JPH0485659A - Binary tree possible solution keeping/restoring processing system - Google Patents

Abstract

PURPOSE:To reduce a using memory capacity at optimizing processing and to shorten the processing time of the whole by keeping the minimum conditions for the restoring of solution information among the information of a node that a possible solution is obtained at optimizing processing by the branch limiting method of a mixed integer programming problem and restoring the solution information based on them. CONSTITUTION:A binary tree 1 that the conditions of the parameter of an inputted mixed integer programming problem are changed and are given to respective nodes in order is prepared, the optimizing of the respective nodes of the binary tree 1 is executed and conditions 2 of a possible solution when all the conditions are satisfied are outputted and kept. Besides, an inverse matrix E is generated based on the conditions 2 of the kept possible solution, the inverse matrix E is operated to the matrix of the original mixed integer programming problem and solution information is restored. Thus, a using memory capacity at optimizing processing can be reduced and the processing time of the whole can be shortened.

Description

[Detailed Description of the Invention] [Summary] Binary tree possible solution storage/restoration processing method for storing and restoring possible solutions of nodes in a binary tree of a mixed integer programming problem During optimization processing using the branch and bound method for a mixed integer programming problem , the minimum information (M
Save the upper limit value, lower limit value, base value, non-base value (value, etc.) of IP variables, and restore the solution state based on this when restoring the solution to reduce the memory capacity used during optimization processing. With the aim of shortening the overall processing time, we created a binary tree in which variable conditions (e.g., upper and lower limits) were changed and sequentially assigned to each node for manually generated mixed integer programming problems. Optimization is performed under the conditions of the variables assigned to each node of this binary tree, and the conditions of the solution (possible solution) when all 27 conditions are satisfied (M I P variable soil, limit value, lower limit value, basis , non-base (value)) and store them, and the solution is configured to be stored with a small memory capacity.

[Industrial application field]

The present invention relates to a binary tree possible solution storage and restoration processing method for storing and restoring possible solutions of nodes of a ternary tree of a mixed integer programming problem. The scope of application of the mathematical generalized one-stroke system is
It has diversified from the traditional focus on low production planning to textile, delivery, and personnel allocation planning. For this reason, from the linear tolerance [3(1,,P problem)] using the Horitai method established in the 1950s, to the mixed integer programming problem (MI
This is showing its development into P problem).

l, )) There is only one solution in the optimization of the company, but
In optimization of the MIP problem, a plurality of solutions (possible solutions) are output to use the branch and bound method. For this reason, it is desired to reduce the memory capacity required for JI1 standardization processing.

[Prior art and problems to be solved by the invention] FIG. 9 shows the flow of a conventional optimization process for an MIP problem. Given M I 1) Divide the problem by the conditions (in the case of integer variables, the upper limit or lower limit -) of each MIP variable (variables that are subject to branching are called MIP variables). Then, perform linear optimization in the area (node) (e
). A branch tree (not shown in FIG. 10) is used to express the branch state. Examine the values of variables defined as VIP variables on this binary tree, and if the linear optimization result satisfies the conditions for MIP variables (19),
Record the linear optimization result as one of the possible solutions to the M I P problem. 2) When all nodes in the branch tree have been searched and completed, the optimization of the problem is terminated. Here, in the optimization process of the MIP problem, the madrisk information of the node is updated.

At this time, a method using base information is known as a method for restoring the solution to the 15P problem. In the Ll) problem, the matrix information was not changed at all, so the solution could be restored using only the base information. However, in the optimization process of MIP problems, matrix information is updated at each node, so it is not possible to restore the solution just by saving the base information, and in conventional systems, a solution fee is charged for each possible solution. It was necessary to output (variable values, ground state, etc.) to a file. When n possible solutions are detected, the size of the solution fee for n possible solutions is, for example, n * (number of all variables (4 hides * 8 + 8 bytes *
6)). This caused the following problems.

(1) Because the cancellation fee is large, it was possible to output small data to memory, but large data had to be output to a file (output to a file on an external storage device). However, there was a problem that it could not be processed quickly.

(2) When optimizing a mixed integer programming problem, even though only the best possible solution is required, it is necessary to output and save all possible solutions to a file.1. There is a problem that 1) the number of 0 times when outputting to a file becomes too large, which makes it difficult to process quickly, and it takes time to process the information because possible solutions are outputted to a file.

The present invention provides the minimum information necessary for solution recovery (the upper limit value of the MIP variable,
To save the base, non-base (base, etc.) and restore the solution state based on this when restoring the solution, thereby reducing the memory capacity used during optimization processing1 and shortening the overall processing time. It is an object.

(Means for solving the problem) One step for solving the problem will be explained with reference to FIG.

In FIG. 1, binary tree 1 is created by changing variable conditions (for example, north limit value, lower limit value, etc.) for an input mixed integer programming problem and sequentially attaching them to each node.

Condition 2 for a possible solution is the condition for a solution when each node of binary tree 1 is optimized and all conditions are satisfied (upper limit value, lower limit value, base, non-base (value), etc. of M I P variables). This is the minimum necessary condition for restoring the previous information.

[Effect]

As shown in FIG. 1, the present invention creates a binary tree 1 for an input mixed integer programming problem by changing variable conditions (for example, upper limit value, lower limit value, etc.) and sequentially assigning them to each node. Optimize each node of this binary tree l, and find the conditions (upper limit, lower limit value, base, non-base (value), etc. of M I P variables) 2 of the solution (possible solution) when all the conditions are satisfied. I am trying to output and save it. Furthermore, an inverse matrix E is generated based on the saved condition 2 of the possible solution, and this inverse matrix E is operated on the madrisk of the original mixed integer programming problem to restore the previous information.

Therefore, when optimizing a mixed integer programming problem using the branch-and-bound method, the minimum conditions necessary for restoring the previous information among the information on the nodes for which possible solutions have been obtained (the upper and lower values of the V I P variables, By storing the base, non-base (values, etc.) 2 and restoring the previous information based on this, it is possible to reduce the memory capacity used during optimization processing, and
It becomes possible to shorten the overall processing time.

〔Example〕

Next, the configuration δ and operation of one embodiment of the present invention will be sequentially explained in detail using FIGS. 1 to 8.

In FIG. 1, binary tree 1 is created by changing one variable condition (for example, upper limit value, lower limit value, etc.) for an input mixed integer programming problem and sequentially assigning it to each node. This is a binary tree for the MIP problem in FIG. 3 (a), which will be described later.

Condition 2 for a possible solution is the condition for a solution (possible solution) when all the conditions are satisfied by optimizing each node of binary tree 1 (
(upper limit, lower limit, base, non-base (value), etc.) of the VIP variable, and is the minimum condition necessary to restore the previous information.

Next, an explanation will be given of the operation of changing one of the variables X and Y of the MIP problem in FIG. 3(a) and creating and optimizing the nodes of the binary tree 1 in FIG. 1 according to the order in FIG. 2.

In FIG. 2, ■ linearly optimizes the MIP problem as a relaxation problem. This is done by relaxing the variable conditions such as, for example, in the MIP problem shown in FIG. 3(A), X is an integer and Y is an integer, and as shown in FIG. 3(C), X is a real number and Y is a real number.

0 changes one condition of the VIP variable and performs linear optimization. This means that for node A of binary tree 1 in Figure 1, the third
The relaxation problem in Figure (C) is further expressed by a tableau (original) in Figure 3 (D), linear optimization is performed on this tableau, and a solution in Figure 4 (E) is obtained (described later).

@ determines whether all MIP conditions are satisfied. This determines whether the solution obtained by linear optimization using @ satisfies all the MIP conditions. Here, the linear optimization result of node A is X=2, as shown in Figure 1.
．． 57, Y=2.86, and the MIP condition (integer here) is not satisfied, so perform O. Furthermore, if the MIP condition is satisfied (when X and Y are all integers), ,
Perform the INVERT process in (2) and calculate possible solutions again.

[Phase] determines whether all VIP conditions are satisfied or not. If satisfied, the node information and base information are updated in [Phase]. If you are not satisfied, perform ■.

■ determines whether all possibilities have been investigated. If there is still some left, repeat steps after ■. If all cases have been investigated, use [Phase] to extract the best possible solution, Madrisk update processing, base restoration processing,
Perform INVERT processing.

Here, the possible solution restoration processing involves searching and finding a node with the possible solution flag on, reading the VIP conditions of that node in order to restore the madrisk information of the possible solution, and updating the madrisk. In order to restore the basis information of the solution, input the basis information of the node into gfc and
V] Execute the RT subroutine and restore °! (Ru.

Taking the above processing into account, for the problem in Figure 3 (A) M I J), after relaxing the variable conditions in the relaxation problem in Figure 3 (C), we can calculate the value of Ii'1) with this relaxation problem. Figure 3 (,-)
Linear optimization based on tableau (original)9)
The results are shown at node 80.
i P g number (here X, Y) (7) change one,
For example, from OshiX≦4 to 0shiX52 (change this)
− to code 13 to perform linear optimization, and similarly convert the result to 1
■Get as shown. , iL2, the obtained result is M I
If all P clauses 4'+' are satisfied, for example, in the case of node I), the Mll) tea fee has been added up to 7, so turn on φ In addition, when restoring the possible solution, output the condition 2 of the IiJ functional solution surrounded by the curve necessary to restore the possible solution. Based on condition 2, the solution information is restored by performing the multi-inverter process. A detailed explanation will be given below.

No. 31'l shows explanation number 1 of the uninvented mitigation problem.

FIG. 3(a) shows the selected MIP problem.

Figure 3 (D) is an illustration of the M I Fi problem of α) in Figure 3 (A).
From , and 2, a possible solution is (direct) of the lattice points of the edges within the shaded area.

The third function (c) is a relaxation problem of the third M (i: l M I )) problem summed with variable conditions 4il. ) l is a real number ♂
・] Change to variable property and relax the problem (2). In optimization of the relaxation problem, we use the revised single/socks method, so in tab 17 and the OGI table, express the problem.

The 31st threshold (J-) does not show a tableau (ori;-yl). This shows the 3rd and 1st (c) relaxation problem in the 3rd + pc ho).)! This is the tab 1) when you change it.

3rd [H] changes the mitigation problem of ③ of 3rd N] (C) to the state in which L7 is applied to the vehicle weight. The mitigation problem of ■ in 31st base J (c) is equivalent to the modified example of (Su゛) in 31st base J (c).

FIG. 4 provides a detailed explanation of the present invention.

Is Figure 4 (a) a tableau? : Ah de 5. It is the same as the 3rd review (d) tableau (original).

FIG. 4 (1:I) shows the calculation of the 4th (A) tab 1]-(original) at the R10 position. In other words, the second row of the tableau (meridinal) in FIG. 4(a) is multiplied by 1/'9, and the result is added to the first and third rows, taking into account the signs.

In Fig. 41 (c), it is further swept out at the R2 position.

I love calculations.

FIG. 4(D) is a representation of FIG. 3(C) with 0 decimal points.

FIG. 4(E) shows the result obtained by extracting the solution of ■ from the M-shaped tab "]-" shown in FIG. 4(5-) with a decimal point.

Here, for example, determining the 0BJO value is shown in Figure 4 (d).
If R1 and R2 take positive values, OBJ
becomes smaller. Also, R1 and R2 are R1≧0, R2≧O
Therefore, it cannot take a negative value, and the value just found is the optimal state (optimal solution for OBJ) that gives the value of OBJ.

FIG. 5 is a diagram illustrating the creation of an inverse matrix according to the present invention.

FIG. 5(a) shows the tableau (optimal state), and FIG. 4(d) shows the tableau (5).

FIG. 5(+-t) does not show the creation of tab 1)-. The optimized tab+:v-72 is obtained by multiplying the inverse matrix E by the tab o −T′ + of the Orisina month. Inverse matrix) Mi is
It consists of two parts (E, , R2 in °).

FIG. 5(c) shows the formation of the inverse matrix F. This inverse matrix l-:, can be seen from Figure 4 (b).
Since the sweep position of itl is 24), enter the element in the second column of the matrix E1 from the left.1. Ru. (2) Y02
Place the reciprocal of element 9 in the row at the position of the arrow shown in the figure.1. Ru.

(3) Multiply the element in the first row of Y by (-1) and enter the number n in the element in the second row at the position indicated by the arrow in the figure. (4) Multiply the element in the third row of Y by (-1) and insert the resulting number into the position of the arrow in the figure. The illustrated inverse matrix E is created using the equations 4) and 4).

FIG. 6(d) shows the creation of the inverse matrix E2. This inverse matrix E2 is 2. It is clear from Figure 4 (c) that the 6th
It was created in the same manner as in Figure (E). Then, the inverse matrix E can be obtained as EtXEI.

Here, if you know which variable is the base and at which position,
El and Et, which are constituent matrices of E, can be easily created.

FIG. 6 shows an example ground state of the present invention. This shows the ground state of node A in FIG. From this ground state example, we create an inverse matrix E as described in Figure 5 (C) and an inverse matrix E2 as described in Figure 5 (D), and calculate the inverse matrix E by calculating both of them. Can be generated.

FIG. 7 shows an illustration of optimization of node B.

FIG. 7(a) shows the problem of node B. This node B
Compared to node A, the variable condition has been changed to 0≦X≦2.

Figure 7 (b) is an illustration of the problem in figure 7 (a). Since XSY is an integer, the possible solutions are:
This is the value of the grid point within the shaded area.

FIG. 7(c) shows a linearly optimized tableau created as described above for the problem of FIG. 7(a).

FIG. 7(d) shows node information after optimization.

The results are (X=2.0, Y=3.1), and since the variable Y is not an integer value and is not a possible solution, the variable conditions are further changed and processing is performed for nodes D and E.

FIG. 8 shows an explanatory diagram of node D.

FIG. 8(a) shows the problem of node D.
Variable condition 20≦X≦2.0≦Y≦2.

FIG. 8(b) is a tableau after optimizing the problem of node D in FIG. 8(a).

FIG. 8(C) shows node information after optimization.

Here, the result is X=2, as shown in node D in Figure 1.
．． Since it is an integer value of 0 and Y=2.0, it is stored as the MIP possible rough flag N.

Also, the possible solution condition 2 surrounded by the curve of node D in FIG. 1 is output and saved.

FIG. 8(d) shows the inverse matrix. This means that node D is
Figure 8 (c) where the base variables are saved as OBJ, R1, and R2
Since it is known from the base information, each is a unit vector, and the inverse matrix E becomes as shown in the figure. Furthermore, since X and Y have an upper limit of 2.00 and are non-basic, this corresponds to the following fact. The simplex method always assumes that the variable has a lower limit value of 0.0 and moves in the positive direction. Now that X has reached the upper limit of 2°00, if X were to move from its current position, it would be in the opposite direction.

Here, regarding X, move the axis and reverse the direction,
Adjust so that normal processing can be performed. Therefore, subtract the upper limit 2.00 from X (axis movement) and then multiply by -1 (direction reversal), so X' = - (X-2) Y' = - (Y-2) replace. Here, -1 times is handled by reversing the sign processing on the summer calculator, so we set X' = X-2 Y = Y-2. Therefore, for both of them, ) and substituting it into the tableau (original) and deleting X and Y, we get
FIG. 8(f) A tableau is obtained.

By further organizing, the tableau in FIG. 8 (g) is obtained.

From this tableau in Figure 8 (g), OBJ is the base in the first row, so 0BJ = 4.0R1 is 2
Since it is the basis in the 3rd line, R1 = 10.00R2 is the basis in the 3rd line, so R2 = 2.00
Therefore, X-2゜Y takes the lower limit value 2.00, and since it is non-basic, Y' = -(Y-2), and since y' = o, oo, Y
=2°O. Furthermore, since Xo and Yo move in opposite directions, the current state is the optimal state where the value of OBJ is maximum.

Similarly, for other nodes C and E, set the variable conditions to the first
By making changes as shown in the figure, the results shown in the figure can be obtained from the tableau after optimization.

〔Effect of the invention〕

As explained above, according to the present invention, the minimum conditions necessary for restoring solution information (MIP L limit value of variable, lower limit I, base, non-base (bond)
etc.) 2, and then restores the solution information based on the information, it is possible to reduce the amount of memory used during optimization processing. By reducing the number of times, the overall processing time can be shortened. Figure 8 is an explanation of node D. Figure 9 is an explanation of the conventional optimization process for the M + 1) problem. 10 is an example of a binary tree. Not shown.

In the figure, ]: Binary tree 2: Possible solution clause A1 A to E: 7 binary trees

Claims (2)

[Claims] (1) Saving and restoring possible solutions of nodes in a binary tree of a mixed integer programming problem In the binary tree possible solution saving and restoring processing method, variable conditions (for example, upper limit value, lower limit value etc.) to create a binary tree (1) in which each node is sequentially assigned, and optimization is performed under the conditions of the variables assigned to each node of this binary tree (1), and all conditions are satisfied. The conditions for the solution (possible solution) (upper limit, lower limit, base, non-base (value) of MIP variables) (2) are output and saved, and the solution information is saved with a small memory capacity. A binary tree solution storage/restoration processing method characterized by the following. (2) An inverse matrix E was generated based on the saved possible solution condition (2) above, and the solution information was restored by calculating this inverse matrix E on the matrix of the original mixed integer programming problem. A binary tree possible solution storage/restoration processing method according to claim (1).