As a criterion for the optimal transport. Optimality criterion for the basic solution of the transport problem

Design parameters. This term denotes independent variable parameters that completely and unambiguously define the design problem being solved. Design parameters are unknown quantities, the values of which are calculated during the optimization process. As design parameters, any basic or derived quantities that serve to quantitatively describe the system can serve. So, it can be unknown values of length, mass, time, temperature. The number of design parameters characterizes the degree of complexity of a given design problem. Usually, the number of design parameters is denoted by n, and the design parameters themselves by x with the corresponding indices. Thus, n design parameters of this problem will be denoted by

X1, X2, X3, ... Xp.

It should be noted that design parameters may be referred to as internal controlled parameters in some sources.

target function. This is the expression whose value the engineer seeks to maximize or minimize. The objective function allows you to quantitatively compare two alternative solutions. From a mathematical point of view, the objective function describes some (n + 1) - dimensional surface. Its value is determined by the design parameters

M = M (x1,x2,…,xn).

Examples of the objective function, often encountered in engineering practice, are cost, weight, strength, dimensions, efficiency. If there is only one design parameter, then the objective function can be represented by a curve on a plane (Fig. 1). If there are two design parameters, then the objective function will be represented by a surface in the space of three dimensions (Fig. 2). With three or more design parameters, the surfaces specified by the objective function are called hypersurfaces and cannot be depicted by conventional means. The topological properties of the objective function surface play an important role in the optimization process, since the choice of the most efficient algorithm depends on them.

Figure 1. One-dimensional objective function.

Figure 2. 2D objective function.

The objective function in some cases can take the most unexpected forms. For example, it is not always possible to express it in a closed mathematical form, in other cases it can be a piecewise linear function. An objective function may sometimes require a technical data table (for example, a steam state table) or it may be necessary to conduct an experiment. In some cases, design parameters take only integer values. An example would be the number of teeth in a gear or the number of bolts in a flange. Sometimes design parameters have only two values - yes or no. Qualitative parameters, such as customer satisfaction, reliability, aesthetics, are difficult to take into account in the optimization process, since they are almost impossible to quantify. However, in whatever form the objective function is presented, it must be a single-valued function of the design parameters.

In a number of optimization problems, the introduction of more than one objective function is required. Sometimes one of them may be incompatible with the other. An example is the design of aircraft, when it is required to provide maximum strength, minimum weight and minimum cost at the same time. In such cases, the designer must introduce a system of priorities and assign some dimensionless multiplier to each objective function. As a result, a “compromise function” appears, which allows one composite objective function to be used in the optimization process.

Finding the minimum and maximum. Some optimization algorithms are adapted for finding the maximum, others for finding the minimum. However, regardless of the type of extremum problem being solved, one and the same algorithm can be used, since the minimization problem can be easily turned into a maximum finding problem by changing the sign of the objective function to the opposite. This technique is illustrated in Figure 3.

Figure 3. When the sign of the objective function is reversed in the minimum task, it turns it into a maximum task.

Design space. This is the name of the area defined by all n, design parameters. The design space is not as large as it might seem, since it is usually limited by a number of conditions related to the physical nature of the problem. The constraints may be so strong that the problem will not have a single satisfactory solution. Constraints are divided into two groups: constraints - equalities and constraints - inequalities.

Equality constraints are dependencies between design parameters that must be taken into account when finding a solution. They reflect the laws of nature, economics, rights, prevailing tastes and the availability of the necessary materials. The number of restrictions - equalities can be any. They look like

C1 (X1, X2, X3, . . ., Xn) = 0,

C2 (X1, X2, X3, . . ., X n) = 0,

..……………………………..

Сj(X1, X2, X 3, . . ., Xn) = 0.

Inequality constraints are a special kind of constraint expressed by inequalities. In the general case, there can be an arbitrarily large number of them, and all of them have the form

z1 ?r1(X1, X2, X3, . . ., Xn) ?Z1

z2 ?r2(X1, X2, X3, . . ., Xn) ?Z2

………………………………………

zk ?rk(X1, X2, X3, . . ., Xn) ?Zk

It should be noted that very often, due to limitations, the optimal value of the objective function is not reached where its surface has a zero gradient. Often the best solution is at one of the boundaries of the design domain.

Direct and functional restrictions. Direct constraints have the form

xni ? xi? xв i at i ? ,

where xнi , xвi are the minimum and maximum allowable values of the i-th controlled parameter; n is the dimension of the space of controlled parameters. For example, for many objects, element parameters cannot be negative: xнi ? 0 (geometric dimensions, electrical resistances, masses, etc.).

Functional restrictions, as a rule, are conditions for the performance of output parameters that are not included in the objective function. Functional restrictions can be:

1) type of equalities
w(X) = 0; (2.1)
2) the type of inequalities

u(X) › 0, (2.2)

where w(X) and u(X) are vector functions.

Direct and functional constraints form the valid scope of the search:

XD = (X | w(X) = 0, u (X) › 0, xi › xni ,

xi ‹ xвi for i ? ).

If constraints (2.1) and (2.2) coincide with the operability conditions, then the admissible area is also called the XP operability area.

Any of the points X belonging to XD is a feasible solution to the problem. Often, parametric synthesis is posed as the problem of determining any of the feasible solutions. However, it is much more important to solve the optimization problem - to find the optimal solution among the feasible ones.

local optimum. This is the name of the point in the design space at which the objective function has the greatest value compared to its values at all other points in its immediate neighborhood. Figure 4 shows a one-dimensional objective function with two local optima. Often the design space contains many local optima and care must be taken not to mistake the first one for the optimal solution to the problem.

Figure 4. An arbitrary objective function can have several local optima.

The global optimum is the optimal solution for the entire design space. It is better than all other solutions corresponding to local optima, and this is what the designer is looking for. The case of several equal global optima located in different parts design space. This allows you to choose the best option from equal best options by objective function. In this case, the designer can choose an option intuitively or based on a comparison of the received options.

Choice of criteria. The main problem of setting extremal problems lies in the formulation of the objective function. The complexity of the choice of the objective function lies in the fact that any technical object initially has a vector character of optimality criteria (multi-criteria). Moreover, the improvement of one of the output parameters, as a rule, leads to the deterioration of the other, since all output parameters are functions of the same controlled parameters and cannot be changed independently of each other. Such output parameters are called conflict parameters.

The objective function must be one (uniqueness principle). The reduction of a multi-criteria problem to a single-criteria one is called the convolution of a vector criterion. The problem of finding its extremum is reduced to the problem of mathematical programming. Depending on how the output parameters are selected and combined, in the scalar quality function, private, additive, multiplicative, minimax, statistical criteria and other criteria are distinguished. The terms of reference for the design of a technical object indicate the requirements for the main output parameters. These requirements are expressed in the form of specific numerical data, their range of change, operating conditions and acceptable minimum or maximum values. The required ratios between output parameters and technical requirements (TT) are called performance conditions and are written as:

yi< TTi , i О ; yi >TTj , j O ;

yr = TTr ± ?yr; r O .

where yi, yj, yr - set of output parameters;

TTi, TTj, TTr - required quantitative values of the corresponding output parameters according to the terms of reference;

Yr - allowable deviation of the r-th output parameter from the value TTr specified in the specification.

The operability conditions are of decisive importance in the development of technical devices, since the design task is to choose a design solution in which all the operability conditions are best met in the entire range of external parameters and when all the requirements of the technical assignment are met.

Particular criteria can be applied in cases where one main parameter yi(X) can be singled out among the output parameters, which most fully reflects the efficiency of the designed object. This parameter is taken as the objective function. Examples of such parameters are: for an energy facility - power, for a technological machine - performance, for a vehicle - carrying capacity. For many technical objects, this parameter is the cost. The conditions for the operability of all other output parameters of the object are referred to as functional limitations. Optimization based on such a statement is called partial criterion optimization.

The advantage of this approach is its simplicity, a significant drawback is that a large working capacity margin can be obtained only by the main parameter, which is taken as the objective function, and other output parameters will not have margins at all.

A weighted additive criterion is used when the performance conditions allow us to distinguish two groups of output parameters. The first group includes output parameters whose values should be increased during optimization y+i(X) (performance, noise immunity, probability of failure-free operation, etc.), the second group includes output parameters whose values should be decreased y-i(X) ( fuel consumption, transient duration, overshoot, offset, etc.). Combining several output parameters, which generally have different physical dimensions, in one scalar objective function requires preliminary normalization of these parameters. Methods for normalizing the parameters will be discussed below. For the time being, we will assume that all y(X) are dimensionless and among them there are no such that correspond to the equality-type performance conditions. Then, for the case of minimizing the objective function, the convolution of the vector criterion will have the form

where aj>0 is a weighting factor that determines the degree of importance of the j-th output parameter (usually aj are chosen by the designer and remain constant during the optimization process).

The objective function in the form (2.1), which expresses the additive criterion, can also be written in the case when all or the main performance conditions have the form of equalities. Then the objective function

determines the root-mean-square approximation yj(X) to the specified specification TTj.

The multiplicative criterion can be applied in those cases when there are no equality-type performance conditions and the output parameters cannot take zero values. Then the multiplicative objective function to be minimized has the form

One of the most significant shortcomings of both the additive and multiplicative criteria is the failure to take into account the technical requirements for the output parameters in the formulation of the problem.

The function form criterion is used when the task of the best match of the given (reference) characteristic yТТ(X, u) with the corresponding output characteristic y(X, u) of the designed object is set, where u is some variable, for example, frequency, time, selected phase variable. These tasks include: designing an automatic control system that provides the required type of transient process for the controlled parameter; determination of the parameters of the transistor model, giving the maximum coincidence of its theoretical current-voltage characteristics with the experimental ones; search for the parameters of the beam sections, the values of which lead to the best match between the given stress diagram and the calculated one, etc.

The use of a particular optimization criterion in these cases is reduced to replacing continuous characteristics with a finite set of nodal points and choosing one of the following objective functions to be minimized:

where p is the number of nodal points uj on the axis of the variable u; aj - weight coefficients, the values of which are the greater, the smaller the deviation y(X, uj) - yTT(X, uj) must be obtained at the j-th point.

Maximin (minimax) criteria make it possible to achieve one of the goals of optimal design - the best satisfaction of performance conditions.

Let us introduce a quantitative estimate of the degree of fulfillment of the jth performance condition, denote it by zj, and call it the performance margin of the parameter yj. The calculation of the margin for the j-th output parameter can be performed in various ways, for example,

where aj - weighting factor; yjnom - nominal value of the j-th output parameter; dj - value characterizing the spread of the j -th output parameter.

It is assumed here that all relations are reduced to the form yi< TТj. Если yi >TTj then -yj< -TТj . Следует принимать аj >1 (recommended values 5 ? aj ? 20), if it is desirable to achieve the j-th technical requirement with a given tolerance, i.e. yj = TTj ± ?yj; aj=l if it is necessary to obtain the maximum possible estimate of zj.

Functional quality technical system is characterized by the vector of output parameters and, consequently, by the vector Z=(zm,zm,…,zm). Therefore, the objective function should be formed as a certain function u(Z) of the estimates vector. For example, if the stock of only that output parameter is considered as the objective function, which at a given point X is the worst from the standpoint of fulfilling the requirements of the TOR, then

where m is the number of health margins.

It is natural now to pose the problem of choosing such a search strategy for X that would maximize the minimum of the reserves, i.e.

where XD is the searchable area.

An optimization criterion with an objective function (2.6) is called a maximin criterion.

Statistical criteria. Optimization under statistical criteria aims to obtain the maximum probability P performance performance. This probability is taken as the objective function. Then we have a task

Rationing of controlled and output parameters. The space of controlled parameters is metric. Therefore, when choosing the directions and values of the search steps, it is necessary to introduce one or another norm, which is identified with the distance between two points. The latter assumes that all controlled parameters have the same dimension or are dimensionless.

Possible various ways rationing. As an example, consider the method of logarithmic normalization, the advantage of which is the transition from absolute increments of parameters to relative ones. In that case i the controlled parameter ui is converted to dimensionless xi as follows:

where oi is a coefficient numerically equal to one of the parameter ui .

Normalization of output parameters can be performed using weight coefficients, as in the additive criterion, or by passing from yj to working capacity margins zj according to (2.5).

The mathematical formulation of the transport problem consists in determining the optimal plan for the transportation of some homogeneous cargo from T departure points A 1 , A 2 , …, A T V P destinations IN 1 , IN 2 , …, IN P . In this case, either the minimum cost of transporting the entire cargo or the minimum time for its delivery is usually taken as an optimality criterion. Let us consider a transport problem, the optimality criterion of which is the minimum cost of transporting the entire cargo.

Denote:

c ij – tariffs for transportation of a unit of cargo from i-th point of departure in j-th destination,

a i - cargo stocks in i-th point of departure,

b j – cargo needs in j– m destination,

x ij – the number of units of cargo transported from i-th point of departure in j-th destination.

Then the mathematical formulation of the problem consists in determining the minimum value of the function: (6.1)

under conditions
(6.2)

(6.3)

Any non-negative solution of systems linear equations(6.2) and (6.3) defined by the matrix X=(x ij ) , is called the transport task plan.

Plan X*=(x*ij) , at which the function (6.1) takes its minimum value, is called the optimal plan of the transport problem.

Usually, the initial data of the transport task is recorded in the form of a table.

There are several methods for determining the baseline: the northwest corner method, the least cost method, the Vogel approximation method, etc.

Northwest corner method

The maximum allowable transportation is entered in the most northwestern cell of the table, while either the entire cargo is exported from the station A 1 and all other cells of the first row are crossed out, or the needs of the first consumer IN 1 are completely satisfied, then all the cells of the first column are crossed out. After that, the northwesternmost cell becomes the cell A 1 IN 2 or IN 2 A 1 . The algorithm continues until the table is filled. Disadvantages - the cost of transportation is not taken into account, and the plan is far from optimal.

Least Cost Method

The method takes into account transportation costs to some extent and is constructed as follows: a matrix is considered and the cell with the lowest cost is found, which is filled with the maximum allowable transportation. In most cases, this method gives a plan close to optimal.

Vogel approximation method

At each iteration, in all columns and in all rows, the difference between the two minimum rates recorded in them is found. These differences are recorded in a row and column specially designated for this purpose in the table of the conditions of the problem. Among these differences, choose the minimum. In the row or column to which this difference corresponds, the minimum tariff is determined.

A widely used method for solving transport problems is the method of potentials.

This method allows you to evaluate the initial reference solution and find the optimal solution using the method of successive improvement.

Theorem 1. If the basic plan X=(x ij ) is optimal, there exists a system of ( t+p) numbers called potentials U i , V j, such that:

U i + V j =C ij , for x ij >0 (basic variables);

U i + V j =C ij , for x ij =0 (free variables).

Thus, to check the optimality of the initial optimal plan, the following conditions must be met:

for each occupied cell, the sum of the potentials is equal to the cost of transporting a unit of cargo in this cell:

U i + V j =C ij

for each free cell, the sum of the potentials is less than or equal to the cost of transporting a unit of cargo in this cell:

U i + V j £ WITH ij

Theorem 2. Any closed transport problem has a solution that is achieved in a finite number of steps of the potential method.

Building a cycle and determining the amount of redistribution of cargo.

A cycle in the transportation table is a polyline with vertices in cells and edges located along the rows or columns of a matrix that meets two requirements:

The broken line must be connected, i.e. from any of its vertices, one can get to another vertex by moving along the edges;

exactly two edges converge at each vertex of the cycle - one along the row, the other along the column.

Free counting cycle A cell is such a cycle, one of the vertices of which is in a free cell, and the rest are in the basic ones.

Let us give examples of some cycles.

Theorem 3. There is one recalculation cycle for each free cell in the transportation table.

Potential Method Algorithm

Let's assign each station A i potential And i, and each station IN j potential v j. To do this, for each filled cell X ij ≠0 write an equation And i + v j = with ij . Let's give And 1 =1 (any other value is possible) and find all other potentials.

Let's check the optimality of the found base plan. To do this, we calculate the sum of potentials for free cells. If this amount is less than the cost of transportation With ij standing in this cell, then the optimal solution is found. If more, then there is a violation in this cell, equal to the difference between this amount and the cost of transportation. Find all such violations (we will write them in the same cells at the bottom right). Let's choose the largest of these violations and construct a free cell recalculation cycle that will start from the marked cell with the greatest violation.

3. The recalculation cycle begins in a free cell, where we put a plus sign, and the rest of the cells are occupied. The signs in these cells alternate. We choose the smallest of the transportations in cells with a minus sign. Then we add this transportation to transportations with a plus sign and subtract from transportations with a minus sign. We get a new optimal solution. Let's check it for optimality.

4. For new potentials, we check the optimality condition. If the optimality conditions are met, then the optimal solution is obtained, if not, then we continue the search for the optimal solution using the potential method.

Example 7.1. Four enterprises of this economic region use three types of raw materials for the production of products. The raw material requirements of each of the enterprises are respectively equal to 120, 50, 190 and 110 units. Raw materials are concentrated in three places of their receipt, and the stocks are respectively equal to 160, 140, 170 units. Raw materials can be delivered to each of the enterprises from any point of its receipt. Freight rates are known values and are given by the matrix
.

Draw up a transportation plan in which the total cost of transportation is minimal.

Solution:

closed task.

Let's draw up the first plan of the transport problem using the northwest corner method. Let's start filling in the cells of the table from the top left cell.

S 1 \u003d 120 7 + 40 8 + 10 5 + 130 9 + 60 3 + 110 6 \u003d 3120

Let's make the first plan using the minimum cost method. We will fill the cells with minimal tariffs.

S 2 \u003d 160 1 + 120 4 + 20 8 + 50 2 + 30 3 + 90 6 \u003d 1530

The cost of such a plan of transportation is almost two times less. Let's start solving the problem with this plan. Let's check it for optimality. Let's introduce the potentials α i – respectively, departures, β j – respectively destinations. Based on the occupied cells, we compose a system of equations α i + β j =c ij:

For free cells of the table, we check the optimality criterion

We will make differences

The plan is not optimal because there is a positive assessment
Let's build a recalculation cycle from it. This is a broken line of links that are located strictly vertically or horizontally, and the vertices are in occupied cells. In a bad cell, put a sign (+). At the other vertices, the signs alternate. Of the negative vertices, select the smallest number and shift it around the cycle. Moved to a new base plan.

S 3 \u003d 70 1 + 90 2 + 120 4 + 20 8 + 50 2 + 120 3 \u003d 1350

The cost of transportation is less, i.e. the plan has been improved. Checking now new plan for optimality. By occupied cells:

For free cells:

The plan is not optimal because there is a positive assessment
We build a recalculation cycle and move on to a new plan.

S 4 \u003d 50 1 + 110 2 + 120 4 + 20 5 + 30 2 + 140 3 \u003d 1330

We check the new plan for optimality.

By occupied cells:

For free cells:

The optimality criterion is met, i.e., the last plan is optimal.

Answer:

When solving a transport problem, the choice of an optimality criterion is important. As you know, the estimate economic efficiency An approximate plan can be determined by one or another criterion, which is the basis for calculating the plan. This criterion is an economic indicator that characterizes the quality of the plan. To date, there is no generally accepted single criterion that comprehensively takes into account economic factors. When solving a transport problem, the following indicators are used as an optimality criterion in various cases:

1) The volume of transport work (criterion - distance in t / km). The minimum mileage is useful for estimating travel plans because the travel distance is easily and accurately determined for any direction. Therefore, the criterion cannot solve transport problems involving many modes of transport. It is successfully used in solving transport problems for road transport. When developing optimal schemes for the transportation of homogeneous goods by cars.

2) Tariff for the carriage of goods (criterion - tariffs for carriage charges). Allows you to get a transportation scheme that is the best in terms of self-supporting indicators of the enterprise. All the surcharges, as well as existing feed-in tariffs, make it difficult to use.

3) Operating costs for the transportation of goods (criterion - the cost of operating costs). More accurately reflects the economy of transportation various types transport. Allows you to draw reasonable conclusions about the feasibility of switching from one mode of transport to another.

4) Terms of delivery of goods (criterion - the cost of time).

5) Reduced costs (taking into account operating costs, depending on the size of the movement and capital investment in the rolling stock).

6) Reduced costs (taking into account the full operating costs of capital investments for the construction of facilities in the rolling stock).

where is the operating cost,

Estimated investment efficiency ratio,

Capital investments coming per 1 ton of cargo throughout the section,

T - travel time,

C - the price of one ton of cargo.

Allows you to more fully evaluate the rationalization of different options for transportation plans, with a fairly complete expression of the quantitative and simultaneous influence of several economic factors.

Let us consider a transport problem, the optimality criterion of which is the minimum cost of transporting the entire cargo. Let us denote by the tariffs for the transportation of a unit of cargo from the i-th point of departure to j-th item destination, through - stocks of cargo in i-th paragraph departure point, through is the demand for cargo at the jth destination, and through is the number of units of cargo transported from the i-th origin to the j-th destination. Then the mathematical formulation of the problem consists in determining the minimum value of the function

under conditions

(2)

(3)

(4)

Since the variables satisfy the systems of linear equations (2) and (3) and the non-negativity condition (4), then the export of the existing cargo from all points of departure, delivery required amount shipment to each of the destinations, as well as return shipments are excluded.

Thus, the T-problem is an LP problem with m*n the number of variables, and m+n the number of restrictions - equalities.

Obviously, the total availability of cargo from suppliers is equal to , and the total need for cargo at destinations is equal to units. If the total demand for cargo at destinations is equal to the cargo stock at origins, i.e.

then the model of such a transport problem is called closed or balanced.

There are a number of practical problems in which the balance condition is not met. Such models are called open. Possible two cases:

In the first case, full satisfaction of demand is impossible..

Such a problem can be reduced to an ordinary transport problem as follows. In case of excess of demand over stock, i.e., a fictitious ( m+1) - th point of departure with cargo stock and tariffs are assumed to be zero:

Then it is required to minimize

under conditions

Consider now the second case.

Similarly, for , a fictitious ( n+1) – th destination with need and the corresponding tariffs are considered equal to zero:

Then the corresponding T-problem can be written as follows:

Minimize

under conditions:

This reduces the problem to an ordinary transport problem, from the optimal plan of which the optimal plan of the original problem is obtained.

In the future, we will consider a closed model of the transport problem. If the model of a particular problem is open, then, proceeding from the above, we rewrite the table of conditions of the problem so that equality (5) is satisfied.

In some cases, you need to specify that products cannot be transported along any routes. Then the costs of transportation along these routes are set so that they exceed the highest costs of possible transportation (in order to make it unprofitable to carry on inaccessible routes) - when solving the problem for a minimum. To the max, it's the other way around.

Sometimes it is necessary to take into account that contracts for fixed supply volumes are concluded between some points of dispatch and some points of consumption, then it is necessary to exclude the volume of guaranteed supply from further consideration. To do this, the guaranteed supply volume is subtracted from the following values:

from the stock of the respective dispatch point;

· from the needs of the corresponding destination.

Example.

Four enterprises of this economic region use three types of raw materials for the production of products. The raw material requirements of each of the enterprises are respectively equal to 120, 50, 190 and 110 units. Raw materials are concentrated in three places of their receipt, and the stocks are respectively equal to 160, 140, 170 units. Raw materials can be delivered to each of the enterprises from any point of its receipt. Freight rates are known values and are given by the matrix

Draw up a transportation plan in which the total cost of transportation is minimal.

Solution. Let us denote by the number of units of raw materials transported from the i-th point of its receipt to the j-th enterprise. Then the conditions for the delivery and export of the necessary and available raw materials are ensured by fulfilling the following equalities:

(6)

With this plan transportation, the total cost of transportation will be

Thus, the mathematical formulation of this transport problem consists in finding such a non-negative solution to the system of linear equations (6), in which the objective function (7) takes the minimum value.

Solution of the transport problem

The main steps in solving the transport problem:

1. Find an initial feasible plan.

2. Choose from non-basic variables the one that will be introduced into the basis. If all non-basic variables satisfy the optimality conditions, then finish the solution, otherwise go to the next. step.

3. Choose a variable to be derived from the basis, find a new basic solution. Return to step 2.

Any non-negative solution of systems of linear equations (2) and (3) determined by the matrix , is called the transport task plan. The reference (basic) plan of the T-problem is any of its feasible, basic solutions.

Usually, the initial data of the transport task is recorded in the form of a table.

Matrix C is called the matrix of transportation costs, the matrix X that satisfies the conditions of the T-problem (2) and (3) is called the transportation plan, and the variables are called transportation. The plan , at which the objective function is minimal, is called optimal.

The number of variables in the transportation problem with m departure points and n destinations equals m*n, and the number of equations in systems (2) and (3) is m+n. Since we assume that condition (5) is satisfied, the number of linearly independent equations is equal to m+n-1. Therefore, the basic plan of the transport task can have no more than m+n-1 non-zero unknowns.

If in the reference design the number of non-zero components is exactly equal to m+n-1, then the design is non-degenerate, and if less, then degenerate.

As for any linear programming problem, the optimal plan of the transport problem is also a base plan.

Construction of an admissible (reference) plan in the transportation problem

By analogy with other problems of linear programming, the solution of the transport problem begins with the construction of an admissible basic plan. There are several methods for constructing initial base plans for the T-problem. Of these, the most common northwest corner method And minimum element method.

The simplest way to find it is based on the so-called northwest corner method. The essence of the method is the sequential distribution of all stocks available at the first, second, etc. points of production, according to the first, second, etc. points of consumption. Each distribution step is reduced to an attempt to completely deplete stocks at the next point of production or to an attempt to fully satisfy the needs at the next point of consumption. At every step q current unallocated reserves are indicated and i (q ), and current unmet needs - b j (q ) . The construction of an acceptable initial plan, according to the northwest corner method, starts from the upper left corner of the transport table, while we assume a i (0) = a i , b j (0) = b j . For the next cell located in the row i and column j , we consider the values of unallocated stock in i -th point of production and unmet need j -th point of consumption, of which the minimum is selected and assigned as the volume of transportation between these points: x i, j =min(a i (q) , b j (q) ) . After that, the values of unallocated stock and unmet demand in the respective points are reduced by this amount:

a i (q+1) = a i (q) - x i , j , b j (q+1) = b j (q) - x i , j

Obviously, at each step, at least one of the equalities is satisfied: and i (q+1) = 0 or b j (q+1) = 0 . If the first is true, then this means that the entire stock of the i-th production point is exhausted and it is necessary to proceed to the distribution of the stock at the production point i+1 , i.e. move to the next cell down the column. If b j (q+1) = 0, This means that the need for j -th point, after which the transition to the cell located to the right of the line follows. The newly selected cell becomes the current one, and all the listed operations are repeated for it.

Based on the condition of the balance of supplies and needs, it is not difficult to prove that in a finite number of steps we will obtain an admissible plan. By virtue of the same condition, the number of steps of the algorithm cannot be more than m+n-1 , so will always remain free (zero) mn-(m+n-1) cells. Therefore, the resulting plan is basic. It is possible that at some intermediate step, the current unallocated stock turns out to be equal to the current unmet need (a i (q) = b j (q)) . In this case, the transition to the next cell occurs in a diagonal direction (the current production and consumption points change simultaneously), which means the “loss” of one nonzero component in the plan, or, in other words, the degeneracy of the constructed plan.

A feature of an acceptable plan constructed by the northwest corner method is that the objective function on it takes a value, as a rule, far from the optimal one. This is because it does not take into account the values c i , j . In this regard, in practice, to obtain the original plan, another method is used - minimum element method, in which, when distributing traffic volumes, the cells with the lowest prices are occupied first.

An example of finding a baseline

F=14 x 11 + 28 x 12 + 21 x 13 + 28 x 14 + 10 x 21 + 17 x 22 + 15 x 23 + 24 x 24 + 14 x 31 + 30 x 32 + 25 x 33 + 21 x 34

The original plan was obtained using the northwest corner method. The problem is balanced (closed).

Table 1

The cost of transportation under this plan is: 1681:

F=14 *27 + 28* 0 + 21*0 + 28*0 + 10 *6 + 17 *13 + 15*1 + 24 *0 + 14 *0 + 30 *0 +25*26 + 21 *17 = 1681

General setting transport task consists in determining the optimal plan for the transportation of some homogeneous cargo from T departure points in P destinations . In this case, either the minimum cost of transporting the entire cargo or the minimum time for its delivery is usually taken as an optimality criterion. Let us consider a transport problem, the optimality criterion of which is the minimum cost of transporting the entire cargo. Denote by the tariffs for the transportation of a unit of cargo from i-th point of departure in j-th destination, through - stocks of cargo in i-th point of departure, via – cargo needs in j – m destination, and through – the number of units of cargo transported from i-th point of departure in j-th destination. Then the mathematical formulation of the problem consists in determining the minimum value of the function

under conditions

(64)

(65)

(66)

Since the variables satisfy the systems of equations (64) and (65) and the condition non-negativity(66), the delivery of the required amount of cargo to each of the destinations, the removal of the existing cargo from all points of departure are ensured, and return shipments are also excluded.

Definition 15.

Any non-negative solution of systems of linear equations (64) and (65) defined by the matrix , is called plan transport task.

Definition 16.

Plan , at which function (63) takes its minimum value, is called optimal plan transport task.

Usually, the initial data of the transport task is recorded in the form of table 21.

Table 21

then the model of such a transport problem is called closed. If the specified condition is not met, then the model of the transport problem is called open.

Theorem 13.

For the solvability of the transport problem, it is necessary and sufficient that the stocks of cargo at the points of departure are equal to the needs for cargo at the destinations, i.e., that the equality (67).

In case of excess stock over demand, i.e., a fictitious ( n+1) – th destination with need and the corresponding tariffs are considered equal to zero: The resulting problem is a transport problem for which equality (67) is satisfied.

Similarly, for , a fictitious ( m+1) - th point of departure with cargo stock and tariffs are assumed to be zero: This reduces the problem to an ordinary transport problem, from the optimal plan of which the optimal plan of the original problem is obtained. In the future, we will consider a closed model of the transport problem. If the model of a specific problem is open, then, proceeding from the above, we rewrite the table of conditions of the problem so that equality (67) is satisfied.

The number of variables in the transportation problem with T departure points and P destinations equals Fri, and the number of equations in systems (64) and (65) is p+t . Since we assume that condition (67) is satisfied, the number of linearly independent equations is equal to p+t – 1. Consequently, the basic plan of the transport task can have no more than P + T – 1 non-zero unknowns.

If in the reference design the number of non-zero components is exactly equal to P +t – 1, then the design is non-degenerate, and if less, then degenerate.

As for any , the optimal plan of the transport task is also a reference plan.

To determine the optimal plan for the transport problem, you can use the above methods.

Example 19.

Draw up a transportation plan in which the total cost of transportation is minimal.

Solution. Denote by the number of units of raw materials transported from i-th point of its receipt on j-th enterprise. Then the conditions for the delivery and export of the necessary and available raw materials are ensured by fulfilling the following equalities:

(68)

With this plan transportation, the total cost of transportation will be

Thus, the mathematical formulation of this transport problem consists in finding such a non-negative solution to the system of linear equations (68), in which the objective function (69) takes the minimum value.

Program for solving the transport problem by the potential method

The general statement of the transport problem is to determine the optimal plan for the transportation of some homogeneous cargo from m departure points (suppliers) A 1 , A 2 , . . ., A m V n points of consumption (consumers) B 1 , B 2 , . . . B n so that:

Take out all goods from suppliers;

Satisfy the demand of each consumer;

Ensure the minimum total transport costs for the transportation of all goods.

Consider the transportation problem as the optimality criterion of which is the minimum cost of transportation of the entire cargo.

Denote:

a i- the presence of cargo in i -th point of departure;

b j- the value of the need for this cargo in j th destination ;

with ij- the cost of transporting a unit of cargo from i -th point of departure to j -th point of consumption (transportation tariff);

xij- the amount of cargo transported from i -th point of departure to j -th destination, destination, xij ≥ 0.

The mathematical formulation of the transport problem consists in finding such a non-negative solution to the system of linear equations, in which the objective function takes the minimum value.

Let us write down the mathematical model of the transport problem.

It is required to determine the matrix ) that satisfies the following conditions:

,
(5.1)

,
(5.2)

,
,
(5.3)

and delivers the minimum value of the objective function

L() =
(5.4)

Since the variables satisfy the system of linear equations (5.1), (5.2) and the non-negativity condition, it ensures the delivery of the necessary cargo to each consumer, the export of the existing cargo from all suppliers, and also excludes return transportation.

Definition 1. Any non-negative solution of the systems of linear equations (5.1) and (5.2) defined by the matrix ) is called valid transport task plan.

Definition 2 . Plan ) at which the function (5.4) takes the maximum value, is called the optimal plan of the transport problem.

Theorem 1. The rank of the matrix, composed of the coefficients for the unknowns of the system of linear equations (5.1) and (5.2) of the transport problem, is one less than the number of equations, i.e. equals (m + n - 1).

Therefore, the number of linearly independent equations is (m + n - 1), they form a basis, and the variables corresponding to them will be basic.

Definition 3. An admissible plan of the transport task, which has no more than (m + n - 1) non-zero values, is called basic or reference.

Definition 4. If the number of non-zero values of variables in the reference design is exactly (m + n – 1), then the design is non-degenerate, and if it is less, then it is degenerate.

Theorem 2. For the solvability of the transport problem, it is necessary and sufficient that the total stocks of cargo at the points of departure were equal to the total needs at the points of destination, i.e.

Definition 5. A transport problem model that satisfies condition (5.5) is called closed. If the specified condition is not met, then the model is open.

To obtain an optimal plan, the open model of the transport problem must be reduced to a closed model.

In case of excess stock over demand, i.e. > , a fictitious (n+ 1) –th destination is entered with a need b n +1 = − , and the corresponding tariffs are considered equal to zero: With i , n + 1 = 0

If< , то вводится фиктивный (m + 1) –ый пункт отправления с потребностью a m +1 = − and tariffs With m+ 1, j = 0

Consider one of the methods for constructing the first basic plan of the transport problem - the method of minimum cost or the best element of the matrix of unit costs.

Definition 6. The best element of the matrix of unit costs (tariffs) will be called the lowest tariff, if the task is set to the minimum of the objective function, the highest tariff, if the task is set to the maximum.

Algorithm for constructing the first reference plan.

1. Among the matrix of unit costs, we find the best tariff.

2. The cell of the distribution table with the selected tariff is filled with the maximum possible volume of cargo, taking into account the restrictions on the row and column. In this case, either the entire cargo is exported from the supplier, or the consumer's need is fully satisfied. The row or column of the table is deleted from consideration and does not participate in the further distribution.

3. From the remaining tariffs, we again choose the best one and the process continues until the entire load is distributed.

If the transport task model is open and a fictitious supplier or consumer is entered, then the distribution is first performed for the actual suppliers and consumers, and lastly, the unallocated cargo is directed from the fictitious supplier or to the fictitious consumer.

Further improvement of the first basic plan of the transport task and obtaining the optimal plan is carried out using the potential method.

Theorem 3. The plan ) of the transport problem is optimal if there is a system (m + n) of numbers u i and v j (called potentials) that satisfies the conditions:

(5.6)

(5.7)

The potentials u i and v j are the variables of the dual problem, composed to the original transport problem, and denote the estimate of the unit of cargo at the points of departure and destination, respectively.

Denote: ) the estimate of the free (unoccupied) cell of the table.

Definition 7. The reference plan of the transport problem is optimal if all estimates of the free cells of the distribution table (the problem is set to a minimum).

Potential Method Algorithm

1. Construction of the first basic plan transportation problem by the minimum cost method.

2. Degeneracy check .

Potentials can only be calculated for a non-degenerate plan. If the number of occupied cells in the reference plan (the number of basic variables) is less than (m+n−1), then we introduce zero into one of the free cells of the table so that total number occupied cells became equal to (m+n−1). Zero is entered into the cell with the best rate The that belongs to the row or column. Simultaneously crossed out when drawing up the first reference plan. In this case, the cell of the table fictitiously occupied by zero should not form a closed rectangular contour with other occupied cells of the table.

3. Calculation of the value of the goal function(5.4) by summing up the products of tariffs (unit costs) by the volume of transported cargo for all occupied cells of the table.

4. Checking the optimality of the plan.

We determine the potentials. For each occupied cell, we write down the equation , as a result we get a system of (m + n−1) equations with (m + n) variables.

Since the number of variables is greater than the number of equations, the resulting system is not defined and has an infinite number of solutions. Therefore, one of the unknown quantities is given an arbitrary value. For simplicity of calculations, we assume , then the remaining potentials are determined uniquely, and their values are entered in an additional row and column of the distribution table.

For each free cell, we determine estimates .

If everything (the problem is solved for the minimum of the objective function), then the optimal plan is found. If at least one estimate of the free cell does not satisfy the optimality condition, then it is necessary to improve the plan by redistributing the load.

5. Building a new baseline.

Of all positive estimates of free cells, we choose the largest one (the task is set to a minimum); of all the negative ones - the largest in absolute value (the task is set to the maximum). The cell, which corresponds to the highest score, should be filled in, i.e. send a load to it. When filling in the selected cell, it is necessary to change the volume of supplies recorded in a number of other occupied cells and associated with the so-called cycle being filled.

A cycle or a rectangular contour in the distribution table of a transport problem is a broken line, the vertices of which are located in the occupied cells of the table, and the links along the rows and columns, and at each vertex of the cycle there are exactly two links, one of which is in the row, the other in the column . If the polyline forming the cycle intersects, then the intersection points are not vertices. For each free cell, a single cycle can be constructed.

To the vertices of the cycle, starting from the vertex located in the selected cell for loading, we assign the signs "+" and "−" in turn. We will call these cells plus and minus.

From the volumes of cargo in minus cells, we choose the smallest one and denote it by θ. We redistribute the value of θ along the contour, adding θ to the corresponding cargo volumes in the plus cells, and subtracting θ from the cargo volumes in the minus cells of the table. As a result, the cell that was free and selected for loading becomes occupied, and one of the occupied cells of the contour becomes free.

The resulting base plan is checked for optimality, i.e. We return to the fourth stage of the algorithm.

Remarks

1. If there are two or more identical minimum values in the minus cells of the constructed cycle, then not one, but two or more cells are released during the redistribution of cargo volumes. In this case, the plan becomes degenerate. To continue the solution, it is necessary to occupy one or several simultaneously freed cells of the table with zero, and preference is given to the cells with the best rate. Zeros are introduced so that in the newly obtained base plan the number of occupied cells (basic variables) is exactly (m + n−1).

2. If in the optimal plan of the transport problem the estimate for some free cell is equal to zero ) , then the problem has a set of optimal plans. For a cell with a zero score, you can build a cycle and redistribute the load. As a result, the resulting plan will also be optimal and have the same value of the objective function.

3. The value of the objective function at each iteration can be calculated as follows:

(the task is set to a minimum),

(the task is set to the maximum),

where is the value of the volume of cargo moved along the contour;

Evaluation of the free cage into which the cargo is sent when moving to a new reference plan;

− value of the goal function at the k-th iteration;

− value of the goal function at the previous iteration.

Example

Three warehouses of the wholesale base have a homogeneous cargo in quantities of 40, 80 and 80 units. This cargo must be transported to four stores, each of which must receive 70, 20, 60 and 60 units, respectively. Unit delivery costs (tariffs) from each warehouse ) to all stores ) are given by the matrix .

Draw up a plan for the transportation of a homogeneous cargo with minimal transport costs (conditional numbers).

Solution.

1. Let us check the necessary and sufficient condition for the solvability of the problem:

40+80+80 = 200,

70+20+60+60 = 210.

As you can see, the total demand for cargo exceeds its stocks in the warehouses of the wholesale base. Consequently, the model of the transport problem is open and has no solution in its original form. To obtain a closed model, we introduce an additional (fictitious) warehouse A 4 with a cargo stock equal to A 4 = 210 - 200 = 10 units we assume that the tariffs for transporting a unit of cargo from warehouse A 4 to all stores are equal to zero.

We enter all the initial data in table 7.

Stocks

IN 1

AT 2

AT 3

AT 4

A 1

A 3

Needs

2. Construction of the first basic plan by the minimum cost method.

Among the tariffs, C 14 =1 is the minimum or best. We send the maximum possible load equal to min(60,40) = 40 into cell A 1 B 4. Then x 14 = 40. From warehouse A 1, all the cargo was taken out, but the need of the fourth store is unsatisfied by 20 units. line A 1 is out of consideration.

Among the remaining tariffs, the minimum element is C 23 = 2. In cell A 2 B 3 we send the cargo min (60.80) = 60. In this case, column B 3 is out of consideration, and 20 units were not taken out of warehouse A 2.

Of the remaining elements, the minimum C 22 \u003d 3. In the cell A 2 B 2 we send the load in the amount of min (20,20) \u003d 20. In this case, the column A 2 and column B 2 are simultaneously crossed out.

We select the minimum element C 31 = 4. In cell A 3 B 1 we send a load equal to min(70.80) = 70. In this case, column B 1 is out of consideration, and 10 units are not taken out of warehouse A 3. The remaining cargo from the third warehouse is sent to the notch A 3 B 4, x 34 = 10. The need of the fourth store is not satisfied by 10 units. send from a fictitious supplier - warehouse A 4 10 units. cargo in a cage A 4 B 4, x 44 = 10.

As a result, the first baseline was obtained, which is valid, since all goods have been removed from the warehouses and the needs of all stores have been satisfied.

3. Degeneracy check

The number of occupied cells or basic variables in the first reference plan is six. the plan of the transport problem is degenerate, since the number of basic variables in the non-degenerate plan is m + n - 1 = 4 + 4 - 1 = 7. To continue solving the problem, the base plan must be supplemented with the introduction of a fictitious transportation, i. occupy one of the free cells with zero.

When constructing the first reference plan, row A 2 and column B 2 were simultaneously crossed out, so the plan degenerated. The right of fictitious transportation is claimed by free cells of row A 2 and column B 2, which have a minimum tariff and do not form a closed rectangular contour with occupied cells. These cells are A 2 B 4 and A 3 B 2 . Zero is sent to the cell A 2 B 4.

4. Calculation of the objective function value

The value of the objective function of the first reference plan is determined by summing up the products of tariffs for the volume of transported cargo for all occupied cells of the table.

L(X 1) = 4∙70 + 3∙20 + 2∙60 + 1∙40 + 3∙0 + 6∙10 + 0∙10 = 560 (thousand rubles).

5. Checking the optimality condition

Calculate the potentials for the occupied cells of the table from the condition: (Since the number of unknown potentials is greater than the number of equations (m + n > m + n - 1), then one of the potentials is taken equal to zero. Let . Then for the occupied cells we can write down the system of equations:

Assuming , we get , , , ,

The calculated potentials are entered in Table 7. Let us calculate the estimates of free cells.

The first reference plan is not optimal, since there are positive estimates of free cells and . We choose the maximum positive evaluation of a free cell - .

6. Building a new baseline

For cell A 3 B 2, we will construct a rectangular closed circuit 0table 7) and redistribute the load to the circuit. To the vertices of the contour, starting from the vertex located in the free cell A 3 B 2, we assign the signs "+" and "-" in turn.

From the volumes of cargo in minus cells, we choose the smallest, i.e. θ = min(20,10) = 10. We add the value θ = 10 to the cargo volumes in the plus cells, subtract from the cargo volumes in the minus cells of the closed loop. As a result, we obtain a new reference plan, shown in Table 8.

Stocks

IN 1

AT 2

AT 3

AT 4

A 1

A 3

3 10

Needs

The second reference plan of the transport problem is non-degenerate, since the number of occupied cells is 7.

L (X 2) \u003d L (X 1) - θ ∙ \u003d 560 - 10 ∙ 3 \u003d 530 (thousand rubles).

We check this plan for optimality. We again find the potentials of suppliers and consumers. To do this, we compose the following system of equations for the occupied cells.