Zlp graphic method examples. Graphical method for solving linear programming problems: diagram and examples

The graphical method for solving the ZLP is based on the statements given in paragraph 2.1. According to Theorem 2, the optimal solution is at the top of the domain of feasible solutions and therefore to solve the ZLP is to find the top of the domain of feasible solutions, the coordinates of which give optimal value objective function.

The graphical method is used to solve a limited class of problems with two variables, sometimes with three variables. It should be noted that for three variables this area is not clear enough.

Algorithm for the graphical method of solving problems

We will consider the implementation of the graphical method for solving the ZLP using examples.

Example 2.2.1. Decide ZLP graphic method:

(2.2.1)

max z=x 1 + 4x 2 (2.2.2)

Solution. To construct a region of feasible solutions, which consists of the intersection of half-planes corresponding to each inequality of the system of constraints (2.2.1), we write the equations of the boundary straight lines:

l 1: x 1 + 5x 2 = 5; l 2: x 1 + x 2 = 6; l 3: 7x 1 + x 2 = 7.

l 1 to the form (2.2.3.) we divide both its parts by 5:
. Thus, straight l 1 cuts on axis Oh 1 5 units, on axis Oh 2 1 unit. Similarly we have for l 2:
And l 3:
.

To determine half-planes that meet the constraints of system (2.2.1), you need to substitute the coordinates of any point that does not lie on the boundary line into the constraints. If we obtain a true inequality, then all points from this half-plane are solutions to this inequality. Otherwise, choose another half-plane.

Thus, the first and second desired half-planes are located in the opposite direction from the origin of coordinates (0 – 5 0 - 5; 7 0 + 0 7), and the second – towards the origin of coordinates (0 + 0 6). The region of feasible solutions in Figure 2.2.1 is shaded.

Figure 2.2.1 – Area of ​​feasible solutions

To find the optimal plan, which will be located at the vertex of the solution polygon, you need to construct a vector of directions
=(With 1 ,With 2), which indicates the direction of the greatest increase in the objective function z=With 1 X 1 +With 2 X 2 .

In this problem, the direction vector
= (1, 4): it starts at the point ABOUT(0,0) and ends at the point N(1, 4).

Next, we construct a straight line that passes through the region of feasible solutions, perpendicular to the vector, and is called target level line functions. We move the level line in the direction of the vector in case of maximizing the objective function z and in the opposite direction, in the case of minimization z, until the last intersection with the region of feasible solutions. As a result, the point or points are determined where the objective function reaches an extreme value, or the unboundedness of the objective function is established z on the set of solutions to the problem.

Thus, the maximum point of the objective function z is the point A line intersections l 2 and l 3 .

To calculate the optimal value of the objective function z find the coordinates of the point A . Since the point A is the point of intersection of the lines l 2 and l 3, then its coordinates satisfy a system of equations composed of the equations of the corresponding boundary lines:

So the point A has coordinates x 1 =1/6, x 2 = 35/6.

To calculate the optimal value of the objective function, you need to substitute the coordinates of the point into it A .

Substituting the coordinates of the point A into the objective function (2.4), we obtain

max z = 1/6 + 4·(35/6) = 47/2.

Example 2.2.2. Construct on the plane the region of feasible solutions of the system of linear inequalities (2.2.4) and find the largest and smallest values ​​of the objective function (2.2.5):

(2.2.4)

z= –2x 1 –x 2 (2.2.5)

Solution. To construct a region of feasible solutions, which consists of the intersection of half-planes corresponding to each inequality of the system of constraints (2.2.4), we write the equations of the boundary straight lines:

l 1: 4x 1 – x 2 = 0; l 2: x 1 + 3x 2 = 6; l 3: x 1 – 3x 2 = 6; l 4: x 2 = 1.

Straight l 1 passes through the point with coordinates (0;0). To construct it, we express x 2 through x 1: x 2 = 4x 1 . Let's find another point through which the line passes l 1, for example (1;4). Through the point with coordinates (0;0) and the point with coordinates (1;4) we draw a straight line l 1 .

To reduce the equation of a straight line l 2 to the form in segments on the axes (2.2.3), we divide both of its parts by 6:
. Thus, straight l 2 cuts on axis Oh 1 6 units, on axis Oh 2 - 2 units. Similarly we have for l 3:
and Direct l 4 parallel to the axis Oh 1 and passes through the point with coordinates (0;1) .

To determine half-planes that meet the constraints of system (2.2.4), it is necessary to substitute the coordinates of any point that does not lie on the boundary line into the constraints. Due to restrictionsX 1 0, X 2 0, the region of admissible solutions of the ZLP lies in the first quarter of the coordinate plane.

ABOUT
the area of ​​feasible solutions in Figure 2.2.2 is shaded.

Figure 2.2.2 – Area of ​​feasible solutions

Let's construct a vector of directions
= (–2,–1). Next, we build a level line perpendicular to the vector .

To find highest value of the objective function, we move the level line in the direction of the vector until the last intersection with the region of feasible solutions. Thus, the maximum point of the objective function z is the point A(intersection of lines l 1 and l 2).

To calculate the optimal value of the objective function z find the coordinates of the point A. Since the point A is the point of intersection of the lines l 1 and l 2, then its coordinates satisfy a system of equations composed of the equations of the corresponding boundary lines:

So the point A has coordinates x 1 =6/13, x 2 = 24/13.

Substituting the coordinates of the point A into the objective function (2.2.5), we obtain the optimal value of the objective function

max z= – 2·(6/13) – (24/13) = – 36/13.

To find the smallest value of the objective function, we move the level line in the direction opposite to the vector until the last intersection with the region of feasible solutions. In this case, the objective function is unlimited in the region of feasible solutions, i.e. ZLP has no minimum.

As a result of the decision of the PPP, the following cases are possible:

The objective function reaches its optimal value at a single vertex of the solution polygon;

The objective function reaches its optimal value at any point on the edge of the solution polygon (the ZLP has alternative reference plans with the same values z );

The PAP does not have optimal plans;

ZLP has an optimal plan in the case of an unlimited range of feasible solutions.

Task. Solve the problem graphically linear programming, having determined the extreme value of the objective function:

under restrictions

Let us construct a region of feasible solutions, i.e. Let's solve the system of inequalities graphically. To do this, we construct each straight line and define the half-planes defined by the inequalities (the half-planes are indicated by a prime).

Let's build the equation 3x 1 +x 2 = 9 at two points.
To find the first point, we equate x 1 = 0. We find x 2 = 9. To find the second point, we equate x 2 = 0. We find x 1 = 3. We connect the point (0;9) with (3;0) with a straight line. Let us define the half-plane defined by the inequality. Having chosen the point (0; 0), we define the inequality sign in the half-plane: 3. 0 + 1 . 0 - 9 ≤ 0, i.e. 3x 1 +x 2 - 9≥ 0 in the half-plane above the straight line.
Let's construct the equation x 1 +2x 2 = 8 at two points.
To find the first point, we equate x 1 = 0. We find x 2 = 4. To find the second point, we equate x 2 = 0. We find x 1 = 8. We connect the point (0;4) with (8;0) with a straight line. Let us define the half-plane defined by the inequality. Having chosen the point (0; 0), we define the inequality sign in the half-plane: 1. 0 + 2 . 0 - 8 ≤ 0, i.e. x 1 +2x 2 - 8≥ 0 in the half-plane above the straight line.
Let's build the equation x 1 + x 2 = 8 at two points.
To find the first point, we equate x 1 = 0. We find x 2 = 8. To find the second point, we equate x 2 = 0. We find x 1 = 8. We connect the point (0;8) with (8;0) with a straight line. Let us define the half-plane defined by the inequality. Having chosen the point (0; 0), we define the inequality sign in the half-plane: 1. 0 + 1 . 0 - 8 ≤ 0, i.e. x 1 +x 2 - 8≤ 0 in the half-plane below the straight line.

The intersection of half-planes will be a region whose point coordinates satisfy the inequalities of the system of constraints of the problem.
Let us denote the boundaries of the area of ​​the solution polygon.

You can check the correctness of the construction of function graphs using a calculator

Consider the objective function of the problem F = 4x 1 +6x 2 → min.
Let's construct a straight line corresponding to the value of the function F = 0: F = 4x 1 +6x 2 = 0. The gradient vector, composed of the coefficients of the objective function, indicates the direction of minimization of F(X). The beginning of the vector is point (0; 0), the end is point (4; 6). We will move this line in a parallel manner. Since we are interested in the minimal solution, we therefore move the straight line until it first touches the designated area. On the graph, this straight line is indicated by a dotted line.

Straight F(x) = 4x 1 +6x 2 intersects the region at point B. Since point B is obtained as a result of the intersection of lines (1) And (2) , then its coordinates satisfy the equations of these lines:
3x 1 +x 2 =9
x 1 +2x 2 =8

Having solved the system of equations, we get: x 1 = 2, x 2 = 3
Where do we find it from? minimum value objective function:
F(X) = 4*2 + 6*3 = 26

Brief theory

Linear programming is a branch of mathematical programming used in developing methods for finding the extremum of linear functions of several variables for linear additional restrictions, imposed on the variables. According to the type of problems solved, his methods are divided into universal and special. By using universal methods Any linear programming problems (LPP) can be solved. Special methods take into account the features of the problem model, its objective function and system of constraints. A feature of linear programming problems is that the objective function reaches an extremum at the boundary of the region of feasible solutions.

Graphical method solving linear programming problems makes it possible to visualize their structure, identify features and opens up ways to study more complex properties. A linear programming problem with two variables can always be solved graphically. However, already in three-dimensional space such a solution becomes more complicated, and in spaces with dimensions greater than three, a graphical solution is, generally speaking, impossible. The case of two variables does not have any particular practical significance, but its consideration clarifies the properties of the LLP constraints, leads to the idea of ​​solving it, and makes the solution methods and ways of their practical implementation geometrically clear.

If the constraints and the objective function contain more than two variables, then it is necessary (or by the method of sequential improvement of the solution) - it is universal and can be used to solve any problem. For some applied problems linear programming, such as, special solution methods have been developed.

Example of problem solution

The task

The enterprise produces two types of products: Product 1 and Product 2. To manufacture a unit of Product 1, it is necessary to spend kg of raw materials of the first type, kg of raw materials second type, kg of raw materials of the third type. To manufacture a unit of Product 2, it is necessary to spend kg of the first type, raw materials of the second type, and raw materials of the third type. Production is provided with raw materials of each type in quantities of kg, kg, kg, respectively. The market price of a unit of Product 1 is thousand rubles, and a unit of Product 2 is thousand rubles.

Required:

• Construct a mathematical model of the problem.
• Draw up a production plan for products that ensures maximum revenue from their sale using a graphical method for solving a linear programming problem.

To ensure that the solution to a linear programming problem is as accurate and correct as possible, many inexpensively order test on that website. You can read more details (how to submit a request, prices, deadlines, payment methods) on the page Buy a test paper on linear programming...

The solution of the problem

Model building

Let and denote the number of manufactured products of the 1st and 2nd types.

Then resource restrictions:

In addition, according to the meaning of the task

The target function of the economic-mathematical model, expressing the revenue received from sales:

We get the following economic and mathematical model:

Construction of the domain of feasible solutions

Let's solve the resulting linear programming problem graphically:

To construct a region of feasible solutions, we construct boundary lines corresponding to these inequalities in the coordinate system:

Let's find the points through which the lines pass:

The solution to each inequality of the ZLP constraint system is a half-plane containing the boundary line and located on one side of it.

To define a half-plane, take any point, for example, not belonging to line (1), and substitute the coordinates (0;0) into the corresponding inequality. Because the inequality is true:

The solution region of the corresponding 1st inequality corresponds to the left half-plane

Let's take any point, for example, that does not belong to line (2), and substitute the coordinates (0;0) into the corresponding inequality. Because the inequality is true:

Let's take any point, for example, that does not belong to line (3), and substitute the coordinates (0;0) into the corresponding inequality. Because the inequality is true:

The solution region of the corresponding 2nd inequality corresponds to the left half-plane

The region of feasible solutions is the figure.

Finding a solution to the LP problem

We construct a vector whose coordinates are proportional to the coefficients of the objective function. Here is the proportionality coefficient.

Draw a level line perpendicular to the constructed vector.

We move the level line in the direction of the vector so that it touches the region of feasible solutions at the extreme point. The solution to the maximum is the point , the coordinates of which are found as the point of intersection of lines (2) and (1).

Answer

Thus, it is necessary to produce 56 products of the 1st type and 64 products of the 2nd type. In this case, revenue from the sale of products will be maximum and amount to 5104 monetary units.

Method graphic solution, if a problem with two variables has linear constraints and the objective function is quadratic, discussed in detail here
The page discusses in detail the solution to the linear programming problem simplex method, in addition, the construction is shown dual problem linear programming and finding its solution by solving a direct problem.

Transport problem and potential method
The transport problem, its mathematical model and solution methods are considered in detail - finding the reference plan by the minimum element method and searching for the optimal solution by the potential method.

Convex programming - graphical method
An example of solving a quadratic convex programming problem using a graphical method is given.

Solving a linear programming problem (LPP) using a graphical method

General formulation of the plot

Find the values ​​of n variables x 1 , x 2 , …, x n that deliver an extremum (minimum or maximum) linear function Z=C 1 x 1 ,+ C 2 x 2+…+ C n x n

and simultaneously satisfying m restrictions of the form

a 1.1 x 1 +a 1.2 x 2 +…+a 1.n x n£ =≥b 1 ,

a 2.1 x 1 +a 2.2 x 2 +…+a 2.n x n£ = ≥b 2 ,

. . . . . . . . . . . . . . . . . . . . . . .,

a m,1 x 1 +a m,2 x 2 +…+a m,n x n£ = ≥b m ,

for given a i,j , b i, C j (i=1,2,…,m; j=1,2,…,n). The relation sign can take any of the three given values.

Example of a linear programming problem

Let's consider the following problem. The manager of a company producing two types of paints described to an operations researcher the situation in the production and marketing of paints. It turned out that the factory produces two types of paints: for interior and external works. Both colors come in wholesale. For the production of paints, two initial products are used - A and B. The maximum possible daily reserves of these products are 6 and 8 tons, respectively. Experience has shown that the daily demand for external paint never exceeds the demand for internal paint by more than 1 ton. In addition, it is found that the demand for exterior paint never exceeds 2 tons per day. Wholesale prices for one ton of paint are as follows: 3 thousand rubles for external paint and 2 thousand rubles for internal paint. How much of each type of paint should the factory produce to maximize sales revenue?

To solve the problem posed to the researcher, it is first necessary to develop a mathematical model of the described situation.

When constructing a mathematical model, the operations researcher asks himself three questions.

• For what quantities should the model be built? In other words, you need to identify the task variables.
• What restrictions must be placed on the variables so that the conditions characteristic of the system being modeled are met?
• What is the goal, to achieve which, from all possible (admissible) values ​​of variables, it is necessary to select those that will correspond to the optimal (best) solution to the problem?

Let's introduce the variables:

x 1 – daily production volume of external paint (in tons),

x 2 – daily production volume of interior paint (in tons).

Considering Wholesale prices per ton of each type of paint, the daily income from the sale of manufactured products is given by the linear objective function Z = 3x 1 + 2x 2.

The goal of production is to obtain maximum profit, which means it is necessary to find the values ​​of x 1 and x 2 that maximize the objective function Z.

Because the paint manufacturer cannot control the values ​​of the variables randomly, so it is necessary to identify the set of possible values ​​of these variables, which is determined by the specific conditions of production and sales. This set is called a region acceptable values.

The first type of constraint is determined by the stocks of products A and B from which paints are produced. From production technology it is known that two parts of product A are used to produce a ton of external paint, and one part is used to produce a ton of internal paint. For product B the relationship is reversed. These technological conditions are described by the inequalities

2x 1 + x 2 £ 6 (6 tons of product A in stock),

x 1 + 2x 2 £ 8 (there are 8 tons of product B in stock).

The last two restrictions mean an obvious circumstance: you cannot use more products A and B for the production of paints than are actually in stock.

The situation with the sale of paints on the market leads to the following restrictions: x 1 – x 2 £ 1 (external paint is sold no more than one ton more than internal paint), x 1 £ 2 (external paint is sold no more than two tons per day).

Summarizing all that has been said, a mathematical model that describes the current production situation can be specified in the following form:

find® max( Z=2× x 1 + 3× x 2 ) at following restrictions on the values ​​of the variables x 1 and x 2

2 × x 1 + x 2 £ 6 limitation (1),

X 1 + 2 × x 2 £ 8 limitation (2),

X 1 - x 2 £ 1 limitation (3),

X 1 £ 2 constraint (4)

and the requirement that the variables x 1 ³ 0 (5), x 2 ³ 0 (6) be non-negative.

Received mathematical model is a linear programming problem.

Graphical method solutions to the problem

The graphical method for solving the problem can only be implemented in the two-dimensional case.

The mathematical model obtained for the formulated standard problem requires research, since it is not known in advance whether it has (how math problem) solution. We will conduct the study using graphic constructions. Simultaneously with such research, we will find (if there is one) a solution.

Stage 1. Construction of the domain of feasible solutions

The goal is to construct a region in which every point satisfies all the constraints.

Each of the six constraints geometrically defines a half-plane. In order to build it, you need:

• · replace the inequality sign in the constraint with equality (we get the equation of a straight line);

• · construct a straight line using two points;

• · determine which half-plane is specified by the inequality sign. To do this, substitute some point into the inequality (for example, the origin of coordinates). If it satisfies the inequality, we paint the half-plane containing it.

We perform these actions for all restrictions. We denote each of the lines by the numbers adopted when numbering the restrictions (see figure).

Area of ​​feasible solutions (satisfying all restrictions) is the set of points of the first quadrant of the coordinate plane (x 1, x 2), which is the intersection of all half-planes defined by the inequalities of the restrictions.

The set of points that satisfy all six constraints of the problem is the polygon AFEDCB.

Stage 2: Construction of target function level lines and determination of the maximum point

The goal is to find in the constructed polygon AFEDCB is the point at which the objective function Z=2x 1 + 3x 2 takes on its maximum value.

Let's draw a straight line 2x 1 + 3x 2 = Const (level line) so that it intersects the polygon AFEDCB (for example, Const = 10). This level line is shown as a dotted line in the figure.

If we consider the values ​​of the linear objective function Z on a set of points (x 1 , x 2) belonging to a segment of the dotted line located inside the hexagon, then they are all equal to the same value (Const = 10).

Let us determine the direction of increase of the function. To do this, we will construct a level line with higher value. This will be a straight line, parallel to the one built, but located to the right. This means that in a given direction the value of the objective function increases, and it is in our interests to move it as far as possible in this direction.

The shift can be continued as long as the moving straight line intersects the polygon of feasible solutions. The last position of the line, when it has one common point with the polygon AFEDCB (point C), corresponds to the maximum value of the objective function Z and is achieved at point C with coordinates x 1 = 4/3 (" 1.333), x 2 = 10/3 (" 3.333). In this case, Z = 38/3 (» 12.667).

The task was completely solved. From the geometric reasoning carried out it is clear that the solution is unique. Let us make some generalizations arising from the geometric interpretation of the problem.

First. The region of feasible solutions is a convex polygon ( Why convex? Can the region of feasible solutions be an empty set? Period? Line segment? Ray? Directly? If yes, give an example of a restriction system).

Second. The maximum of the objective function is achieved at the vertex of the polygon of feasible solutions ( But maybe there is not the only solution? Could there be no solution?)

Task 1 (complete in class, show to teacher)

Solve graphically

A) x1=6 x2=4 F=24

B) x1=2 x2=3 F=26

C) x1О x2=8-x1 F=24

Task 2 (complete in class, show to teacher)

Answer the questions in italics.

Task 3 (homework)

Write a program.

Dan text file kind

2 3 (objective function coefficients)

4 (number of restrictions)

2 2 12 (restrictions)

1 2 8

4 0 16

0 4 12

Construct straight lines so that the polygon of feasible solutions is entirely on the screen (for the definition of scale, see Onegov’s book). Straight lines can be parallel to the axes!

Construct several lines at the level of the objective function (press the key - the straight line moves, the value of the objective function is displayed). Show scale.

An important method of scientific analysis of statistical material is graphic images. The first attempts to use graphical methods in economic research began in the 1780s. However, the graphical method received wider use later - in the middle of the 18th century, especially after the report of the representative of the Berlin statistical bureau Schwabe, “Theory”, made for the first time in the history of statistics. graphic images"at the 8th International Statistical Congress (St. Petersburg, 1872). According to the well-known expression of the German physicist F. Auerbach, the 20th century was marked by the "triumphant advance of the graphical method in science."

What is a schedule? A graph is a form visual representation statistical data on socio-economic phenomena and processes through geometric images, drawings or schematic geographic Maps and explanations for them.

A graph has five main elements general design: field, coordinate grid, graphic signs and their placement in the graphic field, scale and legend (Fig. 10.3).

Rice. 10.3. Basic elements of a chart

Each of these elements has its own purpose and plays a corresponding role in construction and interpretation. The graphic field is the space on which the geometric and other signs that make up the graphic image are placed.

A graphic image is a set of various symbolic signs with the help of which statistical data is reflected. These signs can be depicted in the following forms: lines, dots, geometric, graphic, and sometimes non-geometric figures.

A coordinate grid is a rectangular coordinate system in which time is plotted on the abscissa axis, and quantitative indicators of scale are plotted on the ordinate axis.

Scale is a conditional measure of converting the numerical value of a statistical phenomenon into a graphic one and vice versa. It is used to install numerical values phenomena expressed on the graph.

Explication of a graph is a verbal explanation of its specific content, which usually includes:

1) title with the necessary additional explanations;

2) an exact explanation of the essence, conditionally provided in this chart its graphic signs (geometric, pictorial, background, purely conventional)

3) other explanations, notes, etc.

In addition, you can put some additional information, for example, numerical data that is reflected in some graphic signs and repeats them in digital form exact values, expressed graphically.

Graphs play a particularly important role in studying the complex interrelations of socio-economic phenomena and processes, identifying trends, patterns and changes in dynamics, as well as in ongoing analysis. The main differences and advantages of the graphical method compared to others are: better visibility; the ability to generally cover the data of those being studied; the ability to express some analytical dependencies that are not very clear and difficult to identify with other methods of data presentation.

With the help of schedules, you can exercise operational control over production, sales of products, fulfillment of contractual obligations and assigned tasks. Thus, the schedules are assigned:

To summarize and analyze data;

Data distribution image;

Identification of patterns of development of the studied phenomena and processes in dynamics;

Reflection of the interrelations of indicators;

Monitoring production, implementation of sales contracts, etc.

There are different classifications of graphs - according to shape graphic images, in terms of content and nature of the tasks.

Based on the shape of graphic images, they are distinguished following types graphs:

1) point;

2) linear;

3) planar;

4) volumetric;

5) artistic (visual, conventional).

In scatter plots, the volume of a population is expressed either by a single point or by an accumulation of points. One point can mean one case or several (for example, one plant, 500 workers).

Linear graphs consist of only lines: straight segments, broken lines, stepped, smooth curves (mainly to convey the dynamics of the population). Often straight segments are replaced with strips same width, which also act as graphic signs but with one dimension (length). In such cases, graphs are called bar graphs if the stripes are placed vertically, or strip graphs when the stripes are horizontal.

In turn, column graphs are divided into column diagrams: simple and solid, from groups of columns, etc., and strip diagrams are divided into strip diagrams: simple and stepwise, component-wise, sliding, bilaterally directed (for example, an “age pyramid” of population composition) .

TO special types linear graphs include spiral ones (for phenomena that develop indefinitely in time and in increasing magnitude), radial diagrams(to display patterns of periodically repeating phenomena, their rhythm, seasonality).

Planar graphs are graphs of two dimensions in the form of planes of different geometric shapes. Depending on this, they can be square, circular, sector. It is advisable to use these graphs to compare phenomena represented by absolute and relative values.

Important features of planar charts are the two-dimensional "Warzar sign", the strip or current chart and the balance chart.

The two-dimensional “Varzar sign” (named after its inventor, the Russian statistician V.E. Varzar) is a rectangle with a base a, height b and area Sab, which is useful for graphically expressing fairly common similar relationships between the three quantities a, by S.

A strip, or current, diagram is used to schematically express the volume and composition of cargo flows between two points in one and the second direction.

A balance sheet chart is a two-sided strip chart, the ribbons of which branch in two directions into narrower strips, their width expressing the corresponding values ​​of income and expense items, asset and liability items, and the like.

Volumetric - 3D graphics, which are rarely used because they are less expressive compared to linear and planar ones.

Artistic (visual, conventional) - graphs with conventional graphic signs that reflect the totality or its individual meanings in the form of human figures, animal outlines, schematic drawings of objects, etc.

The classification of graphs according to their content is of great importance. Taking this into account, graphs are divided into two classes - diagrams and statistical maps.

A diagram is a graphic expression of the volumes and characteristics of one or more aggregates using quantitative graphic symbols (geometric, artistic, background, purely conventional).

However, the diagram does not give graphical representation o territorial placement of depicted populations or territorial changes in their characteristics. For this purpose, statistical maps are used, designed to depict the territorial distribution of populations or territorial changes in their characteristics. They are divided into two classes - cartograms and map diagrams.

Cartograms are contour geographic maps on which the quantitative territorial characteristics of a population are presented using graphic symbols.

Map diagrams are contour geographic maps, where individual areas (regions, points) of a territory are plotted with the same type of diagram (one or more), depicting the volume and territorial features of similar populations in these areas. So, for example, the flow of goods, passengers transported, population migrating, and the like are depicted.

Diagrams and statistical maps perform the following important tasks in population research:

General comparison of them;

Study of structure;

Study of dynamics;

Studying the relationships between their characteristics;

Measuring the degree of implementation of economic plans and contractual obligations in economic planning practice.

In turn, both diagrams and cartograms, depending on their purpose, are divided into subclasses, groups and forms (Table 10.27).

When constructing graphs, the following requirements must be observed:

1) rely on reliable numerical data;

2) graphs must be meaningful in design and interesting in content;

3) must be built in accordance with the assigned tasks and their practical purpose;

4) be extremely economical - contain a maximum of information and ideas with a minimum of means of graphic expression, simple, clear, understandable;

5) technically well executed.

Let's take a closer look at the main types and forms of diagrams and statistical maps, which are most often used in the practice of analytical work.

A line diagram is one of the most common types of graphs, which serves to depict the dynamics of the phenomena under study. To construct it, a rectangular coordinate system is used. Equal segments are laid out on the abscissa axis - periods of time (days, months, years, etc.), and on the ordinate axis a scale is adopted that characterizes the units of measurement. Points are drawn on the coordinate field that are equal to the value of the indicator on certain period. Then all the points are connected by straight lines, resulting in a broken line that characterizes the change in the phenomenon being studied over a certain period of time (Table 10.28, Fig. 10.4).

Table 10.28. Investments in fixed capital in housing construction in Ukraine in 2000-2005 pp., in actual prices, million UAH

The graph data shows that the volume of investments in fixed capital in housing construction in Ukraine in actual prices grew from 2000 to 2005

Rice. 10.4. Dynamics of the volume of investments in fixed capital in housing construction in Ukraine in 2000-2005, in actual prices, million UAH

Planned-line graphs are built on a specially designed grid, where time units are laid out horizontally, and research objects are placed vertically. Moreover, each horizontal segment corresponds to 100% completion of the planned task. These segments are divided into 5 equal parts, each of which corresponds to 20% of the planned task.

The degree of plan implementation on the graph is depicted by two lines: a thin dashed line - per unit of time (day, decade) and a solid bold line - for the reporting period as a whole.

Let's look at the procedure for constructing a planned linear graph using an example.

Example. Build line graph fulfillment of the planned task by a team of workers from construction and installation works, using the data in Table. 10.29.

Table 10.29. Fulfillment of the planned task by a team of workers from construction and installation works

The schedule for completing the planned task by the construction team for construction and installation work is presented in Fig. 10.5.

The thin continuous line of the first day corresponds to 90% of the plan and occupies four and a half cells, and the line of the second day - 80% and occupies four cells, the line of the third day stretches exactly five, and the fourth - five cells (100%) plus an additional one the segment below, which occupies 20%, etc.

Depicting the level of plan implementation on an accrual basis requires some additional calculations. So, on the first day, the solid thick line will be the same length as the thin continuous line - 90% and will occupy four and a half cells. Next, the following calculations should be made: in two days, 513 m2 (225 + 288) were actually completed. Of this amount, 250 m2 is credited to the implementation of the plan for the first day. Then, on the second day, 263 m2 will remain, which, according to the plan, on this day is 91% (263,288).

According to the bold line, it occupies five cells of the first day and 91% of the second. In three days, 923 m2 (225 + 288 + 410) were actually completed. 610 m2 is recorded for the completion of the plan for the first two days, and 313 m2 for the third day, which, according to the plan for this day, is 76% (313: 410). The thick line will occupy 5 cells of the first and second days and 76% of the third. All further calculations are carried out similarly. The degree of implementation of the plan for each day is indicated by dots on the thick line.

Column chart- a very common type of graphs in one dimension due to their clarity and simplicity. Statistical data in them is depicted in the form of rectangles of the same width, located vertically along a horizontal line (Fig. 10.6).

The height of the columns should correspond to the magnitude of the phenomena depicted. If the bars are placed horizontally, then such a graph is called a strip graph (Fig. 10.7).

Column and strip charts allow you to compare values different meanings, characterize the same phenomenon in dynamics; characterize the population.

Pie charts (or pie charts) are diagrams designed to display the structure of the phenomena and processes being studied. They are depicted in the form of a circle divided into sectors, the sizes of which correspond to the sizes of the depicted phenomena (Fig. 10.8).

As evidenced by the graph (Fig. 10.8), the main source of financing for leasing operations in Ukraine is bank loans (80.9%), then own funds (16.1%). Borrowed funds legal entities account for only 3.6%.

Rice. 10.6. Dynamics of the volume of investments in fixed capital in housing construction in Ukraine in 2000-2005 pp., in actual prices, million UAH

Rice. 10.7. Dynamics of the volume of investments in fixed capital in housing construction in Ukraine in 2000-2005 pp., in actual prices, million UAH

IN modern conditions development of information and computer systems, it became possible to build graphs using packages computer programs, including electronic EXCEL tables, "Statistica-6", etc. They are easy to use and greatly simplify this work.

Rice. 10.8. Structure of sources of financing for leasing operations in Ukraine at the beginning of 2005 p.,%