Maximization of the Choquet integral over a convex set and its application to resource allocation problems

Abstract

We study the problem of the Choquet integral maximization over a convex set. The problem is shown to be generally non-convex (and non-differentiable). We analyze the problem structure, and propose local and global search algorithms. The special case when the problem becomes concave is analyzed separately. For the non-convex case we propose a decomposition scheme which allows to reduce a non-convex problem to several concave ones. Decomposition is performed by finding the coarsest partition of a capacity into disjunction of totally monotone measures. We discuss its effectiveness and prove that the scheme is optimal for 2-additive capacities. An application of the developed methods to resource allocation problems concludes the article.

Appendix

1.1 A.1 Proofs

Lemma 1

Let $A \subset N, T = \bigcap_{(i,j) \in N^{2}_{T} } (f_{i} < f_{j}) \neq \emptyset$ and exists , j∉A, and C—is the largest subset of A such that m ^β(C)≠0, i.e. $\nexists(i,j) \in N^{2}_{T} \colon i \in C, j \notin C$. Then, for any subset D⊂A holds either

(37)

or

(38)

Proof

Notice that $C = A \setminus\bigcup_{(i,j) \in N^{2}_{T} } i$ for all $(i,j) \in N^{2}_{T}$ such that i∈A,j∉A. Then, all subsets $D \subset A, D \not\subset C$ contain an element i such that j∉D. Hence, m ^β(D)=0. If D does not include such an element, then D⊂C. □

Proof of Theorem 8

We will first show that for all sets A⊂N such that $\nexists (i,j) \in N^{2}_{T}\colon\allowbreak i \in\nobreak A,\ j \notin\nobreak A$,

(39)

entails β ^T(A)=ν(A). Pick an arbitrary set A for which , j∉A holds. In the equation

(40)

there exist two types of sets B⊂A. Either , j∉B, and then m ^β(B)=0, or $\nexists(i,j) \in N^{2}_{T} : i \in B, j \notin B$, and then m ^β(B) is given by (39). Note, that for each set B such that m ^β(B)=0 there exists a unique C such that $B \subset C = B\cup(\bigcup_{(i,j) \in N^{2}_{T} } j) \subset A$ for all , j∉B. In this case the Möbius transform coefficient m ^β(C) is given by (39) and would include m ^ν(B) as a summand. Thus,

(41)

Now we will show that m ^β(A)=0⇒β ^T(A)=⋁ _B⊆̷A β ^T(B) for all A such that , j∉A. Moreover, β ^T(A)=⋁ _B⊆̷A β ^T(B)=β ^T(C)=ν(C), where $C = A \setminus\bigcup_{(i,j) \in N^{2}_{T} } i$, , j∉A. According to Lemma 1, all mass is concentrated on the subsets of the set $C = A \setminus \bigcup_{(i,j) \in N^{2}_{T} } i$, hence β ^T(L)=β ^T(C) for all C⊂L⊂A. At the same time, for any M⊂A, $M \not\subset C$, $C \not\subset M$, m ^β(M)=0 and β ^T(M)=β ^T(D), where D is the largest subset included in C such that D⊂M. Due to monotonicity of ν, ν(D)≤ν(C), hence β ^T(M)≤β ^T(C), and thus β ^T(A)=β ^T(C)=⋁ _B⊆̷A β ^T(B). □

Proof of Theorem 9

A criterion for total monotonicity is the non-negativity of all Möbius transform coefficients. We will demonstrate that m ^β is actually a permutation of m ^T. First, we will show that the number of elements equal to zero is identical:

(42)

Fix the coordinates corresponding to i, j for all $(i,j) \in N^{2}_{T}$ in the characteristic vectors of all elements from 2^N. Apparently, the number of vectors which can be formed by various combinations of values for the remaining coordinates does not depend on whether the fixed values are 0 or 1. Hence, the number of elements equal to zero is identical in both capacities. Next, we will show that the values

(43)

correspond to the values

(44)

It is sufficient to notice that $\tilde{A} = A \cup C$, where C={j|i∈A, $j \notin A, (i,j) \in N^{2}_{T} \rbrace= \lbrace j \vert i \in\tilde{A}$, $j \in\tilde{A}$, $(i,j) \in N^{2}_{T} \rbrace$. □

In order to prove Theorem 10 we would introduce a few notions. Capacities are those from (26) and form a disjunctive decomposition of the capacity ν.

Definition 9

We will call capacities and non-∨_2m -joinable, if the capacity is not 2-monotone, and there do not exist capacities such that is 2-monotone.

Since we analyze 2-additive capacities, their Möbius transform can have negative coefficients only for sets of the form {a,b}, i.e. for sets of power 2. Denote the set of all unordered pairs {a,b} having a negative Möbius coefficient as K. In other words,

(45)

To simplify the notation we will write ab instead of {a,b} and Ya instead of Y∪{a}.

Denote the subset of N comprised of elements that are included in at least one pair from K as N _K . In the following proofs we will denote partial orders over N _K induced by various combinations of orderings (with relation <) on each of the pairs from K as P _i . Note, that not every combination induce a partial order, see Theorem 11 for details. The elements of N not included into at least one pair in K do not influence the 2-monotonicity of the capacity, and therefore will be excluded from the analysis.

We will prove that it is possible to find at least (−1)^p χ(−1) (see Theorem 11) necessity measures which are pairwise non-∨_2m -joinable. In particular, it will be shown that it is possible to pick a necessity measure corresponding to each partial order so that these measures are pairwise non-∨_2m -joinable. We assume here that the graph made of the pairs from K does not contain disconnected parts. Otherwise, the proof below can be applied to each part separately. In this case the number of resulting measures would be equal to product of numbers generated by each part (recall also Theorem 11 and the fact that the chromatic polynomial of a disconnected graph equals to the product of chromatic polynomials of its connected parts (e.g. Read 1968)).

Each partial order P _i corresponding to some combination of orderings on pairs from K can have several linear extensions. For the proofs we will additionally require that the chosen extension complies with the following rule. Index all elements included in pairs from K with numbers from 1 to m. For a given partial order, if two or more elements are incomparable put that with the lowest index first when constructing the linear extension.

Lemma 2

For any two chains , there exists a (possibly empty) set Y⊂N _K and elements a,b∈N _K such that , . If a and b are such that {a,b}∉K, then , and there exist elements c,d∈N _K ∖Y such that {a,c}∈K, {b,d}∈K.

Proof

Fix two arbitrary chains and corresponding to different partial orders P ₁ and P ₂. Set Y=∅. Look at the following elements of each chain: , i=1. Assume they are equal. Then, add an element $c^{1}_{1}$ to the set Y and move on to the next elements , , i=2. Assume, at some step i we have found elements , such that $c^{1}_{i} \neq c^{2}_{i}$. If such elements could not be found the chains are identical.

Assume, $\{c^{1}_{i},c^{2}_{i}\} \in K$. Then, the required elements have been found. In the opposite case $\{c^{1}_{i},c^{2}_{i}\} \notin K$ verify if , . If this holds, the required elements have been found. If this does not hold, two options are possible. Assume , . Then, add $c^{1}_{i}$, $c^{2}_{i}$ to Y and move on to the next elements of the chains (i.e., let i=i+2). Note, that this means that all pairs containing $c^{1}_{i}$ and $c^{2}_{i}$ have the same ordering in partial orders P ₁ and P ₂. If $Yc^{1}_{i}c^{2}_{i}$ is only included in one chain, e.g. , , then look at the elements with index i=i+1, which will have the form , . Add the element $c^{2}_{i}$ to the set Y and repeat the reasoning for the pair , . Again, note that all pairs including $c^{2}_{i}$ have the same ordering in both partial orders. If on some iteration we have reached the end of the chains, and the required combination of elements have not been found, then all pairs of elements from K have the same ordering, hence chains and correspond to the same partial order, which contradicts the initial assumptions.

Now, assume the sets Ya and Yb such that {a,b}∉K, Yab∉C ₁, Yab∉C ₂ have been found. If for either a or b there do not exist elements c and d, c,d∈N _K ∖Y, such that {a,c}∈K and {b,d}∈K, then all elements c _i ,d _j such that {a,c _i }∈K, {b,d _j }∈K belong to Y. Therefore, either a or b must be the element with the lowest index from the set N _K ∖Y (according to the agreement above). It follows that Yab∈C ₁ or Yab∈C ₂ which is again a contradiction. □

Corollary 1

Observe, that by construction all pairs comprising elements from Y have the same ordering in both chains. Therefore, either there exist {w ₁,w ₂}∈K, w ₁,w ₂∈N _K ∖Y, or the chains correspond to the same partial order.

Lemma 3

Capacities and are non-∨_2m -joinable, if there exist , , where , —are maximal chains such that , , {a,b}∈K.

Proof

The 2-monotonicity condition is

$$ \nu(A \cup B) - \nu(A) - \nu(B) + \nu(A \cap B) \geq0, \quad\forall A,B \subset N. $$

(46)

We will write it down for the element Yab:

(47)

Since ν is 2-additive for all A⊂N holds (Grabisch 1997):

$$ \nu(A) = \sum_{i,j \in A} \nu(ij) - \bigl(\vert A \vert- 2\bigr) \sum_{i \in A} \nu(i). $$

(48)

Thus,

(49)

And according to the definition of the measures (see (27)) and the conditions of the lemma .

From the definition of the measures follows and . Therefore, the inequality holds for any combination , i.e. . Therefore, capacities and are non-∨_2m -joinable. □

Note

Since {a,b}∈K, conditions of the lemma imply that and correspond to different partial orders.

Lemma 4

is not 2-monotone if there exist , , , , {a,b}∉K, where , , , , and P ₁ and P ₂ are different partial orders.

Proof

Yab does not belong to maximal chains corresponding to or . Hence, (see (27)). Also, since by construction Y belongs to at least one of the chains. Without loss of generality assume ν(Ya)≥ν(Yb). Therefore,

(50)

Now take sets Ybc and Yc such that {b,c}∈K. According to Lemma 2 such a pair can always be found. Due to monotonicity of ν it holds:

(51)

Since {b,c}∈K, it holds m(bc)=ν(bc)−ν(b)−ν(c)<0. Therefore, ∑ _i∈Y [ν(i)+ν(b)−ν(ib)]−ν(b)<0. Thus the capacity is not 2-monotone. □

Lemma 5

and are non-∨_2m -joinable if there exist , , , , {a,b}∉K, where , , , , and P ₁ and P ₂ are different partial orders.

Proof

According to the result of Lemma 4 the capacity is not 2-monotone. We will show that the capacity is also not 2-monotone. It was shown, that if Yab is included in neither of the chains, corresponding to the measures and , it holds:

(52)

Since Y belongs to at least one of the chains, the only way to reverse the inequality is by increasing the value . It can be done by adding another necessity measure which corresponds to a chain containing Y′ab and not containing Yab, where Y′=Y∖y, and y—is an arbitrary element of Y. At the same time, ν(Y′ab) must be greater than max(ν(Ya),ν(Yb)). However, in this case, the conditions of Lemmata 3 or 4 hold for the added necessity measure and both of the old (consider pairs ((Y′a)b,(Y′a)y), and ((Y′b)a,(Y′b)y)), so that the obtained capacity is still not 2-monotone.

Another option is to add a necessity measure, which corresponds to a chain including Yab. We then look at pairs of sets (Yab,Yax ₁) and (Yab,Ybz ₁), where Yax ₁ and Ybz ₁ are the next elements in chains, corresponding to capacities and (see Fig. 15). The following options are possible

• 1. {x ₁,b}∈K or {z ₁,a}∈K. Lemma 3 applies.

• 2. {x ₁,b}∉K, {z ₁,a}∉K.

  Look at the pairs (Yax ₁,Yab) and (Ybz ₁,Yab) (in the special case x ₁=z ₁ look at the pairs (Yaz ₁,Yab), (Yab,Ybz ₁), and (Yz ₁ a,Yz ₁ b)). The sets Yabx ₁ and Yabz ₁ must belong to the union of the chains corresponding to capacities  . Otherwise, Lemma 4 applies. Therefore, it is necessary to add necessity measures which correspond to chains containing these sets. However, after adding them three new pairs of sets will emerge for which either Lemma 3 or 4 can be applied. These are the pairs (Yax ₁ x ₂,Yabx ₁), (Yabx ₁,Yabz ₁) and (Yabz ₁,Yaz ₁ z ₂). Hence, we would be required to add yet another necessity measures, corresponding to chains containing the joins of sets in every pair. By repeating this process we will obtain a structure, generated by joins of all elements from chains Y⊂Ya⊂⋯⊂N _K (sub-chain of ) and Y⊂Yb⊂⋯⊂N _K (sub-chain of ). It is in fact a distributive lattice (Birkhoff 1967). This entails the existence of two elements YaZw ₁ and YbZw ₂, where Z⊂N _K ∖Yw ₁ w ₂ and {w ₁,w ₂}∈K (see Corollary 1). Thus Lemma 3 applies.

The process is illustrated in Fig. 15. On the left half of the figure we provide the case when elements z _i ,i=1,…,k and x _i ,i=1,…,k are distinct, and on the right a more general one. The boxed sets are those formed when adding new necessity measures.

Fig. 15

Set creation by joining maximal chains

 □

Proof of Theorem 10

Lemmata 2, 3, 5 show that for a 2-additive capacity ν and a set of pairs {i,j}∈K it is possible to find at least one necessity measure for each partial order induced by the combination of orderings on these pairs, such that all thus selected measures are pairwise non-∨_2m -joinable. At the same time, the algorithm presented in Figs. 5, 6, 7 will produce exactly this number of 2-monotone measures (see Theorem 11). Therefore, the obtained decomposition is minimal. □

1.2 A.2 Application example data

Table 9 Protection

Table 10 Technical measures

Table 11 Hosting

Table 12 Office

Table 13 Office network

Table 14 Office servers

Table 15 Staff

Table 16 Staff technical

Table 17 Access control

Table 18 HR