Kidney Exchange with Immunosuppressants∗
Youngsub Chun†
Eun Jeong Heo‡
Sunghoon Hong§
October 2, 2015
Abstract
This paper investigates implications of introducing immunosuppressants (or suppressants) in
kidney exchange problems. We begin with the standard kidney exchange model without suppressants
and define two versions of the top-trading cycles (TTC) solutions satisfying Pareto efficiency. We
then extend the model by introducing suppressants that make a patient compatible with any donor
by relaxing immunological compatibility constraints. We examine the existence of solutions satisfying
Pareto efficiency together with “monotonicity” and “maximal improvement”. Monotonicity requires
that no patient should get worse off after the introduction of suppressants. Maximal improvement
requires that if one set of recipients of suppressants enables more transplantations than the other
(in terms of set inclusion), then the latter should not be chosen as the set of recipients. We propose
two modified versions of TTC solutions satisfying these requirements.
We also provide simulation analyses based on the actual transplantation data to see how much
reduction we can expect on the current practice of using suppressants. In a blood-type restricted environment, we introduce an algorithm that minimizes the use of suppressants, given that all patients
in the data receive transplantations. Our simulation result suggests that the use of suppressants can
be reduced by 55 percent when the recipients of suppressants are chosen as we propose.
JEL classification Numbers: C71; D02; D63; I10
Keywords: immunosuppressants; kidney exchange; top-trading cycles solutions; Pareto efficiency;
monotonicity; maximal improvement
∗
File name: KEI-2015-1001.tex
Seoul National University. E-mail: ychun@snu.ac.kr
‡
Vanderbilt University. E-mail: heoeunjeong@gmail.com
§
Korea Institute of Public Finance. E-mail: sunghoonhong@kipf.re.kr
†
1
Introduction
When a patient suffers from End Stage Renal Disease (ESRD) and needs to receive a kidney transplantation, three options are available depending on the immunological compatibility with her own donor.
If the patient is compatible with the donor, a direct transplantation within the pair can be performed
immediately. Otherwise, she has to list herself in a waitlist to get a transplantation from a deceased
donor, or she has to participate in a kidney exchange program to seek other incompatible patient-donor
pairs. Unfortunately, the transplantations from deceased donors or through exchanges are very limited
in general to cover the increasing number of patients in the waitlists, as the KONOS (Korean Network
for Organ Sharing) data show in Tables 1 and 2.
Recent developments in immunosuppressive protocols brought in the fourth possibility of getting
transplantations. Immunosuppressants, or simply suppressants, relax immunological compatibility constraints, which are mainly determined by ABO blood types, HLA (Human Leukocyte Antigen) typing,
and crossmatching.1 By using a suppressant, a patient becomes compatible with any donor, being
able to receive incompatible kidney transplantations. It is reported that the long-term survival rates of
incompatible kidney transplantations using suppressants are equivalent to those of compatible kidney
transplantations.2
This new protocol has led to the increased practice of incompatible kidney transplantations in a
number of countries including Germany, Japan, Korea, Sweden, and the United States. In Korea, for
example, the proportion of ABO-incompatible living donor kidney transplantations using suppressants
has been increased from 4.7 percent in 2009 to 21.2 percent in 2014 as shown in Table 2.3 Such a sharp
increase is due to the National Health Insurance Service (NHIS) of Korea that covers a large fraction of
the total cost of using suppressants since 2009. Patients are paying a small share, down to 20 percent,
of the total cost, depending on their medical conditions.
As the number of such incompatible kidney transplantations increases, the NHIS expenditure also
increases to subsidize the cost. Since the NHIS has a limited budget assigned for suppressants each year,
we should ask whether suppressants (and the corresponding expenditure) are allocated in an efficient
manner and there is a room for further improvement.
In this paper, we investigate whether it is possible to reduce the use of suppressants while maintaining the same set of patients receiving transplantations. We also investigate how to assign the unused
1
Human Leukocyte Antigens (HLAs) are proteins on the surface of cells that are responsible for immune systems.
There are three subsets of HLAs, namely, HLA-A, HLA-B, and HLA-DR. Within each of these subsets, there are many
different HLA proteins, say HLA-A1, HLA-A2, and so on. If the donor and the patient share the same HLAs, they are
called an identical match. Two siblings are an identical match with probability of 25 percent. A parent and a child share
the half of HLAs. On the other hand, if the crossmatch is positive between the patient’s antibodies and the donor’s HLAs,
the patient’s antibodies will attack the organ transplanted from the donor, and thus, transplantation is incompatible. In
addition to immunological compatibility, kidney size and age may also affect transplantation outcomes.
2
There has been a variety of immunosuppressive protocols reported in the medical literature but most of these protocols
include the use of rituximab (which is an immunosuppressant medicine) and plasmapheresis (which is a procedure to remove
harmful antibodies from the blood of patients). For studies on long-term outcomes of incompatible kidney transplantations,
see Takahashi et al. (2004), Tyden et al. (2007), Montgomery et al. (2012), and Kong et al. (2013).
3
In contrast, the proportion of living donor kidney transplantations through an exchange program had been decreased
from 5.3 percent to near zero percent during the same period.
1
Year
2009
2010
2011
2012
2013
2014
Patients
in waitlists
4,769
5,857
7,426
9,245
11,381
13,612
Deceased donor
transplants
488
491
680
768
750
808
Living donor
transplants
750
796
959
1,020
1,011
1,000
Table 1. Kidney transplantations in Korea
Year
2009
2010
2011
2012
2013
2014
Living donor
transplants
in total
750
796
959
1,020
1,011
1,000
Transplants
within
compatible pairs
675 (90.0%)
689 (86.6%)
828 (86.3%)
827 (81.1%)
795 (78.6%)
783 (78.3%)
Transplants
using
suppressants
35 (4.7%)
78 (9.8%)
113 (11.8%)
193 (18.9%)
212 (21.0%)
212 (21.2%)
Transplants
through
exchanges
40 (5.3%)
29 (3.6%)
18 (1.9%)
0 (0.0%)
4 (0.4%)
5 (0.5%)
Table 2. Living donor kidney transplantations in Korea: percentage of cases to total shown in parentheses
suppressants to the remaining incompatible pairs. For an illustration, consider a case where the immunological compatibility is determined only by ABO blood types. A pair of a patient and a donor
is represented as type X-Y, where the patient’s blood type is X and the donor’s type is Y. Suppose
that there are three pairs of types A-B, B-AB, and O-AB.4 Since each patient is incompatible with her
own donor, no patient can get a transplantation within the pair. In a kidney exchange program, on
the other hand, patients switch their donors and get transplantations across pairs, if possible. Unfortunately, there is no such “trading cycle” among these pairs, since the patient of pair A-B is incompatible
with the donor of pair B-AB and the patient of pair O-AB is incompatible with any other donor.
Now suppose that the NHIS can afford at most two patients to use suppressants. One possibility
is that two pairs, for example, A-B and O-AB, receive suppressants and they perform direct transplantations within the pairs. Note that the “Transplants using suppressants” in the fourth column of
Table 2 represents this case. The remaining pair B-AB can not receive a transplantation. We find that,
however, there exists a more efficient way of using suppressants. Note that the two pairs A-B and B-AB
form a kind of “chain” in that the donor of A-B is compatible with the patient of pair B-AB, although
the remaining patient and donor in these pairs are not compatible. Such a chain can be viewed as a
4
The blood type X specifies the types of antigen and antibody a person has. For example, if a person’s blood type is
A, then her antigen is type A and her antibody is type B; if a person’s blood type is AB, then her antigen is type A and
type B, and she has no antibody. A person with antibody X cannot get a transplantation from a donor with antigen X.
For example, if a person’s blood type is A, then her antibody is type B, and thus, she cannot get a transplantation from
any donor having antigen B, namely, blood types B and AB. Similarly, if a person’s blood type is O, then her antibody
is A and B, therefore, she cannot get a transplantation from any donor of blood types A, B, and AB.
2
trading cycle with a “missing link”. We point out that a suppressant can be used to fill this missing
link to form a trading cycle. That is, it is possible that the patient of pair B-AB gets a transplantation
from the (compatible) donor of pair A-B, while the patient of pair A-B uses a suppressant and gets
a transplantation from the (initially incompatible) donor of pair B-AB. The pair O-AB can use the
remaining suppressant and perform a direct transplantation within the pair. Then, all patients receive
transplantations.
In what follows, we formalize this idea, together with additional considerations on fairness. Before
we proceed, it is worth emphasizing that our analysis assumes the voluntary participation of patientdonor pairs who become compatible with any donor by using suppressants. In the example above, if
the patient of pair A-B uses a suppressant, then she can get a transplantation directly from her own
donor, without having to participate in the exchange program. What we propose, however, is that the
pair A-B participates in the program to benefit the pair B-AB without bearing any welfare loss. As
long as it results in transplantations that are equally successful as those within pairs, it is plausible to
add such “altruistic” pairs to the exchange program, as proposed in the proceeding literature (Sönmez
and Ünver, 2014; Roth et al., 2005; Gentry et al., 2007).
We begin with the standard kidney exchange model without suppressants. We define two TopTrading Cycles (TTC) solutions, without restricting the size of exchanges.5 Both solutions are defined
by an algorithm in which patient-donor pairs form trading cycles at each step, subject to the compatibility constraints. We propose two different ways of choosing trading cycles among multiple cycles,
which define two versions of TTC. We show that both are Pareto efficient for all priority orderings.
We then extend the model by introducing suppressants. To reflect the limited availability of suppressants, we assume that at most k patients can use suppressants. For each problem, we have to decide
(i) the set of at most k recipients of suppressants and (ii) the matching of pairs who participate in the
exchange program.
We first require that all patients should be weakly better off after suppressants are introduced. We
regard it as a fairness requirement: since the use of suppressants is (fully or partially) subsidized by
public health insurance, as in Korea, its benefit should be distributed to everyone. We refer to this
requirement as “monotonicity”.
We next require that the set of recipients of suppressants should be chosen to maximize transplantations. This is an efficiency requirement in assigning suppressants subject to the limited availability.
Consider two sets of potential recipients of suppressants. They can result in two different sets of patients getting transplantations in the end. If one set of recipients enables more transplantations than
the other in terms of set inclusion, then it is reasonable not to choose the latter set. We refer to
this requirement as “maximal improvement”. We can also formulate the same idea, but in terms of
5
Shapley and Scarf (1974) propose the TTC solution in a general model with indivisible goods. Roth et al. (2004)
develop the TTC solution for kidney exchange problems by considering chains formed with patients on waitlists. Note
that Roth et al. (2004) impose no constraint on the size of exchanges, i.e., the length of cycles or chains, when designing
their Top-Trading Cycles and Chains (TTCC) solution. For more discussion on the size of exchanges, see Roth et al.
(2007) and Saidman et al. (2006).
3
total number of transplantations, not in terms of set inclusion. We refer to it as “cardinally maximal
improvement”.
We present two impossibility results. First, there is no solution satisfying Pareto efficiency and a
strong form of monotonicity, which requires that all patients should be weakly better off after suppressants are introduced, no matter who becomes recipients of suppressants. Given this impossibility,
we weaken monotonicity requirement: all patients should be weakly better off for some recipients of
suppressants. In other words, it requires that, for each problem, there exists a “right” set of recipients
who can make all patients weakly better off by using suppressants. Unfortunately, this requirement is
again incompatible with Pareto efficiency and cardinally maximal improvement.
We therefore turn to see if there is any solution satisfying the weak notion of monotonicity, Pareto
efficiency, and maximal improvement. We introduce two TTC solutions for the extended model. Each
of these extended TTC solutions runs in four stages. First, apply the two standard TTC solutions
respectively, assuming that no one uses a suppressant; and identify the sets of patients who receive
no transplantations under these solutions, respectively (this is the only step where the two versions
of extended TTC solution diverge). Second, modify an initial priority ordering so that the patients
identified in the previous step have lower priorities than the remaining patients. Third, select feasible
cycles and chains according to the modified priority ordering, subject to the availability of suppressants;
and assign suppressants to the patients at the end of these selected chains. Finally, obtain a new
compatibility profile and apply the first standard TTC solution associated with the modified priority
ordering. We prove that these extended TTC solutions satisfy the aforementioned requirements.
To see how much we can improve on the current practice of using suppressants, we provide simulation
results based on the actual transplantation data in South Korea. The KONOS data provides the profiles
of blood types of all patient-donor pairs who received living donor kidney transplantations during
the recent four years (2011-2014). Assuming that the ABO blood-types determine immunological
compatibility, we identify seven types of incompatible pairs, A-B, B-A, A-AB, B-AB, O-A, O-B, and
O-AB, who used suppressants to get transplantations.
We propose an algorithm that minimizes the use of suppressants given that all incompatible pairs
in the data receive transplantations. This algorithm runs as follows. We first identify all trading
cycles. We match patients and donors along these cycles and then exclude them from the pool. We
next identify all 3-chains (a k-chain is a chain consisting of k pairs), which is the longest chain in this
setting. We assign suppressants to the tails of these chains, match patients and donors along the chains,
and exclude them from the pool. We repeat this process for 2-chains, and then for 1-chains.
We compare what we obtain from the algorithm with the actual use of suppressants in the KONOS
data. According to our algorithm, 340 pairs need to use suppressants, while 757 patients had used
suppressants in the actual data during the four years. This is a reduction of 55.1 percent in total.
Note that the KONOS data only includes compatible/incompatible pairs that have received transplantations. Therefore, the KONOS data can be biased to represent the whole population of patientdonor pairs (especially, it may underrepresent the population of incompatible pairs). We construct a
4
hypothetical population of 1,001 pairs according to the distribution of ABO blood types in Korea. We
collect all incompatible pairs and run the algorithm. We again obtain a reduction of 55.4 percent.
The rest of this paper is organized as follows. Section 2 introduces the standard kidney exchange
model without suppressants, and defines two versions of top-trading cycles solutions. Section 3 extends the model to include suppressants, and shows two impossibility results. Section 4 provides two
modifications of top-trading cycles solutions, and examines whether both solutions satisfy desirable
properties. Section 5 presents simulation results with an algorithm to minimize the use of suppressants
for incompatible pairs. Section 6 concludes.
2
Standard Kidney Exchange Model without Immunosuppressants
Let N ≡ {1, . . . , n} be the set of patients. Each patient has her donor. For each i, j ∈ N , patient i is
either compatible or incompatible with the donor of patient j. Note that j can be i herself. For each
i ∈ N , let Ni+ ≡ {j ∈ N : patient i is compatible with the donor of patient j} and Ni− ≡ N \ Ni+ . Each
i ∈ N has a weak preference Ri over N as follows: Each patient prefers all elements in Ni+ to all elements
in Ni− . All elements in Ni+ are equally desirable and so are those in Ni− . For patients incompatible with
all donors, without loss of generality, ∅ is preferred to all donors. Note that compatibility of a donor
and a patient is verifiable and determined independently of other pairs’ compatibility. Let Pi and Ii be
the asymmetric and symmetric part of Ri , respectively. Let R be the set of all such weak preferences.
A kidney exchange problem, or simply, a problem is defined as a list (N, R) with R = (Ri )i∈N . Since
we fix N , we represent a problem as R. Let RN be the set of problems.
We introduce a graph whose nodes are patients in N . A directed arc from one node to another, say
i to j, is represented as i → j. We allow j = i, in which case i → i. Let g be the set of such directed
arcs. For each R ∈ RN and each i, j ∈ N , i → j if and only if patient i is compatible with the donor
of patient j at R. Let g(R) be the set of these directed arcs. A problem R is uniquely represented as
graph g(R).
We say that (i1 , . . . , ik , i1 ) forms a cycle in g(R) if i1 , . . . , ik are distinct patients and i1 → i2 →
· · · → ik → i1 . As long as i1 , . . . , ik are ordered, it does not matter who comes first in the list (i1 , . . . , ik ).
Note that a cycle can be composed of a single patient who is compatible with her own donor. Let C(R)
be the set of all such cycles in g(R). Consider a pair of cycles c, c0 ∈ C(R) with c 6= c0 . We say that
c and c0 are feasible if no patient appears both in c and c0 . Let C(R) ⊆ 2C(R) be the collection of all
sets of feasible cycles in C(R). For each C ∈ C(R), let N (C) be the set of patients appearing in the
cycles of C. Let |C| be the number of the cycles in C.
We say that (i1 , . . . , ik ) forms a chain if i1 , . . . , ik are distinct patients and i1 → i2 → · · · → ik and
ik 9 i1 . We call ik the tail of this chain. Let L(R) be the set of all such chains in g(R). Consider a
pair of chains l, l0 ∈ L(R) with l 6= l0 . We say that l and l0 are feasible if no patient appears both in l
and l0 . Let L(R) ⊆ 2L(R) be the collection of all sets of feasible chains in L(R). For each L ∈ L(R), let
N (L) be the set of patients appearing in the chains of L. Let |L| be the number of the chains in L.
5
Consider c ∈ C(R) and l ∈ L(R). We say that c and l are feasible if no patient appears both in
c and l. Let H(R) ≡ C(R) ∪ L(R). Let H(R) ⊆ 2H(R) be the collection of all sets of feasible cycles
and chains in H(R). For each H ∈ H(R), let N (H) be the set of patients appearing in the cycles and
chains of H.
A matching µ specifies which patient is matched to whose donor. That is, for each i, j ∈ N , µ(i) = j
implies that patient i is matched to patient j’s donor. A matching µ is feasible if (i) for each i ∈ N ,
µ(i) ∈ N and |{j ∈ N : µ(j) = i}| = 1 and (ii) for each i ∈ N with µ(i) ∈
/ Ni+ , µ(i) = i. A matching
µ is individually rational at R if for each i ∈ N , µ(i) Ri i. Let M(R) be the set of all feasible and
individually rational matchings at R. For each µ ∈ M(R), let N (µ) be the set of patients receiving
transplantations.
We also introduce a priority ordering over patients.6 Denote a linear ordering over patients by .
For each pair i, j ∈ N , i j if and only if patient i has a higher priority than patient j.
Example 1. Let N = {1, 2, 3}. Suppose that N1+ = {2}, N2+ = {1, 3}, and N3+ = {2, 3}. The
corresponding R and g(R) are given as follows:
R1
2
1, 3
R2
R3
1, 3 2, 3
2
•
1
•
2
•
3
1
The set of cycles in g(R) is C(R) = {c ≡ (1, 2, 1), c0 ≡ (2, 3, 2), c00 ≡ (3, 3)}. Note that c and c00 are
feasible. Since patient 2 appears both in c and c0 , c and c0 are not feasible. The collection of feasible
cycles is C(R) = {{c}, {c0 }, {c00 }, {c, c00 }}. Let C ≡ {c, c00 } ∈ C(R). Then, N (C) = {1, 2, 3}. The set of
chains in g(R) is L(R) = {l ≡ (1), l0 ≡ (2), l00 ≡ (1, 2, 3), l000 ≡ (3, 2, 1)}. Note that l and l0 are feasible.
Since patient 1 appears both in l and l00 , l and l00 are not feasible. The collection of feasible chains
is L(R) = {{l}, {l0 }, {l00 }, {l000 }, {l, l0 }}. Let L ≡ {l, l0 } ∈ L(R). Then, N (L) = {1, 2}. Also, H(R) =
{c, c0 , c00 , l, l0 , l00 , l000 } and H(R) = {{c}, {c0 }, {c00 }, {l}, . . . , {l00 }, {l, l00 }, {c0 , l}, {c00 , l}, {c00 , l0 }, {c00 , l, l0 }}.
A solution is a mapping ϕ : RN ⇒ ∪R∈RN M(R) such that for each R ∈ RN , ϕ(R) ⊆ M(R). We say
that ϕ is essentially single-valued if for each R ∈ RN , each µ, µ0 ∈ ϕ(R), and each i ∈ N , µ(i) Ii µ0 (i).
If ϕ is essentially single-valued, then for each R ∈ RN and each µ, µ0 ∈ ϕ(R), N (µ) = N (µ0 ).7
For each R ∈ RN , we say that µ ∈ M(R) is Pareto efficient at R if there is no other µ0 ∈ M(R)
such that for each i ∈ N , µ0 (i) Ri µ(i) and for some i ∈ N , µ0 (i) Pi µ(i). We say that ϕ is Pareto
efficient if for each R ∈ RN and each µ ∈ ϕ(R), µ is Pareto efficient at R.
We define Top-Trading Cycles (TTC) solutions adapted to this model.8 Without loss of generality,
let 1 2 · · · n. For simplicity, a priority ordering can be written as : 1, 2, . . . , n.
6
For more detailed discussion on these priorities, see Roth et al. (2005).
Suppose, by contradiction, that there are R ∈ RN , i ∈ N , and a pair µ, µ0 ∈ ϕ(R) such that µ(i) ∈ Ni+ and µ0 (i) ∈
/ Ni+ .
+
−
0
Since all patients in Ni are strictly preferred to those in Ni , we have µ(i) Pi µ (i), contradicting that ϕ is essentially
single-valued.
8
The TTC solution is proposed by Shapley and Scarf (1974) in a general model and applied to kidney exchange
problems by Roth et al. (2004).
7
6
Top-trading cycles solution associated with (simply T T C ): For each R ∈ RN , let C0 ≡ C(R).
Step 1. Find a set of feasible cycles C ∈ C0 such that 1 ∈ N (C). If there is no such C, then C1 ≡ C0 .
Otherwise, let C1 be the collection of all such sets of feasible cycles.
Step t ≥ 2. Find C ∈ Ct−1 such that t ∈ N (C). If there is no such C, then Ct ≡ Ct−1 . Otherwise, let
Ct be the collection of all such sets of feasible cycles.
Step n. Obtain Cn . For each C ∈ Cn , each patient in N (C) receives a kidney from the donor of the
patient to whom she is directing and all other patients stay with their own donors. Collect the resulting
matchings for all C’s in Cn .
Note that this solution can also be viewed as a “sequential priority” solution as we first figure out
all matchings that patient 1 receives a transplantation, and then figure out all matchings that patient 2
does, and so on. When only two-way exchanges are allowed between patients and donors, such a sequential priority solution is Pareto efficient and it also maximizes the number of transplantations, which
is another important issue in kidney exchange problems. If we allow more than two-way exchanges,
however, Pareto efficient matchings do not necessarily maximize the number of transplantations. We
thereby propose another version of top-trading cycles solutions.
Top-trading cycles solution maximizing the number of transplantations (simply T T C ): For
each R ∈ RN , let C0 ≡ argmaxC∈C(R) |N (C)|.
Step 1. Find a set of feasible cycles C ∈ C0 such that 1 ∈ N (C). If there is no such C, then C1 ≡ C0 .
Otherwise, let C1 be the collection of all such sets of feasible cycles.
Step t ≥ 2. Find C ∈ Ct−1 such that t ∈ N (C). If there is no such C, then Ct ≡ Ct−1 . Otherwise, let
Ct be the collection of all such sets of feasible cycles.
Step n. Obtain Cn . For each C ∈ Cn , each patient in N (C) receives a kidney from the donor of the
patient to whom she is directing and all other patients stay with their own donors. Collect the resulting
matchings for all C’s in Cn .
Remark. It is straightforward from definitions that T T C and T T C are essentially single-valued.
Proposition 1. For every priority ordering , T T C and T T C are Pareto efficient.
Proof. Suppose, by contradiction, that T T C is not Pareto efficient. Then, there are R ∈ RN , µ ∈
T T C (R), and µ0 ∈ M(R) such that for each i ∈ N , µ0 (i) Ri µ(i) and for some i ∈ N , µ0 (i) Pi
µ(i). This implies that all patients who receive transplantations at µ also receive transplantations at
µ0 . Furthermore, there is at least one patient, say i who additionally receives a transplantation at µ0 .
Among such patient i’s, choose the one with the highest priority under . According to the definition
of T T C , the matchings should be chosen so that patient i receives a transplantation, a contradiction.
Next, suppose, by contradiction, that T T C is not Pareto efficient. Then, there are R ∈ RN , µ ∈
T T C (R), and µ0 ∈ M(R) such that for each i ∈ N , µ0 (i) Ri µ(i) and for some i ∈ N , µ0 (i) Pi µ(i). This
again implies that all patients who receive transplantations at µ also receive transplantations at µ0 and
7
at least one patient additionally receives a transplantation at µ0 . Thus, the number of transplantations
at µ0 is greater than that at µ. Hence, it must be that µ ∈
/ T T C (R), a contradiction.
3
Extended Kidney Exchange Model with Immunosuppressants
We introduce suppressants to the standard model. If patient i receives a suppressant, she gets compatible with all donors, including her own. We denote such preference by Ri∗ . If a set of patients
S receive suppressants, we denote their preferences by RS∗ ≡ (Ri∗ )i∈S . For each R ∈ RN and each
S ⊆ N , let R(S) ≡ (RS∗ , R−S ) denote the preference profile derived from R when patients in S receive
suppressants. Since R(S) ∈ RN , all definitions in the previous section are carried over to this setting
by replacing R with R(S).
Example 2. (Example 1 continued) Consider R and g(R) given in Example 1:
R1
2
1, 3
R2
R3
1, 3 2, 3
2
•
1
•
2
•
3
1
Now suppose that S = {2}. Then, R(S) = (R1 , R2∗ , R3 ) and g(R(S)) are represented as follows:
R1
2
1, 3
R2∗
R3
1, 2, 3 2, 3
•
1
•
2
•
3
1
The set of cycles in g(R2∗ , R−2 ) is C(R2∗ , R−2 ) = {c ≡ (1, 2, 1), c0 ≡ (2, 3, 2), c00 ≡ (3, 3), c000 ≡ (2, 2)}.
Thus, the collection of feasible cycles is C(R2∗ , R−2 ) = {{c}, {c0 }, {c00 }, {c000 }, {c, c00 }, {c00 , c000 }}. The set
of chains in g(R2∗ , R−2 ) is L(R2∗ , R−2 ) = {(1), (1, 2, 3), (3, 2, 1)}.
To accommodate the limited availability of suppressants, we introduce a non-negative integer k to
be an upper bound on the number of patients who can receive suppressants. Let Z+ be the set of
non-negative integers and for each k ∈ Z+ , let 2N (k) be the collection of subsets of at most k patients.
When k = 0, the extended model coincides with the standard model.
A solution specifies for each problem (i) how to assign at most k suppressants to patients, and
(ii) how to match patients to donors at the resulting preference profile. Let σ : RN → 2N (k) be a
S
recipient-selection rule. Let ϕ : RN ⇒ R∈RN M(R) such that for each R ∈ RN , ϕ(R) ⊆ M(R), be a
matching rule. A solution is represented as a pair (σ, ϕ).
The following two requirements are defined for a matching rule, ϕ:
Pareto efficiency: For each R ∈ RN , ϕ(R) is Pareto efficient at R.
8
Our next requirement says whenever some patients receive suppressants – no matter who they are
– everyone should be weakly better off.
Strong monotonicity: For each k ∈ Z+ , each σ : RN → 2N (k), each R ∈ RN , each µ ∈ ϕ(R), each
µ0 ∈ ϕ(R(σ(R))), and each i ∈ N , µ0 (i) Ri µ(i).
Unfortunately, we cannot achieve this requirement together with Pareto efficiency.
Proposition 2. No matching rule satisfies Pareto efficiency and strong monotonicity.
Proof. We present an example of a problem with three patients. Let N ≡ {1, 2, 3}. Let ϕ be a Pareto
efficient matching rule. Consider the following preferences:
R1
R2
R20
R3
2
1
3
2
1, 3
2, 3 1, 2
1, 3
Suppose that k = 0. Let R ≡ (R1 , R2 , R3 ) and R0 ≡ (R1 , R20 , R3 ). By Pareto efficiency, we have
ϕ(R) = {µ = (µ(1) = 2, µ(2) = 1, µ(3) = 3)} and ϕ(R0 ) = {µ0 = (µ0 (1) = 1, µ0 (2) = 3, µ0 (3) = 2)}.
Now suppose that k = 1. Let σ : RN → 2N (k) be such that σ(R) = {2} and σ(R0 ) = {2}. The
new preference profile resulting from σ is (R1 , R2∗ , R3 ) for both problems. To make everyone weakly
better off at the new profile than at R, we should have µ ∈ ϕ(R1 , R2∗ , R3 ). Again, to make everyone
weakly better off at the new profile than at R0 , we should also have µ0 ∈ ϕ(R1 , R2∗ , R3 ). Altogether,
{µ, µ0 } ⊆ ϕ(R1 , R2∗ , R3 ). However, patient 1 is worse off at µ0 than at µ and patient 3 is worse off at µ
than at µ0 , a contradiction to strong monotonicity.
The implication of this result is that, as long as we want to achieve Pareto efficiency, we cannot make
all agents weakly better off by allowing an arbitrary set of patients to receive suppressants. In other
words, sets of patients receiving suppressants should be chosen carefully depending on the specification
of a problem to achieve Pareto efficiency.
Remark. It is usually the case that if a solution is not essentially single-valued, it is harder to satisfy
strong monotonicity. Consider two matchings µ and µ0 chosen by a solution for a problem. Let T be the
set of patients receiving transplantations at µ. Let T 0 be the set of patients receiving transplantations
at µ0 . Suppose that a set of patients receive suppressants. If the solution is essentially single-valued,
T = T 0 and strong monotonicity requires that all patients in T = T 0 receive transplantations at the
new preference profile. If the solution is not essentially single-valued, it may be that T 6= T 0 , and strong
monotonicity requires that all patients in T ∪ T 0 receive transplantations, which is more demanding.
Even if a solution is essentially single-valued, however, Proposition 2 shows that it is impossible to
achieve both requirements. Examples are T T C and T T C as we show below.
Example 3. (T T C and T T C violating strong monotonicity) Let N ≡ {1, 2, 3, 4}, : 1, 2, 3, 4 and
R, R̄ ∈ RN be given as follows:
9
R1
R2
R3
R4
R̄1
R̄2
R̄3
R̄4
2
3
4
2
2
1
4
2
1, 3, 4 1, 2, 4 1, 2, 3 1, 3, 4
1, 3, 4 2, 3, 4 1, 2, 3 1, 3, 4
4
•
3
•
4
•
3
•
•
1
•
2
•
1
•
2
It is easy to check that T T C (R) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)} and T T C (R̄) =
{(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4)}. Now suppose that k = 1 and consider σ : RN → 2N (k) such
that σ(R) = σ(R̄) = {2}. Patient 2’s preference changes to R2∗ as illustrated in the following graph:
4
•
3
•
4
•
3
•
•
1
•
2
•
1
•
2
Let R0 ≡ (R1 , R2∗ , R3 , R4 ) and R̄0 ≡ (R̄1 , R2∗ , R̄3 , R̄4 ). Then,
T T C (R0 ) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4)}
T T C (R̄0 ) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)}.
Under T T C , patients 3 and 4 are worse off while patient 1 is better off and patient 2 remains indifferent.
Under T T C , patient 1 is worse off while patients 3 and 4 are better off and patient 2 remains indifferent.
As shown in the above example, even T T C and T T C , which seem most natural solutions in this
setting, violate strong monotonicity. Having a closer look at this example, however, we observe that it
is possible to make everyone weakly better off by assigning suppressants to carefully chosen patients.
Given R of the above example, consider another assignment of suppressants at which patient 1 receives
a suppressant. While maintaining the cycle (2, 3, 4, 2), we find an additional self-cycle of patient 1,
which allows all patients weakly better off. Similarly, at R̄, suppose that patient 4, instead of patient
2, receives a suppressant. While maintaining the cycle (1, 2, 1), we can find an additional cycle (3, 4, 3),
which allows all patients weakly better off.
Motivated by this example, we propose a weaker requirement than strong monotonicity for a solution
(σ, ϕ). It requires the existence of a set of patients for each problem whose use of suppressants makes
all patients weakly better off.
10
Monotonicity: For each k ∈ Z+ , there is σ : RN → 2N (k) such that for each R ∈ RN , each µ ∈ ϕ(R),
each µ0 ∈ ϕ(R(σ(R))), and each i ∈ N , µ0 (i) Ri µ(i).9
Given the limited availability of suppressants, on the other hand, it makes sense to consider the
most “effective” way of using suppressants. Note that Pareto efficiency of a matching rule only requires
there be no further Pareto improvement from the matchings chosen for a given preference profile. As it
does not say anything about the assignment of suppressants which determines the preference profile, we
propose a new efficiency notion regarding the assignment of suppressants. The following requirement
says that the recipients of suppressants should be chosen to maximize the set of patients getting
transplantations in terms of set inclusion.
Maximal Improvement (MI): For each k ∈ Z+ , each R ∈ RN , and each T, T 0 ∈ 2N (k), if there are
µ ∈ ϕ(R(T )) and µ0 ∈ ϕ(R(T 0 )) such that N (µ0 ) ( N (µ), then T 0 6= σ(R).
Example 4. (Maximal improvement) Let N ≡ {1, 2, 3, 4}. Consider a Pareto efficient solution ϕ
associated with σ. Suppose that R ∈ RN is given as follows:
R1
R2
R3
R4
2
3
4
3
1, 3, 4 1, 2, 4 1, 2, 3 1, 2, 4
4
•
3
•
4
•
3
•
4
•
3
•
•
1
•
2
•
1
•
2
•
1
•
2
S=∅
S 0 = {1}
S 00 = {2}
Consider three cases of assigning suppressants when k = 1. Suppose first that no patient receives a
suppressant. Since the solution is Pareto efficient, the rule selects {µ = (µ(1) = 1, µ(2) = 2, µ(3) =
4, µ(4) = 3)} and then N (µ) = {3, 4}. Suppose instead that patient 1 receives a suppressant. By Pareto
efficiency, the solution selects {µ} and then N (µ) = {1, 3, 4}. Lastly, suppose that patient 2 receives a
suppressant. By the same argument, the solution selects {µ0 = (µ0 (1) = 2, µ0 (2) = 1, µ0 (3) = 4, µ0 (4) =
3)} and then N (µ0 ) = {1, 2, 3, 4}. When patient 2 receives a suppressant, the set of patients receiving
transplantations includes the patients receiving transplantations in the other cases considered above.
Therefore, maximal improvement requires that σ(R) 6= ∅ and σ(R) 6= {1}.
As discussed in Section 2, it is also considered important in kidney exchange problems to maximize
the number of transplantations. We propose the following requirement to accommodate this idea. It
9
Note that any essentially single-valued solutions trivially satisfy monotonicity by setting σ(R) = ∅ for all R ∈ RN . To
avoid such triviality, we impose monotonicity together with an “efficiency” requirement of maximal improvement defined
later.
11
says that the use of suppressants should result in the maximal improvement in terms of the number of
transplantations.
Cardinally Maximal Improvement (CMI): For each k ∈ Z+ , each R ∈ RN , and each T, T 0 ∈ 2N (k),
if there are µ ∈ ϕ(R(T )) and µ0 ∈ ϕ(R(T 0 )) such that |N (µ0 )| < |N (µ)|, then T 0 6= σ(R).
It is straightforward from the definitions that CMI implies MI. Unfortunately, it turns out that we
cannot achieve CMI together with other requirements we consider.
Proposition 3. No monotonic solution satisfies Pareto efficiency and cardinally maximal improvement.
Proof. We present an example of a problem with six patients. Let N ≡ {1, 2, 3, 4, 5, 6} and ϕ be a
monotonic solution satisfying Pareto efficiency and CMI. Then, for each k ∈ Z+ , there is σ : RN →
2N (k) associated to ϕ. Consider the following preferences:
•
1
R1
R2
R3
R4
R5
R6
5
∅
5
1, 2
4
3
N \ {5}
N
4
•
2
•
•
5
•
3
N \ {5} N \ {1, 2} N \ {4} N \ {3}
•
6
•
1
4
•
2
•
•
5
•
3
•
6
Suppose first that k = 0. By Pareto efficiency, ϕ(R) = {µ = (µ(1) = 5, µ(2) = 2, µ(3) = 3, µ(4) =
1, µ(5) = 4, µ(6) = 6)}. Suppose instead that k = 1. To satisfy CMI, it is easy to check that we should
have σ(R) = {2} and all patients except for patient 1 get transplantations at the resulting preference
profile. Since patient 1 gets worse off, monotonicity is violated.
4
Top-Trading Cycles Solutions with Immunosuppressants
We propose two solutions satisfying Pareto efficiency, monotonicity, and maximal improvement. They
are based on the top-trading cycles solutions defined in Section 2, but a major modification is made in
specifying the assignment of suppressants.
The extended top-trading cycles solution, eT T C :
Stage 1. For each R ∈ RN , apply T T C . Denote by N̄ the set of patients who receive no transplantations at any matching selected by T T C . Without loss of generality, let N̄ ≡ {j1 , . . . , jn̄ } and
j1 · · · jn̄ .
Stage 2. Modify into a new priority ordering. All patients in N \ N̄ have higher priories than those
in N̄ , while the relative priorities within N̄ and N \ N̄ remain the same, respectively. Denote this new
12
priority ordering by ∗ (R).
Stage 3. Find a set of feasible cycles and chains. Let
H0 ≡ {H ∈ H(R) : N (H) ⊇ N \ N̄ and the number of chains in H does not exceed k},10 and
for each t ∈ {1, . . . , n̄},

 {H ∈ H
if {H ∈ Ht−1 : jt ∈ N (H)} =
6 ∅,
t−1 : jt ∈ N (H)},
Ht ≡
 Ht−1 ,
otherwise.
Choose any H ∈ Hn̄ and assign suppressants to the tails of all chains in H. Denote the set of these
tails by σ ∗ (R).
Stage 4. Apply to R(σ ∗ (R)) the top-trading cycles rule associated with ∗ (R). For each R ∈ RN ,
let eT T C (R) ≡ T T C∗ (R) (R(σ ∗ (R))).
In the first stage, we apply T T C to the initial preference profile of the standard model and find
the set of patients N̄ who do not receive transplantations. Since T T C is essentially single-valued, the
sets of patients receiving transplantations remain the same across all matchings selected by T T C for
each problem. In the second stage, we replace an initial priority ordering with the modified priority
ordering ∗ (R). Now all patients in N̄ have lower priorities than the rest of patients. In the third
stage, we find sets of feasible cycles and at most k chains consisting of the patients in N \ N̄ and some
selected patients in N̄ in order of priorities. We choose any one of these sets and assign suppressants
to the tails of the chains in the set. In the last stage, we derive the resulting new preference profile and
apply to it T T C associated with the modified priority ordering.
Note that to satisfy monotonicity, all patients in N \ N̄ should receive transplantations after suppressants are introduced. For this, we sort them out and give them higher priorities than those in N̄
when we apply T T C for the new preference profile. As long as they have higher priorities than N̄ ,
the cycles or chains involving these patients will be selected ahead of the rest, guaranteeing that they
still receive transplantations at the new profile. To satisfy maximal improvement, on the other hand,
we figure out at most k chains involving a “largest” set of patients and then assign suppressants to the
tails of these chains. By doing this, we can improve all patients’ welfare in a maximal way. Note that
eT T C is essentially single-valued, which is straightforward from the definition.
Example 5. (eT T C ) Let N ≡ {1, . . . , 9} and : 1, . . . , 9. Let R ∈ RN and the corresponding g(R)
be given as follows:
R1
R2
R3
R4
R5
R6
R7
R8
R9
2
1, 3, 4
5, 7
7
∅
1, 9
2
1
8
N \ {2} N \ {1, 3, 4} N \ {5, 7} N \ {7}
N
N \ {1, 9} N \ {2} N \ {1} N \ {8}
10
Such H0 is always non-empty. This is because at g(R), (i) patients outside N̄ form cycles among themselves and
(ii) patients in N̄ do not form cycles, but only chains, among themselves (if there were any cycle among patients in N̄ ,
such a cycle should have been chosen in Stage 1 of the algorithm, making these patients not belong to N̄ ). Therefore, we
can choose the cycles from (i) and at most k chains from (ii).
13
8
•
1
•
2
•
3
•
•
9
•
6
•
4
•
7
5
•
The set of cycles is C(R) = {c = (1, 2, 1), c0 = (2, 4, 7, 2), c00 = (2, 3, 7, 2)}. Since all these cycles are
not feasible with each other, T T C chooses the cycle involving the patient with the highest priority.
Therefore, T T C (R) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4, . . . , µ(9) = 9)} and N̄ = {3, 4, . . . , 9}.
Now suppose that at most one patient can receive a suppressant, i.e., k = 1. Note that patient 3 has
the highest priority in N̄ . Note also that the collection of sets of feasible cycles and chains including
patients 1, 2, and 3 includes l = (6, 9, 8, 1, 2, 3, 5) and l0 = (6, 9, 8, 1, 2, 3, 7). Among all these sets, we
search for the one involving the patient with the next highest priority in N̄ , who is patient 4. Because
there is no such set, we move on to patient 5. We end up with a single chain {l} and assign the
suppressant to patient 5, who is the tail of l, i.e., σ ∗ (R) = {5}. We also modify to ∗ (R), which
happens to coincide with . By definition, eT T C (R) = T T C∗ (R) (R(σ ∗ (R))). The matching is
(µ(1) = 2, µ(2) = 3, µ(3) = 5, µ(4) = 4, µ(5) = 6, µ(6) = 9, µ(7) = 7, µ(8) = 1, µ(9) = 8).
Next suppose that at most two patients can receive suppressants, i.e., k = 2. Similarly as above, we
search for the set of cycles and at most two chains including patients 1 and 2 as well as other patients
in N̄ in order of their priorities. We end up with two sets of chains, L = {(6, 9, 8, 1, 2, 3, 5), (4, 7)} and
L0 = {(6, 9, 8, 1, 2, 4, 7), (3, 5)}. Thus, σ ∗ (R) = {5, 7}. We can also modify to ∗ (R) as above. The
two resulting matchings of eT T C (R) are µ and µ0 such that
µ = (µ(1) = 2, µ(2) = 3, µ(3) = 5, µ(4) = 7, µ(5) = 6, µ(6) = 9, µ(7) = 4, µ(8) = 1, µ(9) = 8)
µ0 = (µ0 (1) = 2, µ0 (2) = 4, µ0 (3) = 5, µ0 (4) = 7, µ0 (5) = 3, µ0 (6) = 9, µ0 (7) = 6, µ0 (8) = 1, µ0 (9) = 8).
Note that all patients receive transplantations under µ and µ0 , showing that eT T C is essentially
single-valued.
Remark. When multiple suppressants are available, it is also possible to assign them one by one
sequentially, instead of assigning them all at once as in eT T C . Example 5 shows that the resulting matchings may be different from eT T C (R): the only resulting matching is µ in this case, while
eT T C (R) includes both of µ and µ0 .
The extended top-trading cycles solution maximizing the number of transplantations,
eT T C :
Stage 1. For each R ∈ RN , apply T T C . Denote by N̄ the set of patients who receive no transplantations at any matching selected by T T C . Without loss of generality, let N̄ ≡ {j1 , . . . , jn̄ } and
j1 · · · jn̄ .
14
Stage 2. Modify into a new priority ordering. All patients in N \ N̄ have higher priories than those
in N̄ , while the relative priorities within N̄ and N \ N̄ remain the same, respectively. Denote this new
¯ ∗ (R).
priority ordering by Stage 3. Find a set of feasible cycles and chains. Let
H0 ≡ {H ∈ H(R) : N (H) ⊇ N \ N̄ and the number of chains in H does not exceed k}, and
for each t ∈ {1, . . . , n̄},

 {H ∈ H
if {H ∈ Ht−1 : jt ∈ N (H)} =
6 ∅,
t−1 : jt ∈ N (H)},
Ht ≡
 Ht−1 ,
otherwise.
Choose any H ∈ Hn̄ and assign suppressants to the tails of all chains in H. Denote the set of these
tails by σ̄ ∗ (R).
¯ ∗ (R). For each R ∈ RN , let
Stage 4. Apply to R(σ̄ ∗ (R)) the top-trading cycles rule associated with eT T C (R) ≡ T T C¯ ∗ (R) (R(σ̄ ∗ (R))).
Basically we make the same modifications of T T C as we did for T T C with an exception of Stage
1 where we use T T C instead of T T C . Note that, because T T C is essentially single-valued, the sets
of patients receiving transplantations remain the same across all matchings selected by T T C for each
problem. Also, eT T C is essentially single-valued, which is straightforward from the definition.
Example 6. (eT T C ) Consider the problem in Example 5.
8
•
1
•
2
•
3
•
•
9
•
6
•
4
•
7
5
•
Recall that C(R) = {c = (1, 2, 1), c0 = (2, 4, 7, 2), c00 = (2, 3, 7, 2)}. Since all these cycles are not feasible
with each other, T T C chooses a cycle of the largest size; if there are many, it chooses a cycle including
patients with higher priorities in order. Thus, c00 is chosen and the matching is T T C (R) = {(µ(1) =
1, µ(2) = 3, µ(3) = 7, µ(7) = 2, µ(4) = 4, . . . , µ(9) = 9)} with N̄ = {1, 4, 5, 6, 8, 9}.
Suppose that at most one patient can receive a suppressant, i.e., k = 1. Note that patient 1
has the highest priority in N̄ . Among all sets of feasible cycles and single chains, we find the sets
including patients 1, 2, 3, and 7. Note that L ≡ {(6, 9, 8, 1), (2, 3, 7, 2)} and L0 ≡ {(6, 9, 8, 1, 2, 3, 7)}
are among them including the largest set of patients and this is what we obtain at the last step of
¯ ∗ (R) :
the eT T C algorithm. Suppose that we choose L0 . Then, σ̄ ∗ (R) = {7}. We modify to 2, 3, 7, 1, 4, 5, 6, 8, 9. By definition, eT T C (R) = T T C¯ ∗ (R) (R(σ̄ ∗ (R))). The resulting matching is µ
such that (µ(1) = 2, µ(2) = 3, µ(3) = 7, µ(4) = 4, µ(5) = 5, µ(6) = 9, µ(7) = 6, µ(8) = 1, µ(9) = 8).
If we choose L, then σ̄ ∗ (R) = {1}. We have a different matching, but the set of patients receiving
transplantations remains the same as above.
15
Now suppose that at most two patients can receive suppressants, i.e., k = 2. Similarly as above,
we figure out the collection of sets of feasible cycles and at most two chains including patients 2, 3,
and 7 as well as other patients in order of priorities. At the last step of the eT T C algorithm, we
obtain L = {(6, 9, 8, 1, 2, 3, 5), (4, 7)} and L0 = {(6, 9, 8, 1, 2, 4, 7), (3, 5)} that include the largest set of
¯ ∗ (R) as above. The
patients. Suppose that we choose L0 . Then, σ̄ ∗ (R) = {5, 7}. We modify to resulting matching is µ such that (µ(1) = 2, µ(2) = 4, µ(3) = 5, µ(4) = 7, µ(5) = 3, µ(6) = 9, µ(7) =
6, µ(8) = 1, µ(9) = 8). If we choose L, then σ̄ ∗ (R) = {5, 7}. We have a different matching, but the set
of patients receiving transplantations remains the same as above.
Our main result follows.
Theorem 1. eT T C and eT T C satisfy Pareto efficiency, monotonicity, and maximal improvement.
Proof. (Pareto efficiency ) For each R ∈ RN , note that R(σ ∗ (R)), R(σ̄ ∗ (R)) ∈ RN . By definition,
eT T C (R) ≡ T T C∗ (R) (R(σ ∗ (R)) and eT T C (R) ≡ T T C¯ ∗ (R) (R(σ̄ ∗ (R)). By Proposition 1, T T C
and T T C are Pareto efficient for every . Therefore, for each R ∈ RN , eT T C (R) and eT T C (R)
are Pareto efficient at R(σ ∗ (R)) and R(σ̄ ∗ (R)), respectively.
(Monotonicity ) We show that eT T C is monotonic. Let R ∈ RN . Recall that T T C and
eT T C are essentially single-valued. Note also that at each matching, any patient who receives a
transplantation cannot be better off further and any patient who receives no transplantation cannot be
worse off further. Therefore, it is enough to show that all agents who used to receive transplantations at
T T C (R) still receive transplantations at eT T C (R). Denote the set of those patients by N \ N̄ . Note
that they are the only patients involved in feasible trading cycles that generate matchings in T T C (R)
and the preferences of these patients are not affected by the use of suppressants under the eT T C
algorithm. Thus, the feasible trading cycles involving N \ N̄ still remain feasible in g(R(σ ∗ (R))). Given
ordering ∗ (R), these patients have higher priorities than those in N̄ . Consider the algorithm of
T T C∗ (R) applied to R(σ ∗ (R)). The feasible trading cycles involving N \ N̄ at the T T C algorithm
are not ruled out at the step where we figure out the sets of feasible cycles including the patient with
the lowest priority in N \ N̄ . Among these sets, we continue to sort out the set of feasible cycles in
order of priorities among N̄ . This guarantees that there are trading cycles involving all patients in
N \ N̄ . Therefore, we end up with eT T C (R) in which all of these patients receive transplantations,
completing the proof. The same argument shows that eT T C is monotonic.
(Maximal improvement) We show that eT T C satisfies maximal improvement. Suppose, by
contradiction, that there are R ∈ RN , S ⊆ N with |S| ≤ k, µ ∈ eT T C (R), and µ0 ∈ M(R(S)) such
that N (µ) ( N (µ0 ). Let N̄ be the set of patients who receive no transplantations at T T C (R). Let
H ∈ H(R) be the set of cycles and chains that results in µ, that is, a set of cycles and chains chosen in
the last stage of the eT T C algorithm for R. Let H 0 ∈ H(R) be the set of cycles and chains that results
in µ0 when the patients in S use suppressants. According to the definition of eT T C , H should have
been chosen to include all patients in N \ N̄ as well as patients in N̄ in the order of priorities over N̄ ,
16
as long as the number of chains does not exceed k. Since N (µ) ( N (µ0 ), there is i∗ ∈ N (µ0 ) \ N (µ). If
patient i∗ has a higher priority than some patients in N (µ), then the set of cycles and chains should
have been chosen to include i∗ in the eT T C algorithm. Since i∗ ∈
/ N (H), this is a contradiction to
the definition of eT T C . If i∗ has a lower priority than all patients in N (µ), then the set of cycles and
chains should have been chosen to include all patients in N (µ) as well as i∗ , which is possible as in H 0 .
Since i∗ ∈
/ N (H), again, this is a contradiction to the definition of eT T C . The same argument shows
that eT T C satisfies maximal improvement.
5
Simulation Results
We present simulation results to show that the use of suppressants can be significantly reduced from
the current practice. For this analysis, we assume that immunological compatibility is determined only
by ABO blood types. We denote a pair by its type X-Y where the patient’s blood type is X and the
donor’s type is Y. Let #(X-Y) be the number of pairs of type X-Y. A k-cycle is a cycle consisting of k
pairs and a k-chain is a chain consisting of k pairs.
5.1
Simulation based on the KONOS data
We use the KONOS (Korean Network for Organ Sharing) data on living donor kidney transplantations
in the recent four years (2011-2014). The data set collects all patient-donor pairs who receive transplantations during this period and specifies their ABO blood types. As shown in Table 2, almost no pairs in
the set receive transplantations through kidney exchanges. For instance, in 2012, no pair participated
in kidney exchanges. For In other words, almost all transplantations in the data are performed within
pairs: if a pair is compatible, the patient receives a kidney directly from her own donor; otherwise, the
pair uses a suppressant.
In 2012, there are 1,020 living donor kidney transplantations in total. Table 3 summarizes the blood
type profile. For instance, there are 210 compatible pairs of type A-A, who receive the kidneys from their
own donors; there are 35 incompatible pairs of type A-B, who use suppressants to get transplantations
from their donors. In the blood-type restricted setting, there are seven types of incompatible pairs,
A-B, B-A, A-AB, B-AB, O-A, O-B, and O-AB. It turns out that there are 193 incompatible pairs in
2012, as summarized in Table 4.
Since there are 193 incompatible pairs in the data, suppressants are used 193 times for incompatible
kidney transplantations. We propose an algorithm that minimizes the use of suppressants, while all
these incompatible pairs still receive transplantations.
Simulation Algorithm For any set of incompatible pairs and their blood type profile:
Step 1. Find all 2-cycles. Since only pairs of types A-B and B-A can form cycles, the number of all
2-cycles is min {#(A-B),#(B-A)}. Remove the pairs of these 2-cycles from the pool.
17
Type
A
B
O
AB
Total
A
B
O
AB
210
28
38
34
35
172
39
37
80
77
164
20
24
24
5
33
349
301
246
124
Total
310
283
341
86
1020
Table 3. The profile of blood types of living kidney transplantations in 2012 in Korea:
patients’ types are in leftmost column; donors’ types are in top row.
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
35
28
24
24
38
39
5
193
Table 4. Blood-types of incompatible pairs in 2012 in Korea
Step 2. Find all 3-chains, which are the longest chains in the setting. If #(A-B) > #(B-A), these
3-chains must be of the form O-A ← A-B ← B-AB. If #(A-B) < #(B-A), the 3-chains must be of the
form O-B ← B-A ← A-AB. If #(A-B) = #(B-A), no 3-chain can be formed. Remove the pairs of the
3-chains from the pool.
Step 3. Find all 2-chains. These chains must be ones of O-A ← A-B, O-A ← A-AB, O-B ← B-A,
O-B ← B-AB, A-B ← B-AB, and B-A ← A-AB, depending on which types of pairs remain in the pool.
Remove the pairs of the 2-chains from the pool.
Step 4. Each remaining pair forms a 1-chain. Assign suppressants to the tail patient of each chain
and convert each chain to a cycle. Perform transplantations according to these cycles.
In Step 1, given a set of incompatible pairs, a cycle can be formed only by pairs A-B and B-A.
Thus, without the use of suppressants, two pairs of A-B and B-A form 2-cycles and the patient of a
pair receives a transplantation from the donor of the other pair. After all feasible 2-cycles are formed,
at least one of the two types does not remain in the pool.
In Step 2, if there is a pair A-B but no pair B-A in the pool, the longest chain is a 3-chain of the
form O-A ← A-B ← B-AB. On the other hand, if there is a pair B-A but no pair A-B, the longest
chain is a 3-chain O-B ← B-A ← A-AB. If there are no pairs A-B and B-A in the pool, no 3-chain can
be formed. The patient of tail pair O-A (pair O-B) receives a suppressant and a transplantation from
the donor of pair B-AB (pair A-AB respectively). Without the use of suppressants, the patient of pair
B-AB (pair A-AB) receives a transplantation from the donor of pair A-B (pair B-A) and the patient
of pair A-B (pair B-A) from the donor of pair O-A (pair O-B respectively).
In Step 3, if there are no pairs A-B and B-A in the pool of the remaining pairs, the longest chain
is a 2-chain of the forms O-A ← A-AB and O-B ← B-AB. The patient of the tail pair O-A (pair O-B)
receives a transplantation from the donor of pair A-AB (pair B-AB respectively) with a suppressant.
The patient of pair A-AB (pair B-AB) receives a transplantation from the donor of pair O-A (pair O-B
18
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
# pairs
35
28
24
24
38
39
5
193
2-cycle
3-chain
2-chain
2-chain
1-chain
1-chain
1-chain
28 (7)
7 (0)
28 (0)
5 (0)
56
21
48
34
7
22
5
7 (17)
24 (0)
7 (31)
24 (7)
17 (0)
17 (22)
7 (0)
22 (0)
Table 5. Simulation algorithm for the set of incompatible pairs in 2012
2011
2012
2013
2014
# pairs
2-cycle
3-chain
2-chain
1-chain
# Suppressants
131
193
216
217
17
28
29
34
2
7
10
3
30
41
38
48
31
34
52
44
63
82
100
95
Table 6. Simulations for incompatible pairs during 2011-2014
respectively) without the use of suppressants.
After all feasible 2-chains are formed, in Step 4, each pair remaining in the pool forms a 1-chain,
and each remaining patient receives a transplantation from her own donor with a suppressant. Because
we require that all patients receive transplantations, the number of suppressants required is equal to
the number of chains formed by the algorithm.
Remark. The simulation algorithm minimizes the use of suppressants when all incompatible pairs
receive transplantations.11
Table 5 shows the set of cycles and chains formed by our algorithm for the incompatible pairs in
2012. There are 28 2-cycles, 7 3-chains, 41 2-chains, and 34 1-chains, adding up to 82 chains in total.
In Table 5, the numbers in the parentheses represents the number of pairs who still remain in the
pool after cycles or chains are formed and removed. By assigning suppressants to pairs as we propose
and allowing them to exchange, the use of suppressants has been decreased from 193 to 82 while all
incompatible pairs still receive transplantations.
Other years show similar simulation results, as summarized in Table 6. During 2011-2014, 757
recipients of suppressants could have been reduced to 340, given that all incompatible pairs of the data
still receive transplantations.
11
In all cases, the number of suppressants required is at least as large as the number of pairs with blood type O patients.
This is because blood type O patients can receive transplantations only from the type O donors, but there is no pair with
type O donor in the set. Therefore, for all patients to receive transplantations, each pair of types O-A, O-B, and O-AB
should form a chain, and any two pairs of such types cannot form a chain together. Thus, the total number of pairs of
types O-A, O-B, and O-AB is a lower bound for the number of chains.
19
Type
A
B
O
AB
Total
A
B
O
AB
116
92
95
37
92
73
76
30
95
76
78
31
37
30
31
12
340
271
280
110
Total
340
271
280
110
1001
Table 7. The profile of blood types of a hypothetical population: patients’ types are in leftmost column; donors’ types
are in top row.
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
# pairs
92
92
37
30
95
76
31
453
2-cycle
3-chain
2-chain
2-chain
1-chain
1-chain
1-chain
92 (0)
92 (0)
31 (0)
184
0
74
60
58
46
31
37 (0)
37 (58)
30 (0)
30 (46)
58 (0)
46 (0)
Table 8. Simulation for incompatible pairs in a hypothetical population
Hypothetical
population
# pairs
2-cycle
3-chain
2-chain
1-chain
# Suppressants
453
92
0
67
135
202
Table 9. Simulations for incompatible pairs of the hypothetical population
5.2
Simulation based on the hypothetical population
The simulation results presented in the previous section are based on the data of the patient-donor pairs
who already received living kidney transplantations. However, this data set may be biased to represent
the whole population of patient-donor pairs, because there can be incompatible pairs who do not get
transplantations. As they are not included in this data set, the KONOS data may underrepresent
incompatible pairs.
Thus, we construct a hypothetical population of 1,001 patient-donor pairs based on the distribution
of ABO blood types in Korea. Blood type A is about 34.0 percent, type B about 27.1 percent, type
O about 28.0 percent, and type AB about 11.0 percent of the total population. We assume that the
types of patients and donors are identically and independently distributed (i.i.d.) according to this
distribution. Out of 1,001 pairs, it turns out that 453 pairs are incompatible. Table 7 summarizes the
profile of blood types of these pairs.12
From this distribution, we identify all incompatible pairs and run the simulation algorithm. Because
12
The number of pairs of each type is rounded to the nearest integer, with the total equal to 1,001.
20
# pairs
AB-O
2-cycle
4-cycle
3-cycle
2-chain
1-chain
# Suppressants
2011
2012
2013
2014
131
193
216
217
20
20
17
11
17
28
29
34
2
7
10
3
18
13
7
8
12
28
31
40
31
34
52
44
43
62
83
84
Hypothetical
population
453
31
92
0
31
36
135
171
Table 10. Simulations for incompatible pairs when pairs of type AB-O participate.
of the i.i.d. assumption, the pairs of type A-B are as many as those of type B-A, and therefore, there
is no 3-chain. Tables 8 and 9 show the result. For this hypothetical pool of incompatible pairs, the
number of suppressants assigned by our algorithm is equal to the number of pairs of types O-A, O-B,
and O-AB, implying that our simulation algorithm minimizes the use of suppressants.
5.3
Simulation based on the KONOS data with the participation of “altruistic”
compatible pairs
Note that we so far excluded all compatible pairs from simulation analysis and focused on the incompatible pairs. The use of suppressants could be reduced further, however, when these pairs stay in
the exchange pool and “facilitate” the exchanges among other pairs. Such compatible pairs are called
“altruistic” pairs (Sönmez and Ünver, 2014). For example, the participation of type AB-O pairs would
convert 3-chains to 4-cycles AB-O ← O-A ← A-B ← B-AB, and 2-chains to 3-cycles AB-O ← O-A ←
A-AB and AB-O ← O-B ← B-AB, and would reduce the use of suppressants by the number of type
AB-O pairs participating in the exchanges. Table 10 shows simulation results when compatible pairs
of type AB-O participate in the exchange pool.
To generalize this, we may allow all patient-donor pairs – either compatible or incompatible –
participate in the exchange pool. It is quite complicated to analyze the actual KONOS data, but we
expect further reduction of use of suppressants. If we run the simulation algorithm with the hypothetical
population, on the other hand, we achieve full efficiency even without using suppressants: because of
the i.i.d. assumption of types of pairs in the hypothetical population, the number of pairs of type X-Y
will be exactly same as the number of pairs of type Y-X. They can form 2-cycles.
6
Conclusion
In this paper, we took the first step to investigate implications of introducing suppressants. We focus
on fairness and efficiency in assigning suppressants and matching patient-donor pairs. We propose
two extended TTC solutions and show that they satisfy the requirements we had in mind. By using
actual transplantation data in a blood-type restricted environment, we also showed that the use of
suppressants in the current practice can be significantly improved.
21
There still remain several interesting questions. First, it is possible to introduce monetary transfers
to the model and analyze the problem as a local public good problem. Recall that we simplified the
model by assuming that all expenses of suppressants are fully subsidized by the public insurance. This
makes patients indifferent between compatible donors and incompatible donors as long as they can get
transplantations. In practice, however, patients may well prefer compatible donors to incompatible
donors, even if they can use suppressants. This is because they often have to pay a certain fraction of
cost to use suppressants and they also have to take a risk of side effects caused by suppressants. If they
decline to use suppressants, expecting to get transplantations from the deceased donors for example,
we cannot expect a large gain by introducing suppressants any more. To avoid such cases, it seems
plausible to introduce a cost sharing process or a compensating process for them. The problem can
then be viewed as a local public good problem, because the use of suppressants incurs costs for a tail of
a chain, but it benefits all other patients in the chain for free. Therefore, we may consider, for instance,
the equal division of the cost incurred by the tail among all patients in each chain.
On the other hand, a generalization can be made on our assumption of dichotomous preferences as
well, for example, to tri-chotomous preferences. Although ABO blood-type generates only dichotomous
compatibility profiles, we have a wide range of compatibility when tissue-type (or HLA typing) is also
taken into considerations. The treatment of immunosuppressive protocols should also be differentiated.
Since our TTC solutions are applicable only to dichotomous profiles, this will be an interesting and
non-trivial extension of our analysis.
Lastly, we may think of a way to apply the celebrated deferred acceptance (DA) solution to our
setting. Since there is a single priority ordering over patients, the DA solution will be more like a
sequential dictatorship: among all collections of feasible cycles and at most k chains, choose ones
including a patient with the highest priority; among the resulting collections, choose ones including a
patient with the second highest priority; and so on. From what we obtain, we choose the tails of chains
to be recipients of suppressants and let patients receive kidneys from donors along the cycles and chains.
By doing this, we will obtain a “stable” assignment in the end, in that if a patient does not receive a
transplantation, then either (i) all patients with lower priorities do not get transplantations, or (ii) all
available suppressants are assigned to patients with higher priorities. This solution also satisfies Pareto
efficiency and maximal improvement. However, it violates monotonicity: it is easily checked with the
example in the proof of Proposition 3 after switching patients 1 and 6.
Appendix 1. Simulation Algorithm
We prove that the simulation algorithm defined in Section 5 minimizes the number of suppressants
used when all patients receive transplantations.
Proof. Among all sets of feasible chains and cycles that include all incompatible pairs, let H be the
collection of all such sets with the smallest number of chains. Let H ∈ H. We proceed with the
following claims.
22
Claim 1. The number of chains remains the same as in H when all 2-cycles are formed.
Proof of Claim 1. Note that cycles can be formed only by pairs A-B and B-A. Consider A-B and B-A
in two separate chains of H, if any. We show that the number of chains does not change even if these
pairs form a cycle, instead of staying in the chains. For A-B, there are four possible types of chains to
which the pair can belong: A-B, O-A←A-B, A-B←B-AB, O-A←A-B←B-AB. Similarly, there are four
possible types of chains to which B-A can belong: B-A, O-B←B-A, B-A←A-AB, O-B←B-A←A-AB.
Case 1. A-B is a 1-chain. We show that B-A should belong to O-B←B-A←A-AB. Suppose otherwise.
If B-A is a 1-chain, then A-B and B-A can form a cycle, reducing the number of chains by 2, a
contradiction that H ∈ H. If B-A belongs to O-B←B-A, then A-B and B-A can form a cycle while O-B
remains as a 1-chain. This reduces the number of chains by 1, a contradiction again. If B-A belongs to
B-A←A-AB, then the same argument applies. Thus, if A-B is a 1-chain in H, then B-A should belong
to the 3-chain. Let the pairs of A-B and B-A form a cycle and the remaining pairs in the 3-chain, O-B
and A-AB, form two 1-chains. The number of chains remains the same as in H.
Case 2. A-B belongs to O-A←A-B. We show that B-A should belong to either O-B←B-A or O-B←BA←A-AB. Otherwise, we can reduce the number of chains by 1, a contradiction. (For example, if B-A
belongs to B-A←A-AB, then A-B and B-A can form a cycle and the remaining pairs O-A and A-AB
can form a 2-chain.) Suppose that B-A belongs to O-B←B-A. Let A-B and B-A form a cycle and the
remaining pairs O-A and O-B form two 1-chains. The number of chains remains the same as in H.
Suppose that B-A belongs to the 3-chain. Let A-B and B-A form a cycle and the remaining pairs form
two chains O-A←A-AB and O-B. Then the number of chains remains the same as in H.
Case 3. A-B belongs to A-B←B-AB. We show that B-A should belong to either B-A←A-AB or OB←B-A←A-AB. Otherwise, we can reduce the number of chains by 1, a contradiction. The same
argument of Case 2 applies to show that the number of chains remains the same even after A-B and
B-A form a cycle.
Case 4. A-B belongs to O-A←A-B←B-AB. In this case, B-A can belong to any one of the four types
of chains. Even if A-B and B-A form a cycle, we can keep the same number of chains. For example,
consider the case when B-A belongs to O-B←B-A←A-AB. Let A-B and B-A form a cycle and the
remaining pairs form two 2-chains, O-A←A-AB and O-B←B-AB. Then the number of chains remains
the same as in H.
Therefore, if there are two pairs A-B and B-A in separate chains of H, they can form a cycle and
the remaining pairs can reorganize without changing the total number of chains. After doing this, we
again check if there are two other pairs A-B and B-A in separate chains. If any, we repeat the same
procedure to form a cycle and to reorganize the chains without changing the total number of chains.
It ends when at least one type among A-B and B-A disappears from all the chains. This is what Step
1 of the simulation algorithm does, where we form all possible cycles between A-B and B-A and then
exclude them from the pool.
From Claim 1, we see that Step 1 of the simulation algorithm keeps the minimum number of chains
23
as in H. After the process described in the proof of Claim 1 (or after Step 1), at least one type among
A-B and B-A does not remain in the chains of H. Now consider a set H ∈ H such that at most one of
the two types A-B and B-A exists in the chains of H. Since pairs in cycles do not use suppressants, we
can exclude all the cycles and focus on the chains formed by the remaining pairs.
Claim 2. The number of chains remains the same as in H when 3-chains are formed.
Proof of Claim 2. Consider the following cases.
Case 1. A-B and B-A do not exist in any chain of H. Then all the chains of H are formed by pairs
of types O-A, B-AB, O-B, A-AB, and O-AB. However, these pairs can not form a 3-chain among
themselves.
Case 2. Only one type exists in the chains of H. Without loss of generality, suppose that A-B exists
in the chains of H. Then all the chains of H are formed by O-A, A-B, B-AB, together with O-B,
A-AB, and O-AB. We will consider all the chains of H except for O-AB because O-AB can only form
a 1-chain. Note that a 3-chain is the longest among all possible chains, and any 3-chain must be of
the form O-A←A-B←B-AB. We show that forming O-A←A-B←B-AB as many as possible among
those pairs does not change the total number of chains. Let n1 be the number of chains O-A←A-B,
n2 the number of chains A-B←B-AB, n3 , n4 , and n5 , the numbers of 1-chains, O-A, A-B, and B-AB,
respectively. Note that additional 3-chains can be formed only by reorganizing the pairs that belong
to these chains. Since H ∈ H contains the minimum number of chains, we observe the followings:
• Two 1-chains O-A and A-B cannot exist together in H.13
• A 1-chain O-A and a 2-chain A-B←B-AB cannot exist together in H.
• Two 1-chains A-B and B-AB cannot exist together in H.
• A 1-chain B-AB and a 2-chain O-A←A-B cannot exist together in H.
Altogether, we have following subcases, depending on which types of 1-chains exist in H, in terms
of the cardinalities, n3 , n4 , and n5 , respectively. If n4 > 0, it must be that n3 = n5 = 0, because
otherwise the number of chains can be reduced by forming 2-chains, O-A←A-B or A-B←B-AB.
(1) n4 > 0, n3 = n5 = 0 : Let n1 , n2 ≥ 0 with n1 ≥ n2 . Then change 2-chains A-B←B-AB into
3-chains O-A←A-B←B-AB by moving n2 pairs of type O-A from the chains O-A←A-B. Note that this
is the maximum number of 3-chains that can be formed among these pairs. After forming 3-chains,
(n2 + n4 ) of the 1-chains A-B and (n1 − n2 ) of the 2-chains O-A←A-B can be formed among the
remaining patients. Altogether, the total number of chains remains the same, while we maximize the
number of 3-chains. The symmetric argument applies to the case when n2 ≥ n1 .
(2) n4 = 0, n3 > 0, n5 > 0 : Then, n2 = 0 and n1 = 0. There is no additional 3-chain that can be
formed in this case.
(3) n4 = 0, n3 > 0, n5 = 0 : Then, n1 ≥ 0 and n2 = 0. There is no additional 3-chain that can be
formed in this case.
13
If they exist together, these two 1-chains can form a 2-chain, reducing the total number of chains, a contradiction to
H ∈ H.
24
(4) n4 = 0, n3 = 0, n5 > 0 : Then, n1 = 0 and n2 ≥ 0. There is no additional 3-chain that can be
formed in this case.
(5) n4 = n3 = n5 = 0 : Then, n2 ≥ 0 and n1 ≥ 0. The same argument of case (1) applies, except
that n4 has to be set to 0.
All these subcases show that the number of chains remains the same as in H if 3-chains are formed.
From Claim 2, we see that Step 2 of the simulation algorithm keeps the minimum number of chains
as in H. Therefore, we can remove all the 3-chains from the pool and proceed with the remaining
pairs to find 2-chains. As listed in Step 3, these 2-chains must be of the following forms, O-A ← A-B,
O-A ← A-AB, O-B ← B-A, O-B ← B-AB, A-B ← B-AB, and B-A ← A-AB. These 2-chains cannot
be reorganized to form a 3-chain because all possible 3-chains are formed and removed. Because the
number of chains cannot be reduced by converting these 2-chains to other 2-chains and 1-chains, the
simulation algorithm achieves the minimum number of chains as in H. Appendix 2. Simulations based on the KONOS data
We herein present the simulation results of 2011, 2013, and 2014 based on the KONOS data.
Type
A
B
O
AB
Total
A
219
19
76
13
327
B
17
148
73
19
257
O
33
21
195
9
258
AB
34
30
20
33
117
Total
303
218
364
74
959
Table A1. The profile of blood types of living kidney transplantations in 2011 in Korea:
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
# pairs
19
17
13
19
33
21
9
131
2-cycle
17 (2)
17 (0)
3-chain
2 (0)
2-chain
2-chain
34
2 (17)
13 (0)
2 (31)
6
13 (18)
26
17 (0)
1-chain
17 (4)
34
18 (0)
1-chain
18
4 (0)
1-chain
4
9 (0)
Table A2. Simulation for incompatible pairs in 2011
25
9
Type
A
B
O
AB
Total
A
192
39
93
24
348
B
29
138
70
24
261
O
51
44
173
5
273
AB
37
42
17
32
128
Total
309
263
353
85
1010
Table A3. The profile of blood types of living kidney transplantations in 2013 in Korea
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
# pairs
39
29
24
24
51
44
5
216
2-cycle
29 (10)
29 (0)
3-chain
10 (0)
58
10 (14)
2-chain
24 (0)
2-chain
10 (41)
30
24 (17)
48
14 (0)
1-chain
14 (30)
28
17 (0)
1-chain
17
30 (0)
1-chain
30
5 (0)
5
Table A4. Simulation for incompatible pairs in 2013
Type
A
B
O
AB
Total
A
193
37
84
26
340
B
34
143
54
25
256
O
46
40
185
9
280
AB
54
31
11
28
124
Total
327
251
334
88
1000
Table A5. The profile of blood types of living kidney transplantations in 2014 in Korea
26
Type
A-B
B-A
A-AB
B-AB
O-A
O-B
O-AB
Total
# pairs
37
34
26
25
46
40
9
217
2-cycle
34 (3)
34 (0)
3-chain
3 (0)
68
3 (22)
2-chain
26 (0)
2-chain
3 (43)
9
26 (17)
52
22 (0)
1-chain
22 (18)
44
17 (0)
1-chain
17
18 (0)
1-chain
18
9 (0)
9
Table A6. Simulation for incompatible pairs in 2014
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