A Game-theoretic Approach to Multicriteria Problems
Andrew Liapounov
Institute for Economics and Mathematics,
Russian Academy of Science, St.Petersburg,
1, Tchaikovsky st., St. Petersburg, 191187, Russia e-mail address: anlaypunov@yandex.ru
Abstract. A new concept of the single-valued solution for the multicriteria problems using the principles of the consistency and equilibrium from the game theory is introduced. For this solution new equations are constructed and conditions of its Pareto optimality are established.
Examples are considered.
Keywords: Multicriteria optimisation, Pareto optimality, consistency, equilibrium,
game theory.
Introduction
It is known that any “solution” of the multicriteria problem must be Pareto optimal (effective). It means that the improvement of such a solution in any criterion leads to its deterioration in another criterion. The systematical study of the Pareto optimal solutions of the multicriteria problems and of the methods of finding such solution are given in the monograph [Podinovsky and Noghin, 1982]. One of such methods of finding a solution is the method of reducing to one criterion, in particular to the non-negative linear combination of the criteria. This approach contains also in [Kostreva et al., 2004]; [Leskinen et al., 2004]. Another approach to the multicriteria problems consists of an ordering of the criteria by their relative importance. To this approach monograph [Noghin, 2002] and papers [Angilella et al., 2004]; [Doumpos and Zopounidis, 2004] are devoted.
In the papers [Liapounov, 2005a; 2005b; 2005c; 2007] a new concept of the solution the multicriteria problems based on the axiomatic approach with using principles of consistency and equilibrium from the game theory is proposed. The principle of consistency consists of the following: a problem is considered as an element of a class of problems depending on a parameter and the form of dependence of the solution
from this parameter is postulated. In our case (see subsection 2.1) the principle of consistency is reduced to the continuous and monotone dependence of the solution for the segment on the angle of rotation. In our case (see subsection 3.1) the principle of equilibrium consists of the following: the principle of consistency must fulfil in every variable.
On the other hand, the proposed approach is connected with the bargaining problem [Abhinay, 1999]; [Nash, 1950]; [Rubinstein and Osborne, 1990]; [Thomson, 1994].
This paper contains the systematic description of this solution and its properties. Here the equations for this solution are given, its existence and the conditions of its Pareto optimality are proved. Examples are considered.
The section 1 contains definitions. In section 2 the problem with one variable is considered. In subsection 2.1 from the principle (the axiom) of consistency the equations for the solution of the problem with the linear criteria (for the segment) are established. In subsection 2.2 this solution with using of the axiom of the additivity is generalized on the nonlinear problem with one variable (or for the curve). In subsection 2.3 the conditions of the Pareto optimality are established. In subsection 2.4 Liapunov’s function is given.
In section 3 the general problem is considered. In subsection 3.1 from the solution for the problem with one variable with using of the equilibrium axiom the basic equations are established. In subsection 3.2 the existence theorem is proved. In subsection 3.3 the properties of the solution are given, in subsection 3.4 the problem with the linear criteria is considered, in the subsection 3.5 the problem with the concave criteria is considered, in the subsection 3.6 the examples are given and in subsection 3.7 the non-cooperative game connected with the multicriteria problem is considered. In section 4 (conclusion) some features of the solution are discussed.
1. Preliminaries
For x,y G R" we shall write x > y if Xj > yj, j = 1,...,n. Let X C R" be a set.
Definition 1. A point x G X is called Pareto optimal (or effective) in the set X if do not exists a point y G X such that y > x, y = x. The set of Pareto optimal points of the set X we shall denote E(X). If E(X) = X, then we say that the set X is Pareto optimal.
Let X C R" be a set, f : X ^ Rm be a map.
Definition 2. The pair P = (X, f) is called multicriteria problem, or merely problem. The components fi, i = 1,...,m, of the map f are called criteria of the problem
P.
Definition 3. A point x G X is called Pareto optimal (or effective) in the problem P if do not exists a point y G X such that f (y) > f (x), f (y) = f (x). The set of the Pareto optimal points of the problem P will be denoted by E(P) = E(X, f). If E(P) = X, then we say that the problem P is Pareto optimal.
Obviously, £(X) = £(X, f) where f (x) = x.
Definition 4. A map s such that s(P) G X is called a solution of the problem P. If s = s(P) is a solution of the problem P then the vector v = f (s) is called the vector value of the problem P or merely the value of the problem P.
The solution must be Pareto optimal, i.e. the following axiom must be fulfil:
A1. Axiom of Pareto optimality. s(P) G £(P).
It may be required to fulfil a stronger axiom: the solution must be determined only by the Pareto optimal set of the problem. More precise, let P = (X, f) be a problem. Consider the problem P = (£ (P), f), where f is the restriction f to £ (P). Obviously, £(P) = £(P). Then the axiom consists of the following:
A2. Strengthened axiom of Pareto optimality. s(P) = s(P). In the paper the conditions under which the axiom A1 and A2 are satisfied are given.
2. Case of One Variable (Solution for a Curve)
2.1. Solution for a Segment
Let x G Rn. Put x+ = (x+ ,...,x+), where x+ = max(xj, 0). Consider the following function:
— I'*'' II d \
INI ’
where || • || is an arbitrary norm in R".
In l2 norm the function (1) is the cosinus of the angle between the vector x and the positive ortant R+.
Function (1) has the following properties:
F1. The function p is continuous for all x G R", x = 0.
F2. 0 < p(x) < 1 for all x G R", x = 0, p(x) = 1 for x G R^, p(x) = 0 for x G R".
F3. p(x) + p(-x) > 1, and in li-norm p(x) + p(-x) = 1.
F4. For A > 0 p(Xx) = p(x).
Lemma 1. p is the nondecreasing function and for x G R+ and x G R" it is the strictly increasing function.
Proof.
In the lp-norm, where p > 1, we have from (1)
d^xS) =P(x+)p-1J2\xj\p -p|xj-|p“1signxj^(xt)p, j=1 / j j=1 j=1
j = 1,...,n.
Since
(x+ )p-i > \xj |p-1signxjpp(x), (2)
the following inequality holds:
dp(x)
dxj
0.
Moreover, if 0 < p(x) < 1 and xj = 0, from (2) it follows that
dp(x)
dxj
> 0.
The proof for p =1 we get by going to limit as p ^ 1. □
Consider the problem P = (X, f), where
X = [0,1], f (x) = a(1 - x) + bx, a, b G Rm . (3)
Definition 5. The point s(a, b) is called the solution of the problem (3) or the
solution to the segment [a, b] if it fulfils to the following axiom:
A3. Axiom of consistency
s(a, b) = a + p(b — a)(b — a) , (4)
where p is defined by (1) .
Remark 1. Equality (4) expresses the property of the consistency: the solution s continuously and monotonously depends on the rotation angle of the vector b — a around the point a.
Solution (4) has the following properties:
51. The function s is continuous on Rm x Rm.
52. s(a, b) G [a, b].
53. b) — all = ||(b — a)+||.
S4. s(a, b) is Pareto optimal for all a,b G Rm.
Remark 2. The solution defined in (4) is not symmetric: s(a,b) = s(b,a). Instead (4) we may consider the symmetric solution:
s{a, b) = ^(s(a, b) + s(b, a)) = a + —(1 + p(b - a) - p(a ~ b)) . (5)
From (5) it follows that for the symmetric solution function (1) must be replaced by the following function:
ps(x) = ^ (! + p(x) - p(-x)) , (6)
and ||x+|| must be replaced by
n(x) = i(||x|| + ||x+||-||x-||) . (7)
Note that in the l1-norm
ps (x) = p(x), n(x) = ||x+||.
For the sequel we shall use function (4).
Example 1. Let in problem (3)
to = 2, a = 0, \\b\\ = ^b\ + b\ = 1,
and the vector b rotates around origin. Then the solution s = s(0,b) describes the following curve:
S1 + s^ = 1, ii s1
s1 + si = 1, if s1 > 0, s2 > 0,
1 \ 2 1
[ S2 — —j = —, if si < 0, S2 > 0,
1 \ 2 1
si — — j + s\ = if si > 0, S2 < 0.
2.2. Solution for a Curve
Let X = [a, (3\ C R be a segment, f : [a, (3] ^ Rm be a set of criteria. Consider the problem
P =([a,3],f). (8)
Together with problem (8) consider the following curve
L = {y G Rm \ y = f (x), a < x < 3}. (9)
Let [a',3'] C [a, 3]. Put
L(a', 3') = {y G Rm \ y = f (x), a' < x < 3'}. (10)
By l(a',3') denote the length of the curve L(a',3'). It is known that
I'p'
l(a',3')= ||f'(x)lldx. (11)
J a
Let s(a',3') = x*, a' < x* < 3', be a solution for curve (10). Let us require that the solution satisfies the following axiom:
A4. Axiom of additivity. For all a < a' < 7 < 3' < 3 the equality must hold:
l(a\ s(a\ 3)) = l(a\ s(a\ y)) + l(Y, s(Y, 3')) . (12)
Note that equality (8) for the segment holds in view of (4), property F4 of the function p and property S3 of the solution.
Let x0 = a < x1 < ... < xk = 3 be a partition of the segment [a, 3] and let Lk
be a piecewise line corresponding to this partition, i.e.
Lk = [f (xo), f (x1)] U ... U [f (xk-1), f (xk)] .
From (8) and property S3 of the solution for the segment it follows that if sk is the solution for Lk and lk(sk) is the length of the part of Lk from f (a) to sk then the following equality fulfils:
k
lk(sk)=£ ll(f (xi) — f (xi-1))+1. (13)
i=1
Let s = x* be a solution for curve (9), l(s) = l(a,x*) be length (11). Going to the limit in (13) in view of (11) we get the following equation for x*:
rx* fP
/ ||f'(x)||dx = ||(f'(x))+ y dx. (14)
«/ a J a
Equation (14) is the extension of the property S3 of the solution to the segment. Equation (14) has the following interpretation.
The norm of the vector determines the length of the curve L by the formula:
P
' x)
||x| - l(L)=/ Uf'(x)| dx.
a
The norm of the positive part of the vector determines the “quasilength” of the curve by the formula:
rP
l|x+|| - l(L) = H(f'(x))+1 dx.
a
Equation (14) implies the equality
l(a, x*) = l(a, 3).
2.3. Conditions of the Pareto Optimality
Let in problem (8) the functions fi, i = 1,...,rn be strictly concave on [a, 3] and points xi, i = 1,...,m be determined by the following conditions:
fi (xi )= max fi(x), i = 1,...,m. (15)
xE[a,P]
Without loss of generality it can be assumed that
a < x1 < ... < xm < 3 . (16)
Lemma 2. If condition (16) holds then E(P) = [x1,xm].
Proof.
Let x* € [x1,xm]. If x < x* then fm(x) < fm(x*) and if x > x* then f1(x) < f1(x*). Hence, there does not exist x € [a,3] such that f (x) > f (x*), f (x) = f (x*). If x* € [a,x1) then f (x1) > f (x*), and if x* € (xm,3] then f (xm) > f (x*). □
Theorem 1. If the functions fiy i = 1,...,m are strictly concaved then the solution x* of equation (14) satisfies the axiom A2.
Proof.
From the conditions of the theorem and (15), (16) it follows that f '(x) > 0 for x € [a, x1) and f'(x) < 0 for x € (xm, 3]. Hence equation (14) is reduced to
f X* !■ Xm
/ Uf'(x)U dx = ||(f'(x))+ || dx.
J X1 j X1
The theorem follows from lemma 1. □
Corollary 1. If m =1 and f is strictly concave then the solution x* of equation (14) is the maximum of the function f.
Example 2. Consider problem (8), where m = 2, a = 0, f (x) = (f1(x),f2(x)), f (0) = (0, c) and assume that f1 (x) > 0, f2 (x) < 0. Obviously, the all points of this curve are Pareto-optimal. Equation (14) in l1-norm is reduced to the equation
f1(x*) — f2(x* ) = f1(3) — c, (17)
and in l2-norm to the equation
£ ^№))2 + №))2^ = /i(/3)- (is)
Let /i(x) = x, /2(x) = c — ^x2. From equation (17) we get x* = a/2/3 + 1 — 1, and from equation (18) we get the following equation for x* :
x* ■s/1 + x*2 + ln(x* + ■s/1 + x*2 ) = 2/3.
Let us compare this solution with Nash’s solution for bargaining problem (Nash, 1950). For this aim we must solve the following problem:
f1(x)f2(x) ^ max, 0 < x < 3.
Taking into account that c = /32/2 we get x* = /3/a/3.
Example 3. Let in problem (8) X = [0,n/2], f (x) = (cosx, sinx). We have f '(x) = (— sin x, cos x). Equation (14) in the l1-norm is reduced to
p x (‘n/2
/ (sin x' + cos x') dx' = / cos xdx. J0 J0
Hence, x* = 7t/4, f(x*) = (1/a/2, 1/a/2).
In the l2-norm equation (14) is reduced to
p x (‘n/2
/ dx = cos xdx.
00
Hence, x* = 1, f (x*) = (cos1, sin1).
For symmetric solution in the /2-norm (see (5), (6), (7)) we have x* = 7t/4, f(x*) = (l/V2A/V2).
For Nash’s solution we have the problem:
n
sin x cos x —>■ max, 0 < x < —.
, _ _ 2
Solving this problem we get x* = n/4.
2.4. Lyapunov’s function
Consider the following function:
r x r P
F(x)= (x — t)Hf'(t)H dt — (x — a) Kf'^y | dt. (19)
«/ a J a
The equation F'(x*) = 0 is equation (14). Moreover,
F''(x) = ||f'(x)|| > 0.
Therefore, function (19) is Lyapunov’s function for problem (3) (Lyapunov, 1907).
3. The general case
3.1. The basic equation
Let X C M" be a convex compact set such that dim X = n, f : X ^ Rm is a set of criteria. Consider the problem
P =(X,f). (20)
For x € M" we shall write x = (xj, x-j), where xj is the j-th component of x and
x_j is the set of other components. For x € X put
aj(x-j) = min{xj | (xj,x-j) € X}, bj(x-j) = max{xj | (xj,x-j) € X}. (21)
Note that in view of the made assumptions about the set X functions (21) are well defined, continuous and aj (x-j) < bj (x-j) a.e. x € X.
Take an arbitrary point x € X, choose j and fixe x-j. Consider the problem with one variable
Pj(x-j) = ([aj(x-j), bj(x-j)], f (',x-j)).
(22)
Let us require that the solution of problem (20) satisfies the following axiom:
A5. Axiom of equilibrium. The point x* = s(P) € X is a solution of the problem P if the following equalities hold:
s(pj(x* j)), j = 1,...,n,
(23)
i.e. xj is a solution of problem (22), j = 1,...,n.
Remark 3. This definition expresses the principle of the equilibrium: equations (23) are analogous to Nash’s equilibrium points in the non-coopertive games.
Applying equation (14) to each variable xj , we obtain the following system for solution x* :
j(x-j)
dxj
[■bj (x-j) /
)a,j (x* j)
dxj, j =1,...,n.
(24)
3.2. The existence theorem
Theorem 2. If X is a convex compact set such that dimX = n and the functions f are continuously differentiable, then equations (24) have the solution x* € X.
The proof is based on the following lemma. For x € X put
Q(x) = {y € M" | aj (x-j) < yj < bj (x-j), j = 1,...,n},
(25)
where aj(x-j) and bj(x-j), are determined by (21) and let
Q = {x € M" | a < x < b} (26)
be a minimal parallelepiped containing the set X.
Lemma 3. Let $ : X ^ 2R be a multivalued map such that the following conditions hold:
1. The map $ is upper semicontinuous.
2. $(x) is a convex set for all x € X.
3. $(x) C Q(x) for all x € X.
Then the map $ has in X a fixed point x*: x* € $(x*).
Proof.
Let n be the projection operator on the set X. Determine the map ^ : Q ^ 2Q, where Q is the parallelepiped (26), by the equality ^ = $n. Obviously, the map ^
x
+
x
satisfies the conditions 1 and 2 of the lemma and, hence, has a fixed point x* G Q: x* G *(x*). Prove that x* G X.
Assume the converse: x* /X. Then c = n(x*) — x* = 0. Consider the hyperplane
cx = cn(x*).
We have cx* < cn(x*) and for all x G X
cx > cn(x*). (27)
Show that inequalities (27) hold for all x G Q(n(x*)). Since this contradicts the condition 3, the lemma will be proved.
Since n(x*) is the solution of the problem
||x — x* || ^ min, x G X,
for all x G X the following inequalities hold
c(x — x*) > ||c||2.
From this inequalities and also from (25) it follows that for j = 1,...,n
i(x*) =
aj (n-j (x*)), bj (n-j(x*)),
aj (n-j(x*)) < nj(x*) < bj (n-j(x*))
if cj > 0,
if cj < 0,
if cj = 0.
From this conditions the required proposition follows.
Proof of theorem 2. Consider the map $ : X ^ R", where y = $(x) is determined from the system
j-Vj
'a3 (x-j)
dxj =
[■bj (x-j) 'aj (x-j)
dxj ,j = 1,...,n.
Obviously, the map $ satisfies the conditions of lemma 2 and, hence, has a fixed point x* which is a solution of system (24).
3.3. The properties of the solution Property 1. If in problem (20)
f(x) = hj (xj),
(28)
j=i
then from equations (24) it follows that the solution satisfies the following equations:
rx .. f bj (x-j)
/ \\h'j(xj). dxj =
■>aj(x* j) Jaj(x* j)
(hj (xj)) + dxj, j = 1,...,n. (29)
+
Property 2. If in problem (20) for some j = 1,...,n
J^-(x) > 0, (30)
then from equations (24) it follows that
x* = bj (x-j). (31)
Corollary 2. If inequalities (30) are satisfied for all j = 1,..., n, then from (31) it follows that
x G E(X),
and every point of E(X) is the solution of equations (24).
3.4. The problem with linear criteria
Let in problem (20)
"
f (x) = Cx cjxj, (32)
j=i
where C is a m x n matrix and cj = (cij ... Cm.), j = 1,...,n are its columns. Since
f has form (28) equations (29) are reduced to the following ones:
xj = aj (x-j )(1 — p(cj)) + bj (x-j )<p(cj), j = 1,...,n. (33)
If X is parallelepiped (26) then
xj = aj (1 — p(cj)) + bj p(cj), j = 1,...,n. (34)
If X is a unique cube, i.e. in (26) aj = 0, bj = 1, j = 1,...,n, then xj = p(cj),
j = 1,...,n, and
""
v = f (x) =Yj cjxj = Yj s(cj),
j=i j=i
i.e. value of the problem is the sum of the values for the consisting vectors. Example 4. Let X C R2 be a rectangular with the vertex (1,0), (0,1), (2,3), (3,2),
fl(x) = xi + x2, f2(x) = —xi — x2.
Find the solution in the /1-norm. Since c1 = c2 = (1, —1), from (1) it follows that
p(cl) = p(c?) = 1/2, and by (33) we have
xi = i(a i(x2) + b i(x2)), x2 = ^(a2(xi) + 62(xi)).
Obviously, the solutions of this equations are the points of the segment L with
the ends (1,1) and (2,2). Put y = x1 + x2, for x = (x1 ,x2) G L. Then 2 < y < 4.
Put f1(y) = y, f2(y) = —y and solve the problem ([2,4], f) by (34) (or merely by (4) for the segment [(2,-2),(4,-4)]). We get y = 3, x1 = x2 = 3/2, v = (3, —3).
Example 5. Let in problem (20)
"
X = {x £ R” I J2 J- < 1},
j=1 j
where dj > 0, j = 1,...,n, and f are criteia (32). By (21) we have
= °> bj(x-j) = di (i - 53 ;r) ’ j = ''' ’n'
k=j k
From equations (33) we get
k=j dk
Put
"
A = 1
hdk
Then equations (35) may be written in the following form
x
(1 -<p(c?))-£=<p(c?)A, (36)
Case 1. There exists such j, that p(cj) = 1. In this case, obviously, A = 0, e.g.
E!=>'
dj
j=1
Let p(cj) = 1 for j = 1,...,k and p(cj) < 1 for j = k +1,. ..,n, where k > 1. Then from equations (36) it follows that
xj = 0, for j = k + 1, ...,n (37)
and
tXi = l- (38)
dj
j=1
Let k > 1. We get a new problem V = (X, f), where in view of (37), (38)
X = x Rk
k I k
xj
dj
3=1 J 3=1
1.1. Lagrange’s method. Introduce Lagrange’s function for problem (39):
k X
/(x, A) = 53 — ed_)Xj
j=1 dj
where e G Rk, e = (1,..., 1). Put
K = {x G Rk | 0 < xj < dj, j = 1,...,k}
and consider the problem (K, l(-,X)), where X is the parameter. Solving this problem by (34), we get
Xj = djLp(cj - e^). (40)
By property F4 of the function p equation (40) may be written in the following form:
xj = dj p(cj dj — Xe), (41)
where X is determined from condition (38):
k
dj — Xe) = 1. (42)
j=1
From lemma 1 it follows that equation (42) has the unique solution.
1.2. The method of excluding of the variable. In the problem V (39) exclude the variable xk. We have
k-1
d
xk = dk 1 - ^2 -j- . (43)
j=1
Substituting in (39) for xk (43) we get the problem V = X, f, where
k-1
X = < x e R
k-1
k-1
i xj
f = ^2 i xj + dk ck, (45)
j=1
and
? = j = (46)
dj
From (21) we have
If p(cj) < 1 for all j = 1,...,k — 1, then we get case 2 for the problem P ((44), (45), (46) (47)) (with replacing k — 1 by n and cj by cj).
If there exists j such that p(cj) = 1 then the process of excluding of the variable is repeated.
Case 2. For all j = 1,...,n p(cj) < 1. From (36) we get
P(cj )dj
j 1 — P(cj)
where A is determined from the equation
j=1
x1 x^i
Note that the vector ( — ,-p-, A) G R"+1 is the optimal strategy in the diag-
d1 dn
onal matrix game
and A is its value [Vorob’ev, 1985].
3.5. The problem with the concave criteria
Let in problem (20) X = Rn and the criteria satisfy following condition:
Condition C. The functions fi, i = 1,...,m, are strictly concave and have their maxima on Rn.
Lemma 4. If the functions fj, i = 1,...,m satisfy condition C, then for p G R™, p = 0, the function g = pf = ^2,™= 1 Pifi also satisfies condition C.
Proof.
It is sufficient to prove the lemma for g = f1 + f2. Obviously g is strictly concave. Show that g has the maximum. Without loss of generality it can be assumed that f1(x) < 0. Let Sp = {x G Rn | fj(x) > fi}, i = 1,2, and Sp = {x G Rn | g(x) > f} be Lebesgue’s sets, accordingly, of the functions fi, 1 = 1, 2 and g. From condition C it follows that the sets Sp, i = 1, 2, are bounded and non-empty for f < min{max f1, maxf2}. Since f1(x) < 0, g(x) < f2(x) and Sp C Sp,. Hence, Sp is bounded and non-empty and g has the maximum. □
With
m
Am = {p g r™ ^ Pi = 1}
i=1
denote a simplex in Rm.
Let P = (Rn,f) be a problem and condition C holds. It is known [Podinovsky and Noghin, 1982] that for x0 G E(P) it is necessary and sufficient, that for some p G Am x0 is the solution of the following problem:
pf (x) ^ max, x G Rn. (48)
From condition C and lemma 4 it follows that problem (48) has the unique solution
for every p G Am. Denote this solution with x(p). Then E(P) = x(Am). Put
g(p) = f (x(p)) and consider the problem
P = (Am,g). (49)
Introduce Lagrange’s function for problem (49):
l(P, A) = g(p) - e\
where e G Rm, e = (1,..., 1).
Consider, finally, the third problem:
Pl = (K,l), (51)
where K is the unit cube and l is function (50). In problem (51) A is the parameter. Solving problem (51) by (24) we get the solution p(A) as the function of A. The A is found from condition p(A) G Am, that is
m
5>(A) = 1. (52)
i=1
Remark 4. The solution of problem (49), (51) at the same time gives us the weights pi, i = 1,... ,m, of the criteria.
Remark 5. Consider two problems: (49) and P0 = (E(P),f). This problems are not equivalent: if p* is the solution of the problem (49) and x* is the solution of the problem P0 then x* = x(p*).
Example 6. Let
n
fi(x) = -53 (xj - dj>)2, i = 1,...,m.
j = 1
Solving problem (48) we get
i=1
pi - 1
(50)
m
x(p) ^53Pidi,
i=1
where d1 = (d\,..., dn), i = 1,...,m. This means that
E(P) = convjd1, ...,dm}.
Put m = n +1,
n
f (x) = - V" x2, fi (x) = - V" x2 - (xi - ai)2, ai > 0, i = 1,...,n.
j = 1 j=i
We have
E(P) = { x G R Equalities (21) are reduced to
1 aj j=1 J
= bj(x-j) = ai (1 ) > i = i,•••,«•
k=j
Put
Then
Since
A = i-Y±L.
, aj. j=1 J
bj (x—j) = aj A + xj, j = 1,...,n.
1 df
equations (24) are reduced in l1-norm to the following equations:
naj A+Xj
(aj + (n - 1)xj) dxj = / (aj - xj) dxj, j = 1,...
From this equations we find
\/ n2 — n + 1 — 1
n( n - 1)
aj, j 1,.. .,n.
3.6. The examples
Example 7. Let us find a solution for the Pareto optimal part of the sphere. Consider problem (20), where
X = {x G R+ x2j = 1}, fi(x) = xi, i = 1 ,...,n. (53)
j=1
X
0
0
1. The method of the excluding of the variable. We have
n-1
1 -12 4 j=1
Consider the problem V — (X, f), where
n-1
X — {x e R+ xj ^ 1}, fi(x) — Xi, i — 1,...,n- 1, fn =
j=1
From (21) we have
aj (x-j)=0, bj (x-j) Hence, we get
n-1
1-12xj-
j=1
1-
n
2(x)+xj, j =1,...,n - 1. (54)
k=j
- f 0, if ,n, j, — '
1, if i = j-, i —1,.. .,n, j — 1,. • . ,n - 1
dXj j X3 - fn{x)’ if i — n,
Note that
fn(0,x-j ) = bj (x-j).
In the /1-norm equations (24) are reduced to the following equations
r bj (x-j)
1 +
dfn ( , ,
dxj —
dxj,
(55)
xj - fn(xj ,x-j ) + fn(0, x-j) = bj (xj), j =1,...,n - 1.
In view of (55) and finally
xj = fn (x), j = 1,...,n- 1,
x1 — ... — xn —
In the /2-norm equations (24) have the following form
xf2 f bj (x-j)
1 + —---------------dx^ =
/0 v fn(xj ,x-j) j
/0
dxj.
x
n
2
0
or
In view of (54) we have r xj
Jo \
x'2 pxj b . (x •)
1 + —--------dxj = —, 3 3 dxj = bj(x-j) arcsin , . . .
fli^x-j) 3 Jo Jmx_j)-xf 0 b^-i)
Hence,
arcsin , . 3—- = 1, j = I,... ,n — 1.
W*-i)
From this equations we find tg1
^ j — 1,...,n - 1, x
3 \Jl + (n - l)tg2l’ ’ ’ ’ " x/l + (n - l)tg2l
Let us find the symmetric solution in l2-norm (see remark 2 eq. (5), (6), (7)). We have _________________
p x j I x ^ 2
+ ^-----3------dxj
f2(xj ,x-j)
j
1 rbj (x-j) f j x
2 Jo VV fn(.xi>x-j)
j
dxj.
Integrating we find
n 8
or
xj = 6j(x-j)sin-, j = I,... ,n — 1,
Cj = fn{x)tg| = fn(x)(V2 - 1), j = 1, . .., n - 1.
Finally,
V2-1 . , , 1
- =, j = l, xn = --- .
!(n- 1)(3-2a/2) + 1 y(n- 1)(3-2a/2) + 1
2. Lagrange’s method. Introduce Lagrange’s function for problem (53):
1n
l(x, A) = /(x) - -eA(Vi] - 1),
2 j=1 j
and consider the problem V — (K,l). Obviously, the problem V satisfies property 1 (subsection 3.3) and by the symmetry of equations (29) the solution is given by (56).
Example 8. Let us find the solution for the Pareto optimal part of the sphere determined paramertically. Consider problem (20) where
n -
X = {x = (xux2) e R+ | 0 < xi; x2 < —}
1
0
f.(x) — cos x1 cos x2, f2(x) — cos x1 sin x2, f3(x) — sin x1.
In the l1-norm equations (24) have the following form:
/■ X1 i-n/2
I (sin x[ (cos x2 + sin x2) + cos x[) dxj — cos x1 dx1,
00
/■ X2 en/2
/ cos x1(sin x2 +cosx2) dx2 — cos x1 cos x2 dx2.
00
From the second equation it follows that sinx2 = cosx2 = 1/a/2 and /i(x) /2(x) = —-j=—. From the first equation we get
2
sin x1
— a/2 cos xi — 1 — a/2.
Obviously, this equation has a unique solution in [0,n/2]. In the l2-norm equations (24) take the following form:
fXl r-n/2 pX2 t-n/2
/ dx[ —I cos x1 dx1, cos x1 dx2 — cos x1 cos x2 dx2.
0 0 0 0
From this equations we have
x1 — x2 — 1, f1(x) — cos2 1, f2(x) — sin 1 cos 1, f3(x) — sin 1.
For symmetric solution (see remark 2 eq. (6), (7), (8)) in the l2-norm equations (24) are reduced to the following equations:
f'x l 1 f-n/2
dx\ = -Io 1 2 jo
r n/2
(1 + cos x1 - sin x1 ) dx1 ,
0
px2 1 rn/2
COSXi dx'2 = - (cosxi + COSXi COSX2 — COSXi sin x2)dx2.
Jo 2 J0
00 From this equations we have
n 11
Xl= x2 = f i(x) = /2(x) = /3(x) = -J=.
3.7. The non-cooperative game
In the subsection 2.4 Liapunov’s function for the problem with one variable was been constructed. In the general case there may be constructed the non-cooperative game (see [Vorob’ev, 1985]) with n players that is equivalent to the multicriteria problem.
By analogy with function (19) introduce the following functions
Fj (x)
- (xj - aj (x-j))
■Jaj (x-j ) rbj (x-j ) 'aj (x-j )
(xj - t)
df
dt—
(57)
df
dt, j — 1,...,n
and consider the following non-cooperative game:
r — {J, X, {Fj \ j e J}),
(58)
where J — {1, ...,n} is the set of the players, X is the set of the feasible vectors of the strategies, and Fj is the payoff function of the player j e J determined by (57). Note that in game (58) the players want to minimize their payoffs.
Theorem 3. 1. Game (58) with payoff function (57) has an equilibrium point in X.
2. The set of the equilibrium points in game (58) coincides with the set of the solutions of system (24).
Proof. 1. Since
d2F,
dx]
(x)
df ( ,
the functions Fj are convex in xj for j — 1,..., n. Therefore,
r^j (x-j) — argmin.
xj e [aj (x-j) b (x-j)]Fj (xj
^), j — 1,...:
are convex sets. Determine the map $ : X ^ 2R™ by the equality:
$(x) — ^j (x-j).
j=1
Obviously, the map $ satisfies the conditions of lemma 3 and, hence, has in X a fixed point x* — $(x*). By the definition $ we have
Fj(x* ,x*-j)
min
xj :(xj ,x-j)eX
Fj(xj,x*j), j — 1,...,n.
Hence, x* is the equilibrium point in game (58).
2. Since Fj, j — 1,...,n are convex in xj, the equilibrium point x* satisfies the following equations:
OF)
dxj
This system coincides with system (24).
x
+
4. Conclusion
Let us note some features of the proposed solution.
1. The solution of equations (24) may be not Pareto optimal, as it is seen from examples 4, 5 (subsection 3.4). Here two approaches are possible: a) the determining of the conditions for the Pareto optimality of the solution; b) the restriction of the consideration only the Pareto optimal problems.
2. The solution of equation (24) may be not unique, as it is seen from corollary
1 (property 2, subsection 3.3) and examples 4, 5 (subsection 3.4). In this case it is possible to iterate equations (24), as in example 4, 5.
3. The solution of equations (24) depends on the selected norm and also on the method of the solution: excluding the variable or Lagrange’s method.
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