Methods for constructing the initial support solution. Strike-out method Minimum cost method

There are developments in the software implementation of the method. If you are interested in creating an advisor, write.

Here is a description of the method.

Money management is based on the Martingale modification - Labouchere,
also known as the "deletion method". This method is not as extreme as a regular martingale.
What is the principle of transaction management?

At the dawn of the casino, to play on equal terms (for example, red - black), a method of doubling the bet on loss was invented. I will not go into the description in detail, but this method, despite the fact that mathematically, it certainly allows you to win, has negative features. The rates grow exponentially and sooner or later, you will either win, or you will be faced with the lack of the necessary amount in your pocket for the next doubling of the rate, or with the limitation of the maximum bet on the gaming table.

Let me remind you that the mathematical probability of winning when playing classic roulette is 49%. 1% - ZERO, this is the advantage of the casino.

The deletion method is as follows. We divide our deposit into 100 parts.
1% of the deposit is one contract.

We start the game with 1 contract. We take a paper and a pen, write down the rates in a column below each other.

-1
We add 1 more contract to the lost one. The next rate is 2 contracts. For example, we won. We write in a column
-1
+2
In total, we won 1 contract. We cross out everything, start over. The next rate is 1 contract.

Let's consider a more interesting series.

For example, we lost the first bet. Writing it down on paper
-1
We add 1 more contract to the lost one. The next rate is 2 contracts. For example, we lost. We write in a column
-1
-2
Now, to the first rate in the column (-1), add the last rate (-2). Total 3 contracts. Let's say we lost. We write it down in a column.
-1
-2
-3
Now, to the first rate in the column (-1), add the last rate (-3). Total 4 contracts. Let's say we lost again. We write in a column
-1
-2
-3
-4
Now, to the first rate in the column (-1), add the last rate (-4). Total 5 contracts. Let's say we lost again. We write in a column
-1
-2
-3
-4
-5
Five losses in a row. It happens ... The next rate is 6 Contracts.
For example, we won. We write it down in a column.
-1
-2
-3
-4
-5
+6
6 contracts that we won compensated for the loss -1 and - 5 contracts! Now, cross out -1, -5 and +6.
Left:
-2
-3
-4
Now, to the first rate in the column (-2), add the last rate (-4). Total 6 contracts. The next bet is 6 Contracts. Let's say we won again. We write in a column
-2
-3
-4
+6
6 contracts that we won compensated for the loss of -2 and - 4 contracts! Now, cross out -2, -4 and +6.
There are -3 contracts left. Since there is nothing else in the column, add 1.
The next rate is 4 contracts. If we win, then we cross everything out, stay in positive territory by 1 contract and start the streak all over again.

We had such a series
-1
-2
-3
-4
-5
+6
+6
+4

Three winning trades offset 5 losing trades.
I advise you to practice on paper, several times, until the principle comes to automatism.

So, pay attention! In order for the system to function and win, it is necessary to have the number of profitable trades above 33% -40% percent !!!
When in doubt, write your own long series. You can practice at any online casino that has a test game for virtual money. Divide your deposit into 100 parts. Bet only on red or only on black. Keep in mind that such a method of playing can be considered by the casino as dishonest, and after a while the casino computer will start arranging for you a series of the opposite color 10-20-30 in a row, of course, we will not talk about any 33-40 percent ratio and you will lose.

But the principle remains unchanged, 33% of wins compensate for 66% of losses.
Thus, applying such money management in practical Forex trading, we need a trading system that has a 50% probability of winning, and the ratio of possible profit to possible loss is greater than or equal to 1,
those. Profit factor> = 1.

There are a number of methods for constructing an initial reference solution, the simplest of which is the northwest corner method. In this method, the stocks of the next supplier are used to satisfy the requests of the next consumers until they are completely exhausted, after which the stocks of the next supplier by number are used.
Filling in the transport task table starts from the upper left corner and consists of a number of steps of the same type. At each step, based on the stocks of the next supplier and the requests of the next consumer, only one cell is filled and, accordingly, one supplier or consumer is excluded from consideration. This is done in this way:
1) if a i< b j то х ij = а i , и исключается поставщик с номером i ,
x im = 0, m = 1, 2, ..., n, m ≠ j, b j ’= b j - a i
2) if a i> b j then х ij = b j, and the consumer with number j is excluded, x m j = 0, m = 1,2, ..., k, m ≠ i, a i ’= a i - b j,
3) if ai = bj then x ij = ai = bj, either supplier i, x im = 0, m = 1,2, ..., n, m ≠ j, bj '= 0, or the j -th consumer , xmj = 0, m = 1,2, ..., k, m ≠ i, ai '= 0.
It is customary to enter zero transportations into the table only when they enter the cell (i, j) to be filled. If in the next cell of the table (i, j) it is required to put a carriage, and the i-th supplier or j-th consumer has zero stocks or requests, then a carriage equal to zero (basic zero) is placed in the cell, and after that, as usual, the supplier or consumer concerned is excluded from consideration. Thus, only basic zeros are entered into the table, the remaining cells with zero transportation remain empty.
To avoid errors, after constructing the initial support solution, it is necessary to check that the number of occupied cells is equal to k + n- 1 and the vector-conditions corresponding to these cells are linearly independent.
□ Theorem. The solution to the transport problem, built by the northwest corner method, is the reference one.
Proof ... The number of cells in the table occupied by the support solution must be equal to N = k + n-1. At each step of constructing a solution using the northwest corner method, one cell is filled and one row (supplier) or one column (consumer) of the problem table is excluded from consideration. After k + n– 2 steps, k + n– 2 cells will be occupied in the table. At the same time, one row and one column will remain uncrossed, while there is only one unoccupied cell. When this last cell is filled, the number of occupied cells will be
k + n - 2 +1 = k + n– 1.
Let us check that the vectors corresponding to the cells occupied by the support solution are linearly independent. Let's apply the strikeout method. All occupied cells can be crossed out if you do this in the order of their filling. ■
It should be borne in mind that the northwest corner method does not take into account the cost of transportation; therefore, the reference solution constructed by this method may be far from optimal.
Example ... Create an initial support solution using the northwest corner method for a transport problem, the initial data of which are presented in the following table

a i b j	150	200	100	100
100	1	3	4	2
250	4	5	8	3
200	2	3	6	7

Solution. We distribute stocks of the 1st supplier. Since its stocks a 1 = 100 are less than the requests of the 1st consumer b 1 = 150, then in cell (1, 1) we write down the transportation x 11 = 100 and exclude the 1st supplier from consideration. Determine the remaining unsatisfied requests of the 1st consumer b ’= b 1 - a 1 = 150 - 100 = 50.
We distribute stocks of the 2nd supplier. Since its stocks a 2 = 250 are more than the remaining unsatisfied requests of the 1st consumer b 1 '= 50, then in cell (2, 1) we write down the transportation x 21 = 50 and exclude the 1st consumer from consideration. Determine the remaining stocks of the 2nd supplier a 2 = a 2 - b 1 '= 250 -50 = 200. Because and 2 ’= b 2 = 200, then in cell (2, 2) we write x 22 = 200 and exclude, at our discretion, either the 2nd supplier or the 2nd consumer. Let's exclude the 2nd supplier. We calculate the remaining unsatisfied requests of the 2nd consumer b 2 "= b 2 - a 2" = 200 - 200 = 0.
We distribute stocks of the 3rd supplier. Since a 3> b 2 (200> 0), then in the cell (3, 2) we write x 32 = 0 and exclude the 2nd consumer. Stocks of the 3rd supplier did not change a 3 ’= a 3 -b 2’ = 200 - 0 = 200. Compare a 3 "and b 3 (200> 100), write x 33 = 100 in cell (3, 3), exclude the 3rd consumer and calculate a 3" = a 3 "-b 3 = 200 - 100 = 100. Since a 3 "" = b 4, then in cell (3, 4) we write x 34 = 100. In view of the fact that the problem with the correct balance, the stocks of all suppliers are exhausted and the requests of all consumers are satisfied completely and simultaneously.
The results of constructing the reference solution are shown in the table:

	150	200	100	100
100	100
250	50	200
200		0	100	100

We check the correctness of the construction of the reference solution. The number of occupied cells must be equal to N = k + n - 1 = 3 + 4 - 1 = 6. There are six cells in our table. Applying the method of crossing out, we make sure that the found solution is "crossed out":
Consequently, the condition vectors corresponding to the occupied cells are linearly independent and the constructed solution is supporting.

Minimum cost method

The minimum cost method is simple, it allows constructing a support solution close enough to the optimal one, since it uses the cost matrix of the transport problem C = (c ij), i = 1,2, ..., k, j = 1,2, .. ., n. Like the northwest corner method, it consists of a number of steps of the same type, at each of which only one cell of the table is filled, corresponding to the minimum cost min (with ij)), and only one row (supplier) or one column (consumer ). The next cell corresponding to min (with ij) is filled in according to the same rules as in the northwest corner method. A supplier is excluded from consideration if its reserves are fully used. The consumer is excluded from consideration if his requests are fully satisfied. At each step, either one supplier or one consumer is excluded. In this case, if the supplier is not yet excluded, but his stocks are equal to zero, then at the step when the given supplier is required to deliver the cargo, a basic zero is entered in the corresponding cell of the table and only then the supplier is excluded from consideration. Likewise with the consumer.
□ Theorem ... The solution to the transport problem, constructed by the minimum cost method, is the reference one. ■
The proof is similar to the proof of the previous theorem.
Example ... Using the minimum cost method, construct an initial reference solution to the transport problem, the initial data of which are given in the table:

	4 0	6 0	8 0	6 0
60	1	3	4	2
80	4	5	8	3
100	2	3	6	7

Solution ... Let's write down the matrix of costs separately in order to make it more convenient to choose the minimum costs, cross out rows and columns:
Among the elements of the cost matrix, select the lowest cost with 11 = 1, mark it with a circle. This is the cost of transporting goods from the 1st supplier to the 1st consumer. In the corresponding cell (1, 1) we write the maximum possible volume of transportation x 11 = min (a, A,) = min (60, 40) = 40.
Table 6.6

	40	60	80	60
60	40			20
80			40	40
100		60	40

Decrease the stock of the 1st supplier by 40, i.e. a 1 '= a 1 -b 1 = 60 - 40. = = 20. We exclude from consideration the 1st consumer, since his requests are satisfied. In the matrix, C cross out the 1st column.
In the rest of the matrix C, the minimum cost is with 14 = 2. The maximum possible transportation that can be carried out from the 1st supplier to the 4th consumer is x 14 = min (a 1 ', b 4) = min (20.60) = 20. In the corresponding cell of the table, we write down the transportation x 14 = 20 - The supplies of the 1st supplier are exhausted, we exclude it from consideration. In matrix C, delete the first row. We decrease the requests of the 4th consumer by 20, i.e. b 4 "= b 4 - a 1" = 60-20 = 40.
In the rest of the matrix C, the minimum cost is c 24 = c 32 = 3. We fill in one of the two cells of the table (2, 4) or (3, 2). Let in the cell (2, 4) we write x 24 = min (a 2, b 4) = min (80, 40) = 40. The requests of the 4th consumer are satisfied, we exclude him from consideration "we delete the fourth column in the matrix C. We reduce the stocks of the 2nd supplier a 2 '= a 2 - b 4 = 80 - 40 = 40.
In the rest of the matrix C, the minimum cost is min (c ij) = c 32 = 3. We write in the cell of the table (3,2) transportation x 32 = min (a 3 b 2) = min (100, 60) = 60. We exclude from consideration the 2nd consumer, and from the matrix C the second column. Calculate a 3 '= a3-b 2 = 100 - 60 = 40.
In the rest of the matrix C, the minimum cost is min (c ij) = c 33 = 6. We write in the cell of the table (3,3) the transportation x 33 = min (a 3 ", b 3) = min (40, 80) = 40. We exclude from consideration the third supplier, and from the matrix C the third row. Determine b 3 "= b 3 - a 3" = 80 - 40 = 40. In the matrix C there is only one element with 23 = 8. We write in the cell of the table (2, 3) the transportation x 23 = 40.
We check the correctness of the construction of the reference solution. The number of occupied cells in the table is N = k + n- 1 = 3 + 4-1 = 6. We check the linear independence of the vector-conditions corresponding to the positive coordinates of the solution by the method of deletion. The order of deletion is shown in the matrix X:
The solution is “crossed out” and therefore pivotal.

Moving from one pivot solution to another

In a transport problem, the transition from one support solution to another is carried out using a cycle. For some free cell of the table, a cycle is constructed that contains a part of cells occupied by the support solution. For this cycle, traffic volumes are redistributed. The carriage is loaded into the selected free cell and one of the occupied cells is released, a new support solution is obtained.
□ Theorem (on the existence and uniqueness of the cycle). If the table of the transport problem contains a support solution, then for any free cell of the table there is a single cycle containing this cell and some of the cells occupied by the support solution.
Proof ... The support solution occupies N = k + n- 1 cells of the table, which correspond to linearly independent vector-conditions. According to the theorem proved above, no part of occupied cells forms a cycle. If we add one free cell to the occupied cells, then the corresponding k + n vectors are linearly dependent, and by the same theorem there is a cycle containing this cell. Suppose that there are two such cycles (i 1, j 1), (i 1, j 2), (i 2, j 2), ..., (ik, j 1), and (i 1, j 1), (i 2, j 1), (i 2, j 2),…, (il, j 1), -Then, combining cells of both cycles without a free cell (i 1, j 1), we obtain a sequence of cells (i 1, j 1 ), (i 1, j 2), (i 2, j 2), ..., (ik, j 1), (i 1, j 1), (i 2, j 1), (i 2, j 2) ,…, (Il, j 1) which form a cycle. This contradicts the linear independence of the condition vectors that form the basis of the support solution. Therefore, such a cycle is unique.
The designated cycle.
A cycle is called designated if its corner cells are numbered in order and the “+” sign is assigned to the odd cells, and the “-” sign is assigned to the even ones.
A shift along the cycle by the value θ is an increase in traffic volumes in all odd cells of the cycle marked with a “+” sign by θ and a decrease in traffic volumes in all even cells marked with a “-” sign by θ.
□ Theorem ... If the table of the transport problem contains a support solution, then shifting along any cycle containing one free cell by an amount will result in a support solution.
Proof ... In the table of the transport problem containing the support solution, select a free cell and mark it with a "+" sign. By Theorem 6.6, for this cell there is a single cycle that contains some of the cells occupied by the support solution. Let's number the cells of the cycle, starting with the cell marked with the “+” sign. Let's find the shift along the cycle by this amount
In each row and in each column of the table included in the cycle, there are two and only two cells, one of which is marked with a "+", and the other - with a "-". Therefore, in one cell the volume of transportation increases by θ, while in another it decreases by θ, while the sum of all transportation in a row (or column) of the table remains unchanged. Consequently, after the shift in the cycle, the stocks of all suppliers are still completely exported, and the requests of all consumers are fully satisfied. Since the shift in the cycle is carried out by an amount, all traffic volumes will be non-negative. Therefore, the new solution is valid.
If one of the corresponding cells with zero transportation volume is left free, then the number of occupied cells will be equal to N = k + n-1. One cell is loaded (marked with a "+"), one is released. Since the cycle is unique, the removal of one cell from it breaks it. It is impossible to form a cycle from the remaining occupied cells, the corresponding vector-conditions are linearly independent, and the solution is supporting.

Problem number 4. Increase in the number of transactions:

What calls to action can there be? Example: "Call now", "Find out more on our website", "Find out more by calling ...".

P.S. If you just read this article and did not implement any of the above methods of increase in your enterprise, then you have wasted your time.

If you are going to implement 2-3 of your favorite ways to increase sales in your organization, then good results await you.

If you decide to use each of the methods described here, then the problem of warehouse stocks will cease to exist for you. And you will forget that this question was once so urgent for you.

P.P.S. What is a profitable plant? This is an enterprise that realizes what place its products occupies in the market and sells them competently! Sales work is the same lead generation. Sales funnel analysis, online marketing. All the same!

Graphical method

Graphical methods for determining the most effective project are the least accurate, but the most illustrative, therefore they are usually used in various kinds of presentations. The essence of the graphical technique is that no rating is determined for each calculated and analyzed indicator, but the values ​​of the indicators are plotted on the graphical axes. To build symbolic efficiency, as many equidistant axes are plotted on the coordinate plane, according to how many indicators it is extremely important to draw a conclusion, and these indicators should not be less than three, but optimally there should be as many as possible.

Deposition points of exponents on planes for direct exponents are built from 0, and for reciprocal - from the maximum possible value. The maximum values ​​for the opposite indicators are determined based on the average values ​​for projects of different directions. It is important to note that for the creation of industrial enterprises the maximum value of the payback period is 10 years, for residential construction - 6 years, for the creation of enterprises engaged in heavy metallurgy - 12 years.

For such an indicator as the break-even point, 2 aspects should be taken into account:

1. It is not the break-even volume of production in units of production that is graphically reflected, but the indicator of the profitability threshold, which is such a revenue that will completely cover fixed and variable costs and lead the company to the absence of both profit and loss.

2. At point 0, an amount equal to a quarter of the investment costs is deposited and the advance along the axis is carried out with a scale of 1 = 100t.r.

The indicator of the tax burden is constructed from one and a half standards determined by the federal tax service (normal values ​​of the tax burden have been established for all possible branches of activity).

For those industries where the normal tax burden is up to 20%: 1 division step is 1%, and for those industries where more than 20% - 2%.

For direct monetary indicators, the division step is 1/10 of the investment costs in the project. For direct percentages, the division step is 0.1% (except for IRR, where the division step is 5%).

Putting all points for all projects on the coordinate axes, the line closes each project separately. And the most profitable is the project with the greatest distance of points from the center (if there are several such projects, then the one that is closest to the circular value).

It is based on the principle that if it is impossible to choose the best project based on all the available criteria, then it is extremely important to exclude the criteria from the calculation.

Initially, the deletion method uses such criteria as the payback period of the project IDI, IRR and TSP. In order to cross out any indicator, it is extremely important to assess the rating of this criterion. Before the start of deletion, all criteria are equivalent, that is, each criterion is initially assigned, then each criterion is initially assigned 25 rating points.

Calculations begin with the TSP, determining on the basis of which the investor has established the maximum allowable value for the payback period.

If the optimal value of the payback period is established from the extremely important financing of another project, then the importance of the payback period increases by 3 points. And in this regard, it is extremely important to reduce the significance of the remaining 3 indicators by 3 points, that is, a reduction by 1 point for each indicator. If the five-year payback period is set on the basis of average values ​​of the payback period in the industry, then the payback period rating increases by 1.5 points, while the rating of other indicators decreases by 0.5 points for each.

If the payback period is set on a different basis, then the payback period rating and other indicators do not change.

If the GNI indicator is within the sum of the inflation rate and the refinancing rate, then the GNI rating is increased by 6 points. At the same time, the ratings of other indicators are reduced by 2 points each.

If GNI is set higher than the sum of the refinancing rate and inflation, then for every 0.5% excess, the GNI rating is additionally increased by 0.3 points.

The investor then determines how critical it is to adjust the rating of the TSP. If the minimum permissible TSP indicator is determined on the basis of the extremely important return on borrowed funds, then the rating of the TSP is increased by 6 points, while the ratings of the remaining indicators are reduced by 2 points.

If the TSP is established by the investor on the basis of an investment agreement, that is, it is associated with the extremely important investment of the funds received in another investment project, then the rating value of the TSP is increased by 4.5 points. With a simultaneous reduction in the ratings of other indicators by 1.5 points.

If the minimum TSP is set on a different basis, then the TSP rating is reduced by 1.5 points, while others are increased by 0.5 points.

If the IDN indicator is set (if projects have the same implementation period) in the amount of the inflation rate, increased taking into account the number of years of project implementation, then the IDN rating is increased by 3 points. If the IDN is set below this value, then the rating is increased by 4.5 points.

After all recalculations, the investor determines the final number of rating points after all changes are made.

1. The investor crosses out from the list of criteria that are significant for himself the one that scored the least number of points.

3. If it is impossible to single out the most significant criterion, then an additional criterion is introduced into the calculation in the form of a Fisher point. The quantitative indicator of this criterion is not specified, it is taken into account only for equivalence, and the method of deletion is applied again, but only according to three criteria.

If, based on the results of new calculations, it is impossible to choose the criterion that is paramount, then the investor can include other projects in the calculation, or he can use the search for an optimal or ideal solution.

One of the most difficult economic policy issues is inflation management. The ways of managing it are ambiguous, contradictory in their consequences. The range of parameters for such a policy can be very narrow. On the one hand, it is required to restrain the unwinding of the inflationary spiral, and on the other hand, it is necessary to maintain production incentives, to create conditions for saturating the market with goods.

The main forms of stabilization of monetary circulation, depending on the state of inflationary processes, are monetary reforms and anti-inflationary policy.

Monetary reforms were carried out in the conditions of metallic money circulation. From the second half of the twentieth century. stabilization of money circulation as one of the most important ways to restore the economy is carried out using the following methods: nullification, restoration (revaluation), devaluation and denomination.

Nullification means announcing the cancellation of a heavily discounted unit and introducing a new currency.

Denomination- the method of "crossing out zeros", i.e. enlargement of the scale of prices.

Anti-inflationary policy Is a set of measures for state regulation of the economy aimed at combating inflation. In response to the interaction of factors of inflation, demand and inflation of production costs, two main lines of anti-inflationary policy have taken shape - deflationary policy (or demand regulation) and income policy (or cost regulation). In addition, a new direction has appeared - competitive stimulation of production.

Deflationary policy Are methods of limiting money demand through monetary and tax mechanisms by:

• reducing government spending,
• increasing the interest rate for a loan,
• strengthening the tax press,
• restrictions on the money supply, etc.

The peculiarity of the deflationary policy is that it causes a slowdown in economic growth and even crisis phenomena.

Income policy presupposes parallel control over prices and wages by freezing them completely or setting limits on their growth. For social reasons, this type of anti-inflationary policy is rarely used.

Competitive stimulation of production includes measures both to directly stimulate entrepreneurship by significantly reducing taxes on corporations, and to indirectly stimulate the savings of the population by reducing income tax.

The model of anti-inflationary policy in its Russian version consists of two large blocks.

The first block includes economic factors:

• creation of an effective investment program;
• the formation of a stable macroeconomic structure of the market in order to level out the disproportions in production;
• attracting additional capital to the production sector.

The second block forms the financial focus:

• covering the budget deficit by placing government securities and refusing from central bank loans;
• the establishment of the regulatory significance of the functions of taxes (and not just the fiscal one) in the sphere of production;
• a sharp decrease in the emission of money (in excess of the demand for goods turnover) as the most important monetary inflationary factor (although since 1994 the government has not been issuing money to cover the budget deficit);
• an active income policy, which involves the coordination and coordination of the growth rate of wages, incomes and prices under the supervision and mediation of the state (using the experience of the West on the conclusion of agreements between the government, trade unions and enterprises on these issues).

Achieving financial stability, reducing the federal budget deficit, ensuring its financing is possible through a number of non-inflationary sources:

• increasing the collection of tax payments to the budget (including through a significant reduction in ineffective tax and other benefits);
• reducing the list of federal target programs, which will allow concentrating budget funds on the most effective and socially significant projects;
• financing of regions from regional budgetary expenditures with an appropriate income base.

The fulfillment of these conditions of the anti-inflationary strategy contributes to the achievement of such basic social and economic goals as:

• protection of the social interests of the people, primarily of its needy strata;
• preservation of the economic, scientific and technical potential of the country;
• creating incentives for production, primarily investment activities;
• formation of a competitive market environment.

The main task of economic reforms consists in a complex restructuring of the Russian economy and, above all, in the activation of investment policy.