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In document BOLETÍN OFICIAL DEL ESTADO (página 31-35)

When we have determined an inner approximation IP S of P S, some points in IP S may be dominated by other points in IP S. For a decision maker it is not optimal to choose

one of these dominated points. Therefore, we want to determine all points in IP S that are not strongly or weakly dominated by other points in IP S. We refer to these points as non-IP S-dominated points and denote the set of all non-IP S-dominated points by IP SnId. The set conv{IP S, D} and the following lemma and proposition can be used to

determine IP SnId.

Lemma 7.9. Consider z ∈ conv{IP S, D} with z /∈ IP S. Then z is weakly dominated by a point in IP S.

Proof. Let us denote all points in IP SE by z1, . . . , zm. As z ∈ conv{IP S, D}, we know

that z can be written as z = Piλizi+

P

i,jµi,jdj(zi) with

P

iλi +

P

i,jµi,j = 1, λi ≥ 0

and µi,j ≥ 0 for i = 1, . . . , m, j = 1, . . . , k. Furthermore, because z /∈ IP S, we know that

µi,j > 0 must hold for at least one combination of i and j. By definition, each dummy

point dj(zi) ∈ D is weakly dominated by the point zi ∈ IP SE from which it is generated.

Define z∗ = P

iλizi+

P

i,jµi,jzi as the point obtained by replacing each dummy point

dj(zi) with zi in the convex combination that formed z. Then this point z is an element

of IP S and weakly dominates z.

The above lemma thus shows that we should only consider points z ∈ IP S, even when using z ∈ conv{IP S, D} as an inner approximation.

Proposition 7.18. Consider zIP S ∈ IP S. Then zIP S is not strongly dominated by

another point in IP S if and only if zIP S is on a relevant facet of conv{IP S, D}.

Proof. Assume that zIP Sis not strongly dominated by another point in IP S. This implies

that there exists no bz ∈ conv{IP S, D} such that bz < zIP S and that zIP S should thus be

part of a facet F of conv{IP S, D}. As zIP S ∈ IP S, we know that zIP S 5 zub must hold

and—according to Lemma 7.7—facet F should be a relevant facet of conv{IP S, D}. Let us now assume that zIP S is on a relevant facet F of conv{IP S, D} and, by

contradiction, that there exists a point ez ∈ IP S such that ez strongly dominates zIP S. If

H(w, b) is a supporting hyperplane at F , then all points z ∈ conv{IP S, D} must satisfy w>z ≥ b. Furthermore, because of Lemma 7.8, w>zIP S = b and w ≥ 0. So

w>ez < w>zIP S = b.

This inequality shows that ez cannot be an element of IP S. By contradiction, we have thus proven that zIP S cannot be strongly dominated by another point in IP S.

Taking the points on the relevant facets of conv{IP S, D} that are also in IP S, we obtain a set of points in IP S that are not strongly dominated by any other point in IP S. The added dummy points thus make it quite easy to detect and discard the strongly dominated points. However, the set can still contain points that are weakly dominated by a point

in IP S. Consider, for instance, the points z1 = (1, 0, 1)>, z2 = (0, 1, 1)> and z3 =

(0.5, 0.5, 0)>. For an MOP with three objectives, these points could be elements of IP S

and be on a facet of conv{IP S, D}. However, the point ez = 0.5z1+ 0.5z2 = (0.5, 0.5, 1)>

then also has these properties, but is weakly dominated by z3. Therefore, we need a

different criterion to discard the weakly dominated points. The following proposition provides such a criterion.

Proposition 7.19. Consider zIP S ∈ IP S. Let W be the set of inner unit normals of the

relevant facets of conv{IP S, D} containing zIP S and let w+ be the sum of all w ∈ W .

Then zIP S is not weakly dominated by a point in conv{IP S, D} if and only if w+> 0.

Proof. Assume w+ > 0. This implies that for each dimension d = 1, . . . , k, there exists an

inner normal wd ∈ W for which it holds wd

d > 0. Let H(wd, bd) be the corresponding sup-

porting hyperplane of the facet, so that, by definition, wd>z ≥ bd for z ∈ conv{IP S, D}

and wd>zIP S = bd. As wd is an inner normal of a relevant facet, Lemma 7.8 implies

wd≥ 0. If bz is a point that weakly dominates zIP S, then there is a dimension d such that

b

zd< zdIP S. Because wd ≥ 0 and wdd> 0, it follows that wd>bz < wd>zIP S = bd, so bz is not

an element of conv{IP S, D}. Since w+ > 0, there thus exists no bz in conv{IP S, D} that

weakly dominates zIP S.

Let us now assume that zIP S is not weakly dominated by a point in conv{IP S, D}.

This implies that any point ez satisfying ezd < zdIP S and ezi = ziIP S for i 6= d is not in

conv{IP S, D}. As this result holds for any such point ez, there must exist a supporting hyperplane H(wd, bd) of conv{IP S, D} such that wd>zIP S = bd and wd>z < be d. Because

e

zd− zdIP S < 0 and

wdd(ezd− zdIP S) = wd>(ez − zIP S) < bd− bd = 0,

it follows that wd

d > 0. The vector wd must be in W as only relevant facets can contain

points zIP S 5 zub. Furthermore, w ≥ 0 for all w ∈ W and wd

d > 0 implies that wd+

must also be greater than zero. This result holds for every dimension d = 1, . . . , k, so w+> 0.

To determine IP SnId, we can check all points on relevant facets of IP S with the criterion

in Proposition 7.19. However, as the relevant facets of IP S contain infinitely many points, determining IP SnIdrequires a more efficient method. The set W used in Proposition 7.19,

however, is the same for all points on a face of IP S. This implies that we can check all points on a face simultaneously. If IP S is the convex hull of a finite number of points, the number of faces of IP S is also finite. As this is true for all previously discussed sandwich algorithms, we can use the following algorithm, inspired by XNISE2 in Solanki et al. (1993), to determine IP SnId:

1. Set d = k and IP SnId= ∅.

2. Denote by Pd the set of all (d − 1)-faces of conv{IP S, D} having d extreme points

in IP S.

3. For each face in Pd, determine if it is a subset of a multi-dimensional face in IP SnId.

If so, remove the face from Pd.

4. For each remaining face in Pd, calculate the vector w+ by taking the sum of the

inner unit normals of the facets of which this (d − 1)-face is a subset. If w+ > 0,

add the face to IP SnId.

5. Set d = d − 1.

6. If d ≥ 1, return to Step 2. Otherwise, stop.

To illustrate the above algorithm, we use the example in Figure 7.4. When we take IP S = conv{z1, z2, z3, z4} and zub = [1 1 1]>, the figure shows all relevant facets of

conv{IP S, D}. The Pareto optimal points are on the facet with extreme points z1, z2,

and z3, and on the edge with extreme points z1 and z4. In the algorithm, we look at the

2-, 1- and 0-faces, which in three dimensions correspond to facets, edges, and extreme points, respectively. 0 1 2 3 0 1 2 3 0 1 2 3 z2 z3 z1 d3(z1) z4 d3(z4) d1(z1) d1(z4)

Figure 7.4: Example of determining non-IP S-dominated points in IP S.

We start by looking at all facets in the set P3, which in this example contains only the

facet with extreme points z1, z2, and z3. As the inner normal w of this facet satisfies w >

0, we have w+ > 0 and the facet is added to IP SnId in Step 4. Next, we take d = 2 and

continue with the set P2 containing all edges of conv{IP S, D} connecting two extreme

points of IP S. This implies that P2 = {(z1, z2), (z1, z3), (z2, z3), (z1, z4)}. As the first

in Step 3. For the remaining edge (z1, z4), we determine vector w+by summing the inner

normals of the facets defined by {z1, z4, d1(z4), d1(z1)} and {z1, z4, d3(z4), d3(z1)}. The

first facet has an inner unit normal w1 with w1

1 = 0 and w12, w13 > 0. For the inner unit

normal w2 of the second facet it holds that w2

3 = 0 and w12, w22 > 0. For all points on

the edge (z1, z4), the vector w+ is equal to w1+ w2 > 0. The edge (z1, z4) should thus

be added to the set IP SnId according to Proposition 7.19, and this is exactly what the

algorithm does. Finally, we look at all points in P1 = {z1, z2, z3, z4}. As all these point

are already in IP SnId, the set P1 becomes empty in Step 3 and the algorithm stops.

In document BOLETÍN OFICIAL DEL ESTADO (página 31-35)

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