THE METHOD OF BALANCING THE PRODUCTION AND CONSUMPTION MODEL IN THE CASE
OF INDIVISIBLE ARTICLES
Marek Ładyga, Maciej Tkacz
Institute of Mathematics, Czestochowa University of Technology Częstochowa, Poland
marekladyga@im.pcz.pl, maciej.tkacz@im.pcz.pl
Abstract. In this article, a detailed case of the unbalanced production-consumption, namely a model, which is used in the case of indivisible articles, is presented. Also, the method of balancing this model is given, relying on introducing a certain ordering relation in the consumer’s set.
Keywords: balancing, model, indivisible articles
Introduction
In article [1] the unsustainable production and consumption model was defined, in which occur n ≥ contractors, 2 m ≥ 1 producers and n − m ≥ consumers of 1 a certain good. The model assumes that total demand exceeds total supply in a fixed period of time, what we write in the form of so-called uncorrected n-dimensional supply-demand vector, fulfilling the conditions
1) 0
p
i> for every i < m
2) 0
p
i< for every m + ≤ ≤ 1 i n 3)
1
0
n i i
p
=
∑ <
The general case was discussed in [1-4] articles. A detailed case are so-called indivisible articles, for example: washing machines, television sets etc. (you cannot buy 1.5 washing machine).
In this case, instead of uncorrected supply-demand vector, we consider so called
discrete vector, ie. uncorrected supply-demand vector, of which every coordinate
is an integer.
Let there be vectors:
a) discrete vector p = ( p
1, , p
n)
r
K , where p
i> 0 for i = 1, 2, K m < n ( p
i- demand of i
thcontractor), 0
p
j< for j = m + K 1, n ( p
j- supply of j
thcontractor), so that
1 2
0
p + p + K + p
n< ,
b) maximum concession vector u = ( u
1, u
n)
r
K , where ( u ∈ Ν ∪
i{ } 0 , u
i- the size of the maximum possible concession of i
thcontractor), so that
( p
1+ u
1) ( + p
2+ u
2) + K + ( p
n+ u
n) ≥ 0 , c) vector a = ( a
1, a
2, , a
n)
r
K , a
i- concession weight of i
thcontractor.
Taking p
0= p r , a
0= a
r r
for s ≥ we define the size of 1
(
1 1 1)
s s s
r = − p
−+ K + p
n−:
1s j j
i i
j s
N i n a
−r u
≤
= ≤ ≤
∑
(
1, , )
s s s
a a
na = r
K where
( )
11 1
0 for for
s s
i s s s
i
i N a
a
−B
− −i N
∉
=
∈
where
1
1 1
s
s s
j j N
B a
−
− −
∈
= ∑
1
\
s s s
N = N
−N where N
o= { 1, K , n }
(
1, , )
s s s
p p
np = r
K where
1 1
1
for for for
s s s s
i i
s s
i i i
s j
i j s
p r a i N
p p u i N
p i N
− −
−
〈
+ ⋅ ∈
= + ∈
∈ ∪
The iteration process, whose properties were described in [2], we finish for s
thiteration, such that r = . Then vector
s0 p
s−1r balances the model. Let x = ( x
1, , x
n)
r
K be any vector. By E x ( ) r
we understand vector E ( ) x
1, K , E ( ) x
n where E ( ) x
i- integer’s total part x
i. Because p
s−1= p
sr r
, continuing in this article, by the overall vector correction vector p
0r we adopt vector p
sr .
If E ( )
1sE ( )
2sE ( )
s0
p + p + K + p
n= then vector E p ( ) r
sbalances the model in case of indivisible articles. Assume that E ( )
1sE ( )
s0
t = p + K + p
n> , which means that supply on certain good exceeds demand. In order to achieve sustainable modem, some producers should reduce the size of production (supply reduction) or, chosen consumers increase demand. In this situation, you may prefer consumer and use a second variant of balancing. It therefore remains to determine which consumer and by how much the demand should be increased. The solution can be as follows:
add by one unit of good for those consumers who the most resigned.
1. The balancing of model
It is, therefore necessary to prove that:
a) has remained integer of units goods to distribute among consumers,
b) the number of consumers, which can increase demand by one unit of good (in this case, the border in output demand), is not smaller than number of units of goods, which is left to distribute.
It will be proved by the statement below.
Statement 1.
If:
{
0}
1
: and E(
is) 1
iM = i ∈ N m < ≤ i n p − ≥ p and t > 0 then the cardinality of the set M
1is bigger than t and t is an integer.
Proof.
( )
1 n
s i i
t E p
=
= ∑ is, obviously, an integer as a sum of integers.
For every i ≤ m there is an inequality of
s0
p
i≥ , or E p ( )
is− p
is≤ , and for every 0 i > m there is an inequality
s0
p
i≤ , or 0 ≤ E p ( )
is− p
is< . Therefore: 1
( ) ( )
( ( ) ) ( ( ) ) ( ( ) )
1 1 1
1
1 1 1
#
n n n
s s s
i i i
i i i
m n n
s s s s s s
i i i i i i
i i m i m
t E p E p p
E p p E p p E p p M
= = =
= = + = +
= = − =
= − + − < − ≤
∑ ∑ ∑
∑ ∑ ∑
where: M
1= { i ∈ N
0: m < ≤ and i n E p ( )
is≠ p
is} .
It remains to be presented, that M
1⊂ M
1.
Let i ∈ M
1, or E p ( )
is≠ p
is, p
is≥ p
i0, therefore:
( )
0 s s
i i i
p ≤ p < E p , or p
i0< E p ( )
is, but p ,
i0E p ( )
isbelong to the set of integers, so there is inequality p
io≤ E p ( )
is− , therefore 1 i ∈ M
1.
If M
1⊂ M
1, that # M
1≤ #M
1but # M
1> , hence t #M
1> t ( #M
1- is the cardinal- ity of the set M ), which ends the proof of the above statement.
1Statement 1 shows that if t > , then deducting t coordinates from the 0 M
1set of vector ( )
1, , ( )
s s
E p E p
n
K by one, then a vector, which balances the model, will be obtained. It creates the question, which of t coordinates from the M
1set, should deducted by one, in other words, to which of the consumers an additional unit of good should be assigned. The following solution is being proposed: increase the demand by one unit for those consumers who are the closest to its limits. This distance is determined by number of x
i= ( p
i+ u
i− E p ( )
is+ 1 ) u
i−1.
The 100 x
inumber expresses the percentage of concession remaining i
thcon- sumer, if their demand will verify to the amount equal to E ( ) p
is− . In order to 1 indicate a concrete consumer, to which an additional unit of good in M
1set will be assigned, an ordering relations is introduced.
Statement 2.
If for every i j ∈ , M
1, ,
R ,
i j i j
i j
x x x x
i j
i j x x
< ≠
⇔
≥ =
then R relation is ordering rela- tions in M
1set.
Proof
It should be presented that relation R is:
a) reversible, ie. for every i ∈ M
1and R
i,
b) weakly symmetric, ie. for every i j ∈ , M
1i R and j j R i ⇔ i = j , c) transitive, ie. for every i j k ∈ , , M
1i R j and j R k ⇔ i R k , d) consistent, ie. for every i j ∈ , M
1i R j or j R i.
Ad a) R i i , because i ≥ and i x
i= x
i.
Ad b) R i j , ie. x
i< x
jor ( x
i= x
jand i ≥ j ) , R j i , ie. x
j< x
ior ( x
j= x
iand j ≥ . i )
If R i j and j R i , then x
i= x
j, i ≥ j j , ≥ , therefore i and
i j
x = x i = j . Ad c) R i j , therefore x
i< x
jor ( x
i= x
jand i ≥ j ) ,
R j k , therefore x
j< x
kor ( x
j= x
kand j ≥ k ) .
If x
i≠ x
j≠ x
k, then x
i< x
j< x
kor x
i< x
k, therefore R i k . If x
i= x
j≠ x
k, then x
i= x
j< x
kor x
i< x
k, therefore R i k . If x
i≠ x
j= x
k, then x
i< x
j= x
kor x
i< x
k, therefore R i k . If x
i= x
j= x
k, then i ≥ and or j j ≥ or i k ≥ . k
Therefore R i k .
Ad. d) Proof by contradiction.
If R i j does not occur, then x
j> x
ior ( x
j= x
iand i > j ) .
If x
i> x
j, a contradiction that x
j> x
ior x
i= x
j. If x
j> x
i, a contradiction that x
i> x
jor x
j= x
i. If x
i= x
j, a contradiction that i = j .
Must therefore take place one of the relations: i R j or j R i, which ends the proof of the above statement.
Ordered M
1set, according R relations, determines by M . Increasing demand to
1Rthe first t
thconsumers from the M
1Rset, by one unit, leads to sustainable model.
In addition, supply and demand of all contractors will be expressed by integers.
It will be presented in the statement below.
Statement 3.
If t > , 0 M
1+= { i ∈ M
1R: i ≤ s } ,
1
1
( ) 1 for ( ) for
s i
i s
i
E p i M
d
E p i M
+
+
− ∈
=
∉ , d = ( d
1, , d
n)
r
K
then: for i ≤ n d
iis an integer and
1 2... 0
n
d + d + + d = . Proof:
For i ∈ M
1+d
iis an integer as a difference of integers. For i ∉ M
1+d
iis also an integer.
( )
1 1 1 1 1
1 1
E( ) 1 E( ) E( ) ( 1) 0
n n
s s s
i i i i i i
i i M i M i M i M i i M
d d d p p p t t
+ + + + +
= ∈ ∉ ∈ ∉ = ∈