Return to Article Details Approximations of objective function and constraints in bi-criteria optimization problems

Approximations of objective function and constraints in bi-criteria optimization problems

Ionut Traian Luca Dorel I. Duca

June 14, 2018. Accepted: October 23, 2018. Published online: February 7, 2019.

Faculty of Business, Babes-Bolyai University, e-mail: ionut.luca@tbs.ubbcluj.ro

Faculty of Mathematics and Computer Science, Babes-Bolyai University, e-mail: dduca@math.ubbcluj.ro

In this paper we study approximation methods for solving bi-criteria optimization problems. Initial problem is approximated by a new one which has the components of the objective and the constraints replaced by their approximation functions. Components of the objective function are first and second order approximated and constraints are first order approximated. Conditions such that efficient solution of the approximate problem will remain efficient for initial problem and reciprocally are studied. Numerical examples are developed to emphasize the importance of these conditions.

MSC. 90C46, 90C59

Keywords. efficient solution, bi-criteria optimization, η-approximation, invex and incave function

1 Introduction

Bi-criteria optimization problems are quite often used as mathematical models for all kind of phenomena generated by real-world and theoretical situations. As examples we might mention portfolio theory [ 4 ] , energy field [ 5 ] , data analysis [ 3 ] , logistics [ 6 ] .

Among methods widely used to solve bi-criteria optimization problems are “scalarization" methods [ 2 ] (weighting problem, kth objective Lagrangian problem, kth objective ε-constrained problem). Sometimes mathematical models are highly complex and thus using approximation problems might be a more efficient method to solve bi-criteria optimization problems.

This article is analyzing conditions such that efficient solution of a certain approximate problem will remain efficient for the initial problem and reciprocally. Approximate problem consists of replacing components of objective function and also constraints with their approximate functions.

2 Basic concepts

Let X be a set in Rn, x0 an interior point of X, η:X×XX and f:XR. If f is differentiable at x0 then we denote:

F1(x)=f(x0)+f(x0)η(x,x0)

and call it first η-approximation of f, while if f is twice differentiable at x0 then we denote:

F2(x)=f(x0)+f(x0)η(x,x0)+12η(x,x0)T2f(x0)η(x,x0).

and call it second η-approximation of f.

Definition 1

Let X be a nonempty set of Rn, x0 an interior point of X, f:XR a function differentiable at x0 and η:X×XX. Then function f is: invex at x0 with respect to η if for all xX we have:

f(x)f(x0)f(x0)η(x,x0)

or equivalently:

f(x)F1(x);

incave at x0 with respect to η if for all xX we have:

f(x)f(x0)f(x0)η(x,x0)

or equivalently

f(x)F1(x);

avex at x0 with respect to η if it is both invex and incave at x0 w.r.t. η.

If function f is invex, respectively incave or avex we denote invex1, respectively incave1 or avex1.

Definition 2

Let X be a nonempty set of Rn, x0 an interior point of X, f:XR a function twice differentiable at x0 and η:X×XX. Then function f is:
second order invex at x0 with respect to η if for all xX we have:

f(x)f(x0)f(x0)η(x,x0)+12η(x,x0)T2f(x0)η(x,x0)

or equivalently:

f(x)F2(x);

second order incave at x0 with respect to η if for all xX we have:

f(x)f(x0)f(x0)η(x,x0)+12η(x,x0)T2f(x0)η(x,x0)

or equivalently:

f(x)F2(x);

second order avex at x0 with respect to η if it is both second order invex and second order incave at x0 w.r.t. η.

If function f is second order invex, respectively second order incave or second order avex we denote invex2, respectively incave2 or avex2.

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

We consider the bi-criteria optimization problem (P00,0), defined as:

{min(f1,f2)(x)x=(x1,x2,...,xn)Xgt(x)0, tThs(x)=0, sS.

Assuming that functions f1,f2, are differentiable of order i, j{1, 2} and functions gt, (tT), hs, (sS) are first order differentiable, we will approximate original problem (P00,0) by problems (P1i,j):

{min(F1i,F2j)(x)x=(x1,x2,...,xn)XGt1(x)0, tTHs1(x)=0, sS

where (i,j){(1,0), (1,1), (2,0), (2,1), (2,2)} and F10=f1, F20=f2.

We denote by

Fk={xX: Gtk(x)0, tT, Hsk(x)=0, sS, k{0,1}}

the set of feasible solutions for bi-criteria optimization problem (Pki,j), where (i,j){(1,0), (1,1), (2,0), (2,1), (2,2)} and k{0,1}.

3 Approximate problems and relation to initial problem

In this section we will study the conditions such that efficient solution of approximated problems (P11,0), (P12,0), (P12,1) and (P12,2) will remain efficient also for initial problem (P00,0) and reciprocally.

Conditions for the relation (P00,0) vs. (P11,1) have been studied in [ 1 ] so we will not analyze them anymore.

By approximating also the feasible set it is important to determine conditions such that F0F1 and F1F0. These inclusions were studied in [ 1 ] . We will use them in our work, so we will briefly present the Theorems stating these inclusions.

Theorem 3

[ 1 ] . Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, and gt, hs:XR,(tT, sS).

Assume that:

  • for each tT, the function gt is differentiable at x0 and invex1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

then

F0F1.

Theorem 4

[ 1 ] . Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, and gt, hs:XR,(tT, sS).

Assume that

  • for each tT, the function gt is differentiable at x0 and incave1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

then

F1F0.

Theorem 5

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F0,

  • for each tT, the function gt is differentiable at x0 and invex1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and invex2 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P12,0), then x0 is an efficient solution for (P00,0).

Proof â–¼
 x0 being an efficient solution for (P12,0), implies that
xF1 s.t. (F12(x),f2(x))(F12(x0),f2(x0)).

Conditions b) and c) imply that

F0F1

and thus

xF0 s.t. (F12(x),f2(x))(F12(x0),f2(x0)).
1

Let’s assume that x0 is not an efficient solution for (P00,0). Then

yF0 s.t. (f1(y),f2(y))(f1(x0),f2(x0))

which implies that yF0 s.t.

{f1(y)<f1(x0)f2(y)f2(x0)

or

{f1(y)f1(x0)f2(y)<f2(x0).

Because f1 is invex2 at x0 with respect to η we get F12(y)f1(y), yF0.
Because η(x0, x0)=0 we get f1(x0)=F12(x0). Thus from (2) we get that yF0 s.t.

{F12(y)<F12(x0)f2(y)f2(x0)

which contradicts (1) and from (3) we get that yF0 s.t.

{F12(y)F12(x0)f2(y)<f2(x0)

which contradicts (1).
In conclusion x0 is an efficient solution for (P00,0).

Proof â–¼

Theorem 6

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F1,

  • for each tT, the function gt is differentiable at x0 and incave1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and incave2 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P00,0), then x0 is an efficient solution for (P12,0).

Proof â–¼
 x0 being an efficient solution for (P00,0), implies that
xF0 s.t. (f1(x),f2(x))(f1(x0),f2(x0)).

Conditions b) and c) imply that

F1F0

and thus

xF1 s.t. (f1(x),f2(x))(f1(x0),f2(x0)).
4

Let’s assume that x0 is not an efficient solution for (P12,0). Then

yF1 s.t. (F12(y),f2(y))(F12(x0),f2(x0))

which implies that yF1 s.t.

{F12(y)<F12(x0)f2(y)f2(x0)

or

{F12(y)F12(x0)f2(y)<f2(x0).

Because f1 is incave2 at x0 with respect to η we get f1(y)F12(y), yF1.
Because η(x0, x0)=0 we get f1(x0)=F12(x0). Thus from (5) we get that yF1 s.t.

{f1(y)<f1(x0)f2(y)f2(x0)

which contradicts (4) and from (6) we get that yF1 s.t.

{f1(y)f1(x0)f2(y)<f2(x0)

which contradicts (4).

In conclusion x0 is an efficient solution for (P12,0).

Proof â–¼

Theorem 7

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F0,

  • for each tT, the function gt is differentiable at x0 and invex1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is differentiable at x0 and invex1 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P11,0), then x0 is an efficient solution for (P00,0).

Proof â–¼
Proof is similar to Theorem 5.
Proof â–¼

Theorem 8

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F1,

  • for each tT, the function gt is differentiable at x0 and incave1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is differentiable at x0 and incave1 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P00,0), then x0 is an efficient solution for (P11,0).

Proof â–¼
Proof is similar to Theorem 6.
Proof â–¼

Theorem 9

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F0,

  • for each tT, the function gt is differentiable at x0 and invex1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and invex2 at x0 with respect to η,

  • f2 is differentiable at x0 and invex1 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P12,1), then x0 is an efficient solution for (P00,0).

Proof â–¼
Proof is similar to Theorem 5.
Proof â–¼

Theorem 10

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F1,

  • for each tT, the function gt is differentiable at x0 and incave1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and incave2 at x0 with respect to η,

  • f2 is differentiable at x0 and incave1 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P00,0), then x0 is an efficient solution for (P12,1).

Proof â–¼
Proof is similar to Theorem 6.
Proof â–¼

Theorem 11

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F0,

  • for each tT, the function gt is differentiable at x0 and invex1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and invex2 at x0 with respect to η,

  • f2 is twice differentiable at x0 and invex2 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P12,2), then x0 is an efficient solution for (P00,0).

Proof â–¼
Proof is similar to Theorem 5.
Proof â–¼

Theorem 12

Let X be a nonempty set of Rn, x0 an interior point of X,η:X×XX, T and S index sets, f=(f1,f2):XR2 and  gt, hs:XR,(tT, sS) functions.

Assume that:

  • x0F1,

  • for each tT, the function gt is differentiable at x0 and incave1 at x0 with respect to η,

  • for each sS, the function hs is differentiable at x0 and avex1 at x0 with respect to η,

  • f1 is twice differentiable at x0 and incave2 at x0 with respect to η,

  • f2 is twice differentiable at x0 and incave2 at x0 with respect to η,

  • η(x0, x0)=0.

If x0 is an efficient solution for (P00,0), then x0 is an efficient solution for (P12,2).

Proof â–¼
Proof is similar to Theorem 6.
Proof â–¼

4 Numerical examples

In the above theorems, conditions referring to invexity, incavity or avexity of functions are essential to ensure that efficient solution of the initial problem remains efficient for the approximate problem and reciprocally. If those conditions are not fulfill it is possible either that efficient solution of initial problem remains efficient for the approximate problem (and reciprocally) or it does not remain efficient.

Example 1

Let the initial bi-criteria optimization problem (P00,0) be:

{min(x12x2;x1+x2)x1x2+10x1;x20

An efficient solution of problem (P00,0) is x0=(1, 1)F0 and the value of the objective function in x0 is f(1, 1)=(1, 2). First and second approximate functions for the components of the objective function in x0=(1, 1) are:

Fp1(x)=fp(x0)+fp(x0)η(x,x0),p{1,2}

and

Fp2(x)=fp(x0)+fp(x0)η(x,x0)+12η(x,x0)T2fp(x0)η(x,x0),p{1,2},

while first approximate functions for the constraint is:

Gt1(x)=gt(x0)+gt(x0)η(x,x0),t{1,2,3}.

Considering η(x,x0)=xx0 we get:

F1i(x)=F1i(x)=x12x2, i{0,1,2}
F2j(x)=F2j(x)=x1+x2, j{0,1,2}

and

G11(x)=x1x2+2,G21(x)=x1,G31(x)=x2

Consequently, the approximate problems (P1i,j), with (i,j){(1,0),(1,1),(2,0),(2,1),(2,2)} are:

{min(x12x2;x1+x2)x1x2+20x1;x20

Calculating the value of objective function for problem (P1i,j) in x=(0, 2)F1 we obtain:

(F1i,F2j)(0, 2)=(4, 2)<(1, 2)=(F1i,F2j)(1, 1)

where (i,j){(1,0), (1,1), (2,0), (2,1), (2,2)}, which proves that x0=(1, 1)F1 is not an efficient solution for approximate problem (P1i,j). â–¡

Example 2

Let the initial bi-criteria optimization problem (P00,0) be:

{min(x12+(x2π1)2; (x1+110)212(x2+1)2)x1sinx1+x20x15π20x1;x20

An efficient solution of problem (P00,0) is x0=(π2, 1+π2)F0 and the value of the objective function in x0 is f(π2, 1+π2)=(π22; π289π10199100).

To compute the approximate problem (P11,1) in x0 we have to calculate:

Fp1(x)=fp(x0)+fp(x0)η(x,x0), p{1,2}

and

Gt1(x)=gt(x0)+gt(x0)η(x,x0), t{1,2,3,4}

Considering η(x,x0)=xx0 we get:

F11(x)=πx1πx2+π+π22,
F21(x)=(π+15)x1(π2+2)x2π28+π2+1100,
G11(x)=x1+x21,
G21(x)=x15π2, G31(x)=x1, G41(x)=x2

Thus, the approximate problem (P11,1) is:

{min(πx1πx2+π+π22; (π+15)x1(π2+2)x2π28+π2+1100)x1+x210x15π20x1;x20

Calculating the value for the objective function of problem (P11,1) in x=(5π2, 1+5π2)F1 we get:

F1(5π2, 1+5π2)=(π22, 9π289π2199100)<(π22; π289π10199100)=F1(π2, 1+π2)

and thus we have proved that x0=(π2, 1+π2) is not an efficient solution for problem (P11,1). â–¡