Find the extrema of the function using the Lagrange multiplier method. Lagrange method (variation of constant). Linear differential equations of the first order

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Article topic:	Lagrange method.
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Finding a polynomial means determining the values ​​of its coefficient . To do this, using the interpolation condition, you can form a system of linear algebraic equations(SLAU).

The determinant of this SLAE is usually called the Vandermonde determinant. Vandermonde's determinant is not equal to zero when for , that is, in the case when there are no matching nodes in the lookup table. However, it can be argued that the SLAE has a solution and this solution is unique. Having solved the SLAE and determined the unknown coefficients you can construct an interpolation polynomial.

A polynomial that satisfies the interpolation conditions, when interpolated by the Lagrange method, is constructed in the form of a linear combination of polynomials of the nth degree:

Polynomials are usually called basic polynomials. In order to Lagrange polynomial satisfies the interpolation conditions, it is extremely important that the following conditions are satisfied for its basis polynomials:

For .

If these conditions are met, then for any we have:

Moreover, the fulfillment of the specified conditions for the basis polynomials means that the interpolation conditions are also satisfied.

Let us determine the type of basis polynomials based on the restrictions imposed on them.

1st condition: at .

2nd condition: .

Finally, for the basis polynomial we can write:

Then, substituting the resulting expression for the basis polynomials into the original polynomial, we obtain final look Lagrange polynomial:

A particular form of the Lagrange polynomial at is usually called the linear interpolation formula:

.

The Lagrange polynomial taken at is usually called the quadratic interpolation formula:

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Multiplier methodLagrange(in English literature “LaGrange's method of undetermined multipliers”) ˗ is a numerical method for solving optimization problems that allows you to determine the “conditional” extremum objective function(minimum or maximum value)

in the presence of specified restrictions on its variables in the form of equalities (i.e., the area acceptable values)

˗ these are the values ​​of the function argument ( controlled parameters) on the real domain at which the value of the function tends to an extremum. The use of the name “conditional” extremum is due to the fact that the variables are subject to additional condition, which limits the range of acceptable values ​​when searching for the extremum of a function.

The Lagrange multiplier method allows the problem of searching for a conditional extremum of an objective function on a set of admissible values ​​to be transformed into a problem without conditional optimization functions.

In case the functions And are continuous along with their partial derivatives, then there are such variables λ that are not simultaneously equal to zero, under which the following condition is satisfied:

Thus, in accordance with the Lagrange multiplier method, to find the extremum of the objective function on the set of admissible values, I compose the Lagrange function L(x, λ), which is further optimized:

where λ ˗ is a vector of additional variables called indefinite factors Lagrange.

Thus, the problem of finding the conditional extremum of the function f(x) has been reduced to the problem of finding the unconditional extremum of the function L(x, λ).

And

The necessary condition for the extremum of the Lagrange function is given by a system of equations (the system consists of “n + m” equations):

Solving this system of equations allows us to determine the arguments of the function (X) at which the value of the function L(x, λ), as well as the value of the target function f(x) correspond to the extremum.

The magnitude of the Lagrange multipliers (λ) is of practical interest if the constraints are presented in the form with a free term in the equation (constant). In this case, we can consider further (increase/decrease) the value of the objective function by changing the value of the constant in the equation system. Thus, the Lagrange multiplier characterizes the rate of change in the maximum of the objective function when the limiting constant changes.

There are several ways to determine the nature of the extremum of the resulting function:

First method: Let be the coordinates of the extremum point, and be the corresponding value of the objective function. A point close to the point is taken and the value of the objective function is calculated:

If , then there is a maximum at the point.

If , then there is a minimum at the point.

Second method: A sufficient condition from which the nature of the extremum can be determined is the sign of the second differential of the Lagrange function. The second differential of the Lagrange function is defined as follows:

If in given point minimum, if , then the objective function f(x) has a conditional maximum.

Third method: Also, the nature of the extremum of the function can be determined by considering the Hessian of the Lagrange function. The Hessian matrix is ​​a symmetrical square matrix second partial derivatives of a function at the point at which the matrix elements are symmetrical about the main diagonal.

To determine the type of extremum (maximum or minimum of a function), you can use Sylvester’s rule:

1. In order for the second differential of the Lagrange function to be of positive sign it is necessary that the angular minors of the function be positive. Under such conditions, the function at this point has a minimum.

2. In order for the second differential of the Lagrange function to be negative in sign , it is necessary that the angular minors of the function alternate, and the first element of the matrix must be negativesv. Under such conditions, the function at this point has a maximum.

By angular minor we mean the minor located in the first k rows and k columns of the original matrix.

The main practical significance of the Lagrange method is that it allows you to move from conditional optimization to unconditional optimization and, accordingly, expand your arsenal available methods solving the problem. However, the problem of solving the system of equations to which this method is reduced is general case no easier original problem searching for an extremum. Such methods are called indirect. Their use is explained by the need to obtain a solution to an extremal problem in analytical form (for example, for certain theoretical calculations). When solving specific practical problems, direct methods are usually used, based on iterative processes of calculating and comparing the values ​​of the functions being optimized.

Calculation method

1 step: We determine the Lagrange function from the given objective function and system of constraints:

Forward

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The method for determining a conditional extremum begins with constructing an auxiliary Lagrange function, which in the region of feasible solutions reaches a maximum for the same values ​​of variables x 1 , x 2 , ..., x n , which is the same as the objective function z . Let the problem of determining the conditional extremum of the function be solved z = f(X) under restrictions φ i ( x 1 , x 2 , ..., x n ) = 0, i = 1, 2, ..., m , m < n

Let's compose a function

which is called Lagrange function. X , - constant factors ( Lagrange multipliers). Note that Lagrange multipliers can be given an economic meaning. If f(x 1 , x 2 , ..., x n ) - income consistent with the plan X = (x 1 , x 2 , ..., x n ) , and the function φ i (x 1 , x 2 , ..., x n ) - costs of the i-th resource corresponding to this plan, then X , is the price (estimate) of the i-th resource, characterizing the change in the extreme value of the objective function depending on the change in the size of the i-th resource (marginal estimate). L(X) - function n+m variables (x 1 , x 2 , ..., x n , λ 1 , λ 2 , ..., λ n ) . Determining the stationary points of this function leads to solving the system of equations

It's easy to see that . Thus, the task of finding the conditional extremum of the function z = f(X) reduces to finding the local extremum of the function L(X) . If a stationary point is found, then the question of the existence of an extremum in the simplest cases is resolved on the basis sufficient conditions extremum - study of the sign of the second differential d 2 L(X) at a stationary point, provided that the variable increments Δx i - connected by relationships

obtained by differentiating the coupling equations.

Solving a system of nonlinear equations in two unknowns using the Find Solution tool

Settings Finding a solution allows you to find a solution to the system without linear equations with two unknowns:

Where
- nonlinear function of variables x And y ,
- arbitrary constant.

It is known that the couple ( x , y ) is a solution to system of equations (10) if and only if it is a solution to the following equation with two unknowns:

WITH on the other hand, the solution to system (10) is the intersection points of two curves: f ] (x, y) = C And f 2 (x, y) = C 2 on surface XOY.

This leads to a method for finding the roots of the system. nonlinear equations:

Determine (at least approximately) the interval of existence of a solution to the system of equations (10) or equation (11). Here it is necessary to take into account the type of equations included in the system, the domain of definition of each of their equations, etc. Sometimes the selection of an initial approximation of the solution is used;

Tabulate the solution to equation (11) for the variables x and y on the selected interval, or construct graphs of functions f 1 (x, y) = C, and f 2 (x,y) = C 2 (system(10)).

Localize the supposed roots of a system of equations - find several minimum values from the table, tabulate the roots of equation (11), or determine the intersection points of the curves included in system (10).

4. Find the roots for the system of equations (10) using the add-on Finding a solution.

Classification of mathematical programming problems

PROGRAMMING

METHODS FOR SOLVING NONLINEAR PROBLEMS

Control questions to section 4

Solution diagram transport problem

Let us list the main stages of solving the transport problem.

1. Check the closed condition. If the task is open, the transport table is supplemented with either a column of a fictitious point of consumption or a row of a fictitious supplier.

2. Build a reference plan.

3. Check the support plan for non-degeneracy. If there is not enough occupied cell to satisfy the non-degeneracy condition, one of the cells of the transport table is filled with a supply equal to zero. If necessary, it is permissible to record zero deliveries in several cells.

4. The plan is checked for optimality.

5. If the optimality conditions are not met, move on to the next plan by redistributing supplies. The computational process is repeated until the optimal plan is obtained.

1. What is the meaning of the objective function in the mathematical model of the transport problem?

2.What is the meaning of restrictions in the mathematical model of the transport problem?

3. Is it possible to apply the potential method to solve an open (unclosed) transport problem?

4.What changes need to be made to the original transport table so that the problem can be solved by the potential method?

5.What is the essence of the minimum element method? What stage of solving the transport problem will be completed as a result of applying this method?

6. How do you know if the transportation plan is optimal?

7. In what case and how is it necessary to redistribute supplies in terms of transportation?

8. Suppose the constructed transportation plan is degenerate. Is it possible to continue solving the problem using the potential method and what needs to be done for this?

General task mathematical programming was formulated in section 1.1. Depending on the type of functions included in the model (1.1)-(1.3), the problem is classified as one or another type of mathematical programming. There are linear programming (all functions are linear), integer (the solution is represented by integers), quadratic (the objective function is a quadratic form), nonlinear (at least one of the functions of the problem is nonlinear) and stochastic programming (parameters that are probabilistic in nature are included).

The task class is not linear programming wider class linear models. For example, production costs in most cases are not proportional to the volume of output, but depend on it nonlinearly, income from the sale of production products turns out to be a nonlinear function of prices, etc. The criteria in optimal planning problems are often maximum profit, minimum cost, and minimum capital costs. As variables output volumes are various types products. Constraints include production functions that characterize the relationship between product output and labor and labor costs. material resources, the volume of which is limited.

Unlike linear programming, which uses universal method solutions (simplex method), for solving nonlinear problems there is a whole range of methods depending on the form of the functions included in the model. Of the variety of methods, we will consider only two: the Lagrange method and the dynamic programming method.

WITH The essence of the Lagrange method is to reduce the problem to conditional extremum to solving the unconditional extremum problem. Consider the nonlinear programming model:

(5.2)

Where – known functions,

A – given coefficients.

Note that in this formulation of the problem, the constraints are specified by equalities, and there is no condition for the variables to be non-negative. In addition, we believe that the functions are continuous with their first partial derivatives.

Let us transform conditions (5.2) so that on the left or right sides of the equalities there is zero:

(5.3)

Let's compose the Lagrange function. It includes the objective function (5.1) and the right-hand sides of the constraints (5.3), taken respectively with the coefficients . There will be as many Lagrange coefficients as there are constraints in the problem.

The extremum points of function (5.4) are the extremum points of the original problem and vice versa: the optimal plan of problem (5.1)-(5.2) is the global extremum point of the Lagrange function.

Indeed, let a solution be found problems (5.1)-(5.2), then conditions (5.3) are satisfied. Let's substitute the plan into function (5.4) and verify the validity of equality (5.5).

Thus, in order to find the optimal plan for the original problem, it is necessary to examine the Lagrange function for the extremum. The function has extreme values ​​at points where its partial derivatives are equal zero. Such points are called stationary.

Let us define the partial derivatives of the function (5.4)

,

.

After equalization zero derivatives we get the system m+n equations with m+n unknown

, (5.6)

In the general case, system (5.6)-(5.7) will have several solutions, which will include all the maxima and minima of the Lagrange function. In order to highlight the global maximum or minimum, the values ​​of the objective function are calculated at all found points. The largest of these values ​​will be the global maximum, and the smallest will be the global minimum. In some cases it turns out possible use sufficient conditions for a strict extremum continuous functions (see Problem 5.2 below):

let the function be continuous and twice differentiable in some neighborhood of its stationary point (i.e. )). Then:

A) If ,(5.8)

then is the strict maximum point of the function;

b) If ,(5.9)

then is the strict minimum point of the function;

G ) If ,

then the question of the presence of an extremum remains open.

In addition, some solutions of system (5.6)-(5.7) may be negative. Which is inconsistent with the economic meaning of the variables. In this case, you should consider replacing negative values ​​with zero values.

Economic meaning of Lagrange multipliers. Optimal multiplier value shows how much the criterion value will change Z when the resource increases or decreases j by one unit, since

The Lagrange method can also be used in the case where the constraints are inequalities. Thus, finding the extremum of the function under conditions

,

performed in several stages:

1. Determine stationary points of the objective function, for which they solve a system of equations

.

2. From the stationary points, select those whose coordinates satisfy the conditions

3. Using the Lagrange method, solve the problem with equality constraints (5.1)-(5.2).

4. Examine the points found in the second and third stages for the global maximum: compare the values ​​of the objective function at these points - highest value corresponds to the optimal plan.

Problem 5.1 Let us solve problem 1.3, considered in the first section, using the Lagrange method. Optimal distribution water resources are described by a mathematical model

.

Let's compose the Lagrange function

Let's find the unconditional maximum of this function. To do this, we calculate the partial derivatives and equate them to zero

,

Thus, we obtained a system of linear equations of the form

The solution to the system of equations represents an optimal plan for the distribution of water resources across irrigated areas

Values ​​are measured in hundreds of thousands of cubic meters. - the amount of net income per one hundred thousand cubic meters of irrigation water. Therefore, the marginal price of 1 m 3 of irrigation water is equal to den. units

The maximum additional net income from irrigation will be

160·12.26 2 +7600·12.26-130·8.55 2 +5900·8.55-10·16.19 2 +4000·16.19=

172391.02 (den. units)

Problem 5.2 Solve a nonlinear programming problem

Let's represent the limitation in the form:

.

Let's compose the Lagrange function and determine its partial derivatives

.

To determine the stationary points of the Lagrange function, its partial derivatives should be set equal to zero. As a result, we obtain a system of equations

• Tutorial

Good afternoon everyone. In this article I want to show one of graphic methods construction mathematical models For dynamic systems which is called bond graph(“bond” - connections, “graph” - graph). In Russian literature, I found descriptions of this method only in Tomsky’s Textbook Polytechnic University, A.V. Voronin “MODELING OF MECHATRONIC SYSTEMS” 2008 Also show the classical method through the Lagrange equation of the 2nd kind.

Lagrange method

I will not describe the theory, I will show the stages of calculations with a few comments. Personally, it’s easier for me to learn from examples than to read theory 10 times. It seemed to me that in Russian literature, the explanation of this method, and indeed mathematics or physics in general, is very rich in complex formulas, which accordingly requires a serious mathematical background. While studying the Lagrange method (I study at the Polytechnic University of Turin, Italy), I studied Russian literature to compare calculation methods, and it was difficult for me to follow the progress of solving this method. Even remembering the modeling courses at the Kharkov Aviation Institute, the derivation of such methods was very cumbersome, and no one bothered themselves in trying to understand this issue. This is what I decided to write, a manual for constructing mathematical models according to Lagrange, as it turned out it is not at all difficult, it is enough to know how to calculate derivatives with respect to time and partial derivatives. For more complex models, rotation matrices are also added, but there is nothing complicated in them either.

Features of modeling methods:

• Newton-Euler: vector equations based on dynamic equilibrium force And moments
• Lagrange: scalar equations based on state functions associated with kinetic and potential energies
• Bond Count: flow based method power between system elements

Let's start with simple example. Mass with spring and damper. We ignore the force of gravity.

Fig 1. Mass with spring and damper

First of all, we designate:

• initial system coordinates(NSK) or fixed sk R0(i0,j0,k0). Where? You can point your finger at the sky, but by twitching the tips of the neurons in the brain, the idea passes through to place the NSC on the line of movement of the M1 body.
• coordinate systems for each body with mass(we have M1 R1(i1,j1,k1)), the orientation can be arbitrary, but why complicate your life, set it with minimal difference from the NSC
• generalized coordinates q_i(the minimum number of variables that can describe the movement), in in this example one generalized coordinate, movement only along the j axis

Fig 2. We put down coordinate systems and generalized coordinates

Fig 3. Position and speed of body M1

Then we will find the kinetic (C) and potential (P) energies and the dissipative function (D) for the damper using the formulas:

Fig 4. Complete formula for kinetic energy

In our example there is no rotation, the second component is equal to 0.

Fig 5. Calculation of kinetic, potential energy and dissipative function

The Lagrange equation has the following form:

Fig 6. Lagrange Equation and Lagrangian

Delta W_i This virtual work perfected by the applied forces and moments. Let's find her:

Fig 7. Calculation of virtual work

Where delta q_1 virtual movement.

We substitute everything into the Lagrange equation:

Fig 8. The resulting mass model with spring and damper

This is where Lagrange's method ended. As you can see, it’s not that complicated, but it’s still a very simple example, for which most likely the Newton-Euler method would even be simpler. For more complex systems, where there will be several bodies rotated relative to each other at different angles, the Lagrange method will be easier.

Bond graph method

I’ll show you right away what the model looks like in bond-graph for an example with a mass, a spring and a damper:

Fig 9. Bond-graph masses with spring and damper

Here you will have to tell a little theory, which will be enough to build simple models. If anyone is interested, you can read the book ( Bond Graph Methodology) or ( Voronin A.V. Modeling of mechatronic systems: tutorial. – Tomsk: Tomsk Polytechnic University Publishing House, 2008).

Let us first determine that complex systems consist of several domains. For example, an electric motor consists of electrical and mechanical parts or domains.

bond graph based on the exchange of power between these domains, subsystems. Note that power exchange, of any form, is always determined by two variables ( variable power ) with the help of which we can study the interaction of various subsystems within a dynamic system (see table).

As can be seen from the table, the expression of power is almost the same everywhere. In summary, Power- This work " flow - f" on " effort - e».

An effort(English) effort) in the electrical domain this is voltage (e), in the mechanical domain it is force (F) or torque (T), in hydraulics it is pressure (p).

Flow(English) flow) in the electrical domain it is current (i), in the mechanical domain it is speed (v) or angular velocity (omega), in hydraulics it is the flow or flow rate of fluid (Q).

Taking these notations, we obtain an expression for power:

Fig 10. Power formula through power variables

In the bond-graph language, the connection between two subsystems that exchange power is represented by a bond. bond). That's why this method is called bond-graph or g raf-connections, connected graph. Let's consider block diagram connections in a model with an electric motor (this is not a bond-graph yet):

Fig 11. Block diagram of power flow between domains

If we have a voltage source, then accordingly it generates voltage and transfers it to the motor for winding (this is why the arrow is directed towards the motor), depending on the resistance of the winding, a current appears according to Ohm’s law (directed from the motor to the source). Accordingly, one variable is an input to the subsystem, and the second must be exit from the subsystem. Here the voltage ( effort) – input, current ( flow) - exit.

If you use a current source, how will the diagram change? Right. The current will be directed to the motor, and the voltage to the source. Then the current ( flow) – input, voltage ( effort) - exit.

Let's look at an example in mechanics. Force acting on a mass.

Fig 12. Force applied to mass

The block diagram will be as follows:

Fig 13. Block diagram

In this example, Strength ( effort) – input variable for mass. (Force applied to mass)
According to Newton's second law:

Mass responds with speed:

In this example, if one variable ( force - effort) is entrance into the mechanical domain, then another power variable ( speed - flow) – automatically becomes exit.

To distinguish where the input is and where the output is, it is used vertical line at the end of an arrow (connection) between elements, this line is called sign of causality or causation (causality). It turns out: applied force is the cause, and speed is the effect. This sign is very important for the correct construction of a system model, since causality is a consequence of the physical behavior and exchange of powers of two subsystems, therefore the choice of location of the causality sign cannot be arbitrary.

Fig 14. Designation of causality

This vertical line shows which subsystem receives the force ( effort) and as a result produce a flow ( flow). In the example with mass it would be like this:

Fig 14. Causal relationship for the force acting on the mass

It is clear from the arrow that the input for mass is - force, and the output is speed. This is done so as not to clutter the diagram with arrows and systematize the construction of the model.

Next important point. Generalized impulse(amount of movement) and moving(energy variables).

Table of power and energy variables in different domains

The table above introduces two additional physical quantities used in the bond-graph method. They're called generalized impulse (R) And generalized movement (q) or energy variables, and they can be obtained by integrating power variables over time:

Fig 15. Relationship between power and energy variables

In the electrical domain :

Based on Faraday's law, voltage at the ends of the conductor is equal to the derivative of the magnetic flux through this conductor.

A Current strength - physical quantity, equal to the ratio the amount of charge Q passing through the cross section of the conductor in some time t to the value of this time interval.

Mechanical domain:

From Newton's 2nd law, Force– time derivative of impulse

And correspondingly, speed- time derivative of displacement:

Let's summarize:

Basic elements

All elements in dynamic systems can be divided into two-pole and four-pole components.
Let's consider bipolar components:

Sources
There are sources of both effort and flow. Analogy in the electrical domain: source of effort – voltage source, stream source – current source. Causal signs for sources should only be like this.

Fig 16. Causal connections and designation of sources

Component R – dissipative element

Component I – inertial element

Component C – capacitive element

As can be seen from the figures, different elements of the same type R,C,I described by the same equations. ONLY there is a difference for electrical capacitance, you just need to remember it!

Quadrupole components:

Let's look at two components: a transformer and a gyrator.

The last important components in the bond-graph method are the connections. There are two types of nodes:

That's it with the components.

The main steps for establishing causal relationships after constructing a bond-graph:

1. Give causal connections to everyone sources
2. Go through all the nodes and put down causal relationships after point 1
3. For components I assign an input causal relationship (effort is included in this component), for components C assign output causality (effort comes out of this component)
4. Repeat point 2
5. Insert causal connections for R components

This concludes the mini-course on theory. Now we have everything we need to build models.
Let's solve a couple of examples. Let's start with electrical circuit, it is better to understand the analogy of constructing a bond-graph.

Example 1

Let's start building a bond-graph with a voltage source. Just write Se and put an arrow.

See, everything is simple! Let's look further, R and L are connected in series, which means the same current flows in them, if we speak in power variables - the same flow. Which node has the same flow? The correct answer is 1-node. We connect the source, resistance (component - R) and inductance (component - I) to the 1-node.

Next, we have capacitance and resistance in parallel, which means they have the same voltage or force. 0-node is suitable like no other. We connect the capacitance (component C) and resistance (component R) to the 0-node.

We also connect nodes 1 and 0 to each other. The direction of the arrows is chosen arbitrarily; the direction of the connection affects only the sign in the equations.

You will get the following connection graph:

Now we need to establish causal relationships. Following the instructions for the sequence of their placement, let's start with the source.

1. We have a source of voltage (effort), such a source has only one causality option - output. Let's put it.
2. Next there is component I, let’s see what they recommend. We put
3. We put it down for 1-node. Eat
4. A 0-node must have one input and all output causal connections. We have one day off for now. We are looking for components C or I. We found it. We put
5. Let's list what's left

That's all. Bond graph is built. Hurray, Comrades!

All that remains is to write the equations that describe our system. To do this, create a table with 3 columns. The first will contain all the components of the system, the second will contain the input variable for each element, and the third will contain the output variable for the same component. We have already defined the input and output by causal relationships. So there shouldn't be any problems.

Let's number each connection for ease of recording the levels. We take the equations for each element from the list of components C, R, I.

Having compiled a table, we define the state variables, in this example there are 2 of them, p3 and q5. Next you need to write down the equations of state:

That's it, the model is ready.

Example 2. I would like to immediately apologize for the quality of the photo, the main thing is that you can read

Let's solve another example for mechanical system, the same one that we solved using the Lagrange method. I will show the solution without comment. Let's check which of these methods is simpler and easier.

In Matbala, both mathematical models with the same parameters were compiled, obtained by the Lagrange method and bond-graph. The result is below: Add tags