Linear programming graphic online. The simplex method of solving problems of linear programming. Graphical method for decoupling LP tasks

The graphical method is simple and basic to complete the task of LP from two changes. Vіn foundations on geometric submission of admissible solutions and digital filter tasks.

The skin of the irregularities of the LP control is indicated on the coordinate plane (X 1 ,X 2 ) deyak a half-plane (Fig. 1), and a system of irregularities in a halo - a ridge of circular planes. Nameless point acceptablesolution(ODR). ODR zavzhd є opuklu figure, tobto. if power can come: if two points A and B lie on this figure, then the entire line of AB lies on it. ODR can be graphically represented by a swollen bagatokutnik, a non-surrounded swollen rich-cut area, a vіdrіzk, a change, one point. At the time of the inconsistency of the system, the boundary of the ODR head is an empty multiplicity.

Note № 1. All the foregoing can be seen and to the point, if the system of obmezheniya (1.1) includes equality, shards be-like equality

a il x 1 + a i 2 x 2 =b

possible at the sight of a system of two irregularities (Fig. 1)

A i 2 x 2<Ь 1э +a i 2 x 2 >bj.

ZF L(x)= s1x1 + s2x2 with a fixed value L(x)=L is drawn on the plane as a straight line s1x1 + c2x2 = L. By changing the value of L, we take away the family of parallel lines, titles lines of equal.

It is due to the fact that the change in the value of L will cause the change to be less long-term, that it will show the line on the x2 axis (cob ordinate), and the top coefficient of the straight line tgа = - will become permanent (Fig. 1).

For that, it is enough to induce one of the lines of the line, enough to turn the value of L.

Vector C = (c1; c2) with coordinates of CF coefficients at x1 and x2 perpendicular to the skin line (Fig. 1). straight aheadvector Z zbіgaєtsya from straight ahead growth station ZF, which is an important moment for the accomplishment of tasks. straight ahead decay CF oppositelystraight vector C.

The essence of the graphical method is the y axis. Directly (against) the vector C in the ODS, a search for the optimal point X = (x1; x2 ). The optimal point is considered to be the one to pass the line L max (L min), which gives the highest (least) value of the function L (x). The optimal solution is always to be found on the border of the ODR, for example, in the remaining top of the ODR bagatokutnik, which passes the entire straight line, or on the entire side.

When looking for the optimal solution of LP problems, it is possible to: іsnuє impersonal solution (alternative optium); ZF is not fenced; the range of admissible solutions is a single point; There is no solution.

Permissible area - nap_vploschina

Malyunok 1

1.2. Technique for solving problems with the graphic method

I. Replace the signs of unevenness with signs of exact evenness at the interchanges and stay straight.

II. Know and shade in the areas allowed by the skin borders-irregularities of the plant. For which one, substantiate the coordinates of any point [for example, (0; 0)] into a specific unevenness, that reverses the truth of the taken away unevenness.

Yakscho unevenness is true, then it is required to shade the napіvploshchina, to avenge that point; otherwise (Nerіvnіst khibne) it is required to shade in a flat plane, so as not to avenge the given point.

Shards x1 and x2 must be non-negative, their admissible values ​​will always be higher than axis x 1 and right than axis x2, then. at the 1st quadrant.

Obezhennya-equalities allow only those points, if they lie on a straight line, so you can see such straight lines on the graph.

Appoint ODR as a part of the area, which should be applied to all permissible areas at once, that see it. For the daytime of the ODR not maє decision, about what to build a vіdpovіdny vysnovok.

Yakshcho ODR - do not be empty without a face, then stay straight ahead, tobto. be-yaku z line equal z 1 x 1 + z 2 x 2 = L de L - more number, for example, a multiple of 1 i z 2, then. better than rozrahunkiv. The method of inducing a similar one to inducing direct delimitation.

V. Make a vector C = (c 1, z 2), which starts at point (0; 0), ends at point (c 1, z 2). Yakshcho tsіlova is straight and vector perpendicular.

VI. When asking for max ZF, move the whole line straight ahead of the vector C with a joke min CF - anti straightening vector C. rest in the course of the move, the top of the GDR will be the point max or min ZF. If such a point (point) is not available, then make a visnovok nezamezhenosti CF on impersonal plans to the top (pіd hour poshuk shah) chi below (pіd h poshuku min).

Set the coordinates of the point max(min) CF X = (х1*; х2* ) and calculate the values ​​of the digital filter l(x *). To calculate the coordinates of the optimal point X *, untie the system of alignments of straight lines, on the periphery of which there are X * .

Head 1

We know the optimal solution of the problem, the mathematical model can be seen

L(X) = 3x1 + 2x2 → max

x 1 + 2x 2< 6, (1)

2x 1 + x 2< 8, (2)

x 1 + x 2<1, (3)

x 2< 2, (4)

x 1 >0, x 2 >0.

Let's create a straight line, for which we calculate the coordinates and points of intersection of these lines with the coordinate axes (Fig. 2).

x 1 + 2x 2 = 6, (1)

2x1+ x2 = 8, (2)

(1) x1 = 0, x1 = 6, x2 = 3, x2 = 0,

(2) x1 = 0, x1 = 4, x2 = 8, x2 = 0,

(3) x1 = 0, x1 = -1, x2 = 1, x2 = 0,

The straight line (4) passes through the point x 2 = 2 in parallel to the axis L(X).

Rice. 2. Graphical breakdown of tasks

Significantly ODR. For example, let's put a point (0; 0) at the exit of the exchange (3), take 0< 1, что является истинным неравенством, поэтому стрелкой (или штрихованием) обозначим полуплоскость, take revenge point (0; 0), then. ruffled to the right and lower straight (3). Similarly, it is significant that the admissible napіvshchini for the other borders and, for example, the arrows at the other straight lines (Fig. 2). spіlnoyu region, permitted by usima obezhennyami, tobto. ODR є bagatokutnik ABCDEF.

Tsіlova can be directly encouraged for jealousy

We will be a vector of 3 points (0; 0) to the point (3; 2). Point E-ce is the remaining peak of the bagatokutnik of admissible solutions ABCDEF, through the yak pass the whole straight line, collapsing off straight vector C. To that E tse point to the maximum of the digital filter. Significantly, the coordinates of the point E from the system of equalization of straight lines (1) and (2)

X1+2x2=6, (1) x1=10/3=3 1/3, x2=4/3=1 1/3

2 X1 + x 2 = 8, (2) E 3 1/3; 1 1/3

The maximum value of the digital filter is more L(E) = 3*10/3+2*4/3 = 12 2/3

Let's take a look at the simplest ways, if the ZLP includes exactly two changes:

The skin of irregularities (a)-(b) of the system of boundary problems (3.8) geometrically expands the plane in a similar way to the boundary lines , X 1 =0 and X 2 =0. The skin from the boundary straight lines subdivides the area x 1 x 2 on two pivploschini. The solution to the visual unevenness is to lie at one of the illuminated half-planes (all points of the half-plane) and, then, when the coordinates are set, whether or not the points appear to be the same. With a glance at the target, that half-plane appears, in which lie the unraveling of nervousness, tobto. a way to choose whether it is a point, whether it is a flat surface, or a substitution of її coordinates at a certain unevenness. As unevenness is vikonuetsya for the tsієї point, it is vikonuєtsya and for any other point from the tsієї w napіvploschiny. In a different direction, the breakdown of unevenness lies in a different flat.

In times, since the system of irregularities (a)-(b) is coherent, then the area of ​​її rozvyazkіv є impersonal point, which should lie with the assigned half-planes. If the point of the border of these flat surfaces is convex, then the area of ​​admissible expansions of the problem (3.8) is the swell of the absence, as it is called the rich cuter of the pink tongues (the earlier introductions of the term "the rich border of the pink tongues" sound like n 3). The sides of this bagatokutnik lie on straight lines, evenly they walk exit system obmezhen by replacing the signs of irregularities with signs of exact evenness.

In this way, the output of the ZLP is determined by the significance of such a point of the bagatokutnik solution, in which the target function F gains the maximum (minimum) value.

This point is correct, if the bagatokutnik of roses is not empty and on a new purpose the function is surrounded by the beast. For the significance of the minds of one of the vertices of the bagatokutnik, the solution of the goal function gains the maximum value. To define the given vertices, the line L will be line L: c 1 x 1 + c 2 x 2 \u003d h (de h is the constant), perpendicular to the vector-gradient i pass through the bagatokutnik solution, i pass її parallel to the vector-gradient doti, while still out do not pass through the remaining її corner point of the crossbar with a bagatokutnik solution (when the vector-gradient is called, add a point (z 1; z 2) at the plane x 1 Ox 2 and draw the cob of coordinates of the straight lines of the vertices to it). The coordinates of the assigned points determine the optimal plan for the given task.

Let's create an algorithm for the graphical method of solving the ZLP.

Algorithm for the graphical method of untying the ZLP

1. Encourage a bagatokutnik solution that is set by the system of encirclement of the external ZLP.

2. As if prompted by a bagatokutnik of roses - an empty lot, then the ZLP of roses cannot be released. In a different way, induce the vector-gradient and draw an additional line of the line L, shifting to the maximum at the direction of the vector (or at the reverse direction for the task to the minimum), designate the extreme point of the boundary of the solution, and reach the maximum (min.functional) function

3. Calculate the coordinates of the found optimal point by arranging the system of alignments of two boundary lines that overlap into it.

4. Substituting the found optimal solution for the goal function of the task, calculate the optimal її value, tobto: .

With a graphical impersonality of admissible solutions of the ZLP (rozv'yazkiv) such situations are possible.

The development of problems of linear programming by the graphical method is reviewed. Description of the method. Apply rozvyazannya tasks.

Div. also: Problem solving by the simplex method

Description of the method

As for the tasks of linear programming, there are only two changes, and you can solve it using a graphical method.

Let's look at the problem of linear programming with two changes:
(1.1) ;
(1.2)
There are quite a few numbers here. The task is possible both for the value of the maximum (max), and for the value of the minimum (min). At the system, the boundary can be like signs, і signs.

Pobudova area of ​​acceptable solutions

Graphical method of solving problems (1) offensive.
On the back, we draw the coordinate axes and select the scale. The skin of the irregularities of the system bordering (1.2) defines a flat plane, surrounded by a straight line.

So, first nervousness
(1.2.1)
I signify a flat plane, I will border it with a straight line. From one side in the direction of the straight line, from the other side. On the very straight line. In order to find out from which side the unevenness (1.2.1) is pointed out, we choose a sufficient point that does not lie on a straight line. Next, we present the coordinates of the point (1.2.1). If the unevenness is overcome, then the flat is to avenge the chosen point. As if the unevenness does not vanish, the flat area is expanded from the other side (do not avenge the point). Shaded on the surface, unevenness is drawn on the yak (1.2.1).

The same is true for other irregularities of the system (1.2). So we take away the shading of the infill areas. The points of the region of admissible solutions satisfy all the inconsistencies (1.2). Therefore, graphically, the area of ​​​​permissible solutions (ODR) is the limit of all stimuli. Shaded ODR. Vaughn є opuklim with a bagatokutnik, the edges of which lie straight. So ODR can be an uncircumcised swollen figure, a vіdrіzk, we will exchange it or a straight one.

You can blame and such a fall that the flats do not avenge the sleeping points. Todi the area of ​​admissible solutions is impersonal. There is no such decision.

You can ask the method. It is possible not to shade the skin on the surface, but on the back of the head to sound all straight
(2)
Dali choose a sufficient point, so as not to lie down any of these straight lines. Submit the coordinates of the point y for the system of irregularities (1.2). Since all the irregularities are fixed, then the area of ​​admissible solutions is surrounded by straight lines and including the chosen point. Shading the area of ​​​​admissible solutions beyond the boundaries of the straight lines so that it includes the selected point.

If we want one unevenness not to be overcome, then we choose another point. And so far, the docks will not find one point, the coordinates of which satisfy the system (1.2).

The value of the extremum of the goal function

Also, maybe the area of ​​​​admissible decisions (ODR) is shaded. Vaughn is surrounded by laman, which is made up of windings and changes, which should be straight (2). ODR zavzhdi є opuklim impersonality. It can be like a fringed impersonality, so it’s not a fringed vzdovzh of such direct ones.

Now we can find the extremum of the goal function
(1.1) .

For whom we choose, be it the number that will be direct
(3) .
For the sake of clarity of the distant week, it is important that I go straight through the ODR. On this direct target function is constant and equal. such a direct line of function is called. Tsya directly breaks the flat into two flats. On one pіvploschinі
.
On the other pіvploschіnі
.
So from one side in a straight line (3) the target function grows. I scho gave me a speck in the straight line (3), then it will be more important. From the lower side in the straight line (3) the target function changes. If you gave me a point in the straight line (3) in the next book, then it will be less important. If we draw a straight line, parallel to the straight line (3), then the new straight line will also be the line of the target function, but with other values.

In this way, in order to know the maximum value of the goal function, it is necessary to draw a straight line, parallel to the straight line (3), as far as possible in the direction of its growth, and pass through one point of the ODR. To know the minimum value of the target function, you need to draw a straight line, parallel to the straight line (3) and as far as possible in the direction of the change in the value , and want to pass through one point of the ODR.

For example, the ODR is not bordered, you can win a vipadok, if you can’t carry out such a straight line. Because we didn’t see a straight line in the line (3) at the bik growth (change), then we’ll go straight through the ODR. In this vipadka, you can be as always great (malim). There is no maximum (minimum) value for that. There is no solution.

We can look at the slope, if it is extremely straight, parallel to the fairly straight view (3), to pass through one peak of the ODR bagatokutnik. From the graphics, the coordinates of the vertex are visible. Then the maximum (minimum) value of the target function is assigned to the following formula:
.
Problem solving є
.

Also, there can be a sharp drop, if the straight line is parallel to one of the faces of the ODR. Todi is straight to pass through two peaks of the ODR bagatokutnik. The coordinates of the vertices are visible. To determine the maximum (minimum) value of the target function, you can select the coordinates of any of these vertices:
.
The manager may be an impersonal decision. To the decisions, whether it be a point, it is sorted into a vіrіzku between the points ta, including the points themselves ta.

An example of solving problems of linear programming by a graphical method

The company manufactures fabrics of two models A and B. With which fabrics are made on three types. On the preparation of one cloth model A, 2 m of fabric of the first type, 1 m of fabric of another type, 2 m of fabric of the third type are required. For the preparation of one cloth model, 3 m of fabric of the first type, 1 m of fabric of another type, 2 m of fabric of the third type are required. Stocks of textiles of the first type become 21 m, another type - 10 m, third type - 16 m. one, one virobu type B - 300 den. one.

Put together a plan of production that will secure the maximum income for the company. The task is to scribble in a graphical way.

Let them change and mean the number of spun cloth models A and B are obvious. To the same number of stained fabrics of the first kind we become:
(m)
The number of stained fabrics of another type becomes:
(m)
Number of stained fabrics of the third type in stock:
(m)
Oscilki viroblen kіlkіst cloth cannot be negative, then
that .
Dokhіd vіd vіroblennyh cloth stock:
(Den. od.)

Then the economic-mathematical model of the task can be seen:

Virishuemo in a graphic way.
Conducted axes of coordinates i .

We will be direct.
At .
At .
We draw a straight line through the points (0; 7) and (10.5; 0).

We will be direct.
At .
At .
We draw a straight line through the points (0; 10) and (10; 0).

We will be direct.
At .
At .
We draw a straight line through the points (0; 8) and (8; 0).

Shaded area, so that the speck (2; 2) squandered to the shaded part. Take the OABC cutter.

(P1.1) .
At .
At .
We draw a straight line through the points (0; 4) and (3; 0).

Further, we note that the scores of the coefficients with the target function are positive (400 and 300), then they will increase with the increase. We draw a straight line, parallel to the straight line (A1.1), as far as possible in the direction of it at the b_k growth of the station, i want to pass through one point of the chotirikutnik OABC. Such a straight line passes through the point C. З prompt її coordinates.
.

Solution for task: ;

.
To get the highest income, it is necessary to prepare 8 cloths of model A. Dokhid with a warehouse of 3200 den. one.

butt 2

Solve the problem of linear programming by the graphical method.

Virishuemo in a graphic way.
Conducted axes of coordinates i .

We will be direct.
At .
At .
We draw a straight line through the points (0; 6) and (6; 0).

We will be direct.
Zvіdsi.
At .
At .
We draw a straight line through the points (3; 0) and (7; 2).

We will be direct.
Be straight (all abscissa).

The region of admissible solutions (SDR) is surrounded by prompted straight lines. In order to recognize, from any side, respectfully, that the point is to lay the ODR, the shards are satisfied with the system of irregularities:

Shade the area beyond the boundaries of the prompted lines, so that the speck (4; 1) has squandered to the shaded part. We take the tricot ABC.

There will be a sufficient line of equal function, for example,
.
At .
At .
We draw a straight line through the points (0; 6) and (4; 0).
Oscilki tsіlova funktsіya zbіlshuєtsya with zbіlshennі, then it is carried out straight, parallel to the line of the line and as far as possible in the direction of it at the bіk growth, and if you want to pass through one point of the tricot ABC. Such a straight line passes through the point C. З prompt її coordinates.
.

Solution for task: ;

An example of the validity of the solution

Razv'yazati graphically zavdannya linear programming. Find the maximum and minimum values ​​of the target function.

Solve the problem using a graphical method.
Conducted axes of coordinates i .

We will be direct.
At .
At .
We draw a straight line through the points (0; 8) and (2.667; 0).

We will be direct.
At .
At .
We draw a straight line through the points (0; 3) and (6; 0).

We will be direct.
At .
At .
We draw a straight line through the points (3; 0) and (6; 3).

Straight lines with coordinate axes.

The area of ​​admissible solutions (ODR) is surrounded by lines and coordinate axes. In order to recognize, from any side, respectfully, that the point is to lay the ODR, the shards are satisfied with the system of irregularities:

Shaded area, so that the speck (3; 3) squandered to the shaded part. We take away the unbounded area, I will be surrounded by the laman ABCDE.

There will be a sufficient line of equal function, for example,
(P3.1) .
At .
At .
We draw a straight line through the points (0; 7) and (7; 0).
Oskіlki koefitsіenti with positive, then growth with zbіlshennі і.

In order to know the maximum, it is necessary to draw a parallel line, as far as possible at the bik growth and pass through one point of the region ABCDE. However, the oskіlki region is not bordered on the side of great values, so it is impossible to draw such a straight line. I used to conduct a straight line, there will always be points in the region, more distant kills and more. There is no such maximum. you can robiti like a truly great one.

Shukaemo minimum. We draw a straight line, parallel to the straight line (A3.1) and as far as possible in the direction of the change, i want to pass through one point of the area ABCDE. Such a straight line passes through the point C. З prompt її coordinates.
.
Minimum value of the target function:

There is no maximum value.
Minimum value
.

The most simple and straightforward method of solving the problem of linear programming (LLP) is the graphic method. The reasons for the foundations on the geometric interpretation of the problem of linear programming and zastosovuєtsya with the violation of the ZLP from two nevidomimi:

Let's look at the rozv'yazannya tsgogo zavdannya on the square. Skin unevenness of the system of functional borders geometrically defines the surface area with the boundary line a p x, + + a j2 x 2 = b n i = 1, those. Wash away the invisibility and signify the flat surfaces with the boundary lines X (= 0, x 2= 0 is reasonable. If the system is spіlna, then the nap_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p_p, merge, utvoryuyuyut zagal’nuyu part, yak є swollen impersonal and є suupnіstyu point; coordinates of the skin s tsikh point є rozvyazkom tsієї system. Sukupnіst tsikh dot call rich cuter of roses. Vіn can be a point, vіdrіzkom, exchange, fringe it with a non-fenced bagatoknik.

Geometrically ZLP є to see such an apex point of the bagatokutnik solution, the coordinates of which deliver the maximum (minimum) value of the linear target function, moreover, the admissible solutions are all points of the bagatokutnik of roses.

A linear alignment describes an impersonal point that lies on one straight line. Linear unevenness defines the region on the flat.

Significantly, as part of the area is signified by unevenness 2x (+ Zx 2 12.

First, let's go straight 2x, + Zx 2 = 12. Won pass through the points (6; 0) and (0; 4). In a different way, it is significant that the surface is pleased with the unevenness. For which we choose whether a point on the graph, so that there is no straight line, and we substitute these coordinates into the unevenness. If there is some inconsistency, then given pointє admissible solutions and napіvploshchina, to avenge the point, satisfying the unevenness. To substitute for nerivnist, manually tweak the cob of coordinates. Imagine x (= x 2 = 0 y unevenness 2x, + Зх 2 12. Take 2 0 + 3 0

Similarly, you can graphically depict all the tasks of linear programming.

To solve the skin unevenness of the system, the border of the ZLP is the nap_plyshchina, which avenges the boundary line and is lined from one side in front of it. Peretin on the surface, the skin of which is characterized by a type of nervousness of the system, is called area of ​​acceptable solutions(ODR) or the area of ​​appointment.

It is necessary to remember that the area of ​​acceptable solutions satisfies the minds of negativity (Xj > 0, j = 1, e). Coordinates of any point, which should be assigned to the designated area, is a valid assignment of the task.

For the value of the extreme value of the target function with a graphical solution of the ZLP, we use gradient Vector, coordinates of some private target function:

This vector shows directly the most obvious change of the target function. Straight c [ x l + c 2 x 2 \u003d f (x 0), perpendicular to the gradient vector, є the line of equal goal function (Fig. 2.2.2). Whether the point of the line is equal, the function of the same value is the same. It is equal to the goal function of a constant value a. Changing values a, we take away the family of parallel lines, the skin line of the line of equal function.

Rice. 2.2.2.

The power of the line is important, the linear function of the field is in the fact that with a parallel line shifted in one row, the line is less grow up, and when shifted to the next book - only change.

The graphical method of completing the ZLP consists of several stages:

• 1. There will be an area of ​​acceptable solutions (ODR) of the LLP.
• 2. There will be a vector-gradient of the integer function (CF) from the cob at the point x 0(0; 0): V = (s, h 2).
• 3. Line of river CjXj+ h 2 x 2 \u003d a (a - constant value) - a straight line, perpendicular to the vector-gradient V, - shifts at the direction of the vector-gradient in order to maximize the target function f(x v x 2) doti, deprive the docks between ODR. When minimizing /(*, x 2) the straight line moves straight ahead, to the opposite vector-gradient. The extreme point (or points) of the ODR with which Russia is the maximum (minimum) point f(x p jc 2).

If it’s straight, it’s a clear line, with its own Russia it doesn’t fill the ODR, then the minimum (maximum) of the function f(x p x 2) no.

Якщо лінія рівня цільової функції паралельна функціональному обмеження задачі, на якому досягається оптимальне значення ЦФ, то оптимальне значення ЦФ буде досягатися в будь-якій точці цього обмеження, що лежить між двома оптимальними кутовими точками, і, відповідно, кожна з цих точок є оптимальним рішенням ZLP.

4. The coordinates of the point are assigned to the maximum (minimum). For which it is sufficient to diverge the system of straight lines, to give a point to the maximum (minimum) at the crossbar. Value f(x ( , x 2), known in the minimum point to the maximum (minimum) values ​​of the target function.

Possible situations of a graphic solution of the ZLP are presented in Table. 2.2.1.

Table 2.2.1

Example 2.2.1. Planning for the release of products for the sewing of enterprises (zavdannya about costumes).

We are launching the release of two types of costumes - human and female. For a woman's costume, 1 m of wool, 2 m of lavsan and 1 person's day of labor are required; for a person - 3.5 m outside, 0.5 m lavsan and 1 person day of labor. Usyi є 350 m of outdoors, 240 m of lavsan and 150 man-days of labor costs.

It is necessary to designate how many costumes of a skin type it is necessary to sew, in order to secure the maximum allowance, so that the allowance for a woman's suit becomes 10 den. one, like a person - 20 den. one. With any trace of the mother in the country, it is necessary to sew at least 60 human costumes.

Economic and mathematical model of the problem

Changes: X, - number of women's costumes; x 2 - number of human costumes.

Purpose function:

Exchange:

First obmezhennya (vovna) may look x (+ 3.5 x 2 x (+ 3.5 x 2 = 350 to pass through the points (350; 0) and (0; 100). 2x (+ 0.5 x 2 2x x + 0.5 x 2 \u003d 240 pass through the points (120; 0) and (0; 480). Third obmezhennya (on pratsi) may look x y + x 2150. Straight x (+ x 2 = 150 pass through the points (150; 0) and (0; 150). A quarter of an exchange (for a quantity of human costumes) may look x 2> 60 x 2 = 60.

As a result, the peretina of the chotirokh napіvploschins is taken to be a bagatok, which is the area of ​​​​admissible decisions of our plant. Whether or not the point of the bagatokutnik satisfies all the functional inconsistencies, and for whether or not the point of the position of the bagatokutnik, wanting to, one nerіvnіst will be destroyed.

On fig. 2.2.3 shaded area of ​​acceptable solutions (ODR). For a straight forward direction to the optimum, we will use the gradient vector V, the coordinates of which are private, similar target functions:

To induce such a vector, it is necessary to set the point (10; 20) from the cob of coordinates. For clarity, you can use a vector proportional to the vector V. So, in fig. 2.2.3 shows the vector (30; 60).

Let's wait for the equal line 10xj + 20x2 = a. It is equal to the goal function of a constant value a. Changing values a, we take into account the family of parallel lines, skin lines of equal function:

We gave a transfer line to the її exit from the ODR. In our viewpoint (when maximizing the goal function), the flow of the line is similar to the straight line of the gradient. At the extreme apex point, the maximum function of the target is reached. For the significance of the coordinates of the center of the point, we develop a system of two equal lines, which gives a point to the maximum at the crossbar:

Acceptable For these values

Vidpovid. To get the maximum allowance (2300), it is necessary to sew 70 women (xj 1 = 70) and 80 men (x 2 = 80) suits.

Rice. 2.2.3. Krapka (70; 80) - the optimal solution to the problem. Example 2.2.2. Know the maximum and minimum f(X):

at dewatering

Solution. At the cherry this butt to the maximum blame the situation, if the line is equal to Zx + 3 x 2 = a parallel to the first interchange: x x + x 2 8. The target function reaches the maximum value at two points: BUT(3; 5) that AT(6; 2) - that one takes on a vіdrіzka AB one and the same meaning, which is good 24:

With the deviation of the butt for the minimum function of the line 3xj + 3 x 2 - a next to the collapse of the straight line, the reverse direction of the vector-gradient. The target function reaches the minimum value at the point D (0,5; 0):

The graphic solution for the butt is shown in fig. 2.2.4.

Rice. 2.2.4.

Value: max /(2Q = 24; min /( x)= 1.5. Stock 2.2.3. Know maximum / ( X):

at dewatering

Solution. There is no solution, the shards of the ZF are not surrounded by animals on the ODR (Fig. 2.2.5).

Linear programming uses a graphical method, for the help of which they designate multiples (multiple solutions). Since the main task of linear programming can be an optimal plan, then the goal function gains the value of one of the vertices of the solution bug-hedron (div. little ones).

Service assignment. For help this service you can in online mode solve the problem of linear programming using the geometric method, and find the solution of a two-sided task (evaluate the optimality of the resource selection). Dodatkovo create a solution template in Excel.

Instruction. Choose the number of rows (number of borders). If there are more than two replacements, it is necessary to bring the system to SZLP (div. butt No. 2). For example, 1 ≤ x 1 ≤ 4, it is divided into two: x 1 ≥ 1, x 1 ≤ 4 (so the number of rows increases by 1).
Promote the area admissible solution(ODR) can also be used for additional services.

At the same time with the sim calculator, you can also use the following:
Simplex method of solving ZLP

Dispatch of the transport department
Matrix solution
For additional service in online mode, you can determine the price of the matrix edge (lower and upper boundary), check the presence of the saddle point, know the solution of the mixed strategy by methods: minmax, simplex method, graphic (geometric) method, Brown's method.
Extremum of a function of two variables
Calculation between

Solve problems of linear programming by graphical method including next steps:

1. On the area X10X2 there will be straight lines.
2. The surfaces are indicated.
3. They signify a bagatokutnik rozvyazkіv;
4. There will be a vector N(c 1 ,c 2), which will directly indicate the target function;
5. Transfer direct target function c 1 x 2 + c 2 x 2= 0 for the straight line of the vector N to the extreme point of the cut-out loop.
6. Calculate the coordinates of the point and the value of the function of the qiy point.

Butt. The company produces two types of products - P1 and P2. There are two kinds of syrovina for vyrobnitstva - C1 and C2. Wholesale prices for a single defective product are expensive: CU 5 for P1 and 4 d.o. for P2. Vitrata syrovini per unit of production type P1 and type P2 is given in the table.
Table - Product Vitrata

1. Formulate a mathematical model of the problem of linear programming.
2. Solve the problem of linear programming in a graphic way(For two men).

Resource exchange
6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6

shodo drink
x 1 +1 ≥ x 2
x2 ≤ 2

Purpose function
5x1 + 4x2 → max

Todi otrimuemo come ZLP:
6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6
x 2 - x 1 ≤ 1
x2 ≤ 2
x 1 , x 2 ≥ 0
5x1 + 4x2 → max

If there are more than two changing tasks of linear programming, then the task should be brought up to the standard ZLP in advance.
→ max at defrost:
x1 + x2 + x3 = 12
2x1 - x2 + x4 = 8
- 2x1+2x2+x5=10
F(X) = 3x 1 - 2x 2 + 5x 3 - 4x 5
Transition to SZLP.

Peculiarities of the development of problems of linear programming by the graphical method

f=x 1 +13x 2 -x 3 +2x 4 +3x 5
-x 2 +x 3 -x 5 =-3
x 1 -4x 2 +3x 3 -x 4 +2x 5 =3
4x 2 -x 3 +x 4 -x 5 =6

From the first level, it can be shown x 5:
x5 = -x2 +x3 +3

f=x 1 +13x 2 -x 3 +2x 4 +3(-x 2 +x 3 +3)
x 1 -4x 2 +3x 3 -x 4 +2(-x 2 +x 3 +3)=3
4x 2 -x 3 +x 4 -(-x 2 +x 3 +3)=6
or
f=x 1 +10x 2 +2x 3 +2x 4 +9
x 1 -6x 2 +5x 3 -x 4 =-3
5x2 -2x3 +x4 =9

On the other hand, it can be shown x 4:
x4 = 9-5x2 +2x3
i can be represented in usi virazi:
f=x 1 +6x 3 +27
x1 -x2 +3x3 = 6

Change x 2 is accepted as an additional change and robably replace it with the “≥” sign:
f=x 1 + 6x 3 + 27
x 1 + 3x 3 ≥6

Butt #2

x1 + x2 + x3 = 12
2x1 - x2 + x4 = 8
- 2x1+2x2+x5=10
F(X) = 3x 1 - 2x 2 + 5x 3 - 4x 5
Transition to SZLP.
The matrix of the system of fencing-evennesses of this plant has been expanded:

Example number 3. Put together a mathematical model of a linear programming problem and find solutions in a geometric way.

• Fold the system of mathematical deposits (irregularities) and the goal function.
• Draw a geometric interpretation of the problem.
• Find the optimal solution.
• Carry out an analytical review.
• Signify the sutts and nesuttєve resources and their superfluous.
• Calculate the value of the target function.
• Calculate the objectively estimated scores.
• Fold down the strength of durability.