscillation amplitudes:

where A 2 and A 7 are the amplitudes of the second and seventh oscillations, respectively.

a) diagram b) measurement result

Figure 2.1 - Vibration method

ADO-2M measures oscillation amplitudes of 0.01 ... 2.0 mm with a frequency of 1 ... 3 Hz.

The greater the degree of corrosion, the faster the vibrations decay. The disadvantage of the method is that the oscillation decrement largely depends on the parameters of the soil, the method of embedding the support, deviations in the manufacturing technology of the support, and the quality of concrete. A noticeable effect of corrosion is manifested only with a significant development of the process.

The task is to choose the Xo value of the parameter X in such a way that for X>Xo a decision is made to replace the support, and for X<Хо не проводили управляющего воздействия.

. (2.2)

The support oscillation decrement depends not only on the degree of corrosion, but also on many other factors. Therefore, we can talk about a certain area in which the value of the decrement can be located. The distributions of the vibration decrement for a serviceable and corroded bearing are shown in fig. 2.2.

Figure 2.2 - Probability density of the support oscillation decrement

It is significant that the areas of serviceable D 1 and corrosive D 2 states intersect, and therefore it is impossible to choose x 0 in such a way that rule (2.2) would not give erroneous solutions.

Type I error- making a decision about the presence of corrosion (defect), when in reality the support (system) is in good condition.

Type II error- making a decision about the serviceable condition, while the support (system) has corroded (contains a defect).

The probability of an error of the first kind is equal to the product of the probabilities of two events: the probability of having a good state and the probability that x > x 0 in a good state:

, (2.3)

where P(D 1) \u003d P 1 - a priori probability of finding the support in good condition (it is considered known based on preliminary statistical data).

Type II error probability:

, (2.4)

If the costs of errors of the first and second kind c and y are known, respectively, then we can write an equation for the average risk:

Let us find the boundary value x 0 for rule (2.5) from the condition of minimum average risk. Substituting (2.6) and (2.7) into (2.8), differentiating R(x) with respect to x 0 , we equate the derivative to zero:

= 0, (2.6)

. (2.7)

This is a condition for finding two extremums - a maximum and a minimum. For the existence of a minimum at the point x = x 0, the second derivative must be positive:

. (2.8)

This leads to the following condition:

. (2.9)

If the distributions f(x/D 1) and f(x/D 2) are unimodal, then for:

(2.10)

condition (4.58) is satisfied.

If the distribution densities of the parameters of a healthy and faulty (system) are subject to the Gauss law, then they have the form:

, (2.11)

. (2.12)

Conditions (2.7) in this case takes the form:

. (2.13)

After transformation and logarithm, we get the quadratic equation

, (2.14)

b= ;

c= .

By solving equation (2.14), one can find such a value x 0 at which the minimum risk is achieved.

Initial data:

Working condition:

Expected value:

The probability of a good system state:

Standard deviation:

The given costs for good condition:

Faulty state:

Expected value: ;

Minimum risk method. This method was developed in connection with the problems of radar, but can be quite successfully used in problems of technical diagnostics.

Let the parameter x be measured (for example, the vibration level of the product) and, based on the measurement data, it is required to make a conclusion about the possibility of continuing operation (diagnosis - good condition) or about sending the product for repair (diagnosis - faulty condition).

On fig. 1 shows the values ​​of the probability density of the diagnostic parameter x for two states.

Let the control norm for the level of vibrations be set.

In accordance with this norm, they accept:

The sign means that an object with a vibration level x is assigned to a given state.

From fig. 1 it follows that any choice of value is associated with a certain risk, since the curves intersect.

There are two types of risk: the risk of "false alarm", when a serviceable product is considered faulty, and the risk of "missing the target", when a faulty product is considered good.

In the theory of statistical control, they are called the risk of the supplier and the risk of the receiver, or errors of the first and second kind.

Given the probability of a false alarm

and the probability of missing the target

The task of the theory of statistical decisions is to choose the optimal value

According to the minimum risk method, total cost risk

where is the “price” of a false alarm; - "price" of missing the target; - a priori probabilities of diagnoses (conditions), determined by preliminary

Rice. 1. Probability density of a diagnostic feature

statistical data. The value represents the "average value" of the loss in an erroneous decision.

From the necessary minimum condition

we get

It can be shown that for unimodal distributions condition (23) always ensures the minimum of the value If the cost of erroneous decisions is the same, then

The last relation minimizes the total number of erroneous decisions. It follows also from the Bayes method.

Neumann-Pearson method. This method proceeds from the condition of the minimum probability of skipping a defect at an acceptable level of false alarm probability.

Thus, the probability of a false alarm

where is the allowable false alarm level.

In the one-parameter problems under consideration, the minimum probability of missing the target is achieved when

The last condition determines the boundary value of the parameter (value

When assigning a value, take into account the following:

1) the number of decommissioned products must exceed the expected number of defective products due to the inevitable errors in the condition assessment method;

2) the accepted false alarm value should not, unless absolutely necessary, disrupt normal operation or lead to large economic losses.

Example 2.5. For the consequence matrix shown in example 2.1, select the best option solutions based on the Hurwitz criterion for λ =1/2.

Solution. Considering the consequence matrix Q row by row, for each i we calculate the values ​​ci= 1/2minqij + 1/2maxqij. For example, c1=1/2*2+1/2*8=5; similarly, c2=7; c3=6.5; c4= 4.5. The largest is c2=7. Therefore, the Hurwitz criterion for a given λ =1/2 recommends choosing the second option ( i=2).

2.3. Analysis of a coupled decision group under conditions of partial

uncertainty

If, when making a decision, the decision maker knows the probabilities pj that the real situation can develop according to option j, then we say that the decision maker is in conditions of partial uncertainty. In this case, you can be guided by one of the following criteria (rules).

Criterion (rule) for maximizing the average expected income. This criterion is also called criterion for the maximum average payoff. If the probabilities are known pj options for the development of a real situation, then the income received from the i-th solution is a random variable Qi with a distribution series

Expected value M[qi] of the random variable Qi is the average expected income, also denoted by :

= M[qi ] = .

For each i-th variant of the solution, the values ​​are calculated, and in accordance with the criterion under consideration, the variant is selected for which

Example 2.6. Let, for the initial data of Example 2.1, the probabilities of the development of a real situation for each of the four options that form a complete group of events are known:

p1=1/2, p2=1/6, p3=1/6, p4=1/6. Find out which solution provides the greatest average income and what is the value of this income.

Solution. Let's find for each i-th solution the average expected income: =1/2*5+1/6*2+1/6*8+1/6*4= 29/6, = 25/6, = 7, = 17/6. The maximum average expected return is 7 and corresponds to the third solution.

Average expected risk minimization rule (other name - minimum average loss criterion).

Under the same conditions as in the previous case, the decision maker's risk when choosing the i-th solution is a random variable Ri with a distribution series

Expected value M and is the average expected risk, also denoted by : = M = . . The rule recommends making a decision that entails the minimum average expected risk: .

Example 2.7 . The initial data are the same as in example 2.6. Determine which solution option achieves the smallest average expected risk, and find the value of the minimum average expected risk (loss).

Solution. For each i-th solution, we find the value of the average expected risk. Based on the given risk matrix R, we find: = 1/2*3+1/6*3+1/6*0+1/6*8=20/6, = 4, = 7/6, = 32/6.

Therefore, the minimum average expected risk is 7/6 and corresponds to the third solution: = 7/6.

Comment. When talking about the average expected return (gain) or the average expected risk (loss), they mean the possibility of repeating the decision-making process according to the described scheme or the actual repeated repetition of such a process in the past. The conditionality of this assumption lies in the fact that the actual required number of such repetitions may not be.

Criterion (rule) of Laplpas of equal opportunity (indifference). This criterion is not directly related to the case of partial uncertainty, and it is applied under conditions of complete uncertainty. However, it is assumed here that all states of the environment (all variants of the real situation) are equally probable - hence the name of the criterion. Then the calculation schemes described above can be applied, considering the probabilities pj identical for all variants of the real situation and equal to 1/n. So, when using the criterion for maximizing the average expected income, a solution is chosen that achieves . And in accordance with the criterion of minimizing the average expected risk, a solution option is selected for which .

Example 2.8. Using the Laplace criterion of equal opportunity for the initial data of example 2.1, choose the best solution based on: a) the rule for maximizing the average expected income; b) rules for minimizing the average expected risk.

Solution. a) Taking into account the equiprobability of variants of the real situation, the average expected income for each of the solution options is = (5+2+8+4)/4=19/4, = 21/4, = 26/4, = 15/4. Therefore, the third solution is the best, and the maximum average expected return is 26/4.

b) For each solution option, we calculate the average expected risk based on the risk matrix, taking into account the equiprobability of the situation options: = (3+3+0+8)/4 = 14/4, = 3, = 7/4, = 18/4 . It follows that the third option will be the best, and the minimum average expected risk will be 7/4.

2.4. Pareto optimality of two-criteria financial

operations under conditions of uncertainty

It follows from the above that each decision (financial transaction) has two characteristics that need to be optimized: the average expected return and the average expected risk. Thus, the choice of best solution is an optimization two-criteria problem. In multicriteria optimization problems, the main concept is the concept Pareto optimality. Let's consider this concept for financial operations with two specified characteristics.

Let every operation a has two numbers E(a),r(a)(for example, effectiveness and risk); when optimizing E strive to increase r decrease.

There are several ways to formulate such optimization problems. Let's consider this problem in a general way. Let BUT - some set of operations, and different operations necessarily differ in at least one characteristic. When choosing the best operation, it is desirable that E was more and r was less.

We will say that the operation a dominates operation b, and designate a > b if E(a) ≥ E(b) and r(a) ≤ r(b) and at least one of these inequalities is strict. At the same time, the operation a called dominant, and the operation b-dominated. Obviously, no dominated operation can be recognized the best. Therefore, the best operation must be sought among non-dominated operations. The set of non-dominated operations is called set (domain) Pareto or set of Pareto optimality.

For the Pareto set, the statement is true: each of the characteristics E,r is a single-valued function of the other, i.e., on the Pareto set, one characteristic of the operation can uniquely determine another.

Back to analysis financial solutions under conditions of partial uncertainty. As shown in section 2.3, each operation is characterized by an average expected risk and the average expected income. If we introduce a rectangular coordinate system, on the x-axis of which we plot the values , and on the y-axis - values ​​, then each operation will correspond to a point ( , ) on the coordinate plane. The higher this point on the plane, the more profitable the operation; the more to the right the point, the more risky the operation. Therefore, when searching for non-dominated operations (Pareto sets), you need to choose points above and to the left. Thus, the Pareto set for the initial data of examples 2.6 and 2.7 consists of only one third of the operation.

To determine the best operation in some cases, you can apply some weighing formula, in which the characteristics and enter with certain weights, and which gives a single number that specifies the best operation. Let, for example, for the operation i with characteristics ( , ) the weighting formula has the form f(i) = 3 - 2, and the best operation is chosen by the maximum value f(i). This weighting formula means that the decision maker agrees to increase the risk by three units if the income of the operation increases by at least two units. Thus, the weighting formula expresses the ratio of decision makers to indicators of income and risk.

Example 2.9. Let the initial data be the same as in examples 2.6 and 2.7, i.e. for the consequences and risk matrices of example 2.1, the probabilities of the real situation development options are known: p1 =1/2, p2=1/6, p3=1/6, p4=1/6. Under these conditions, the decision maker agrees to increase the risk by two units, if at the same time the income of the operation increases by at least one unit. Determine the best operation for this case.

Solution. The weighting formula has the form f(i) = 2 - . Using the results of calculations in examples 2.6 and 2.7, we find:

f(1) = 2*29/6 – 20/6 = 6,33; f(2) = 2*25/6 – 4 = 4,33;

f(3) = 2*7 – 7/6 = 12,83; f(4) = 2*17/6 – 32/6 = 0,33

Therefore, the third operation is the best, and the fourth is the worst.

Topic 3. Measures and indicators of financial risks

Quantitative risk assessment. The risk of a single operation. General measures of risk.

This topic discusses criteria and decision-making methods in cases where it is assumed that the probability distributions of possible outcomes are either known or can be found, and in the latter case it is not always necessary to explicitly specify the distribution density.

3.1. General Methodological Approaches to Quantitative Risk Assessment

Risk is a probabilistic category, therefore the methods of its quantitative assessment are based on a number of the most important concepts of probability theory and mathematical statistics. So, the main tools of the statistical method of risk calculation are:

1) expected value m, for example, such a random variable as the result of a financial transaction k: m = E{k};

2) dispersion as a characteristic of the degree of variation of the values ​​of a random variable k around the grouping center m(recall that variance is the mathematical expectation of the squared deviation of a random variable from its mathematical expectation );

3) standard deviation ;

4) the coefficient of variation , which has the meaning of risk per unit of average income.

Comment. For a small set n values ​​- small sample! - discrete random variable Strictly speaking, it is only estimates listed risk measures .

So, average (expected) value of the sample, or selective analogue of mathematical expectation , is the quantity , where Ri- the probability of realizing the value of a random variable k. If all values ​​are equally likely, then the expected value of a random sample is calculated by the formula .

Likewise, sample variance (sample variance ) is defined as the standard deviation in the sample: or

. In the latter case, the sample variance is biased estimate of the theoretical variance . Therefore, it is preferable to use an unbiased estimate of the variance , which is given by the formula .

It is obvious that the estimate can be calculated as follows or .

It is clear that the estimate coefficient of variation now takes the form .

In economic systems under risk, decision-making is most often based on one of the following criteria.

1. expected value (profitability, profit or expenses).

2. Sample variance or standard (rms) deviation .

3. Expected Value Combinations and dispersion or sample standard deviation .

Comment . under a random variable k in each specific situation, the indicator corresponding to this situation is understood, which is usually written in the accepted notation: mp – portfolio return securities, IRR - (Internal Rate of Return) internal (rate) of return etc.

Let's consider the stated idea on specific examples.

3.2. Probability distributions and expected returns

As has been said more than once, risk is associated with the likelihood that the actual return will be lower than its expected value. Therefore, probability distributions are the basis for measuring the risk of an operation. However, it must be remembered that the resulting estimates are probabilistic in nature.

Example 1. Suppose, for example, that you intend to invest $100,000. for a period of one year. Alternative investment options are given in Table. 3.1.

Firstly, these are GKO-OFZs with a maturity of one year and an income rate of 8%, which can be purchased at a discount, i.e. at a price below par, and at the time of redemption their face value will be paid.

Table 3.1

Estimated return on four investment alternatives

Note. The yield corresponding to different states of the economy should be considered as an interval of values, and its individual values ​​- as points within this interval. For example, a 10% yield on a corporate bond in a slight downturn is most probable return value at a given state of the economy, and the point value is used for convenience of calculations.

Secondly, corporate securities (blue chips), which are sold at par with a coupon rate of 9% (i.e., for 100,000 dollars of invested capital, you can receive 9,000 dollars per annum) and a maturity of 10 years. However, you intend to sell these securities at the end of the first year. Therefore, the actual yield will depend on the level of interest rates at the end of the year. This level, in turn, depends on the state of the economy at the end of the year: rapid economic development are likely to cause higher interest rates, which will reduce the market value of blue chips; in the event of an economic downturn, the opposite situation is possible.

Third, Investment Project 1, with a net worth of $100,000. Cash flow during the year is zero, all payments are made at the end of the year. The amount of these payments depends on the state of the economy.

And, finally, alternative investment project 2, which coincides in all respects with project 1 and differs from it only probability distribution of payments expected at the end of the year .

Under probability distribution , we will understand the set of probabilities of possible outcomes (in the case of a continuous random variable, this would be the density of the probability distribution). It is in this sense that the data presented in Table 1 should be interpreted. 3.1 four probability distributions corresponding to four alternative investment options. The yield on GKO-OFZ is known exactly. It is 8% and does not depend on the state of the economy.

Question 1 . Can the GKO-OFZ risk be considered unconditionally equal to zero?

Answer: a) yes; b) I think that not everything is so unambiguous, but I find it difficult to give a more complete answer; c) no.

The correct answer is c).

For any answer, see Help 1.

Help 1 . Investments in GKO-OFZs are risk-free only in the sense that they nominal returns do not change over a given period of time. At the same time their real the return contains a certain amount of risk, since it depends on the actual rate of inflation during the period of holding this security. Moreover, GKOs can present a problem for an investor who has a portfolio valuable papers in order to obtain continuous income: when the GKO-OFZ payment period expires, it is necessary to reinvest funds, and if interest rates decrease, the portfolio income will also decrease. This type of risk, which is called risk reinvestment rate , is not taken into account in our example, since the period during which the investor owns GKO-OFZ corresponds to their maturity. Finally, we note that relevant yield For any investment, this is the after-tax return, so the return values ​​used to make a decision should reflect income after taxes.

For the other three investment options, the real, or actual, returns will not be known until the end of the respective asset holding periods. Because returns are not known with certainty, these three types of investments are risky .

The probability distributions are discrete or continuous . Discrete distribution probabilities has a finite number of outcomes; so, in table. 3.1 discrete probability distributions of profitability of various investment options are given. The GKO-OFZ yield takes only one possible value, while each of the three remaining alternatives has five possible outcomes. Each outcome is assigned a probability of its occurrence. For example, the probability that GKO-OFZs will yield 8% is 1.00, while the probability that corporate securities will yield 9% is 0.50.

If we multiply each outcome by the probability of its occurrence, and then add the results, we get a weighted average of the outcomes. The weights are the corresponding probabilities, and the weighted average is expected value . Since the outcomes are internal rates of return (Internal Rate of Return, abbreviation IRR), the expected value is expected rate of return (Expected Rate of Return, abbreviation ERR), which can be represented as follows:

ERR = IRRi, (3.1)

where IRRi , - i-th possible Exodus; pi- probability of occurrence of the i-th outcome; P - the number of possible outcomes.

Assume that the decision maker (decision maker) considers several possible solutions: i = 1,…,m. The situation in which the decision maker operates is uncertain. It is only known that there is one of the options: j = 1,…, n. If i -e decision is made, and the situation is j -i, then the firm headed by the decision maker will receive income q ij . The matrix Q = (q ij) is called the matrix of consequences (possible solutions). What decision needs to be made by the LPR? In this situation of complete uncertainty, only some preliminary recommendations can be made. They will not necessarily be accepted by the decision maker. Much will depend, for example, on his risk appetite. But how to assess the risk in this scheme?
Let's say we want to estimate the risk that the i -e decision bears. We do not know the real situation. But if they knew it, they would choose the best solution, i.e. generating the most income. Those. if the situation is j-th, then a decision would be made that gives income q ij .
This means that when making the i -e decision, we risk getting not q j , but only q ij , which means that the adoption of the i -th decision carries the risk of not getting r ij = q j - q ij . The matrix R = (r ij) is called the risk matrix.

Example #1. Let the consequence matrix be
Let's create a risk matrix. We have q 1 = max(q i 1) = 8, q 2 = 5, q 3 = 8, q 4 = 12. Therefore, the risk matrix is

Decision making under complete uncertainty

Wald's rule(rule of extreme pessimism). Considering the i -e solution, we will assume that in fact the situation is the worst, i.e. yielding the smallest income a i But now let's choose a solution i 0 with the largest a i0 . So, Wald's rule recommends making a decision i0 such that
So, in the above example, we have a 1 \u003d 2, a 2 \u003d 2, a 3 \u003d 3, a 4 \u003d 1. Of these numbers, the maximum number is 3. Hence, the Wald rule recommends making the 3rd decision.

Savage's rule(rule of minimum risk). When applying this rule, the risk matrix R = (rij) is analyzed. Considering the i -e solution, we will assume that in fact there is a situation of maximum risk b i = max
But now let's choose a solution i 0 with the smallest b i0 . So, Savage's rule recommends making a decision i 0 such that
In the example under consideration, we have b 1 = 8, b 2 = 6, b 3 = 5, b 4 = 7. The minimum of these numbers is the number 5. I.e. Savage's rule recommends making the 3rd decision.

Hurwitz rule(weighing pessimistic and optimistic approaches to the situation). A decision is made i , on which the maximum is reached
, where 0 ≤ λ ≤ 1 .
The value of λ is chosen from subjective considerations. If λ approaches 1, then Hurwitz's rule approaches Wald's rule, as λ approaches 0, Hurwitz's rule approaches the "pink optimism" rule (guess what that means). In the above example, for λ = 1/2, the Hurwitz rule recommends the 2nd solution.

Decision making under partial uncertainty

Laplace's rule

Example #2. Consider an example of solving a statistical game in an economic problem.
An agricultural enterprise can sell some products:
A1) immediately after cleaning;
A2) during the winter months;
A3) in the spring months.
Profit depends on the selling price in a given period of time, storage costs and possible losses. The amount of profit calculated for different states-ratios of income and costs (S1, S2 and S3), during the entire implementation period, is presented in the form of a matrix (million rubles)

1. Bayes criterion (maximum mathematical expectation)

it means that according to this criterion, strategy A3 is optimal - to sell in the spring months.

2. Laplace's Insufficient Reason Criterion (IUT)

3. Wald's maximin criterion (MM)

4. Criterion of pessimism-optimism Hurwitz (P-O)

5. Hodge-Lehmann criterion (Kh-L)

5. Savage's minimax risk criterion

Conclusion:

1. Strategy A1 (sell immediately after harvest) is not optimal in any of the criteria.
2. Strategy A2 (sell in the winter months) is optimal according to the criteria of insufficient Laplace reason, Wald's maximin criterion and Savage's minimax criterion.
3. Strategy A3 (sell in the spring months) is optimal according to the criteria of Bayes, Hurwitz pessimism-optimism, Hodge-Lehmann.

Example #2. In an ordinary strategic game, each player takes exactly those actions that are most beneficial to him and less beneficial to the enemy. It is assumed that the players are reasonable and antagonistic opponents. However, very often there is uncertainty, which is not associated with the conscious opposition of the enemy, but depends on some objective reality.
The agricultural enterprise has three plots of land: wet, medium moisture and dry. One of these plots is supposed to be used for growing potatoes, the rest - for sowing green mass. To obtain a good potato crop, a certain amount of moisture is required in the soil during the growing season. With excessive moisture, planted potatoes in some areas may rot, and with insufficient rainfall, they will develop poorly, which leads to a decrease in yield. Determine in which area to sow potatoes in order to get a good harvest of it, if the average potato yield in each area is known, depending on weather conditions. Location on A 1 the yield is 200, 100 and 250 centners per 1 hectare with a normal amount of precipitation, respectively, more and less than the norm. Similarly in the area A2- 230, 120 and 200 c, and on the site A 3- 240, 260 and 100 c.
Let's use a game approach. Agricultural enterprise - player A, which has three strategies: A 1- sow potatoes in a moist area, A2- in an area of ​​medium humidity, A 3- in a dry area. Player P- nature, which has three strategies: P 1 corresponds to less than normal rainfall, P 2- norm, P 3- more than normal. The payoff of an agricultural enterprise for each pair of strategies ( Ai, P j) is given by the potato yield per 1 ha.

Example 9 The company produces popular children's dresses and suits, the sale of which depends on the state of the weather. The company's costs during August-September per unit of production amounted to: dresses - 7 den. units, costumes - 28 den. units The selling price is 15 and 50 den. units respectively. According to observations over several previous years, the company can sell 1,950 dresses and 610 suits in warm weather, and 630 dresses and 1,050 suits in cool weather.
Create a payment matrix.
Solution. The firm has two strategies: A 1: release products, assuming that the weather will be warm; A2: release products, assuming that the weather will be cool.
Nature has two strategies: B1: the weather is warm; B2: the weather is cool.
Let's find the payoff matrix elements:
1) a 11 - the company's income when choosing a strategy A 1 on condition B1:
a 11 \u003d (15-7) 1950 + (50-28) 610 \u003d 29020.
2) a 12 - the income of the firm when choosing A 1 on condition B2. The company will produce 1,950 dresses and sell 630, income from the sale of dresses
(15-7) 630-7 (1950-630)=5040-9240
a 12 \u003d 5040-9240 + 22 610 \u003d 9220.
3) similarly for the strategy A2 in conditions B1 the firm will produce 1,050 suits and sell 610;
a 21 =8 630+22 610-28 (1050-610)=6140
4) a 22 \u003d 8 630 + 22 1050 \u003d 28140
Payment matrix:

Example 2 . The association carries out mineral exploration at three deposits. The pool of means of association makes 30 den. units Money in the first deposit M1 can be invested in multiples of 9 den. unit, second M2– 6 den. unit, third M3– 15 den. units Mineral prices at the end of the planning period may be in two states: C1 and C2. The experts found that in the situation C1 profit from the mine M1 will be 20% of the amount invested den. units for development, for M2– 12% and M3- fifteen %. In a situation C1 at the end of the planned period, the profit will be 17%, 15%, 23% at the fields M1, M3, M3 respectively.
Player A- an association. Player P(nature) - a set of external circumstances that determine one or another profit in the fields. Player A there are four possibilities that make full use of available funds. The first strategy A 1 is that A will invest in M 1 9 days units, in M 2 - 6 den. units, in M 3 - 15 den. units Second strategy A 2: in M 1 - 18 den. units, in M 2 - 12 den. units, in M 3 do not invest money. The third strategy A 3: 30 den. units invest in M 3 . The fourth strategy A four:. 30 den. units invest in M 2. Briefly, one can write A 1 (9, 6, 15), A 2 (18, 12, 0), A 3 (0, 0, 30), A 4 (0, 30, 0).
Nature can realize one of its two states, characterized by different prices for minerals at the end of the planning period. Denote the states of nature P 1 (20 %, 12 %, 15 %), P 2 (17 %, 15 %, 23 %).
The elements a ij of the payoff matrix have the meaning of the total profit obtained by combining into different situations (Ai, P j) (i=1, 2, 3, 4, j= 1, 2). For example, let's calculate a 12 corresponding to the situation ( A 1, P 2), i.e., the case when the association invests in deposits M 1 , M 2 , M 3, respectively 9 den. units, 6 den. units, 15 den. units, and at the end of the planning period, prices were in the state C2:
a 12\u003d 9 0.17 + 6 0.15 + 15 0.23 \u003d 5.88 den. units

Example 3 . Floods are expected to come, which may have a category from the first to the fifth. Flood damage:

Games with "nature"

• the first pure strategy assumes that the entire piece of land will be sown with crop A 1 ;
• the second pure strategy assumes that the entire plot of land will be sown with crop A 2 ;
• the third pure strategy assumes that the entire area will be planted with crop A 3 .

• dry weather, which corresponds to the first pure strategy B 1 ;
• normal weather, which corresponds to the second pure strategy B 2 ;
• rainy weather, which corresponds to the third pure strategy B 3 .

Example. Planning of production output under different states of nature - demand market.
An enterprise can produce 4 types of products: A 1, A 2, A 3, A 4, while making a profit. Its value is determined by the state of demand (the nature of the market), which can be in one of four possible states: B 1 , B 2 , B 3 , B 4 . The dependence of the amount of profit on the type of product and the state of the market is presented in the table:

The task of the second player loss minimizationV	The task of the first player payoff maximizationV
objective function
F / = x 1 +x 2 +x 3 +x 4 =→ max	F = y 1 + y 2 + y 3 + y 4 =→ min
Functional limitations
Direct Restrictions
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, x 4 ≥ 0	y 1 ≥ 0, y 2 ≥ 0, y 3 ≥ 0, y 4 ≥ 0

Example. The company plans to sell its products on the markets, taking into account possible options for consumer demand P j , j=1.4 (low, medium, high, very high). The company has developed three strategies for selling goods A 1 , A 2 , A 3 . The volume of trade (money units), depending on the strategy and consumer demand, is presented in the table.

Solution find with a calculator.
Bayes criterion.
According to the Bayes criterion, the (pure) strategy A i is taken as optimal, at which the average gain a is maximized or the average risk r is minimized.
We consider the values ​​∑(a ij p j)
∑(a 1,j p j) = 33 0.3 + 10 0.2 + 20 0.4 + 26.5 0.1 = 22.55
∑(a 2,j p j) = 50 0.3 + 67 0.2 + 11.5 0.4 + 25 0.1 = 35.5
∑(a 3,j p j) = 23.5 0.3 + 35 0.2 + 40 0.4 + 58.5 0.1 = 35.9

The management of the company decides to place the production of a new product in a certain place. To form an idea of ​​the situation on the market of a new product at the time of production development, it needs to take into account the costs of delivery finished products to the consumer, the development of the transport and social infrastructure of the region, competition in the market, the ratio of supply and demand, exchange rates and much more. Possible solutions, the investment attractiveness of which is defined as the percentage of income growth in relation to the amount of capital investments, are presented in the table.
Choose:
1) a place to locate production, if the head of the enterprise is sure that situation 4 will develop on the market;
2) a place to locate production, if management estimates the probability of situation 1 at 0.2; situations 2 in 0.1; situations 3 in 0.25;
3) select a variant under conditions of uncertainty according to the criterion: maximax, maximin, Laplace's criterion, Savage's criterion, Hurwitz's criterion (y = 0.3);
4) will the best solution according to the Hurwitz criterion change if the value of a is increased to 0.5?
5) assuming that these tables represent the costs of the enterprise, determine the choice that the enterprise will make when using each of the following criteria: maximin; maximax; Hurwitz criterion (? = 0.3); Savage's criterion; Laplace criterion

Typical tasks

1. Select the optimal project for construction using the criteria of Laplace, Wald, maximum optimism, Savage and Hurwitz at a=0.58. The cost matrix looks like:

   0.07	0.26	0.11	0.25	0.1	0.21
   68	45	54	79	47	99
   56	89	42	56	74	81
   72	87	56	40	62	42
   65	48	75	89	52	80
   69	93	93	56	45	43
   73	94	79	68	67	46
   66	100	64	89	94	49
   70	42	97	42	42	50

2. A retailer has developed several options for a plan to sell goods at an upcoming fair, taking into account the changing market conditions and customer demand, the profits resulting from their possible combinations are presented in the form of a payoff matrix. Determine the best plan for selling goods.
   x=0.7
3. The company plans to sell its products on the markets, taking into account possible options for consumer demand Пj, j=1͞,4͞ (low, medium, high, very high). The company has developed three strategies for selling goods A 1 , A 2 , A 3 . The volume of trade (money units), depending on the strategy and consumer demand, is presented in the table.

   A j	P j
   	P 1	P 2	P 3	P 4
   A 1	30+N	10	20	25 + N/2
   A 2	50	70-N	10 + N/2	25
   A 3	25-N/2	35	40	60-N

   
   Where N=3
   Possible states of consumer demand are known, which, respectively, q 1 =0.3, q 2 =0.2, q 3 =0.4, q 4 =0.1. It is necessary to find a sales strategy that maximizes the average turnover of the firm. In this case, use the criteria of Wald, Hurwitz, Savage, Bayes.
   Solution
4. The cost of the factory during April - May per unit of production was: dresses - 8 monetary units, suits - 27, and the selling price is 16 and 48, respectively. According to past observations, the factory can sell during these months in warm weather conditions 600 suits and 1975 dresses, and in cool weather - 625 dresses and 1000 suits.