Abstract
In a multi-dimensional Euclidean space there are several ways to classify elements of a given set of vectors.
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Keywords
- Multi-dimensional Euclidean Space
- Maximal Efficient Faces
- Pareto Maximal Points
- Efficient Solution Set
- Relative Interior Point
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
In a multi-dimensional Euclidean space there are several ways to classify elements of a given set of vectors. The componentwise order relation introduced in the very beginning of the second chapter seems to be the most appropriate for this classification purpose and leads to the concept of Pareto optimality or efficiency, a cornerstone of multiobjective optimization that we are going to study in the present chapter.
1 Pareto Maximal Points
In the space \(\mathbb {R}^k\) with \(k>1\) the componentwise order \(x\geqq y\) signifies that each component of x is bigger than or equal to the corresponding component of y. Equivalently, \(x\geqq y\) if and only if the difference vector \(x-y\) has non-negative components only. This order is not complete in the sense that not every couple of vectors is comparable, and hence the usual notion of maximum or minimum does not apply. We recall also that \(x>y\) means that all components of the vector \(x-y\) are strictly positive, and \(x\ge y\) signifies \(x\geqq y\) and \(x\not =y\). The following definition lays the basis for our study of multiobjective optimization problems.
Definition 4.1.1
Let Q be a nonempty set in \(\mathbb {R}^k\). A point \(y\in Q\) is said to be a (Pareto) maximal point of the set Q if there is no point \(y'\in Q\) such that \( y' \geqq y \) and \( y'\not =y.\) And it is said to be a (Pareto) weakly maximal point if there is no \(y'\in Q\) such that \( y' > y .\)
The sets of maximal points and weakly maximal points of Q are respectively denoted Max(Q) and WMax(Q) (Figs. 4.1 and 4.2). They are traditionally called the efficient and weakly efficient sets or the non-dominated and weakly non-dominated sets of Q. The set of minimal points Min(Q) and weakly minimal points WMin(Q) are defined in a similar manner. When no confusion likely occurs between maximal and minimal elements, the set Min(Q) and WMin(Q) are called the efficient and weakly efficient sets of Q too. The terminology of efficiency is advantageous in certain circumstances in which we deal simultaneously with maximal points of a set as introduced above and maximal elements of a family of subsets which are defined to be maximal with respect to inclusion. Thus, given a convex polyhedron, a face of it is efficient if it consists of maximal points only. When we refer to a maximal efficient face, it is understood that that face is efficient and maximal by inclusion which means that no efficient face of the polyhedron contains it as a proper subset. In some situations one is interested in an ideal maximal point (called also a utopia point), which is defined to be a point \(y\in Q\) that satisfies
Such a point is generally unattainable, and if it exists it is unique and denoted by IMax(Q) (Fig. 4.3).
Max and Min
WMax and WMin
IMax
Geometrically, a point y of Q is an efficient (maximal) point if the intersection of the set Q with the positive orthant shifted at y consists of y only, that is,
and it is weakly maximal if the intersection of Q with the interior of the positive orthant shifted at y is empty, that is,
Of course, maximal points are weakly maximal, and the converse is not true in general. Here are some examples in \({\mathbb {R}}^2\).
Example 4.1.2
Let Q be the triangle of vertices \(a=\left( \begin{array}{rr} 0\\ 0\end{array}\right) \), \(b=\left( \begin{array}{rr} 1\\ 0\end{array}\right) \) and \(c=\left( \begin{array}{rr} 0\\ 1\end{array}\right) \) in \({\mathbb {R}}^2\). Then \( \mathrm{{ Max}} (Q) = \mathrm{{ WMax}}(Q)=[b,c]\), \(\mathrm{{ Min}}(Q)= \{a\}\) and \(\mathrm{{WMin}}(Q)= [a,b]\cup [a,c].\)
Example 4.1.3
Let Q be the polytope in the space \({\mathbb {R}}^3\), determined by two inequalities
Then \(\mathrm{{ Max}}(Q) = \mathrm{{WMax}}(Q)=\emptyset \), \(\mathrm{{ Min}} (Q)= \emptyset \) and \(\mathrm{{ WMin}} (Q)= Q\setminus \mathrm{{ int}}(Q).\)
Existence of pareto maximal points
As we have already seen in Example 4.1.3, a polyhedron may have no weakly maximal points. This happens when some components of elements of the set are unbounded above. Positive functionals provide an easy test for such situations.
Theorem 4.1.4
Let Q be a nonempty set and let \(\lambda \) be a nonzero vector in \({\mathbb {R}}^k\). Assume that \(y\in Q\) is a maximizer of the functional \(\langle \lambda , .\rangle \) on Q. Then
-
(i)
y is a weakly maximal point of Q if \(\lambda \) is a positive vector;
-
(ii)
y is a maximal point of Q if either \(\lambda \) is a strictly positive vector, or \(\lambda \) is a positive vector and y is the unique maximizer.
In particular, if Q is a nonempty compact set, then it has a maximal point.
Proof
Assume \(\lambda \) is a nonzero positive vector. If y were not weakly maximal, then there would exist another vector \(y'\) in Q such that the vector \(y'-y\) is strictly positive. This would yield \(\langle \lambda , y'\rangle > \langle \lambda , y\rangle \), a contradiction.
Now, if \(\lambda \) is strictly positive, then for any \(y' \geqq y\) and \(y'\not = y\), one has \(\langle \lambda , y'\rangle > \langle \lambda , y\rangle \) as well. Hence y is a Pareto maximal point of Q.
When \(\lambda \) is positive (not necessarily strictly positive) and not zero, the above inequality is not strict. Actually, we have equality because y is a maximizer. But, in that case \(y'\) is also a maximizer of the functional \(\langle \lambda ,.\rangle \) on Q, which contradicts the hypothesis.
When Q is compact, any strictly positive vector \(\lambda \) produces a maximizer on Q, hence a Pareto maximal point too. \(\Box \)
Maximizers of the functional \(\langle \lambda ,.\rangle \) with \(\lambda \) positive, but not strictly positive, may produce no maximal points as seen in the following example.
Example 4.1.5
Consider the set Q in \({\mathbb {R}}^3\) consisting of the vectors \(x=(x_1,x_2,x_3)^T\) with \(x_3\leqq 0.\) Choose \(\lambda =(0,0,1)^T\). Then every element x of Q with \(x_3=0\) is a maximizer of the functional \(\langle \lambda , .\rangle \) on Q, hence it is weakly maximal, but not maximal, for the set Q has no maximal element.
Given a reference point a in the space, the set of all elements of a set Q that are bigger than the point a forms a dominant subset, called a section of Q at a. The lemma below shows that maximal elements of a section are also maximal elements of the given set.
Lemma 4.1.6
Let Q be a nonempty set in \(\mathbb {R}^k\). Then for every point a in \({\mathbb {R}}^k\) one has
Proof Let y be a Pareto maximal point of the section \(Q\cap (a+\mathbb {R}^k_+)\). If y were not maximal, then one would find some \(y'\) in Q such that \(y'\geqq y\) and \(y'\not =y\). It would follow that \(y'\) belongs to the section \(Q\cap (a+\mathbb {R}^k_+)\) and yield a contradiction. The second inclusion is proven by the same argument. \(\Box \)
For convex polyhedra existence of maximal points is characterized by position of asymptotic directions with respect to the positive orthant of the space.
Theorem 4.1.7
Let Q be a convex polyhedron in \({\mathbb {R}}^k\). The following assertions hold.
-
(i)
Q has maximal points if and only if
$$Q_\infty \cap {\mathbb {R}}^k_+=\{0\}.$$ -
(ii)
Q has weakly maximal points if and only if
$$Q_\infty \cap \mathrm{int}({\mathbb {R}}^k_+)=\emptyset .$$
In particular, every polytope has a maximal vertex.
Proof
Let y be a maximal point of Q and let v be any nonzero asymptotic direction of Q. Since \(y+v\) belongs to Q and \(Q\cap (y+\mathbb {R}^k_+) =\{y\}\), we deduce that v does not belong to \(\mathbb {R}^k_+\). Conversely, assume Q has no nonzero asymptotic direction. Then for a fixed vector y in Q the section \(Q\cap (y+\mathbb {R}^k_+)\) is bounded; otherwise any nonzero asymptotic direction of that closed convex intersection, which exists due to Corollary 2.3.16, should be a positive asymptotic vector of Q. In view of Theorem 4.1.4 the compact section \(Q\cap (y+\mathbb {R}^k_+)\) possesses a maximal point, hence, in view of Lemma 4.1.6, so does Q.
For the second assertion, the same argument as above shows that when Q has a weakly maximal point, no asymptotic direction of it is strictly positive. For the converse part, by the hypothesis we know that \(Q_\infty \) and \({\mathbb {R}}^k_+\) are two convex polyhedra without relative interior points in common. Hence, in view of Theorem 2.3.10 there is a nonzero vector \( \lambda \in {\mathbb {R}}^k\) separating them, that is
In particular, for \(v=0\) and for d being usual coordinate unit vectors, we deduce from the above relation that \(\lambda \) is positive. Moreover, the linear function \(\langle \lambda , .\rangle \) is then non-positive on every asymptotic direction of Q. We apply Theorem 3.1.1 to obtain a maximum of \(\langle \lambda , .\rangle \) on Q. In view of Theorem 4.1.4 that maximum is a weakly maximal point of Q.
Finally, if Q is a polytope, then its asymptotic cone is trivial. Hence, by the first assertion, it has maximal points. To prove that it has a maximal vertex, choose any strictly positive vector \(\lambda \in \mathbb {R}^k\) and consider the linear problem of maximizing \(\langle \lambda , .\rangle \) over Q. In view of Theorem 3.1.3 the optimal solution set contains a vertex, which, by Theorem 4.1.4, is also a maximal vertex of Q. \(\Box \)
In Example 4.1.5 a positive functional \(\langle \lambda ,.\rangle \) was given on a polyhedron having no maximizer that is maximal. This, however, is impossible when the polyhedron has maximal elements.
Corollary 4.1.8
Assume that Q is a convex polyhedron and \(\lambda \) is a nonzero positive vector in \({\mathbb {R}}^k\). If Q has a maximal point and the linear functional \(\langle \lambda ,.\rangle \) has maximizers on Q, then among its maximizers there is a maximal point of Q.
Proof
Let us denote by \(Q_0\) the nonempty intersection of Q with the hyperplane \(\{y\in \mathbb {R}^k: \langle \lambda , y\rangle = d\}\) where d is the maximum of \(\langle \lambda , .\rangle \) on Q. It is a convex polyhedron. Since Q has maximal elements, in view of Theorem 4.1.7 one has \(Q_\infty \cap \mathbb {R}^k_+=\{0\}\), which implies that \((Q_0)_\infty \cap \mathbb {R}^k_+=\{0\}\) too. By the same theorem, \(Q_0\) has a maximal element, say \(y_0\). We show that this \(y_0\) is also a maximal element of Q. Indeed, if not, one could find some \(y\in Q\) such that \(y\geqq y_0\) and \(y\not =y_0\). Since \(\lambda \) is positive, we deduce that \(\langle \lambda , y\rangle \geqq \langle \lambda , y_0\rangle =d \). Moreover, as y does not belong to \(Q_0\), this inequality must be strict which is a contradiction. \(\Box \)
We say a set Q in the space \({\mathbb {R}}^k\) has the domination property if its elements are dominated by maximal elements, that is, for every \(y\in Q\) there is some maximal element a of Q such that \(a\geqq y\). The weak domination property refers to domination by weakly maximal elements.
Corollary 4.1.9
A convex polyhedron has the domination property (respectively weak domination property) if and only if it has maximal elements (respectively weakly maximal elements).
Proof
The “only if” part is clear. Assume a convex polyhedron Q has maximal elements. In view of Theorem 4.1.7, the asymptotic cone of Q has no nonzero vector in common with the positive orthant \({\mathbb {R}}^k_+\). Hence so does the section of Q at a given point \(a\in Q\). Again by Theorem 4.1.7 that section has maximal points that dominate a and by Lemma 4.1.6 they are maximal points of Q. Hence Q has the domination property. The weak domination property is proven by the same argument. \(\Box \)
We learned in Sect. 2.3 how to compute the normal cone at a given point of a polyhedron. It turns out that by looking at the normal directions it is possible to say whether a given point is maximal or not.
Theorem 4.1.10
Let Q be a convex polyhedron in \(\mathbb {R}^k\). The following assertions hold.
-
(i)
\(y\in Q\) is a maximal point if and only if the normal cone \(N_Q(y)\) to Q at y contains a strictly positive vector.
-
(ii)
\(y\in Q\) is a weakly maximal point if and only if the normal cone \(N_Q(y)\) to Q at y contains a nonzero positive vector.
Proof
Let y be a point in Q. If the normal cone to Q at y contains a strictly positive vector, say \(\lambda \), then by the definition of normal vectors, the functional \(\langle \lambda , .\rangle \) attains its maximum on Q at y. In view of Theorem 4.1.4, y is a maximal point of Q. The proof of the “only if” part of (i) is based on Farkas’ theorem. We assume that y is a maximal point of Q and suppose to the contrary that the normal cone to Q at that point has no vector in common with the interior of the positive orthant \(\mathbb {R}^k_+\). We may assume that Q is given by a system of inequalities
The active index set at y is denoted I(y). By Theorem 2.3.24, the normal cone to Q at y is the positive hull of the vectors \(a^i, i\in I(y)\). Its empty intersection with int\((\mathbb {R}^k_+)\) means that the following system has no solution
where \(A_{I(y)}\) denotes the matrix whose columns are \(a^i, i\in I(y)\) and e is the vector whose components are all equal to one. By introducing artificial variables z, the above system is equivalent to the system
Apply Farkas’ theorem (Theorem 2.2.3) to obtain a nonzero positive vector v such that
The inequalities (4.1) corresponding to the inactive indices at y being strict, we may find a strictly positive number t such that
In other words, the point \(y+tv\) belongs to Q. Moreover, \(y+tv\geqq y\) and \(y+tv\not =y\) which contradicts the hypothesis. This proves (i).
As to the second assertion, the “if" part is clear, again, Theorem 4.1.4 is in use. For the converse part, we proceed the same way as in (i). The fact that the intersection of \(N_Q(y)\) with the positive orthant \(\mathbb {R}^k_+\) consists of the zero vector only, means that the system
has no nonzero solution. Applying Corollary 2.2.5 we deduce the existence of a strictly positive vector v such that
Then, as before, the vector \(y+tv\) with \(t>0\) sufficiently small, belongs to Q and \(y+tv >y\), which is a contradiction. \(\Box \)
Example 4.1.11
Consider a convex polyhedron Q in \({\mathbb {R}}^3\) determined by the system
We analyze the point \(y=(1/3, 1/3, 1/3)^T\in Q\). Its active index set is \(I(y)=\{1\}\). By Theorem 2.3.24 the normal cone to Q at that point is generated by the vector \((1,1,1)^T\). According to Theorem 4.1.10 the point y is a maximal point of Q. Now, take another point of Q, say \(z=(-1, 0, 1)^T\). Its active index set consists of two indices 2 and 5. The normal cone to Q at z is generated by two directions \((0,1,1)^T\) and \((0,0,1)^T\). It is clear that this normal cone contains no strictly positive vector, hence the point z is not a maximal point of Q because \(z^T\le (0,0,1)^T\). It is a weakly maximal point, however, because normal directions at z are positive. Finally, we choose a point \(w=(0,0,0)^T\) in Q. Its active index set is \(I(w)=\{4\}\). The normal cone to Q at w is the cone generated by the direction \((0,0,-1)^T\). This cone contains no positive vector, hence the point w is not weakly maximal. This can also be seen from the fact that w is strictly dominated by y.
Scalarizing vectors
In remaining of this section we shall use the terminology of efficient points instead of (Pareto) maximal points in order to avoid possible confusion with the concept of maximal element of a family of sets by inclusion. Given a family \(\{A_i: i\in I\}\) of sets, we say that \(A_{i_0} \) is maximal (respectively minimal) if there is no element \(A_i\) of the family such that \(A_i\not = A_{i_0}\) and \(A_{i_0}\subset A_i\) (respectively \(A_{i_0}\supset A_i\)). Another formulation of Theorem 4.1.10 is viewed by maximizing linear functionals on the set Q.
Corollary 4.1.12
Let Q be a convex polyhedron in \(\mathbb {R}^k\). Then the following statements hold.
-
(i)
\(y\in Q\) is an efficient point if and only if there is a strictly positive vector \(\lambda \in \mathbb {R}^k\) such that y maximizes the functional \(\langle \lambda , .\rangle \) on Q.
-
(ii)
\(y\in Q\) is a weakly efficient point if and only if there is a nonzero positive vector \(\lambda \in \mathbb {R}^k\) such that y maximizes the functional \(\langle \lambda , .\rangle \) on Q.
Proof
This is immediate from the definition of normal cones and from Theorem 4.1.10. \(\Box \)
The vector \(\lambda \) mentioned in this corollary is called a scalarizing vector (or weakly scalarizing vector in (ii)) of the set Q. We remark that not every strictly positive vector is a scalarizing vector of Q like not every strictly positive functional attains its maximum on Q. Moreover, an efficient point of Q may maximize a number of scalarizing vectors that are linearly independent, and vice versa, a scalarizing vector may determine several maximizers on Q. For a given polyhedron Q that has efficient elements, the question of how to choose a vector \(\lambda \) so that the functional associated with it furnishes a maximizer is not evident. Analytical choice of positive directions such as the one discussed in Example 4.1.11 is conceivable and will be given in details later. Random generating methods or uniform divisions of the standard simplex do not work in many instances. In fact, look at a simple problem of finding efficient points of the convex polyhedral set given by the inequality
in the two-dimensional space \({\mathbb {R}}^2\). Except for one direction, every positive vector \(\lambda \) leads to a linear problem of maximizing \(\langle \lambda , x\rangle \) over that polyhedron with unbounded objective. Hence, using positive vectors \(\lambda _i\) of a uniform partition
of the simplex \([(1,0)^T, (0,1)^T]\) of the space \({\mathbb {R}}^2\) for whatever the positive integer p be, will never generate efficient points of the set.
Any nonzero positive vector of the space \(\mathbb {R}^k\) is a positive multiple of a vector from the standard simplex \(\varDelta \). This combined with Corollary 4.1.12 yields the following equalities
where \(\mathrm{argmax}_Q \langle \lambda ,.\rangle \) is the set of all maximizers of the functional \(\langle \lambda ,.\rangle \) on Q. Given a point \(y\in Q\) denote
The set \(\varDelta _Q\) is called the weakly scalarizing set of Q and \(\varDelta _y\) is the weakly scalarizing set of Q at y (Fig. 4.4). By Corollary 4.1.12 the set \(\varDelta _y\) is nonempty if and only if the point y is a weakly efficient element of Q. Hence when Q has weakly efficient points, the set \(\varDelta _Q\) can be expressed as
in which every set \(\varDelta _y\) is nonempty. By definition a vector \(\lambda \in \varDelta \) belongs to \(\varDelta _y\) if and only if
The latter inequality signifies that \(\lambda \) is a normal vector to Q at y, and so (4.2) becomes
Scalarizing set at y
Let \(\mathcal {F}=\{F_1,\cdots ,F_q\}\) be the collection of all faces of Q and let \(N(F_i)\) be the normal cone to \(F_i\), which, by definition, is the normal cone to Q at a relative interior point of \(F_i\). Since each element of Q is a relative interior point of some face, the decomposition (4.2) produces the following decomposition of \(\varDelta _Q\):
where \(\varDelta _i = N(F_i)\cap \varDelta \) and I is the set of those indices i from \(\{1,\cdots , q\}\) such that the faces \(F_i\) are weakly efficient. We note that when a face is not weakly efficient, the normal cone to it does not meet the simplex \(\varDelta \). Remember that a face of Q is weakly efficient if all elements of it are weakly efficient elements of Q, or equivalently if a relative interior point of it is a weakly efficient element. A face that is not weakly efficient may contain weakly efficient elements on its proper faces.
We say a face of Q is a maximal weakly efficient face if it is weakly efficient and no weakly efficient face of Q contains it as a proper subset. It is clear that when a convex polyhedron has weakly efficient elements, it does have maximal weakly efficient faces. Below we present some properties of the decompositions (4.2) and (4.3) of the weakly scalarizing set.
Lemma 4.1.13
If P and Q are convex polyhedra with \(P\cap Q\not =\emptyset \), then there are faces \(P'\subseteq P\) and \(Q'\subseteq Q\) such that \(P\cap Q= P'\cap Q'\) and \(\mathrm{ri}(P')\cap \mathrm{ri }(Q') \not =\emptyset .\) Moreover, if the interior of Q is nonempty and contains some elements of P, then \(\mathrm{ri }(P) \cap \mathrm{int} (Q) \not =\emptyset \) and \(\mathrm{ri }(P \cap Q ) = \mathrm{ri }(P) \cap \mathrm{int} (Q). \)
Proof
Let x be a relative interior point of the intersection \(P\cap Q\). Let \(P'\subseteq P\) and \(Q'\subseteq Q\) be faces that contain x in their relative interiors. These faces meet the requirements of the lemma. Indeed, it suffices to show that every point y from \(P\cap Q\) belongs to \(P'\cap Q'\). Since x is a relative interior point of \(P\cap Q\), the segment \([x-\varepsilon (x-y), x+\varepsilon (x-y)]\) belongs to that intersection when \(\varepsilon >0\) is sufficiently small. Moreover, as \(P'\) is a face, this segment must lie in \(P'\) which implies that y lies in \( P'\). The same argument shows that y lies in \(Q'\), proving the first part of the lemma.
For the second part it suffices to observe that P is the closure of its relative interior. Hence it has relative interior points inside the interior of Q. The last equality of the conclusion is then immediate. \(\Box \)
Theorem 4.1.14
The weakly scalarizing set \(\varDelta _Q\) is a polytope. Moreover, if \(\varDelta _Q\) is nonempty, the elements of the decomposition (4.2) and (4.3) are polytopes and satisfy the following conditions:
-
(i)
If \(\varDelta _y=\varDelta _z\) for some weakly efficient elements y and z, then there is \(i\in I\) such that \(y,z\in F_i\) and \(\varDelta _y=\varDelta _z=\varDelta _i\).
-
(ii)
If \(F_i\) is a maximal weakly efficient face of Q, then \(\varDelta _i\) is a minimal element of the decomposition (4.3). Conversely, if the polytope \(\varDelta _i\) is minimal among the polytopes of the decomposition (4.3), then there is a maximal weakly efficient face \(F_j\) such that \(\varDelta _j=\varDelta _i\).
-
(iii)
For all \(i,j\in I\) with \(i\not =j\), one has either \(\varDelta _i=\varDelta _j\) or \(\mathrm{{ri}}(\varDelta _i) \cap \mathrm{{ri}}(\varDelta _j)=\emptyset \).
-
(iv)
Let \(F_i\) and \(F_j\) be two weakly efficient adjacent vertices (zero-dimensional faces) of Q. Then the edge joining them is weakly efficient if and only if \(\varDelta _{i}\cap \varDelta _{j}\not = \emptyset \).
Proof
Since \(\varDelta _y\) is empty when y is not a weakly efficient point of Q, we may express \(\varDelta _Q\) as
which proves that \(\varDelta _Q\) is a bounded polyhedron because the normal cone \(N_Q\) is a polyhedral cone. Likewise, the sets \(\varDelta _y= N_Q(y)\cap \varDelta \) and \(\varDelta _i= N(F_i)\cap \varDelta \) are convex polytopes.
To establish (i) we apply Lemma 4.1.13 to the intersections \(\varDelta _y= N_Q(y)\cap \varDelta \) and \(\varDelta _z=N_Q(z)\cap \varDelta \). There exist faces \(N\subseteq N_Q(y)\), \(M\subseteq N_Q(z)\) and \(\varDelta ^y, \varDelta ^z\subseteq \varDelta \) such that
Choose any vector \(\xi \) from the relative interior of \(\varDelta _y\). Then it is also a relative interior vector of the faces N, M, \(\varDelta ^y\) and \(\varDelta ^z\). This implies that \(N=M\) and \(\varDelta ^y=\varDelta ^z\). Using Theorem 2.3.26 we find a face \(F_i\) of Q such that \(N(F_i)=N\). Then \(F_i\) contains y and z and satisfies
For (ii) assume \(F_i\) is a maximal weakly efficient face. Assume that \(\varDelta _j\) is a subset of \(\varDelta _i\) for some \(j\in I\). We choose any vector \(\xi \) from \(\varDelta _j\) and consider the face \(F'\) consisting of all maximizers of \(\langle \xi , .\rangle \) on Q. Then \(F'\) is a weakly efficient face and contains \(F_j\) and \(F_i\). As \(F_i\) is maximal, we must have \(F'=F_i\). Thus, \(F_j\subseteq F_i\) and
Conversely, let \(\varDelta _i\) be a minimal element among the polytopes \(\varDelta _j, j\in I\). If \(F_i\) is maximal weakly efficient face, we are done. If it is not, we find a maximal weakly efficient face \(F_j\) containing \(F_i\). Then \(\varDelta _j=N(F_j)\cap \varDelta \subseteq N(F_i)\cap \varDelta =\varDelta _i\) and \(\varDelta _j=\varDelta _i\) by hypothesis.
We proceed to (iii). Assume that the relative interior of \(\varDelta _i\) and the relative interior of \(\varDelta _j\) have a vector \(\xi \) in common. In view of Lemma 4.1.13 one can find four faces: \(N^i\) of \(N(F_i)\), \(N^j\) of \(N(F_j)\), \(\varDelta ^i\) and \(\varDelta ^j\) of \(\varDelta \) such that
According to Theorem 2.3.26 there are faces \(F_\ell \) and \(F_m\) of Q which respectively contain \(F_i\) and \(F_j\) with \(N(F_\ell )=N^i\) and \(N(F_m)=N^j\). Then \(\xi \) is a relative interior vector of the faces \(N(F_\ell )\), \(N(F_m)\), \(\varDelta ^i\) and \(\varDelta ^j\). We deduce that the face \(\varDelta ^i\) coincides with \(\varDelta ^j\), and \(F_\ell \) coincides with \(F_m\). Consequently, \(\varDelta _i=\varDelta _j\).
To prove the last property we assume \(F_i\) and \(F_j\) are adjacent vertices (zero-dimensional faces) of Q. Let a one-dimensional face \(F_l\) be the edge joining them. According to Corollary 2.3.28 we have \(N(F_l)=N(F_i)\cap N(F_j)\). Then \(\varDelta _l=\varDelta _i\cap \varDelta _j\) which shows that \(F_l\) is weakly efficient if and only if the latter intersection is nonempty. \(\Box \)
Note that two different faces of Q may have the same weakly scalarizing set. For instance the singleton \(\{(0,0,1)^T\}\) is the weakly scalarizing set for all weakly efficient faces of the polyhedron \(\mathbb {R}^2_+\times \{0\}\) in \(\mathbb {R}^3\).
In order to treat efficient elements of Q we need to work with the relative interior of \(\varDelta \). Corresponding notations will be set as follows
The set \(\varDelta ^r_Q\) is called the scalarizing set of Q. It is clear that \(y\in Q\) is efficient if and only if \(\varDelta ^r_y\) is nonempty, and it is weakly efficient, but not efficient if and only if \(\varDelta _y\) lies on the border of \(\varDelta \). The decompositions of the weakly scalarizing set induce the following decompositions of the scalarizing set
and
where \(I_0\) consists of those indices i from \(\{1,\cdots ,q\}\) for which \(F_i\) are efficient.
Theorem 4.1.15
Assume that the scalarizing set \(\varDelta ^r_Q\) is nonempty. Then
Moreover the decompositions (4.4) and (4.5) of \(\varDelta ^r_Q\) satisfy the following properties:
-
(i)
If \(\varDelta ^r_y=\varDelta ^r_z\) for some efficient elements y and z, then there is \(s\in I_0\) such that \(y,z\in \mathrm{ri }(F_s)\) and \(\varDelta ^r_y=\varDelta ^r_z=\varDelta ^r_s\).
-
(ii)
For \(i\in I_0\) the face \(F_i\) is a maximal efficient face if and only if \(\varDelta ^r_i\) is a minimal element of the decomposition (4.5).
-
(iii)
For all \(i,j\in I_0\) with \(i\not =j\), one has \(\mathrm{{ri}}(\varDelta ^r_i) \cap \mathrm{{ri}}(\varDelta ^r_j)=\emptyset \).
-
(iv)
Let \(F_i\) and \(F_j\) be two efficient adjacent vertices (zero-dimensional efficient faces) of Q. Then the edge joining them is efficient if and only if \(\varDelta ^r_{i}\cap \varDelta ^r_{j}\not = \emptyset \).
Proof
Since the set \(\varDelta ^r_Q\) is nonempty, the set \(\varDelta _Q\) does not lie on the border of \(\varDelta \). Being a closed convex set, \(\varDelta _Q\) is the closure of its relative interior. Hence the relative interior of \(\varDelta _Q\) and the relative interior of \(\varDelta \) have at least one point in common and we deduce
The converse inclusion being evident, we obtain equality \(\varDelta _Q = \mathrm{{cl}} (\varDelta ^r_Q).\)
To prove (i) we apply the second part of Lemma 4.1.13 to have
If y and z were relative interior points of two different faces, in view of Theorem 2.3.26 we would have \(\mathrm{ri}[N_Q(y)]\cap \mathrm{ri}[N_Q(z)]=\emptyset \) that contradicts the hypothesis. Hence they are relative interior points of the same face, say \(F_s\). By definition \(N(F_s)= N_Q(y)\) and we deduce \(\varDelta ^r_s=\varDelta ^r_y.\)
For (ii) assume \(F_i\) is a maximal efficient face. If for some \(j\in I_0\) one has \(\varDelta _j^r\subseteq \varDelta _i^r\), then by Lemma 4.1.13 there is some strictly positive vector that lies in the relative interior of the normal cone \(N(F_j)\) and in the normal cone \(N(F_i)\). We deduce that either \(F_i=F_j\), or \(F_i\) is a proper face of \(F_j\). The last case is impossible because \(F_j\) is also an efficient face and \(F_i\) is maximal. Conversely, if \(F_i\) is not maximal, then there is a face \(F_j\) that is efficient and contains \(F_i\) as a proper face. We have \(\varDelta _j\subseteq \varDelta _i\). This inclusion is strict because the relative interiors of \(N(F_i)\) and \(N(F_j)\) do not meet each other. Thus, \(\varDelta _i\) is not minimal.
We proceed to (iii). If \(\mathrm{{ri}}(\varDelta ^r_i) \cap \mathrm{{ri}}(\varDelta ^r_j)\not =\emptyset \), in view of Theorem 4.1.14 one has \(\varDelta _i=\varDelta _j\), and hence \(\varDelta _i^r=\varDelta _j^r\). By (i), there is some face that contains relative interior points of \(F_i\) and \(F_j\) in its relative interior. This implies \(F_i=F_j\) a contradiction.
For the last property we know that the normal cone to the edge joining the vertices \(F_i\) and \(F_j\) satisfies \(N([F_i,F_j]) = N(F_i)\cap N(F_j)\). Hence the edge \([F_i,F_j]\) is efficient if and only if the normal cone to it meets the set \(\mathrm{ri}(\varDelta )\), or equivalently \(\varDelta _i^r\cap \varDelta _j^r\) is nonempty. \(\Box \)
A practical way to compute the weakly scalarizing set is to solve a system of linear equalities when the polyhedron Q is given by a system of linear inequalities.
Corollary 4.1.16
Assume the polyhedron Q in \(\mathbb {R}^k\) is determined by the system
Then for every \(y\in Q\), the set \(\varDelta _y\) consists of all solutions z to the following system
In particular the weakly scalarizing set \(\varDelta _Q\) is the solution set to the above system with \(I=\{1,\cdots ,m\}.\)
Proof
According to Theorem 2.3.24 the normal cone to Q at y is the positive hull of the vectors \(a^i, i\in I(y)\). Hence the set \(\varDelta _y\) is the intersection of the positive hull of these vectors and the simplex \(\varDelta \), which is exactly the solution set to the system described in the corollary. For the second part of the corollary it suffices to observe that the normal cone of Q is the polar cone of the asymptotic cone of Q (Theorem 2.3.26) which, in view of Theorem 2.3.19, is the positive hull of the vectors \(a^i, i=1,\cdots ,m\). \(\Box \)
Example 4.1.17
Consider the polyhedron defined by
By Corollary 4.1.16 the weakly scalarizing set \(\varDelta _Q\) is the solution set to the system
This produces a unique solution z with \(z_1= 2/3, z_2=0\) and \(z_3=1/3.\) Then \(\varDelta _Q\) consists of this solution only. The scalarizing set \(\varDelta ^r_Q\) is empty, which shows that Q has no efficient point. Its weakly efficient set is determined by the problem
It follows from the second inequality determining Q that the maximum value of the objective function is zero and attained on the face given by the equations \(2y_1+y_3 =0 \) and \( y_1-y_2-y_3 \leqq 1.\)
In the next example we show how to compute the scalarizing set when the polyhedron is given by a system of equalities (see also Exercise 4.4.13 at the end of this chapter).
Example 4.1.18
Let Q be a polyhedron in \(\mathbb {R}^3\) determined by the sytem
We consider the solution \(\overline{y}=(1/2, 1/2, 0)^T\) and want to compute the scalarizing set at this solution if it exists. As the proof of Theorem 4.1.14 indicates, a vector \(\lambda \in \varDelta \) in \(\mathbb {R}^3\) is a weakly scalarizing vector of Q at \(\overline{y}\) if and only if it is normal to Q at that point. Since the last component of \(\overline{y}\) is zero, a vector \(\lambda \) is normal to Q at \(\overline{y}\) if and only if there are real numbers \(\alpha , \beta \) and a positive number \(\gamma \) such that
We deduce \(\lambda \in \varDelta _{\overline{y}}\) if and only if
and hence \(\varDelta _{\overline{y}}\) consists of vectors \(\lambda \) whose components satisfy \(0\leqq \lambda _3 \leqq 1/3\), \(\lambda _1+\lambda _2= 1-\lambda _3\) and \(\lambda _1,\lambda _2\geqq 0\).
To obtain the scalarizing vectors, it suffices to choose \(\lambda \) as above with an additional requirement that \(\lambda _i >0\) for \( i=1,2,3.\)
Structure of the set of efficient points
Given a convex polyhedron Q in the space \(\mathbb {R}^k\), the set of its efficient elements is not simple. For instance, it is generally not convex, and an edge of it is not necessarily efficient even if its two extreme end-points are efficient vertices. Despite of this, a number of nice properties of this set can be scrutinized.
Corollary 4.1.19
Let Q be a convex polyhedron in \(\mathbb {R}^k\). The following statements hold.
-
(i)
If a relative interior point of a face of Q is efficient or weakly efficient, then so is every point of that face.
-
(ii)
If Q has vertices, it has an efficient vertex (respectively a weakly efficient vertex) provided that it has efficient (respectively weakly efficient) elements.
Proof
Since the normal cone to Q at every point of a face contains the normal cone at a relative interior point, the first statement follows directly from Theorem 4.1.10.
For the second statement let y be an efficient point of the polyhedron Q. By Theorem 4.1.10 one can find a strictly positive vector \(\lambda \) such that y is a maximizer of the linear functional \(\langle \lambda , .\rangle \) on Q. The face which contains y in its relative interior maximizes the above functional. According to Corollary 2.3.14 there is a vertex of Q inside that face and in view of Theorem 4.1.10 this vertex is an efficient point of Q. The case of weakly efficient points is proven by the same argument. \(\Box \)
A subset P of \(\mathbb {R}^k\) is called arcwise connected if for any pair of points y and z in P, there are a finite number of points \(y^0,\cdots ,y^\ell \) in P such that \(y^0=y\), \(y^\ell =z\) and the segments \([y^i, y^{i+1}], i=0,\cdots ,\ell -1\) lie all in P.
Theorem 4.1.20
The sets of all efficient points and weakly efficient points of a convex polyhedron consist of faces of the polyhedron and are closed and arcwise connected.
Proof
By analogy, it suffices to prove the theorem for the efficient set. According to Corollary 4.1.19, if a point \(\bar{y}\) in Q is efficient, then the whole face containing y in its relative interior is a face of efficient elements. Hence, Max(Q) consists of faces of Q if it is nonempty. Moreover, as faces are closed, their union is a closed set.
Now we prove the connectedness of this set by assuming that Q has efficient elements. Let y and z be any pair of efficient points of Q. We may assume without loss of generality that y is a relative interior point of a face \(Q_y\) and z is a relative interior point of a face \(Q_z\). Consider the decomposition (4.5) of the scalarizing set \(\varDelta ^r_Q\). For a face F of Q, the scalarizing set \(N(F)\cap \mathrm{ri}(\varDelta )\) is denoted by \(\varDelta ^r_{i(F)}\). Let \(\lambda _y\) be a relative interior point of the set \(\varDelta ^r_{i(Q_y)}\) and \(\lambda _z\) a relative interior point of \(\varDelta ^r_{i(Q_z)}\). Then the segment joining \(\lambda _y\) and \(\lambda _z\) lies in \(\varDelta ^r_Q\). The decomposition of the latter set induces a decomposition of the segment \([\lambda _y, \lambda _z]\) by \([\lambda _i, \lambda _{i+1}], i=0,\cdots ,\ell -1\) where \(\lambda _0=\lambda _y\), \(\lambda _\ell =\lambda _z\). Let \(Q_1,\cdots ,Q_\ell \) be faces of Q such that
For every j, we choose a relative interior point \(y^j\) of the face \(Q_j\). Then \(\lambda _j\) belongs to the normal cones to Q at \(y^j\) and \(y^{j+1}\). Consequently, the points \(y^j\) and \(y^{j+1}\) lie in the face argmax\(_Q\langle \lambda _j, .\rangle \) and so does the segment joining them. As \(\lambda _j\in \varDelta ^r_Q\), by Theorem 4.1.10 the segment \([y^j,y^{j+1}]\) consists of efficient points of Q. Moreover, as the vector \(\lambda _0\) belongs to the normal cones to Q at y and at \(y_1\), we conclude that the segment \([y,y^1]\) is composed of efficient points of Q. Similarly we have that \([y^{\ell -1}, z]\) lies in the set Max(Q). Thus, the union \([y,y^1]\cup [y^1,y^2]\cup \cdots [y^{\ell -1}, z]\) forms a path of efficient elements joining y and z. This completes the proof. \(\Box \)
We know that every efficient point of a convex polyhedron is contained in a maximal efficient face. Hence the set of efficient points is the union of maximal efficient faces. Dimension of a maximal efficient face may vary from zero to \(k-1\).
Corollary 4.1.21
Let Q be a convex polyhedron in \(\mathbb {R}^k\). The following statements hold.
-
(i)
Q has a zero-dimensional maximal efficient face if and only if its efficient set is a singleton.
-
(ii)
Every \((k-1)\)-dimensional efficient face of Q, if any exists, is maximal. In particular in the two dimensional space \(\mathbb {R}^2\) every efficient edge of Q is maximal if the efficient set of Q consists of more than two elements.
-
(iii)
An efficient face F of Q is maximal if and only if the restriction of the decomposition of \(\varDelta ^r_Q\) on \(\varDelta _F\) consists of one element only.
Proof
The first statement follows from the arcwise connectedness of the efficient set of Q. In \(\mathbb {R}^k\) a proper face of Q is of dimension at most \(k-1\). Moreover, a k-dimensional polyhedron cannot be efficient, for its interior points are not maximal. Hence, if the dimension of an efficient face is equal to \(k-1\), it is maximal.
The last statement follows immediately from Theorem 4.1.15. \(\Box \)
Example 4.1.22
Let Q be a polyhedron in \(\mathbb {R}^3\) defined by the system
Since Q is bounded, it is evident that the weakly scalarizing set \(\varDelta _Q\) is the whole standard simplex \(\varDelta \) and the scalarizing set is the relative interior of \(\varDelta \). Denote
Applying Corollary 4.1.16 we obtain the following decomposition of \(\varDelta ^r_Q\):
-
(i)
\(\mathrm{ri}[q^4, q^5]\) is the scalarizing set of the face determined by the equalities \(x_1+x_3=1\) and \(x_2+x_3=1\);
-
(ii)
\(\mathrm{ri(co(}[q^3,q^4, q^5]\mathrm{))}\) is the scalarizing set of the face determined by the equali ties \(x_1+x_3=1\), \(x_2+x_3=1\) and \(x_1=x_2=0\);
-
(iii)
\(\mathrm{ri(co(}[q^1, q^2, q^4, q^5]\mathrm{))}\) is the scalarizing set of the face determined by the equalities \(x_1+x_3=1\), \(x_2+x_3=1\) and \(x_3=0\).
In view of Corollary 4.1.21 the one dimensional face (edge) determined by \(x_1+x_3=1\) and \(x_2+x_3=1\) is a maximal efficient face.
2 Multiobjective Linear Problems
The central multiobjective linear programming problem which we propose to study throughout is denoted (MOLP) and written in the form :
where X is a nonempty convex polyhedron in \(\mathbb {R}^n\) and C is a real \(k\times n\)-matrix. This problem means finding a Pareto efficient (Pareto maximal) solution \(\bar{x}\in X\) such that \(C\bar{x} \in \) Max(C(X)). In other words, a feasible solution \(\bar{x}\) solves (MOLP) if there is no feasible solution \(x\in X\) such that
The efficient solution set of (MOLP) is denoted \(S(\textit{MOLP})\). When x is an efficient solution, the vector Cx is called an efficient or maximal value of the problem. In a similar manner one defines the set of weakly efficient solutions \(\textit{WS}(\textit{MOLP})\) to be the set of all feasible solutions whose image by C belong to the weakly efficient set WMax(C(X)). It is clear that an efficient solution is a weakly efficient solution, but not vice versa as we have already discussed in the preceding section. When the feasible set X is given by the system
where A is a real \(m\times n\)-matrix and b is a real m-vector, we say that (MOLP) is given in standard form, and it is given in canonical form if X is determined by the system
The matrix C is also considered as a linear operator from \(\mathbb {R}^n\) to \(\mathbb {R}^k\), and so its kernel consists of vectors x with \(Cx=0\).
Theorem 4.2.1
Assume that the problem (MOLP) has feasible solutions. Then the following assertions hold.
-
(i)
(MOLP) admits efficient solutions if and only if
$$C(X_\infty ) \cap \mathbb {R}^k_+=\{0\}.$$ -
(ii)
(MOLP) admits weakly efficient solutions if and only if
$$C(X_\infty ) \cap \mathrm{int}({\mathbb {R}}^k_+)=\emptyset .$$
In particular, if all asymptotic rays of X belong to the kernel of C, then (MOLP) has an efficient solution.
Proof
By definition, (MOLP) has an efficient solution if and only if the set C(X) has an efficient point, which, in virtue of Theorem 4.1.7, is equivalent with
Now the first assertion is deduced from this equivalence and from the fact that the asymptotic cone of C(X) coincides with the cone \(C(X_\infty )\) (Corollary 2.3.17). The second assertion is proven by a similar argument. \(\Box \)
Example 4.2.2
Assume that the feasible set X of the problem (MOLP) is given by the system
It is nonempty and parametrically presented as
Its asymptotic cone is given by
Consider an objective function C with values in \({\mathbb {R}}^2\) given by the matrix
Then the image of \(X_\infty \) under C is the set
that has only the zero vector in common with the positive orthant. In view of Theorem 4.2.1 the problem has maximal solutions.
Now we choose another objective function \(C'\) given by
Then the image of \(X_\infty \) under \(C'\) is the set
that has no common point with the interior of the positive orthant. Hence the problem admits weakly efficient solutions. It has no efficient solution because the intersection of \(C'(X_\infty )\) with the positive orthant does contain positive vectors.
Definition 4.2.3
The objective function of the problem (MOLP) is said to be bounded (respectively weakly bounded) from above if there is no vector \(v\in X_\infty \) such that
We shall simply say that (MOLP) is bounded if its objective function is bounded from above. Of course, a bounded problem is weakly bounded and not every weakly bounded problem is bounded. A sufficient condition for a problem to be bounded is given by the inequality
where a is some vector from \(\mathbb {R}^k\). This condition is also necessary when \(k=1\), but not so when \(k>1.\)
Example 4.2.4
Consider the bi-objective problem
The feasible set and its asymptotic cone are given respectively by
and
Then for every asymptotic direction \(v=(t,t,0)^T\in X_\infty \) one has
By definition the objective function is bounded. Nevertheless the value set of the problem consists of vectors
for which no vector \(a\in {\mathbb {R}}^2\) satisfies \(Cx\leqq a\) for all \(x\in X\).
Corollary 4.2.5
The problem (MOLP) has efficient solutions (respectively weakly efficient solutions) if and only if its objective function is bounded (respectively weakly bounded).
Proof
This is immediate from Theorem 4.2.1. \(\Box \)
The following theorem provides a criterion for efficiency in terms of normal directions.
Theorem 4.2.6
Let \(\bar{x} \) be a feasible solution of (MOLP). Then
-
(i)
\(\bar{x}\) is an efficient solution if and only if the normal cone \(N_X(\bar{x})\) to X at \(\bar{x}\) contains some vector \(C^T\lambda \) with \(\lambda \) a strictly positive vector of \({\mathbb {R}}^k\);
-
(ii)
\(\bar{x}\) is a weakly efficient point if and only if the normal cone \(N_X(\bar{x})\) to X at \(\bar{x}\) contains some vector \(C^T\lambda \) with \(\lambda \) a nonzero positive vector of \({\mathbb {R}}^k\).
Proof
If the vector \(C^T\lambda \) with \(\lambda \) strictly positive, is normal to X at \(\bar{x}\), then
which means that
By Theorem 4.1.4 the vector \(C \bar{x}\) is an efficient point of the set C(X). By definition, \(\bar{x}\) is an efficient solution of (MOLP).
Conversely, if \(C \bar{x}\) is an efficient point of C(X), then by Theorem 4.1.10, the normal cone to C(X) at \(C \bar{x}\) contains a strictly positive vector, denoted by \(\lambda \). We deduce that
This shows that the vector \(C^T\lambda \) is normal to X at \(\bar{x}\). The second assertion is proven similarly. \(\Box \)
Example 4.2.7
We reconsider the bi-objective problem given in Example 4.2.2
Choose a feasible solution \(\overline{x}=(9/2, 1/2, 0)^T\) corresponding to \(t= 1/2\). The normal cone to the feasible set at \(\overline{x}\) is the positive hull of the hyperplane of basis \(\{(1,1,-1)^T, (1,-1,0)^T\}\) (the row vectors of the constraint matrix) and the vector \((0,0,-1)^T\) (the constraint \(x_3\geqq 0\) is active at this point). In other words, this normal cone is the half-space determined by the inequality
The image of the positive orthant of the value space \({\mathbb {R}}^2\) under \(C^T\) is the positive hull of the vectors
Using inequality (4.6) we deduce that the vector \(v_2\) lies in the interior of the normal cone to the feasible set at \(\overline{x}\). Hence that normal cone does contain a vector \(C^T\lambda \) with some strictly positive vector \(\lambda \). By Theorem 4.2.6 the solution \(\overline{x}\) is efficient. It is routine to check that the solution \(\overline{x}\) is a vertex of the feasible set.
If we pick another feasible solution, say \(\overline{z}= (5,1,1)^T\), then the normal cone to the feasible set at \(\overline{z}\) is the hyperplane determine by equation
Direct calculation shows that the vectors \(v_1\) and \(v_2\) lie in different sides of the normal cone at \(\overline{z}\). Hence there does exist a strictly positive vector \(\lambda \) in \({\mathbb {R}}^2\) such that \(C^T\lambda \) is contained in that cone. Consequently, the solution \(\overline{z}\) is efficient too.
3 Scalarization
We associate with a nonzero k-vector \(\lambda \) a scalar linear problem, denoted (LP\(_\lambda \))
This problem is referred to as a scalarized problem of (MOLP) and \(\lambda \) is called a scalarizing vector. Now we shall see how useful scalarized problems are in solving multiobjective problems.
Theorem 4.3.1
The following statements hold.
-
(i)
A feasible solution \(\bar{x}\) of (MOLP) is efficient if and only if there is a strictly positive k-vector \(\lambda \) such that \(\bar{x}\) is an optimal solution of the scalarized problem (LP\(_\lambda \)).
-
(ii)
A feasible solution \(\bar{x}\) of (MOLP) is weakly efficient if and only if there is a nonzero positive k-vector \(\lambda \) such that \(\bar{x}\) is an optimal solution of the scalarized problem (LP\(_\lambda \)).
Proof
If \(\bar{x}\) is an efficient solution of (MOLP), then, in view of Theorem 4.2.6, there is a strictly positive vector \(\lambda \) such that \(C^T\lambda \) is a normal vector to X at \(\bar{x}.\) This implies that \(\bar{x}\) maximizes the linear functional \(\langle \lambda , C(.)\rangle \) on X, that is, \(\bar{x}\) is an optimal solution of (LP\(_\lambda \)).
Conversely, if \(\bar{x}\) solves the problem (LP\(_\lambda \)) with \(\lambda \) strictly positive, then \(C^T\lambda \) is a normal cone to X at \(\bar{x}\). Again, in view of Theorem 4.2.6, the point \(\bar{x}\) is an efficient solution of (MOLP). The proof of the second statement follows the same line. \(\Box \)
We notice that Theorem 4.3.1 remains valid if the scalarizing vector \(\lambda \) is taken from the standard simplex, that is \(\lambda _1+\cdots +\lambda _k=1.\) Then another formulation of the theorem is given by equalities
where S(LP\(_\lambda \)) denotes the optimal solution set of (LP\(_\lambda \)). It was already mentioned that a weakly efficient solution is not necessarily an efficient solution. Consequently a positive, but not strictly positive vector \(\lambda \) may produce weakly efficient solutions which are not efficient. Here is an exception.
Corollary 4.3.2
Assume for a positive vector \(\lambda \), the set consisting of the values Cx with x being optimal solution of (LP\(_\lambda \)) is a singleton, in particular when (LP\(_\lambda \)) has a unique solution. Then every optimal solution of (LP\(_\lambda \)) is an efficient solution of (MOLP).
Proof
Let x be an optimal solution of (LP\(_\lambda \)) and let y be a feasible solution of (MOLP) such that \(Cy\geqq Cx. \) Since \(\lambda \) is positive, one has
Actually we have equality because x solves (LP\(_\lambda \)). Hence y solves (LP\(_\lambda \)) too. By hypothesis \(Cx=Cy\) which shows that x is an efficient solution of (MOLP). \(\Box \)
Equalities (4.7) and (4.8) show that efficient and weakly efficient solutions of (MOLP) can be generated by solving a family of scalar problems. It turns out that a finite number of such problems are sufficient to generate the whole efficient and weakly efficient solution sets of (MOLP).
Corollary 4.3.3
There exists a finite number of strictly positive vectors (respectively positive vectors) \(\lambda ^i, i=1,\cdots ,p\) such that
Proof
It follows from Theorem 3.1.3 that if an efficient solution is a relative interior of a face of the feasible polyhedron and an optimal solution of (LP\(_{\lambda }\)) for some strictly positive vector \(\lambda \), then the whole face is optimal for (LP\(_{\lambda }\)). Since the number of faces is finite, a finite number of such vectors \(\lambda \) is sufficient to generate all efficient solutions of (MOLP). The case of weakly efficient solutions is treated in the same way. \(\Box \)
Corollary 4.3.4
Assume that (MOLP) has an efficient solution and (LP\(_\lambda \)), where \(\lambda \) is a nonzero positive vector, has an optimal solution. Then there is an efficient solution of (MOLP) among the optimal solutions of (LP\(_\lambda \)).
Proof
Apply Theorem 4.3.1 and the method of Corollary 4.1.8. \(\Box \)
Corollary 4.3.5
Assume that the scalarized problems
where \(c^i, i=1,\cdots , k\) are the columns of the matrix \(C^T\), are solvable. Then (MOLP) has an efficient solution.
Proof
The linear problems mentioned in the corollary correspond to the scalarized problems (LP\(_\lambda \)) with \(\lambda =(0,\cdots , 1, \cdots ,0)^T\) where the one is on the ith place, \(i=1,\cdots ,k\). These problems provide weakly efficient solutions of (MOLP). The linear problem whose objective is the sum \(\langle c^1, x\rangle +\cdots + \langle c^k, x\rangle \) is solvable too. It is the scalarized problem with \(\lambda =(1,\cdots , 1)^T\), and hence by Theorem 4.3.1, (MOLP) has efficient solutions. \(\Box \)
Decomposition of the scalarizing set
Given a feasible solution x of (MOLP) we denote the set of all vectors \(\lambda \in \varDelta \) such that x solves (LP\(_\lambda \)) by \(\Lambda (x)\), and the union of all these \(\Lambda (x)\) over \(x\in X\) by \(\Lambda (X)\). We denote also
The sets \(\Lambda ^r(X)\) and \(\Lambda (X)\) are respectively called the scalarizing and weakly scalarizing sets of (MOLP). The decomposition results for efficient elements (Theorems 4.1.14 and 4.1.15) are easily adapted to decompose the weakly scalarizing and scalarizing sets of the problem (MOLP). We deduce a useful corollary below for computing purposes.
Corollary 4.3.6
The following assertions hold for (MOLP).
-
(i)
A feasible solution \(x\in X\) is efficient (respectively weakly efficient ) if and only if \(\Lambda ^r(x)\) (respectively \(\Lambda (x)\)) is nonempty.
-
(ii)
If X has vertices, then the set \(\Lambda ^r(X)\) (respectively \(\Lambda (X)\)) is the union of the sets \(\Lambda ^r(x^i)\) ( respectively \(\Lambda (x^i)\)) where \(x^i\) runs over the set of all efficient (respectively weakly efficient ) vertices of (MOLP).
-
(iii)
If X is given by system
$$\langle a^i, x\rangle \leqq b_i, i=1,\cdots ,m$$and x is a feasible solution, then the set \(\Lambda (x)\) consists of all solutions \(\lambda \) to the following system
$$\begin{aligned} \lambda _1+\cdots +\lambda _k= & {} 1\\ \sum _{i\in I(x)} \alpha _i a^i= & {} \lambda _1 c^1+\cdots +\lambda _k c^k\\ \lambda _i\geqq 0, i=1,\cdots ,k, \alpha _i\geqq & {} 0, i\in I(x). \end{aligned}$$In particular the weakly scalarizing set \(\Lambda (X)\) is the solution set to the above system with \(I=\{1,\cdots ,m\}.\)
Proof
The first assertion is clear from Theorem 4.3.1. For the second assertion we observe that when X has vertices, every face of X has vertices too (Corollary 2.3.6). Hence the normal cone of X is the union of the normal cones to X at its vertices. Moreover, by writing the objective function \(\langle \lambda , C( . ) \rangle \) of (LP\(_\lambda \)) in the form \(\langle C^T\lambda , .\rangle \), we deduce that
Consequently,
The proof for efficient solutions is similar. The last assertion is derived from (4.9) and Corollary 4.1.16. \(\Box \)
Example 4.3.7
We reconsider the bi-objective problem
We wish to find the weakly scalarizing set of this problem. According to the preceding corollary, it consists of positive vectors \(\lambda \) from the standard simplex of \(\mathbb {R}^2\), solutions to the following system:
Solving this system we obtain \(\lambda = \left( \begin{array}{c} t\\ 1-t \end{array}\right) \) with \(7/8 \leqq t\leqq 1\). For \(t=1\), the scalarized problem associated with \(\lambda \) is of the form
It can be seen that x solves this problem if and only if \(x_1+x_2=1\) and \(x_1 \leqq 0\). These solutions form the set of weakly efficient solutions of the multiobjective problem.
For \(t=7/8\), the scalarized problem associated with \(\lambda = (7/8, 1/8)^T\) is of the form
Its optimal solutions are given by \(3x_1+2x_2=2\) and \(x_1\geqq 0\). Since \(\lambda \) is strictly positive, these solutions are efficient solutions of the multiobjective problem. If we choose \(\lambda = (1/2,1/2)^T\) outside of the scalarizing set, then the associated scalarized problem has no optimal solution.
Structure of the efficient solution set
We knew in Chap. 3 that the optimal solution set of a scalar linear problem is a face of the feasible set. This property, unfortunately, is no longer true when the problem is multiobjective. However, a few interesting properties of the efficient set of a polyhedron we established in the first section are still valid for the efficient solution set and exposed in the next theorem.
Theorem 4.3.8
The efficient solutions of the problem (MOLP) have the following properties.
-
(i)
If a relative interior point of a face of X is an efficient or weakly efficient solution, then so is every point of that face.
-
(ii)
If X has a vertex and (MOLP) has an efficient (weakly efficient) solution, then it has an efficient (weakly efficient) vertex solution.
-
(iii)
The efficient and weakly efficient solution sets of (MOLP) consist of faces of the feasible polyhedron and are closed and arcwise connected.
Proof
Since the normal cone to X at every point of a face contains the normal cone at a relative interior point, the first property follows directly from Theorem 4.2.6.
Further, under the hypothesis of (ii) there is a strictly positive vector \(\lambda \in \mathbb {R}^k\) such that the scalarized problem (LP\(_\lambda \)) is solvable. The argument in proving (ii) of Corollary 4.1.19 is applicable to obtain an optimal vertex of (LP\(_\lambda \)) which is also an efficient vertex solution of (MOLP).
The proof of the last property is much similar to the one of Theorem 4.1.15. We first notice that in view of (i) the efficient and weakly efficient solution sets are composed of faces of the feasible set X, and as the number of faces of X is finite, they are closed. We now prove the arcwise connectedness of the weakly efficient solution set, the argument going through for efficient solutions too. Let x and y be two weakly efficient solutions, relative interior points of efficient faces \(X_x\) and \(X_y\) of X. Let \(\lambda _x\) and \(\lambda _y\) be relative interior vectors of the weakly scalarizing sets \(\Lambda (X_x)\) and \(\Lambda (X_y)\). The decomposition of the weakly scalarizing set \(\Lambda (X)\) induces a decomposition of the segment joining \(\lambda _x\) and \(\lambda _y\) by
with \(\lambda _1=\lambda _x\), \(\lambda _\ell =\lambda _y\) and \([\lambda _i,\lambda _{i+1}]\subseteq \Lambda (X_i)\) for some face \(X_i\) of X, \(i=1,...,\ell -1\). Since \(\lambda _i\) belongs simultaneously to \(\Lambda (X_i)\) and \(\Lambda (X_{i+1})\), there is some common point \(x^i\in X_i\cap X_{i+1}\), \(i=1,...,\ell -1\). It is clear that \([x, x_1]\cup [x_1,x_2]\cup \cdots \cup [x_{\ell -1}, y]\) is an arcwise path joining x and y and each member segment \([x_i,x_{i+1}] \) is efficient because being in the face \(X_i\), \(i=1,...,\ell \) with \(x_\ell =y\). \(\Box \)
4 Exercises
4.4.1
Find maximal elements of the sets determined by the following systems
4.4.2
Find maximal and weakly maximal elements of the following sets
\(Q_1= \left\{ \left( \begin{array}{c} x_1\\ x_2\\ x_3 \end{array} \right) \in \mathbb {R}^3: x_1\geqq 0, x_2\geqq 0, x_3\geqq 0, x_2^2+x_3^2\leqq 1 \right\} \)
\(Q_2= \)co(A, B) with \(A= \left\{ \left( \begin{array}{c} 1\\ 0\\ s \end{array} \right) \in \mathbb {R}^3: 0\leqq s\leqq 1 \right\} \)
and \(B= \left\{ \left( \begin{array}{c} 0\\ x_2\\ x_3 \end{array} \right) \in \mathbb {R}^3: x_2\geqq 0, x_3\geqq 0, x_2^2+x_3^2\leqq 1 \right\} .\)
4.4.3
We say a real function g on \( \mathbb {R}^k\) is increasing if \(x,y\in \mathbb {R}^k\) and \(x\ge y\) imply \(g(x)>g(y)\), and it is weakly increasing if \(x>y\) implies \(g(x)>g(y)\). Prove that g is increasing (respectively weakly increasing) if and only if for every nonempty subset Q of \(\mathbb {R}^k\), every maximizer of g on Q is an efficient (respectively weakly maximal) element of Q.
4.4.4
Let Q be a closed set in \(\mathbb {R}^k\). Prove the following statements.
-
(i)
The set WMax(Q) is closed.
-
(ii)
The set Max(Q) is closed provided that \(k=2\) and \(Q-\mathbb {R}^2_+\) is convex.
-
(iii)
Max\((-Q)=-\) Min(Q) and Max\((\alpha Q)=\alpha \) Max(Q) for every \(\alpha > 0.\)
4.4.5
Let P and Q be two convex polyhedra in \(\mathbb {R}^k\).
-
(i)
Prove that Max\((P+Q) \subseteq \) Max\((P)+\) Max(Q).
-
(ii)
Find conditions under which equality holds in (i).
4.4.6
Prove that the set of maximal elements of a convex polytope is included in the convex hull of the maximal vertices. Is the converse true?
4.4.7
An element x of a set P in \(\mathbb {R}^k\) is said to be dominated if there is some \(x'\in P\) such that \(x'\ge x\). Prove that the set of dominated elements of a convex polyhedral set is convex and if a face contains a dominated element, its relative interior points are dominated too.
4.4.8
A diet problem. A multiobjective version of the diet problem in hospital consists of finding a combination of foods for a patient to minimize simultaneously the cost of the menu and the number of calories under certain nutritional requirements prescribed by a treating physician. Assume a menu is composed of three main types of foods: meat with potatoes, fish with rice and vegetables. The nutrition facts, calories in foods and price per servings are given below
Using three variables: x= number of servings of meat, y= number of servings of fish and z= number of servings of vegetables, formulate a bi-objective linear problem whose objective functions are the cost and the number of calories of the menu while maintaining the physician’s prescription of at least one unit and at most one and half unit for each nutritional substance. Discuss the menus that minimize the cost and the number of calories separately.
4.4.9
An investment problem . An investor disposes a budget of 20,000 USD and wishes to invest into three product projects with amounts x, y and z respectively. The total profit is given by
and the total sale is given by
Find x, y and z to maximize the total profit and total sale.
4.4.10
Bilevel linear programming problem . A typical bilevel programming problem consists of two problems: the upper level problem of the form
and the lower level problem for which y is an optimal solution:
Here \(c, p, d, b_1\) and \(b_2\) are vectors of dimension \(n_1, n_2, n_2, m_1\) and \(m_2\) respectively; \(A_1, A_2 \) and \(A_3\) are matrices of dimension \(m_1\times n_1\), \(m_2\times n_1\) and \(m_2\times n_2\) correspondingly.
Consider the following multiobjective problem
where e is the vector whose components are all equal to one. Prove that \((\overline{x}, \overline{y})\) is an efficient solution of this latter problem if and only if it is a feasible solution of the upper level problem described above.
4.4.11
Apply Theorem 4.1.15 to find a decomposition of the scalarizing set for the polyhedron defined by the system
4.4.12
Find the weakly scalarizing set of a polyhedron in \(\mathbb {R}^k\) determined by the system
and apply it to find the weakly scalarizing set of a multiobjective problem given in standard form.
4.4.13
Scalarizing set at a vertex solution. Consider the problem (MOLP) in standard form
where C is a \(k\times n\)-matrix, A is an \(m\times n\)-matrix and b is an m-vector. Assume \(\overline{x}\) is a feasible solution associated with a non-degenerate basis B. The non-basic part of A is denoted N, the basic and non-basic parts of C are denoted \(C_B\) and \(C_N\) respectively. Prove the following statements.
-
(a)
A vector \(\lambda \) belongs to \(\Lambda (\overline{x})\) if and only if it belongs to \(\varDelta \) and solves the following system
$$[C_N^T-(B^{-1}N)^TC_B^T]\lambda \leqq 0.$$ -
(b)
If the vector on the left hand side of the system in (a) is strictly negative for some \(\lambda \), then \(\overline{x}\) is a unique solution of the scalarized problem
$$\begin{aligned} \mathrm{{maximize}}\;&\langle \lambda , C x\rangle \\ \mathrm{{subject\; to}}\;&Ax=b\\&x\geqq 0. \nonumber \end{aligned}$$In particular, if in addition \(\lambda \) is positive, then \(\overline{x}\) is an efficient solution of (MOLP).
4.4.14
Pascoletti-Serafini’s method. Let \(\lambda \in \mathbb {R}^k\) be a strictly positive vector and \(Cx\geqq 0\) for every feasible solution \(x\in X\) of (MOLP). Show that if \((\overline{x}, \overline{\alpha })\) is an optimal solution of the problem
then \(\overline{x}\) is a weakly efficient solution of (MOLP).
4.4.15
Weighted constraint method. Prove that a feasible solution \(\overline{x}\in X \) is a weakly efficient solution of (MOLP) if and only if there is some strictly positive vector \( \lambda \in \mathbb {R}^k\) such that \(\overline{x}\) solves
for \(\ell =1,\cdots , k\).
4.4.16
Constraint method. Choose \(\ell \in \{1, \cdots , k\}, L_j \in \mathbb {R}, j = 1, \cdots , k, j \ne \ell \) , and solve the scalar problem \((P_\ell )\):
Note that if \(L_j\) are big, then \((P_\ell )\) may have no feasible solution. A constraint \(\langle c^j, x\rangle \ge L_j\) is called binding if equality \(\langle c^j, x\rangle = L_j\) is satisfied at every optimal solution of \((P_\ell )\). Prove that
-
(a)
every optimal solution of \((P_\ell )\) is a weakly efficient solution of (MOLP);
-
(b)
if an optimal solution of \((P_\ell )\) is unique or all constraints of \((P_\ell )\) are binding, then it is an efficient solution of (MOLP);
-
(c)
a feasible solution \(x^0\) of (MOLP) is efficient if and only if it is optimal for all \((P_\ell )\), \( \ell =1,...,k\) and
$$L_\ell =( \langle c^1, x^0\rangle ,\cdots , \langle c^{\ell -1}, x^0\rangle , \langle c^{\ell +1}, x^0\rangle , \cdots , \langle c^k, x^0\rangle ).$$
4.4.17
Let d be a k-vector such that \(Cx\geqq d\) for some feasible solution x of (MOLP). Consider the problem (P)
where e is the vector of ones in \(\mathbb {R}^k\). Show that
-
(a)
a feasible solution \(x^0\) of (MOLP) is efficient if and only if the optimal value of (P) with \(d= Cx^0\) is equal to zero;
-
(b)
(MOLP) has efficient solutions if and only if the optimal value of (P) is finite.
4.4.18
Let \(\bar{x}\) be a feasible solution of the problem
Show that the following statements are equivalent.
-
(i)
\(\bar{x}\) is a weak Pareto maximal solution.
-
(ii)
The system
$$\left\{ \begin{array}{ rcl} Ax &{}\leqq &{} b \\ Cx &{}>&{} C\overline{x} \end{array} \right. $$is inconsistent.
-
(iii)
For every \(t >0\), the system
$$\left\{ \begin{array}{rcl} Ax &{} \leqq &{} b - A\overline{x} \\ Cx &{}\geqq &{} te \end{array} \right. $$is inconsistent, where e is the vector of ones.
-
(iv)
For every \(t >0\), the system
$$\left\{ \begin{array}{rcl} C^T\lambda -A^T\mu &{}=&{} 0\\ \langle A\bar{x} - b, \mu \rangle + t\langle e, \lambda \rangle &{}= &{} 1\\ \lambda , \mu &{}\geqq &{} 0 \end{array} \right. $$is consistent.
4.4.19
Consider the multiobjective problem described in the preceding exercise. Assume that the cone \(\mathrm{pos}\{c^1,\cdots , c^k\}\) contains the origin in its relative interior. Prove that if the interior of the feasible set is nonempty, then every feasible solution of (MOLP) is an efficient solution.
4.4.20
Let X denote the feasible set of the problem (MOLP) given in Exercise 4.4.18. Consider the following function
Prove that \(\overline{x}\) is a weakly maximal solution of (MOLP) if and only if \(h(\overline{x}) = 0\).
4.4.21
Geoffrion’s proper efficient solutions. Let X be a nonempty set in \(\mathbb {R}^n\) and let f be a vector-valued function from \(\mathbb {R}^n\) to \(\mathbb {R}^k\). Consider the following multiobjective problem
A feasible solution \(\overline{x}\) of this problem is said to be a proper efficient solution if there exists a constant \(\alpha >0\) such that for every \(i\in \{1,\cdots , k\}\) and \(x\in X\) satisfying \(f_i(x) >f_i(\overline{x}) \) there exists some \(j\in \{1,\cdots , k\}\) for which \(f_j(x)< f_j(\overline{x})\) and
-
(i)
Justify that every proper efficient solution is efficient. Give an example of efficient solutions that are not proper.
-
(ii)
Prove that when f is linear and X is a polyhedral set, every efficient solution is proper.
4.4.22
Maximality with respect to a convex cone. Let C be a convex cone in \(\mathbb {R}^k\) with \(C\cap (-C)=\{0\} \) (one says C is pointed). For \(y, z\in \mathbb {R}^k\) define \(y\geqq _Cz\) by \(y-z\in C\). A point z of a set A is called C-maximal if there is no \(y\in A\) such that \(y\geqq _C z\) and \(y\not =z\). Prove the following properties:
-
(i)
A point \(z\in A\) is C-maximal if and only if \((A-a) \cap \mathbb {R}^k =\{0\}\);
-
(ii)
If \(\mathbb {R}^k_+\subseteq C\), then every C-maximal point is Pareto maximal, and if \(\mathbb {R}^k_+\supseteq C\), then every Pareto maximal point is C-maximal;
-
(iii)
If A is a polyhedral set, then there is a polyhedral cone C satisfying \(\mathbb {R}^k_+ \subseteq \mathrm{int} (C) \cup \{0\}\) such that a point of A is C-maximal if and only if it is Pareto maximal. Find such a cone for the sets in Exercise 4.4.1 (a) and (b).
4.4.23
Lexicographical order. The lexicographical order \(\geqq _{lex}\) in \(\mathbb {R}^k\) is defined as \(y\geqq _{lex} z\) for \(y,z\in \mathbb {R}^k\) if and only if either \(y=z\) or there is some \(j\in \{1,\cdots , k\}\) such that \(y_i=z_i\) for \(i < j\) and \(y_{j}>z_{j}\). A point z of a nonempty set A in \(\mathbb {R}^k\) is called lex-maximal if there is no \(y\in A\) such that \(y\geqq _{lex} z\) and \(y\not =z\).
-
(i)
Show that the lexicographical order is total in the sense that for every \(y,z\in \mathbb {R}^k\) one has either \(y\geqq _{lex} z\) or \(z\geqq _{lex} y\).
-
(ii)
Find a convex cone C such that \(y\geqq _{lex} z\) if and only if \(y-z\in C\).
-
(iii)
Prove that every lex-maximal element of a set is Pareto maximal.
Do the same for the colexicographical order: \(y\geqq _{colex} z\) if and only if either \(y=z\) or there is some \(j\in \{1,\cdots , k\}\) such that \(y_i=z_i\) for \(i > j\) and \(y_{j}>z_{j}\).
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Luc, D.T. (2016). Pareto Optimality. In: Multiobjective Linear Programming. Springer, Cham. https://doi.org/10.1007/978-3-319-21091-9_4
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