JP4636609B2 - Material allocation method, apparatus and program - Google Patents

Description

  The present invention relates to a material allocation method, apparatus, and program. More specifically, the present invention relates to a calculation method for allocating materials without surplus according to orders, and an apparatus and a program for realizing the calculation method.

  2. Description of the Related Art Conventionally, in the case of a build-to-order manufacturing in which a product is manufactured according to an order from a customer, a technique and a system for optimizing a product material and an order allocation method are known at the time of allocation of an order for product material. In particular, in the steel industry, orders from customers are usually made by specifying the weight (unit weight) of each delivered product with a margin separately from the total order quantity. A method that attempts to optimize the allocation of materials has been proposed.

For example, Patent Document 1 discloses a production management system capable of collecting a material unit so as not to lose a single weight when collecting a product in a final process and collecting a custom product. Further, Patent Document 2 discloses a method of grouping a plurality of types of products and a plurality of types of surplus materials in advance based on characteristics, and determining whether allocation is possible based on the characteristics for each group. .
Japanese Patent Laid-Open No. 5-19809 Japanese Patent Laid-Open No. 6-149850

  In the steel industry, one material is usually cut into small pieces and allocated to several orders (FIG. 2 shows an example in which one material is divided into three orders and allocated). Each order has an order quantity, upper and lower margins (allowable amount allowance range), and unit weight upper and lower limits. For each order, the material is cut to a size within the upper and lower limits of the unit weight, and allocated so that the allocated amount falls within the allowable range. At this time, it is a problem to allocate so that there is no surplus in the material (a surplus not allocated to any order).

  In steelworks, in principle, materials (slabs, coils, etc.) are produced in response to orders from customers. Therefore, the order and the materials are related in each production process. However, as the material passes through each process, the relationship between the order and the material may be broken due to various causes. In such a case, a system for reallocating materials not related to the order to the order again is called a material allocation system.

  In the material allocation system, materials are basically allocated in accordance with the order quantity. For example, if there are two 10t materials for an order of 15t, one 10t material is allocated, and the other 10t material is allocated by cutting 5t. However, the order has a single weight upper limit and a single weight lower limit, which are the upper and lower limits of the weight of each allocated material. Therefore, when allocating a material, the material must be cut and allocated so as to be within this single weight range.

  For example, if there is a single weight limit of 4t to 6t in an order of 14t, it is possible to allocate 10t material using one 4t material (10t material is two for each 5t). However, it is not possible to satisfy the order of 14t using only two 7t materials (7t material can be cut and allocated without surplus) Cannot reach 14t). Since there is a single weight limit, if a material is allocated so as to be the same as the order quantity, many small surpluses (less than about 5 t) are generated in the material. A small surplus is likely to be discarded and is a loss for the steel mill, so there is a certain margin in the order, and only if there is no surplus, the order will be placed within the upper and lower limits. It is allowed to adjust the total allowance. On the contrary, when a small surplus occurs when allocation is made in accordance with the order quantity, it is desirable to use an order margin so that no surplus occurs.

  Optimization of the material allocation system optimizes which material is allocated to which order under the above conditions. Each material is not homogeneous, and there are desirable and non-relevant combinations of individual materials and orders, so not only the amount of the provision is matched to the order quantity, but also the various provision combinations are examined to find the optimum provision There is a need to. However, even if there are only a few orders and materials that are candidates for combination, there are a huge number of possible allocation methods, and it is a reality to find the optimal allocation method covering all possibilities Not right. This is because it is difficult in terms of calculation time to find out all cases of the order quantity and unit weight upper and lower limits given in kilograms while changing the provision amount by 1 kilogram.

  Therefore, candidates for allocation to be examined using some rule of thumb are limited. In practice, provisions are made according to the order volume in principle, and cases where surpluses are generated are often solved by an exceptional method. In other words, in principle, provisions are made according to the order volume, and the margin of the order volume is checked only when a small surplus occurs. Algorithm. For example, if there is one order for 5t, one order for 6t, and one material for 12t, the material is first cut by 5t and allocated to one order, and the remaining 7t is allocated to the other order. Think about it. At this time, if a margin of 1 t or more is set for the order, the generation of surplus is eliminated by allocating 7 t instead of 6 t of the order quantity.

  However, such a solution may cause a surplus even when a provision for eliminating the surplus is possible. For example, consider a case where there are two orders for 11t material and 5t. Assume that the upper and lower limits of the provision amount are 4.5 t to 5.5 t and the upper and lower limits of the unit weight are 2.7 t to 5.1 t for both orders. At this time, in the above solution, first, one order is allocated at 5t, and then the remaining 6t material is allocated to the 5t order. However, in this case, a margin is insufficient (when the above algorithm is applied), a surplus is generated. However, in actuality, it is possible to reserve without surplus by allocating 5.5 t to two orders.

  Therefore, when materials can be allocated without surplus, it is a natural idea to add such allocation methods to search candidates. However, under the single weight restriction, it is difficult to determine whether one material can be allocated to a plurality of orders without surplus, and an efficient algorithm is not known. However, since a single material can usually be allocated to at most about three orders, limiting the number of orders will reduce the difficulty of the problem, but it will still be possible to allocate to eliminate the surplus. Efficient judgment is not a simple problem.

  The problem to be solved by the present invention is to determine whether or not an allowance for eliminating this surplus can be determined, and to provide a method for allocating if it can be reserved. As described above, it is assumed that an enormous number of times is performed in the process of searching for a provision that does not generate surplus. Therefore, it is necessary that the algorithm is sufficiently fast.

  In the method of calculating the weight of allocated material with margin, input the weight of the given material, and input the unit weight, unit weight upper limit, unit weight lower limit, margin upper limit value, margin lower limit value for each order. A step of receiving, a step of determining whether a material can be allocated without surplus based on a predetermined calculation formula for the input order, and a material can be allocated without surplus A method including the step of calculating the weight of the material allocated according to the determination, and an apparatus and a program for realizing the method are provided.

  In the determining step, when the number of orders is one, the product of the maximum integer not exceeding the quotient obtained by dividing the weight of the material by the lower limit of the unit weight and the upper limit of the unit weight is a weight equal to or more than the weight of the material. It is determined based on a calculation formula that

  Further, in the determining step, when there are a plurality of orders, the input order can be allocated based on the weight of the given material, the single weight upper limit and the single weight lower limit of each order. A combination of the allocation amounts is represented by a grid point in the Euclidean space region, and whether or not the material can be allocated without a surplus is determined by detecting the presence or absence of the grid point.

  According to the above invention, it is determined whether or not there is a provision for eliminating the surplus, and when it exists, a high-speed process for providing the provision can be realized. The high speed of processing is due to the fact that when there is one order, it can be determined by one inequality. This is because, in the case where there are a plurality of orders, it has become possible to determine in a practical time whether or not it is possible to make provisions to eliminate surplus quickly and reliably by realizing a combination of techniques such as binary search.

  Hereinafter, the present invention will be described through embodiments of the invention. The following embodiments do not limit the invention according to the claims. Moreover, not all the combinations of the constituent elements described in the embodiments are essential for the solving means of the invention.

  FIG. 1 is a schematic diagram of a material allocation apparatus 100 having the above function. The material allocation apparatus 100 includes an input unit 110 that receives input from a user, an execution unit 115 that performs arithmetic processing, an algorithm storage unit 130, a function storage unit 140, and an output unit 170.

The input unit 110 may be a device that receives an output from another program as input data in addition to an input device that accepts an operation from a human such as a keyboard or a mouse. In addition, a communication function may be provided to receive data from other remote systems. Through the input unit 110, the execution unit 115 is given the weight of the material, the number of orders, the upper limit value of the order quantity for each order, the lower limit value of _the order quantity, the upper limit value of the single weight, and the lower limit value of the single weight. It is done.

The execution unit 115 includes an algorithm selection unit 120 that selects an algorithm to be applied according to the number of orders, an allocation determination unit 150 that determines whether the material can be allocated without surplus according to the execution result of the selected algorithm, and the allocated material It is comprised by the weight calculation part 160 which calculates the weight of. The execution unit 115 determines whether or not the material can be allocated without surplus for each order by the input given from the input unit 110, and calculates the allocation amount for each order when the material can be allocated.

  The algorithm storage unit 130 stores a plurality of algorithms selected according to the number of orders. In addition, each algorithm is connected to the function storage unit 140 in order to call one or a plurality of functions necessary for performing an operation. Each algorithm and function can be called recursively, as described below.

  The output unit 170 displays the result obtained by the execution unit 115, that is, whether or not it is possible to reserve without surplus, and when it is determined that allocation is possible, the calculation result of the allocated weight for each order. The output unit 170 has a function of outputting the result of the execution unit 115, may be a display device such as a CRT or a liquid crystal display, and may be a communication unit that transfers data to other systems or programs.

  In the embodiment of the present invention, when a single material can be allocated to a plurality of orders so as not to generate surplus, the allocation pattern can be limited to a small number of patterns. It is determined whether there is a provision that cannot be issued. For each allocation pattern, a search for an allocation with no surplus was realized by combining the following two algorithms. One is an algorithm that determines at a high speed whether or not the provision amount for an order satisfies the unit weight limit of the order. The other is that when a material is allocated to a plurality of orders, a combination of allocation amounts that can be allocated is represented by lattice points in a polyhedron (including a plane) in the Euclidean space (in the embodiment, on a two-dimensional plane). An algorithm that expresses and finds lattice points in the polyhedron at high speed. The algorithm for finding the lattice points in the polyhedron is realized by using an algorithm that counts the number of lattice points in a triangle mapped to a two-dimensional plane at high speed and a binary search method. Hereinafter, these algorithms will be described in detail.

[Problem to be solved]
Hereinafter, for the sake of simplicity, the case where the number of orders is 1 to 3 will be described, but the method according to the present invention can be extended even when the number of orders is 4 or more.
Input: The number of orders n (1 ≦ n ≦ 3) is given, and for each order O _i (1 ≦ i ≦ n), the following restrictions are given.
1) Maximum order quantity ru _i
2) Lower limit of order quantity rl _i
3) Upper limit of unit weight uu _i
4) Lower limit of unit weight ul _i
For materials, the weight of the material w is given.
Output: Provision AQ _{i for} each order O _i (1 ≦ i ≦ n) that satisfies the constraint condition. If the constraint is not satisfied, a message to that effect is output.
Constraints: The allocation amount AQ _i for each order O _i (1 ≦ i ≦ n) must satisfy the following constraints.
1) rl _i ≦ AQ _i ≦ ru _i (restriction of provision amount)
2) There exists an integer k _i and ul _i · k _i ≦ AQ _i ≦ uu _i · k _i (single weight limit)
3) w = AQ ₁ + AQ ₂ + ... + AQ _n (the surplus must be zero)

[algorithm]
In order to solve the above constraints, the applied algorithm differs depending on the number of orders. Algorithm 1 described later is applied when the number of orders is one, algorithm 2 described later is applied when the number of orders is two, and algorithm 3 described later is applied when the number of orders is three. In addition, functions commonly used regardless of the number of orders are described below.

と表記する。 [Function without order quantity]
<Floor function, ceiling function>
Hereinafter, for real number x, floor (x) (floor function) is defined as the largest integer that does not exceed x, and ceil (x) (ceiling function) is defined as the smallest integer that does not fall below x. In the figure, for simplicity,

Is written.

<Function 1 (O _i , w)>
Input: Order O _i and material weight w available
if w ≦ floor (w / ul _i ) ・ uu _i then Judge that the order O _i can be allocated with the allocation amount w
else
FIG. 11 shows a flowchart of the function 1 algorithm.

<Function 2 (O _a , O _b , a, b, c, x _u , x _l )>
Input: Order O _a , O _b and integer a, b, c and real number x _u , x _l
Find the greatest common divisor d of a and b and integers x ₀ and y ₀ that satisfy ax ₀ + by ₀ = d using the extended Euclidean algorithm
if c is divisible by d, and ceil ((x _l- (c / d) x ₀ ) · (d / b)) ≦ floor ((x _u- (c / d) x ₀ ) · (d / b )) Then
Order O _a a reserve amount a ((c / d) x 0 + (b / d) ceil ((x l - (c / d) x 0) · (d / b)))
Order O _b in reserve amount ca ((c / d) x 0 + (b / d) ceil ((x l - (c / d) x 0) · (d / b)))
Judge that it is possible to reserve in
else
FIG. 12 shows a flowchart of the function 2 algorithm.

<Function countT (a, b, m)>
if m <0 then return 0
if a <b then return countT (b, a, m)
Set p = floor (m / a).
if a = b then return p (p-1) / 2
Let k = floor ((a-1) / b).
Let q = floor ((c-ap) / q).
return countT (a-bk, b, cb (kp + q)) + kp (p-1) / 2 + pq
FIG. 13 shows a flowchart of the algorithm of the function countT.

[Algorithm when there is one order]
<Algorithm 1 (O _i , w)>
Input: Order O _i and material weight w available
if rl _i ≦ w ≦ ru _i then
Returns the result of function 1 (O _i , w)
else
FIG. 3 shows a flowchart of algorithm 1 in which it is impossible to reserve to eliminate the surplus.

The following shows that the above algorithm operates correctly. In order to show the correctness of the algorithm, the following lemma is used. In the algorithm 1, first, it is checked whether the weight of the material is within the upper and lower limits of the order quantity for the order O _i . When the order quantity is within the upper and lower limits, it is next necessary to determine whether or not the single weight constraint is satisfied. The fact that the allocation amount x satisfies the unit weight constraint is that the unit weight upper limit uu and unit weight lower limit ul can be written as n · ul ≦ x ≦ n · uu using a certain positive integer n. From Lemma 1, this is equivalent to satisfying the relationship x ≦ floor (x / ul) · uu. In algorithm 1, the determination is made using function 1. Since it is correctly checked whether the allocation amount is within the upper and lower limits of the order amount and whether the single weight constraint is satisfied, it can be seen that Algorithm 1 operates correctly.

[Algorithm when there are two orders]
<Algorithm 2 (O _i , O _j , w)>
Input: Order O _i , O _j and the weight of material available w
if function 1 (O _i , ru _i ) = allowable and algorithm 1 (O _j , w-ru _i ) = allowable then
Allowance amount ru _i in order O _i, provision can be a determination in the reserve amount w-ru _i in order O _j
else if function 1 (O _i , rl _i ) = can be allocated and algorithm 1 (O _j , w-rl _i ) = can be allocated then
It is determined that the allocation amount rl _i can be allocated to the order O _i and the allocation amount w-rl _i can be allocated to the order O _j.
else if function 1 (O _j , ru _j ) = allowable and algorithm 1 (O _i , w-ru _j ) = allowable then
Order O _j is determined to be available with provision amount ru _j and order O _i is determined with provision amount w-ru _j
else if function 1 (O _j , rl _j ) = can be allocated and algorithm 1 (O _i , w-rl _j ) = can be allocated then
Order O _j is determined to be allocated with allocation amount rl _j and order O _i is allocated with allocation amount w-rl _j
else if function 3 (O ₁ , O ₂ , w, uu ₁ ) = can be allocated, or function 3 (O ₁ , O ₂ , w, ul ₁ ) = can be allocated then
Judgment can be made using the judgment result of function 3
else if function 3 (O ₂ , O ₁ , w, uu ₂ ) = can be allocated, or function 3 (O ₂ , O ₁ , w, ul ₂ ) = can be allocated then
Judgment can be made using the judgment result of function 3
else
It is determined that it is not possible to reserve to eliminate the surplus. FIGS.

<Function 3 (O _a , O _b , w, u _a )>
Input: u _{a of} material weight w and order O _a , O _b , order O _a single weight upper and lower limit
x _l = ceil (max {rl _a , w-ru _b } / u _a )
x _u = floor (min {ru _a , w-rl _b } / u _a
if x _l > x _u then
Judgment that it is impossible to eliminate surplus
else if ceil ((wx _l・ u _a ) / uu _b ) ≦ floor ((wx _l・ u _a ) / ul _b ) then
It is determined that the order O _a can be allocated with the allocation amount x _l · u _a and the order O _b with the allocation amount wx _l · u _a
else if ceil ((wx _u・ u _a ) / uu _b ) ≦ floor ((wx _u・ u _a ) / ul _b ) then
It is determined that the order O _a can be allocated with the allocation amount x _u · u _a and the order O _b with the allocation amount wx _u · u _a
else if function 2 (O _a , O _b , u _a , ul _b , w, x _u , x _l ) = can be allocated then
Judge that allocation is possible using the allocation of function 2
else if function 2 (O _a , O _b , u _a , uu _b , w, x _u , x _l ) = can be allocated then
Judge that allocation is possible using the allocation of function 2
else
A flowchart of the function 3 is shown in FIG. 14 which returns the determination of the function 4 (O _a , O _b , w, u _a , x _u , x _l ) as it is.

<Function 4 (O _a , O _b , w, u _a , x _u , x _l )>
Input: Weight of material w, unit weight of order O _a (upper or lower limit) u _a and range to search binary x _l ≤ x ≤ x _u
if x _u -x _l <2 then
else
Let x _p = floor ((x _l + x _u ) / 2)
if ceil ((wx _p・ u _a ) / uu _b ) ≦ floor ((wx _p・ u _a ) / ul _b ) then
It is determined that the order O _a can be allocated with the provision amount x _p · u _a and the order O _b with the provision amount wx _p · u _a
else
k _p = countT (u _a , ul _b , wu _a・ x _p ) -countT (u _a , uu _b , wu _a・ x _p )
k _u = countT (u _a , ul _b , wu _a・ x _u ) -countT (u _a , uu _b , wu _a・ x _u )
k _l = countT (u _a , ul _b , wu _a・ x _l ) -countT (u _a , uu _b , wu _a・ x _l )
if k _p -k _u > 0 then return function 4 (O _a , O _b , w, u _a , x _u , x _p )
else if k _l -k _p > 0 then return function 4 (O _a , O _b , w, u _a , x _p , x _l )
else
FIG. 15 shows a flowchart of the algorithm of the function 4 in which it is impossible to allocate the surplus.

  Hereinafter, an algorithm in the case of two orders will be described. From Lemma 2 (discussed later), if it is possible to reserve to eliminate the surplus, the amount of provision for all other provisions, except for one, will be a multiple of the upper or lower limit of the unit weight, or the upper and lower limits of the provision amount You can make provisions that match. Therefore, except for one provision, the provision amount of all the remaining provisions is considered to be a multiple of one of the upper and lower limits of the unit weight, or to obtain an provision that matches the upper and lower limits of the provision amount.

  In Algorithm 2, first, a case where one allocation matches one of the upper and lower limits of the order quantity is processed. This is because if the amount of provision for one order is matched with one of the upper and lower limits of the order amount, it can be determined using the algorithm 1 whether the provision for eliminating the surplus can be made for the other order.

Next, the case where one allocation matches one multiple of the upper and lower unit weights is processed. In order to examine each of the upper and lower unit weights for the two orders, function 3 is called up to 4 (= 2 × 2) times to examine all possibilities. The function 3 determines whether or not the provision for eliminating the surplus can be made by limiting the provision amount of the order O _a to _a multiple of the single weight u _a .

In Function 3, first, the upper and lower limits of the reserve amount of O _a are obtained. When allocating to O _a by x times the unit weight u _a , it is necessary to satisfy max {rl _a , w-ru _b } ≦ u _a · x ≦ min {ru _a , w-rl _b } (inversely, satisfies the inequality, note also that can Hikiateru by O _b in reserve amount wu _a · x). From this, max {rl _a , w-ru _b } / u _a ≦ x ≦ min {ru _a , w-rl _b } / u _a . If x _l = ceil (max {rl _a , w-ru _b } / u _a ), x _u = floor (min {ru _a , w-rl _b } / u _a ), then x _l > x _u Since there is no integer x that satisfies the above inequality, it is determined that there is no provision for eliminating the surplus.

FIG. 22 is a diagram illustrating regions and grid points when there are two orders. When there is an integer x that satisfies x _l ≤ x ≤ x _u , the four straight lines on the xy plane x = x _l , x = x _u , P: u _a・ x + ul _b · y = w, Q: u _a · x + uu This is equivalent to the existence of a grid point (including a straight line) in an area (area ADFC in FIG. 22) surrounded by _b · y = w. (Lattice point (x ', y' if there is), O _a a reserve amount u _a · x ', the O _b reserve amount wu _a · x' can be Hikiateru in.) And the straight line x = x _l The intersections of the straight lines u _a・ x + ul _b・ y = w, u _a・ x + uu _b・ y = w are (x _l , (wu _a・ x _l ) / ul _b ), (x _l , (wu _a・ x _l ) / uu _b ), so that when floor ((wu _a・ x _l ) / ul _b ) ≧ ceil ((wu _a・ x _l ) / uu _b ) is satisfied, on the straight line x = x _l There are grid points. In this case, allowance amount u _a · x _l in order O _a, it is possible provision in provision amount wu _a · x _l in order O _b. Similarly, it can also be determined whether there is a grid point on the straight line x = x _u .

Next, it is confirmed whether or not there is a grid point on the straight line P: u _a · x + ul _b · y = w, and the straight line Q: u _a · x + uu _b · y = w. From Lemma 5, it can be seen that the correct decision was made. The remaining case is to determine whether there is a grid point in the region ADCF. In the function 4, the discrimination is realized by dividing the region and performing a binary search.

In the function 4, first, it is examined whether or not there is a lattice point on the straight line x = floor ((x _l + x _u ) / 2) between the straight line x = x _l and the straight line x = x _u . If there are no grid points on this straight line, check the number of grid points in each of the left and right areas (area ADEB and area BEFC), and if there is a grid point in either area, recursively for that area. A lattice point is searched by applying function 4. If there is no grid point in any region, it is determined that it is impossible to reserve to eliminate the surplus.

  The number of grid points in the region can be calculated using the function countT from Lemma 6 (described later). For example, for the region ADEB, ((number of grid points in AGJ)-(number of grid points in DGJ))-((of grid points in BHJ)-(of grid points in ΔEHJ) Number)). In function 4, this is used to determine the number of grid points.

[Algorithm when there are three orders]
<Algorithm 3 (O _i , O _j , O _k , w)>
Input: Order O _i , O _j , O _k and the weight of material available w
if function 1 (O _i , rl _i ) = allowable and algorithm 2 (O _j , O _k , w-rl _i ) = allowable then
It is determined that the allocation amount rl _i and the orders O _j and O _k can be allocated to the order O _i with the allocation amount output by the algorithm 2.
else if function 1 (O _j , rl _j ) = can be allocated and algorithm 2 (O _i , O _k , w-rl _j ) = can be allocated then
It is determined that the allocation amount rl _j and the orders O _i and O _k can be allocated to the order O _j with the allocation amount output by the algorithm 2.
else if function 1 (O _k , rl _k ) = can be allocated and algorithm 2 (O _i , O _j , w-rl _k ) = can be allocated then
Order O _k in reserve amount rl _k, the order O _i, O _j is possible provision in provision amount Algorithm 2 has output determination
else if function 1 (O _i , ru _i ) = can be allocated and algorithm 2 (O _j , O _k , w-ru _i ) = can be allocated then
It is determined that the allocation amount ru _i and the order O _j and O _k can be allocated to the order O _i with the allocation amount output by the algorithm 2.
else if function 1 (O _j , ru _j ) = can be allocated and algorithm 2 (O _i , O _k , w-ru _j ) = can be allocated then
It is determined that the provision amount ru _j and the order O _i , _Ok are available for the order O _j with the provision amount output by the algorithm 2.
else if function 1 (O _k , ru _k ) = allowable and algorithm 2 (O _i , O _j , w-ru _k ) = allowable then
Order O _k in reserve amount ru _k, the order O _i, O _j is possible provision in provision amount Algorithm 2 has output determination
else if function 5 (O _i , O _j , O _k , w, uu _i , uu _j ) = can be allocated, or function 5 (O _i , O _j , O _k , w, ul _i , uu _j ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else if function 5 (O _i , O _j , O _k , w, uu _i , ul _j ) = can be allocated, or function 5 (O _i , O _j , O _k , w, ul _i , ul _j ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else if function 5 (O _j , O _k , O _i , w, uu _j , uu _k ) = can be allocated, or function 5 (O _j , O _k , O _i , w, ul _j , uu _k ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else if function 5 (O _j , O _k , O _i , w, uu _j , ul _k ) = can be allocated, or function 5 (O _j , O _k , O _i , w, ul _j , ul _k ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else if function 5 (O _k , O _i , O _j , w, uu _k , uu _i ) = can be allocated, or function 5 (O _k , O _i , O _j , w, ul _j , uu _k ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else if function 5 (O _k , O _i , O _j , w, uu _k , ul _i ) = can be allocated, or function 5 (O _k , O _i , O _j , w, ul _j , ul _k ) = can be allocated then
Judgment that allocation is possible using judgment result of function 5
else
Determining that it is impossible to allocate the surplus, FIGS.

<Function 5 (O _a , O _b , O _c , w, u _a , u _b )>
Input: material weight w, order O _{a unit} weight (upper or lower limit) u _a and order O _{b unit} weight (upper or lower limit) u _b
x _l = ceil (rl _a / u _a ), x _u = floor (ru _a / u _a )
if x _l > x _u then Judge that it is impossible to reserve to eliminate surplus
else
_{y l = ceil (rl b /} u b), placing a _{y u = floor (ru b /} u b)
if y _l > y _u then it is determined that it is not possible to allocate the surplus
else if function 1 (O _c , wx _l・ u _a -y _l・ u _b ) = Allowable then
O _a a reserve amount x _l · u _a, O _b in reserve amount y _l · u _b, the O _c enabling provision in provision amount _{wx l · u a -y l ·} u b determines
else if function 1 (O _c , wx _l・ u _a -y _u・ u _b ) = Allowable then
O _a a reserve amount x _l · u _a, O _b in reserve amount y _u · _{u b,} the O _c enabling provision in provision amount _{wx l · u a -y l ·} u b determines
else if function 1 (O _c , wx _u・ u _a -y _l・ u _b ) = Allowable then
O _a a reserve amount x _u · _{u a,} O _b in reserve amount y _l · u _b, the O _c enabling provision in provision amount _{wx u · u a -y l ·} u b determines
else if function 1 (O _c , wx _u・ u _a -y _u・ u _b ) = Allowable then
O _a reserve x _u・ u _a , O _b reserve y _u・ u _b , O _c reserve wx _u・ u _a -y _u・ u _b
else
z _l = ceil (rl _c / ul _c ), z _u = floor (ru _c / ul _c )
z = z _l
while z ≦ z _u
s _u = min {floor ((wz ・ uu _c -y _l・ u _b ) / u _a ), x _u }
s _l = max {ceil ((wz ・ uu _c -y _u・ u _b ) / u _a ), x _l }
if function 2 (O _a , O _b , u _a , u _b , wz · uu _c , s _u , s _l ) = can be allocated then
The order O _a and O _b are allocated using the allocation amount of the function 2, and the order O _c is determined to be allocated with the allocation amount wz · uu _c.
else
t _x = max {floor ((wz ・ ul _c -y _u・ u _b ) / u _a ), x _l -1}
if s _l ≤t _x then
Order O _a in order amount t _x · u _a, orders O _b in order quantity y _u · _{u b,} and can reserve the order O _c in _{wt x · u a -y u ·} u b determines
else
t _y = max {floor ((wz ・ uu _c -x _u・ u _a ) / u _b ), y _l -1}
if s _u = x _u and (w-uu _c・ zx _u・ u _a ) / u _b ≦ t _y then
Order O _a in order quantity x _u · _{u a,} orders O _b in order amount t _y · u _b, determines the possible provision to apply O _c in _{wx u · u a -t y ·} u b
else if function 6 (O _a , O _b , O _c , w, u _a , u _b , uu _c・ z, ul _c・ z) = Allowable then
Return the judgment of function 6 as it is
else
z = z + 1
while end
16 to 19 show flowcharts of the algorithm of the function 5.

<Function 6 (O _a , O _b , O _c , w, u _a , u _b , r _u , r _l )>
Input: Material weight w, Order O _{a unit} weight (upper or lower limit) u _a , Order O _{b unit} weight (upper or lower limit) u _b , O _c allowance upper limit r _u and lower limit r _l
r _p = floor ((r _u + r _l ) / 2)
s _u = min {floor ((wr _p -y _l・ u _b ) / u _a ), x _u }
s _l = max {ceil ((wr _p -y _u・ u _b ) / u _a ), x _l }
if r _p is divisible by d, and function 2 (O _a , O _b , u _a , u _b , wr _p , s _u , s _l ) = can be allocated then
Order O _a, O _b provision with provision of function 2 determines that allow provision in provision amount wr _p in order O _c
else
k _p = countT (u _a , u _b , wr _p -x _l・ u _a -y _l・ u _b ) -countT (u _a , u _b , wr _p -x _u・ u _a -y _l・ u _b ) -countT (u _a , u _b ,
wr _p -x _l・ u _a -y _u・ u _b )
k _u = countT (u _a , u _b , wr _u -x _l・ u _a -y _l・ u _b ) -countT (u _a , u _b , wr _u -x _u・ u _a -y _l・ u _b ) -countT (u _a , u _b ,
wr _u -x _l・ u _a -y _u・ u _b )
k _l = countT (u _a , u _b , wr _l -x _l・ u _a -y _l・ u _b ) -countT (u _a , u _b , wr _l -x _u・ u _a -y _l・ u _b ) -countT (u _a , u _b ,
wr _l -x _l・ u _a -y _u・ u _b )
if k _p -k _u > 0 then
Returns the judgment of function 6 (O _a , O _b , O _c , w, u _a , u _b , r _p , r _u ) as it is
else if k _l -k _p > 0 then
Returns the judgment of function 6 (O _a , O _b , O _c , w, u _a , u _b , r _l , r _p ) as it is
else
Determining that it is not possible to allocate the surplus, FIGS. 20 to 21 show flowcharts of the algorithm of the function 6.

  Hereinafter, an algorithm in the case of three orders will be described. As in the case of two orders, from Lemma 2, if the allowance to eliminate the surplus is possible, the provision amount of all other provisions, except for one provision, is either a multiple of the upper or lower unit weight or the provision Provisions can be made to match the volume limits. Therefore, except for one provision, the provision amount of all the remaining provisions is considered to be a multiple of one of the upper and lower limits of the unit weight, or to obtain an provision that matches the upper and lower limits of the provision amount.

  In Algorithm 3, first, a case where at least one allocation matches one of the upper and lower limits of the order quantity is processed. This is because if the amount of provision for one order is matched with one of the upper and lower limits of the order amount, it can be determined using the algorithm 2 whether or not provision for eliminating the surplus can be made for the remaining orders.

Next, the case where the two allocations match one multiple of the upper and lower unit weights is handled. There are three ways to select orders that do not match the multiple of the single weight upper and lower limits. In order to examine each of the single weight upper and lower limits for two orders, function 5 is called a maximum of 12 (= 3 × 2 × 2) times. Examine all possibilities. Function 5 determines whether the provision of order O _a multiple of unit weight u _a, by limiting the provision of order O _b a multiple of unit weight u _b, eliminating the surplus reserve can.

In function 5, firstly, provision amount x · u _a to order O _a, when the provision of the order O _b was _{y · u b, x, y} ( both integers) lower x _l ≦ x ≦ on the , seeking y _l ≦ y ≦ y _u. Next, when the amount of provision for order O _{a is set} to x _l · u _a and x _u · u _a respectively, the amount of provision for order O _{b is set} to y _l · u _b and _yu · u _b It is determined using function 1 whether the surplus can be eliminated for the four ways.

Then, using the integer z, it is checked whether or not the allocation to the order O _c is limited to the range of z · ul _{c to} z · uu _c (the integer z And change to different values to cover all cases). FIG. 23 is a diagram illustrating regions and grid points when there are three orders. When it is possible to reserve to eliminate the surplus, there are lattice points in the area BDFGEC (including on the side) in FIG. Whether or not there is a grid point in the region is determined by function 6.

  What function 6 does is a combination of an algorithm for counting the number of grid points and a binary search, as in the case of two orders. In the binary search, except for the fact that the method of dividing in half is changed to a straight line parallel to the straight line P and the straight line Q, the determination is almost the same as that of the function 4, and the detailed description is omitted.

  Here, the lemma used in the description of the algorithm will be described.

[Lemma 1]
For a real number x> 0, b ≧ a> 0, the existence of a natural number n satisfying an ≦ x ≦ bn is equivalent to x ≦ floor (x / a) · b.
(Proof)
When there is a natural number n satisfying an ≦ x ≦ bn, n ≦ x / a from an ≦ x.
Since n is a natural number, n ≦ floor (x / a) also holds. From x ≦ bn, x ≦ floor (x / a) · b.
Next, suppose that x ≦ floor (x / a) · b holds. If n = floor (x / a), an ≦ a · floor (x / a) ≦ a · x / a ≦ x. Bn ≧ b · floor (x / a) ≧ x (the last inequality is assumed). Therefore, an ≦ x ≦ bn holds.
(End of proof)

[Lemma 2]
For each order O _i , assuming that the amount of provision is p _i , it is assumed that the single weight constraint is satisfied and that the amount is within the upper and lower limits of the amount of provision. At this time, except for one provision, the provision amount p ′ _i is obtained such that all the provision amounts of the other provisions coincide with either the integral multiple of the unit weight (upper limit or lower limit) or the upper and lower limits of the provision amount. be able to.
(Proof)
When both of the two provisions do not match an integer multiple of the upper and lower limits of the unit weight or the upper and lower limits of the provision amount, one of the two provisions is changed to the single weight without changing the provision amount of the other provisions. It is shown that it can be made to be an integral multiple of or an upper and lower limit of the provision amount. It can be easily understood that a desired allocation can be obtained by repeating this operation.
Suppose that p ₁ and p ₂ do not match an integer multiple of unit weight or the upper and lower limits of the provision amount. here,
d ₁ = min {ceil (p ₁ / b ₁ ) ・ b ₁ , floor (p ₁ / a ₁ ) ・ a ₁ , u ₁ -p ₁ -p ₁ -l ₁ },
d ₂ = min {ceil (p ₂ / b ₂ ) ・ b ₂ , floor (p ₂ / a ₂ ) ・ a ₂ , u ₂ -p ₂ , p ₂ -l ₂ }
far. Note that d ₁ > 0, d ₂ > 0.
If d ₁ <d ₂ then (p ′ ₁ , p ′ ₂ ) = (p ₁ -d ₁ , p ₂ + d ₁ ) or (p ′ ₁ , p ′ ₂ ) = (p ₁ + d ₁ , p ₂ -d ₁ ), p ′ ₁ matches an integer multiple of the unit weight or the upper and lower limits of the provision amount. If d ₁ ≧ d ₂ , (p ′ ₁ , p ′ ₂ ) = (p ₁ -d ₂ , p ₂ + d ₂ ) or (p ′ ₁ , p ′ ₂ ) = (p ₁ + d ₂ , p ₂ -d ₂ ), p ′ ₂ is equal to an integral multiple of the unit weight or the upper and lower limits of the provision amount. In either case, it can be easily confirmed that (p ′ ₁ , p ′ ₂ ) is within the upper limit and lower limit of the unit weight limit / allocation amount.
(End of proof)

[Lemma 3]
Let d = GCD (a, b) be the greatest common divisor of integers a and b. At this time, a necessary and sufficient condition for ax + by = c to have a solution is that c is divisible by d. If a solution exists, if another set of solutions is (x, y) = (x ₀ , y ₀ ), another arbitrary solution is generally given by the following equation.
x = x ₀ + (b / d) k, y = y _0- (a / d) k
(Proof)
The first half is almost self-explanatory. Since a and b are divisible by d, ax + by is also divisible by d. Therefore, c must be divisible by d.
In the second half, first, a ′ = a / d and b ′ = b / d. Since a′x + b′y = c / d and a′x ₀ + b′y ₀ = c / d, a ′ (xx ₀ ) + b ′ (yy ₀ ) = 0. Since a ′ and b ′ are relatively prime, they can be expressed as a ′ (xx ₀ ) = − b ′ (yy ₀ ) = a ′ · b ′ · k using the integer k. Therefore, x = x ₀ + b ′ · k, y = y ₀ −a ′ · k.
(End of proof)

[Lemma 4]
Find the greatest common divisor d = GCD (a, b) of a and b for positive integers a and b, and at the same time, set x = x ₀ , y = y of integer x and y satisfying ax + by = d There is an algorithm for finding a set of ₀ (extended Euclidean algorithm). The amount of calculation is the same as the normal Euclidean algorithm, O (max {log a, log b}).

[Lemma 5]
a, b, and c are positive integers. The fact that the lattice point (x ′, y ′) exists on the straight line ax + by = c and the relationship of x _l ≦ x ′ ≦ x _u is satisfied is that the following conditions (1) and (2 ) Is equivalent to both. Where d is the greatest common divisor of a and b.
(1) c is divisible by d
(2) ceil ((x _l -c / d ・ x ₀ ) (d / b)) ≦ floor ((x _u -c / d ・ x ₀ ) (d / b)
(Proof)
From Lemma 4, the extended Euclidean algorithm can be used to obtain a set of integers x ₀ and y ₀ satisfying the greatest common divisor d of a and b and ax ₀ + by ₀ = d. Since a and b are divisible by d, ax + by is also divisible by d. Therefore, c must be divisible by d ((1) holds). Since x = (c / d) x ₀ , y = (c / d) y ₀ satisfies ax + by = c, from Lemma 3, x ′ = (c / d) x ₀ + (b / d) It can be expressed as k. In order to satisfy x _l ≦ x ′ ≦ x _u , (x _l − (c / d) x ₀ ) (d / b) ≦ k ≦ (x _u − (c / d) x ₀ ) (d / b An integer k that satisfies) is required. This shows that (2) needs to hold. Conversely, if (1) and (2) hold, it can be easily understood that the lattice point (x ′, y ′) can be found.
(End of proof)

[Lemma 6]
Let a and b be relatively prime positive integers. The number of lattice points contained in the region T (a, b, c) = {(x, y) ∈R ² │ax + by ≦ c, x> 0, y> 0} is N (a, b, c) far. Here, R ² represents a secondary Euclidean space. N (a, b, c) can be calculated in O (max {log a, log b}) time.

(Proof)
Let p = floor (c / a) and q = (c-ap) / b. When p = 0, clearly N (a, b, c) = 0, so p> 0.
Consider representing the area of a triangle T (a, b, c) using N (a, b, c). The number of squares with a side length of 1 in T (a, b, c) is equal to N (a, b, c). Therefore, consider dividing the remaining region into a plurality of trapezoids and one triangle (see Equation 3). For each 0 <j ≦ p, the trapezoid S _j is defined as follows. For real number x, {x} is defined by x-floor (x).

S _j = {(x, y) ∈R ² | ax + by ≦ c, j-1 ≦ x ≦ j, y ≧ floor ((c-aj) / b)}
Next, the triangle R is defined as follows.
R = {(x, y) ∈R ² | ax + by ≦ c, p ≦ x, y ≧ 0}
At this time, the area of the triangle T (a, b, c) can be expressed as follows.

Since | T (a, b, c) | = c ² / (2ab) and | R | = q (c / ap) / 2 hold, the following relationship can be derived.

Next, consider the area of the trapezoid S _j . The area | S _j | of the trapezoid S _j can be expressed as follows.

Therefore, N (a, b, c) can be expressed as follows.

The second equation consists of the relationship q = (c-ap) / b. Here, since N (a, b, c) = N (b, a, c) holds, it is assumed that a ≧ b without losing generality. Further, when a = b, it can be easily shown that N (a, b, c) = floor (c / a) (floor (c / a) -1) / 2. Therefore, the case of a> b may be considered. If a is divisible by b, k = a / b-1, otherwise k = floor (a / b). If c ′ = cb (kp + floor (q)), the straight line (a-bk) x + by = c ′ passes through two points (0, c ′ / b) and (p, {q}). From this, N (a-bk, b, c ′) can be expressed as follows.

Organize

Therefore, N (a, b, c) can be calculated with the following algorithm.
Algorithm calc (a, b, c)
a, b, c is a positive integer such that a ≧ b
begin
if (a mod b) = 0 then k: = a / b-1; else k: = floor (a / b);
p: = floor (c / a);
q: = floor ((c-ap) / b);
if p = 0 then
return 0;
else if a ＝ b then
return p (p-1) / 2;
else
return calc (b, a-bk, cb (kp + q)) + kp (p-1) / 2 + pq;
endif
end

The calculation time of the algorithm calc is proportional to the number of recursive calls of calc. Since the number of recursive calls of calc is equal to the number of recursive calls when Euclidean algorithm is applied to a and b, the calculation time of the algorithm calc is O (max {log a, log b}).
(End of proof)

  Hereinafter, specific examples of the calculation method described above will be shown.

  Example 1: When order quantity 10.0t (acceptable range 8.0t-12.0t), single weight limit 4.5t-5.5t order and 10.0t material are given, algorithm 1 is applied It is determined that 10.0 t of material can be allocated.

  Example 2: When order 10.0t (allowable range 8.0t to 12.0t), single weight limit 4.5t to 5.5t and 8.0t material are given, algorithm 1 is applied Since the relationship of 8.0 ≦ floor (8.0 / 4.5) * 5.5 is not satisfied in the function 1, it is determined that it is impossible to allocate 8.0 t of material without surplus.

  Example 3: Order amount 9.0t (allowable range 1.45t to 10.0t), single weight limit 1.45t to 1.55t order A and order amount 3.0t (allowable range 2.5t to 3.5t) ), When order B with a weight limit of 2.5t to 2.9t and material of 12.0t are given, apply algorithm 2 to allocate 9.3t for order A and 2.7t for order B Thus, it is determined that it is possible to reserve without surplus.

  Example 4: Order amount 9.0t (allowable range 2.5t to 10.0t), single weight limit 1.5t to 1.6t order A and order amount 4.0t (allowable range 2.5t to 5.5t) ) When the order B of the single weight limit of 1.55 t to 1.7 t and the material of 12.0 t are given, it is determined that it is impossible to allocate without surplus if the algorithm 2 is applied.

  Example 5: Order quantity 1.9t (allowable range 1.5t to 5.0t), single weight limit 1.5t to 1.9t order A and order quantity 4.8t (allowable range 2.5t to 5.0t) ), Order B with single weight limit 2.5t to 3.0t and order quantity 4.0t (allowable range 3.5t to 8.0t), Order C with single weight limit 1.2t to 1.5t and 11.0t When the algorithm 3 is applied, it is determined that the order A can be allocated without surplus by allocating 1.5 t to the order A, 5.0 t to the order B, and 4.5 t to the order C. .

  Example 6: Order amount 4.3t (allowable range 1.5t to 10.0t), single weight limit 1.5t to 1.9t order A and order amount 4.8t (allowable range 4.5t to 9.0t) ), Order B with single weight limit 1.5t to 1.7t and order quantity 4.8t (allowable range 4.5t to 9.0t), Order C with single weight limit 1.5t to 1.7t and 9.0t When algorithm 3 is applied, it is determined that it is impossible to reserve without surplus.

  According to the embodiment of the present invention, it is determined whether or not there is a provision for eliminating the surplus, and when it exists, high-speed processing for providing the provision can be realized. The high speed of processing is due to the fact that it is possible to determine with one inequality for a single order. In the conventional method, it is usually necessary to check whether the weight of the material is in an interval satisfying the single weight constraint by multiplying the upper limit and lower limit of the single weight twice or three times. The speed of this part not only contributed to speeding up the judgment when there are multiple orders, but also speeded up the entire material provisioning program, such as the part to determine whether a certain provision meets the single weight limit. Greatly contribute. In the case where there are a plurality of orders, it is possible to determine in a practical time whether or not the provision for eliminating the surplus can be made quickly and reliably by realizing by combining techniques such as binary search.

  As mentioned above, although this invention was demonstrated using embodiment and a specific example, the technical scope of this invention is not limited to the range as described in the said embodiment. Various modifications or improvements can be added to the above embodiment. It is apparent from the scope of the claims that the embodiments added with such changes or improvements can be included in the technical scope of the present invention. As can be understood from the above embodiments, the method and apparatus of the present invention can be applied regardless of the weight. For example, the present invention can be applied to a confectionery having a single weight restriction on the weight of the material or to allocate a small material without departing from the essence of the present application.

  As one embodiment, the present invention can be realized by a computer program on the material allocation apparatus 100 (FIG. 1). The storage medium storing the program can be an electronic, magnetic, optical, electromagnetic, infrared, or semiconductor system (or apparatus or device) or a propagation medium. Examples of computer readable media include semiconductor or solid state storage, magnetic tape, examples of removable computer readable media include semiconductor or solid state storage, magnetic tape, removable computer diskettes, Random access memory (RAM), read only memory (ROM), rigid magnetic disk and optical disk are included. Current examples of optical disks include compact disk-read only memory (CD-ROM), compact disk-read / write (CD-R / W) and DVD.

It is a material allocation apparatus schematic. In this example, one material is allocated to three orders. 3 is a flowchart of Algorithm 1. It is a flowchart (1/3) of algorithm 2. It is a flowchart (2/3) of algorithm 2. It is a flowchart (3/3) of the algorithm 2. It is a flowchart (1/4) of algorithm 3. It is a flowchart (2/4) of the algorithm 3. It is a flowchart (3/4) of the algorithm 3. It is a flowchart (4/4) of the algorithm 3. It is a flowchart of function 1. It is a flowchart of function 2. It is a flowchart of function countT. 10 is a flowchart of Function 3. 10 is a flowchart of function 4. It is a flowchart (1/4) of function 5. It is a flowchart (2/4) of the function 5. It is a flowchart (3/4) of the function 5. It is a flowchart (4/4) of function 5. It is a flowchart (1/2) of the function 6. It is a flowchart (2/2) of the function 6. It is a figure showing the area | region and lattice point when there are two orders. It is a figure showing the area | region and lattice point when an order is three.

Explanation of symbols

DESCRIPTION OF SYMBOLS 100 Material allocation apparatus 110 Input part 115 Execution part 120 Algorithm selection part 130 Algorithm storage part 140 Function storage part 150 Allocation determination part 160 Weight calculation part 170 Output part

Claims (6)

A method of calculating a weight of a provision material with a margin by a computer including an input unit and an arithmetic processing unit ,
Receiving the input of the weight of the material given from the input unit and receiving the input of the unit weight, unit weight upper limit, unit weight lower limit, margin upper limit value, margin lower limit value for each order for two or more orders ; ,
Determining whether the material can be allocated without surplus based on a predetermined calculation formula for the order input by the arithmetic processing unit ;
In response to determining that the material can be allocated without surplus, the arithmetic processing unit calculates the weight of the allocated material;
Only including,
In the determining step,
A combination of allocations that can be allocated for the input order is represented by polyhedral grid points in Euclidean space, the weight of the given material, the unit weight upper limit of each of the orders, and the unit A method of determining whether or not the material can be allocated without surplus by detecting the presence or absence of the lattice points in the region formed by the lower and upper limit . The area in the determining step is in response to the number of orders being two,
The weight of the given material is w, the unit weight upper limit of one order uu _a , the unit weight lower limit ul _a , the margin upper limit value ru _a , the margin lower limit value rl _a , and the unit weight upper limit of another order. uu _b , the unit weight lower limit ul _b , margin upper limit value ru _b , margin lower limit value rl _b , u _a = uu _a or ul _a , u _b = uu _b or ul _b , ceil (x) is the ceiling function of x, Let floor (x) be the floor function of x
u _a・ x + ul _b・ y ≦ w, u _a・ x + uu _b・ y ≧ w,
ceil (max {rl _a , w-ru _b } / u _a ) ≦ x ≦ floor (min {ru _a , w-rl _b } / u _a )
The method according to claim 1, wherein the given material can be allocated by calculating the presence / absence of a lattice point in a region formed in step ( 1 ). The area in the determining step is in response to the number of orders being three,
The weight of the given material is w, the unit weight upper limit of one order is uu _a , the unit weight lower limit ul _a , the margin upper limit value ru _a , the margin lower limit value rl _a , and the unit weight upper limit of the second order is uu _b , the unit weight lower limit ul _b , the margin upper limit value ru _b , the margin lower limit value rl _a , the unit weight upper limit of the third order uu _c , the unit weight lower limit ul _c , the margin upper limit value ru _c , and the margin lower limit value rl _c , u _a = uu _a or u _a = ul _a , u _b = uu _b or u _b = ul _b , ceil (x) is the ceiling function of x, floor (x) is the floor function of x,
xl = ceil (rl _a / u _a ), xu = floor (ru _a / u _a ),
yl = ceil (rl _b / u _b ), yu = floor (ru _b / u _b ),
zl = ceil (rl _c / uu _c ), zu = floor (ru _c / ul _c ),
As
For each integer value z satisfying zl ≦ z ≦ zu,
u _a・ x + u _b・ y ≦ w-ul _c・ z, u _a・ x + u _b・ y ≦ w-uu _c・ z
max {ceil ((wz ・ uu _c -yu ・ u _b ) / u _a ), xl} ≦ x ≦ min {floor ((w-zu ・ l _c -yl ・ u _b ) / u _a ), xu},
max {ceil ((wz ・ uu _c -xu ・ u _a ) / u _b ), yl} ≦ y ≦ min {floor ((wz ・ ul _c -xl ・ u _a ) / u _b ), yu}
By detecting the presence or absence of lattice points in the region formed by, for the given material to be possible provision method according to claim 1. The method according to any one of claims 1 to 3 the presence of the grid points in the step of determining, detected using a binary search. A device for calculating the weight of the allocated material with a margin,
An input unit that receives input of the weight of a given material and receives input of a unit weight, a unit weight upper limit, a unit weight lower limit, a margin upper limit value, and a margin lower limit value for each order for two or more orders ;
A determination unit that determines whether it is possible to reserve the material without surplus based on a predetermined calculation formula for the input order;
In response to determining that the material can be allocated without surplus, a calculating unit that calculates the weight of the allocated material;
Including
The determination unit
A combination of allocations that can be allocated for the input order is represented by polyhedral grid points in Euclidean space, the weight of the given material, the unit weight upper limit of each of the orders, and the unit An apparatus for determining whether or not it is possible to reserve the material without surplus by detecting the presence or absence of the lattice points in the region formed by the lower and upper limit . A computer program for calculating the weight of a provisioned material with a margin, the computer comprising an input unit and an arithmetic processing unit ,
It receives the weight of the material provided by the input unit, and the unit weight for each of the orders for two or more orders, unit weight limit, unit weight lower, margin upper limit value, the step of subjecting the input margin lower limit value When,
For the input order, a step of determining whether the material can be allocated without surplus based on a predetermined calculation formula by the arithmetic processing unit ;
In response to determining that the material is capable of Hikiateru without excess, a step of causing calculate the weight of the reserve material by the processing unit,
Execute
In the step of determining,
A combination of allocations that can be allocated for the input order is represented by polyhedral grid points in Euclidean space, the weight of the given material, the unit weight upper limit of each of the orders, and the unit A computer program that makes it possible to determine whether or not the material can be allocated without surplus by detecting the presence or absence of the lattice points in the region formed by the lower and upper limit .

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