CAPÍTULO 2 MODELADO Y SIMULACIÓN EN MATLAB
2.1 INTRODUCCIÓN AL MODELAMIENTO DEL INTERCAMBIADOR DE CALOR
LetB be the set of maximal elements inF with respect to set inclusion, i.e., the set of all hospital-feasible contracts. To prove the claim, it is enough to show that B satisfies the following property: For allY, Z ∈ Band y∈Y \Z, there exists z ∈ Z \Y such that Y −y+z ∈ B (e.g., see Theorem 39.6 of Schrijver (2003)).
LetY, Z ∈ B,y∈Y\Z andy∈Yhfor some hospitalh∈H. If|Yh| ≤ |Zh|
then there exists z ∈ Zh \Yh which also satisfies Y −y +z ∈ B because
ξ(Y) =ξ(Y −y+z). In the rest of the proof, we assume that |Yh| > |Zh|.
Since |Y| = |Z|, there exist a hospital h′ ∈ H and a contract z such that
|Yh′| <|Zh′| and z ∈ Zh′ \Yh′. Let Y′ = Y −y+z. Obviously, Y′ satisfies |Y′|=n and |Y′
i| ≤qi for all i∈H. If δ is the Chebyshev distance function,
Y′ is hospital-feasible because|ξ
h(Y′)−ξh∗| ≤max{|ξh(Y)−ξh∗|,|ξh(Z)−ξh∗|}
and |ξh′(Y′)−ξh∗′| ≤max{|ξh′(Y)−ξh∗′|,|ξh′(Z)−ξh∗′|}. For the remainder of
the proof, we assume δrepresents the Manhattan distance. If |Yh|> ξh∗ then
Y′ is hospital-feasible because|ξ
h(Y′)−ξh∗|=|ξh(Y)−ξh∗| −1. Let us assume
that ξ∗
h ≥ |Yh| > |Zh|. If |Yh′| < ξh∗′ then Y′ is hospital-feasible because
δ(ξ(Y′), ξ∗) =δ(ξ(Y), ξ∗). We finally assume that for eachi∈Hif|Y
i|<|Zi|
thenξ∗
{i:|Yi|< ξ∗i} ∪ {h} ⊆ {i:|Zi|< ξ∗i} and δ(ξ(Y), ξ∗) = 2 ∑ i:|Yi|<ξ∗ i (ξi∗− |Yi|)≤2 ∑ i:|Yi|<ξ∗ i (ξi∗− |Zi|) ≤2 ∑ i:|Zi|<ξ∗i (ξi∗− |Zi|) = δ(ξ(Z), ξ∗),
where the first equality follows from the assumption |Y| = ∑iξ∗
i, the first
inequality results from the assumption that |Yi| ≥ |Zi| for all i ∈ H with
ξ∗
i > |Yi|, the second inequality follows from the relation {i : |Yi| < ξi∗} ∪
{h} ⊆ {i:|Zi|< ξi∗}, and the last equality results from the assumption that
|Z| = ∑iξ∗
i. Also, note that at least one of the above inequalities holds
strictly. This impliesδ(ξ(Y′), ξ∗) = δ(ξ(Y), ξ∗) + 2≤δ(ξ(Z), ξ∗), that is, the
hospital-feasibility of Y′.
B.5
Proof of Proposition 5
Proof. If R is a laminar family, the family of hospital-feasible contracts is given as: {X′ ⊆ X | |X′
r| ≤ qr (∀r ∈ R),|Xh′| ≤ qh (∀h ∈ H)}. This is a
laminar matroid (Definition 10).
IfR is not a laminar family, there exist two regionsr,r′ such that each of
r∩r′, r\r′, andr′\r is non-empty. Let us chooseh
1, h2, andh3 fromr∩r′,
r\r′, and r′ \r, respectively. Also, let us choose q
r = 1, qr′ = 1, (d, h1) ∈ Xh1, (d ′, h 2) ∈ Xh2, and (d ′′, h 3) ∈ Xh3. Then, X ′ = {(d′, h 2),(d′′, h3)} and X′′ = {(d, h
1)} are hospital-feasible. However, there exists no x ∈ X′ \X′′
such that x+X′′ becomes hospital-feasible.
B.6
Proof of Proposition 6
Proof. We prove the first assertion by induction onj. Clearly,Bbk={{x1, x2. . . , xk}}
satisfies (5). We assume thatj > kandBbj−1 satisfies (5). Let us consider the
situation just before Step 5. For each pair of Z ∈Bbj−1 and Y ∈ Y, (6) holds
forBbj−1∪Y. Thus, for eachY ∈ Y, there existsZ ∈Bbj−1with|Z∩Y|=k−1.
It is sufficient to show that for all Y1, Y2 ∈ Y with |Y1\Y2| ≥ 2, and for all
xi ∈ Y1 \Y2, there exists xi′ ∈ Y2 \Y1 with Y1 −xi +xi′ ∈ Y. We suppose
From symmetry, it is sufficient to prove the case where i = i1. Let us
choose Z1, Z2 ∈ Bbj−1, such that Z1 =Y1−xj +xi0,Z2 =Y2−xj +xi′0, and
|Z1∩Y1|=|Z2∩Y2|=k−1 hold. We consider the following three cases.
Case 1: i0 ∈ {i′1, . . . , ih′}. Without loss of generality, we assume i0 = i′h.
Since Z1 ∈Bbj−1, Y2 ∈ Y, and xi1 ∈Z1\Y2, there existsxi′ ∈Y2\Z1 =
{xi′
1, . . . , xi′h−1, xj} such that Z ′
1 =Z1−xi1 +xi′ ∈Bbj−1 ∪ Y by (5). If
i′ = j then Z′
1 = Y1 −xi1 +xi′h ∈ Y satisfies the required condition.
Thus, we assume that i′ ∈ {i′
1, . . . , i′h−1}. Without loss of generality,
we assume i′ = i′
1. Since Y1 ∈ Y, Z1′ ∈ Bbj−1, and xj ∈ Y1 \Z1′, there
exists xℓ ∈ Z1′ \Y1 such that Z1′ +xj −xℓ ∈ Bbj−1∪ Y by (6). Since ℓ
is either i′
1 or i′h,Z1′ +xj−xℓ is either Y1−xi1 +xi′1 orY1−xi1+xi′h,
each of which is in Y and satisfies the required condition.
Case 2: i0 = i′0. Since Z1, Z2 ∈ Bbj−1, and xi1 ∈ Z1 \ Z2, there exists
xi′ ∈ Z2 \Z1 such that Z1′ = Z1 −xi1 +xi′ ∈ Bbj−1 by (5). Without
loss of generality, we assume i′ = i′
h. Furthermore, since Z1′ ∈ Bbj−1, Y1 ∈ Y, and xi0 ∈ Z ′ 1 \ Y1, there exists xℓ ∈ Y1 \ Z1′ = {xj, xi1} such that Z′′ 1 = Z1′ − xi0 + xℓ ∈ Bbj−1 ∪ Y by (5). If ℓ = j then Z′′
1 =Y1−xi1 +xi′h ∈ Y satisfies the required condition; otherwise we
haveZ′′
1 =Z1−xi0+xi′h =Y1−xj+xi′h ∈Bbj−1, and hence, we induce Case 1.
Case 3: i0 ̸=i′0 andi0 ̸∈ {i′1, . . . , i′h}. Since Z1, Z2 ∈Bbj−1, and xi0 ∈Z1\Z2,
there exists xi′ ∈Z2\Z1 such that Z1′ =Z1−xi0 +xi′ ∈Bbj−1 by (5),
and hence, we induce Case 1 orCase 2.
To show the maximality ofBb, assume for contradiction that there exists
B′ ⊋ Bb, such that B′ ⊆ B and B′ forms a matroid. Let Y′ be the first
element ofB′\Bbremoved by Algorithm 1, andx
h be the element ofY′ with
the maximum subscript. Then there existsZ ∈Bbh−1 ⊆ B′ such that (6) does
not hold for Y′, Z and Bb
h−1 ∪ Y′, where Y′ is Y in Algorithm 1 just before
Y′ is removed. By the definition ofY′, if Y ⊊{x
1, . . . , xh} is an element of
B′, it is also an element of Bb
h−1∪ Y′. Thus (6) does not hold for Y′,Z and