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Solution to the problem of locating production with discontinuous functions at zero transportation costs Sultankul k. A.
Решение задачи размещения с разрывными в нуле функциями затрат Султанкул к. А.
Султанкул кызы Айнура /Sultankul kyzy Ajnura - старший преподаватель, кафедра кибернетики и информационных технологий, факультет математики, информатики и кибернетики,
Кыргызский национальный университет им. Ж. Баласагына, г. Бишкек, Кыргызская Республика
Abstract: it has been proved a sufficient condition for the application of the sequential calculations method for a nonlinear problem of locating production with discontinuous functions at zero transportation costs.
Аннотация: доказано достаточное условие применимости метода последовательных расчетов для нелинейной задачи размещения с разрывными в нуле функциями затрат.
Keywords: fixed co-payment, raw materials extraction item, fixed costs, transport costs, sequential calculations. Ключевые слова: фиксированная доплата, пункт добычи сырья, фиксированные затраты, транспортные расходы, последовательные расчеты.
Statement of the problem: Suppose a large company in the region has m possible stations for extraction of raw materials such that Д, i = 1,2,.. .m , with the desired production volumes of x. > 0. Produced raw material, in turn, is been delivered to n number of enterprises B., j = 1,2,..n for further processing. It is assumed that the feedstock volume y , on each B., j = 1,2,...n bounded above by the value q , i.e. 0 < у j < q, j = 1,2,..., n, as well as amount of transported raw material from possible extraction stations Д to B, in a quantity of a. i.e. 0 < x <a. , i = 1,2,...,m, j = 1,2,...,n.
J ‘J ‘J ‘J
Given
V (X ^
Bj .
the feedstock need i = 1,2,...m, j = 1,2,...n
for the region in a volume of Q and in a function
A
determining expenditures for the transportation of raw materials from i to
In addition, for each Д, i = 1,2, ..., m and B ., j = 1,2, .. n, it is known that their functions are V (x.), if j (у j ) which determine the cost of raw materials extraction and its processing.
For each enterprise it is required to determine the optimal amount of raw materials extracted x. > 0, transported
X > 0 and processed raw materials у. > 0, for which the total cost would be minimal cost.
The above stated problem can be written as the following extreme problem:
Find the minimum
m n m n
L(x) = ^Lv (xj) + z V (xi) + (У]) (1)
i=1 j=1 i=1 j=1
under the constraints
n
Xxp = x,, i = 1,2,...,m, (2)
j=1
m
Xxj = уj <qj, j = l^..^^ (3)
i=1
n
X у j = Q, (4)
j=1
0<xtj <a, i = 1,2,..,m, j = 1,2,..,n, (5)
x, > 0, i = 1,2,..., m, yj 0, j = 1,2,.., n, (6)
where x = x .
I У lm,n
It is assumed that
n m m n
Q -Eq , Ex -EEa (7)
j=1 i=1 i=l J=1
Consider the solution to the problem from (1) to (6), when
V,(X) = c,x,, i = 1,2,...,m, Wj(у) = cjyj, j =1,2,...n, а Vj(хг:/.)
i = 1,2,.., m, j = 1,2,.., n,
where a. is a fixed surcharge.
j
cjxj + aj, XJ > 0,
0, x,j = 0
The problem from (1) to (6) in this case can be written as:
Find the minimum
m n ____
L( x)=EE cvxv+ aP( xj) (8)
i=1 J =1
under conditions
n
Ex,j = xt, i = 1,2,...m, (9)
j=1
m
Exj - Vj , j = 1,2,. n (10)
i=1
m n
EExi=Q, <“>
i=1 j =1
0 - xj - Qjj, i = 1,2,...,m, j = 1,2,...,n, (12)
where cy = cy + c, + Cj, i = 1,2,...,m, j = 1,2,...,n.
To solve the problem (8)-(12) we’ll use the method of successive calculations [1]. Therefore, we’ll introduce some notations and transformations.
Consider that every possible extraction station of raw materialsД, i = 1,2,...,m, has many extraction stations A , i = 1,2,...,m, k = 1,2,...,n. Then each extraction station A corresponds to a certain volume of extraction (production) of raw materials
xik, 0 - xik - aik, i = 1,2,...,m, к = 1,2,...,n,
xikj - volume of products transported from A to B, and fixed costs
aikj=aijSkj , i = 1,2,..., m, к, j = 1,2,..., n, as well as production and transportation costs
ск = CjSj + M(1 -5j ), i = l^..^m k,j = l^..^n, where
aft = a^kj , i = 1,2,.m, k, j = 1,2,..,n, М is sufficiently large positive number (barring rate).
Let G denote the set of pairs of indices {ikJ, i = 1,2,..m, к = 1,2,..,n.
Then the problems (8)-(12) can be written as Find the minimum
__ n
L(x, G) = EE ckjxkj + E U,ksign(x,k ) (13)
ikGG j=1 ikGG
under conditions
n
E xk = xk - ak, ik gG, (14)
j=1
E xkj - Vj, j = ^^..^n (15)
ikGG
n
EExtj- = Q, (16)
ikGG j =1
xkj > 0 , ik g G, j = 1,2,...,n,
(17)
___ n
where x = |*J П* = Xai , ik e G.
h j=i
We introduce a conditional clause A0 with a volume of raw material extraction equal to the value of
n
Xqj ~Q having transport costs c00j = 0, j = 1,2,...,n,
j=i
with the fixed costs П00 = 0.
This extraction station, having the index {00}, is considered as element of any subset of indices © c G . Then each subset © c G may be determined by the function:
___ n
L( x ©) = XX + X (18)
ikea j=1 ike©
under conditions
X xikj = q, j = n, (19)
ike©
n
Xxk <a,k, ike© (20)
j=1
__ n
X X xk = Q, (21)
ike©\{00} j=1
Xkj ^ 0, ik e©, j = 1,2,..., n. (22)
Denote p(o), having the minimum value of L(x, ©), under the conditions (19) - (22). Then the problem can be formulated as follows:
It is needed to define a subseto c G so that p(o) reaches its lowest valuep(o*) , i. e.
p (©*) = min {p (o)}. (23)
©cG ^ *
We’ll prove a sufficient applicability of the method of successive calculations [2] to the problem, i. e., that for any subsets ф, ©2 c G there is condition
s(©, ©2) = p(o) + p(o) -p(d)-p(P) < 0, (24)
where a = ©1lu©2, 3 = ©1ni©2,
and p(o), p(©2), p(a), p(P) is minimum value of the function L(x,©) under conditions (19)-(22),
and replacement set ©respectively to sets ©, ф, a, Д
Then the condition (24) for this problem takes the following form:
s(© ,©) = min { X X cikjxik) + ХПЛ + m-in 1 X X Cikjxikj + X Пik f
x 1 ike© j =1 ike© x ike<©2 i=\ ike<©2 1 2 J (25)
- min {xX ck ixikj + X Пik - min { n XX j k + Xn, f< 0
iikea j=1 ikea j i x i ike3 j=1 ikep J
Looking at inputted a and 3 it is easy to see that Xw+Xn,k = Xn,k +Xn,k.
ike© ike©2 ikea ike3
Suppose that the problems (13) - (17) on the set©, ф , have valid plans |x© |, |x©2| that satisfy the following conditions:
x©j = xa, ik e a\©2, J = 1,2,...,n, (26)
x©j = x^, ik ea\©, j = 1,2,..., n, (27)
x©j + x© = xl + x3, ik e3, J = 1,2,...,n (28)
n
Xx© < ak, ik e©\ {°°}, (29)
j=1
Xxa —aik, ikeafi{00}, (30)
j=1
where | x^ | is optimal plan of the problems (13) - (17) on the set a , and | xfi | - on the set fi .
Then to prove the condition (25) it is sufficient to show that
_ n n n n
s(ai,a2) = X Ёcikjxikj + X X cikjxikj “X Xcikjxik “X X cikjxfkj — 0 (31)
ikea j=1 ikea2 j=1 ikea j=1 ikefi j=1
Obviously, since \x“S , x5 are not optimal solutions of the corresponding problems, then the inequality takes
place:
min
x
X Xc~x~ +
cijkxikj 1 X Пik Г — X X
[ikeal j=1 ikeal I ikea j=1
c xWi +
cikjxikj +
Xn*.
ike a
m_in <
x
XX
ciilrx.,. +
Xn. —XX
c^x2 +
'ijk^ikj 1 X ik \ — X X ikjikj
ikea j=1 ikea 1 I ikea 1=1 ike
Xn »■
Consequently, s(o ,o2) — 0, then solving the condition (25) is followed.
It is easy to establish that for feasible plans (26)-(30) the inequality (25) takes place. Obviously, from (26)-(28) it follows:
X cikjx<ikj + X
ike a
cikjxikj + X cikj X cikxikj- 1 X cikjxikj
+ X cikjxikj +X cikj (xikj' +xikj )
ike o
ikea\a
X c xa
ikea\a ikefi
+ X cikjxikj +X cikjxikj = X cikjXikj +X
ikefi ikea ikefi
j = 1,2,..., n.
Summing obtained equality for all j, j = 1,2,..., n, we get
xa + v c xa
cikjxikj ^ X cikjxikj
ikea\o2 ikea\a ikefi
X X cikjxikj + X X cikjxikj = X X c*jxikj + X X
j=1 ikefi
ikj ikj X X ikj ikj
j=1 ikea j=1 ik j=1 ikea
c xfi
cikjxikj.
Consequently, s(a ,a2 ) — 0.
Thus, the condition (25) is proved. Therefore, for the problem (23) a sufficient applicability of the method of successive calculations (24) holds under the assumption of the existence of feasible plans |x0 |, ^x^2 | , which satisfy the conditions (26)-(30).
In the end we need to determine whether there exist such feasible plans for solving this problem. Proof of the existence of these feasible plans |x0 | and jx^21 , satisfying (26) - (30) in this problem, are analogous to that given in [3].
Further, we note that the proof of sufficient applicability of the method of successive calculations (24) allows the use of the algorithm of successive calculations in the formulation of V.P. Cherenin in [2], for the tasks (13)-(17), with additional cull as follows:
Before starting to the calculation of p(a) = m_in {L(x, a)}, for each variant a, the condition X aik - Q > 0 needs to be tested. (32)
ikea\{00}
Calculation of p(a) is carried out only for variants that satisfy condition (32), however, for options a that do not satisfy the condition (32), calculation of p(a) is not carried out and is excluded from further consideration of all fi c a .
n
n
n
n
References
1. Cherenin V. P. The solution of some combinatorial problems of optimal planning using the method of successive calculations //Scientific and methodological materials from Economic and Mathematical Seminar of Laboratory of Economic and Mathematical Methods at Academy of Sciences USSR. - M., 1962.
2. Cherenin V. P., Hachaturov V. R. Solving the same type problems by method of successive calculations regarding the location of production. Economic and mathematical methods. - M.: Science, 1965. - issue 2.
3.
Zhusupbaev A. Problem about the location of production using nonlinear function from a vector space into its underlying scalar field. // Application of mathematical methods in economic research. - Frunze: Ilim, 1976. - Pp.
30-41.
