Kidney Exchange with Immunosuppressants - Boston College · Kidney Exchange with Immunosuppressants Youngsub Chuny Eun Jeong Heoz Sunghoon Hongx October 2, 2015 Abstract This paper

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 en-

vironment, 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: [email protected]‡Vanderbilt University. E-mail: [email protected]§Korea Institute of Public Finance. E-mail: [email protected]

1 Introduction

When a patient suffers from End Stage Renal Disease (ESRD) and needs to receive a kidney transplan-

tation, 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 con-

straints, 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 maintain-

ing the same set of patients receiving transplantations. We also investigate how to assign the unused

1Human 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 manydifferent HLA proteins, say HLA-A1, HLA-A2, and so on. If the donor and the patient share the same HLAs, they arecalled an identical match. Two siblings are an identical match with probability of 25 percent. A parent and a child sharethe 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. Inaddition to immunological compatibility, kidney size and age may also affect transplantation outcomes.

2There has been a variety of immunosuppressive protocols reported in the medical literature but most of these protocolsinclude the use of rituximab (which is an immunosuppressant medicine) and plasmapheresis (which is a procedure to removeharmful 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).

3In contrast, the proportion of living donor kidney transplantations through an exchange program had been decreasedfrom 5.3 percent to near zero percent during the same period.

1

YearPatients

in waitlistsDeceased donor

transplantsLiving donortransplants

2009 4,769 488 750

2010 5,857 491 796

2011 7,426 680 959

2012 9,245 768 1,020

2013 11,381 750 1,011

2014 13,612 808 1,000

Table 1. Kidney transplantations in Korea

YearLiving donortransplants

in total

Transplantswithin

compatible pairs

Transplantsusing

suppressants

Transplantsthrough

exchanges

2009 750 675 (90.0%) 35 (4.7%) 40 (5.3%)

2010 796 689 (86.6%) 78 (9.8%) 29 (3.6%)

2011 959 828 (86.3%) 113 (11.8%) 18 (1.9%)

2012 1,020 827 (81.1%) 193 (18.9%) 0 (0.0%)

2013 1,011 795 (78.6%) 212 (21.0%) 4 (0.4%)

2014 1,000 783 (78.3%) 212 (21.2%) 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 im-

munological 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. Unfortu-

nately, 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 trans-

plantations 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

4The blood type X specifies the types of antigen and antibody a person has. For example, if a person’s blood type isA, 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 andtype 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 fromany donor having antigen B, namely, blood types B and AB. Similarly, if a person’s blood type is O, then her antibodyis 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 patient-

donor 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 (Sonmez

and Unver, 2014; Roth et al., 2005; Gentry et al., 2007).

We begin with the standard kidney exchange model without suppressants. We define two Top-

Trading 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 compat-

ibility 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 sup-

pressants, 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 transplan-

tations. 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 pa-

tients 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

5Shapley 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. Notethat Roth et al. (2004) impose no constraint on the size of exchanges, i.e., the length of cycles or chains, when designingtheir 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 sup-

pressants 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 trans-

plantations. Therefore, the KONOS data can be biased to represent the whole population of patient-

donor 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 ex-

tends 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 N+i ≡ {j ∈ N : patient i is compatible with the donor of patient j} and N−i ≡ N \N

+i . Each

i ∈ N has a weak preference Ri over N as follows: Each patient prefers all elements in N+i to all elements

in N−i . All elements in N+i are equally desirable and so are those in N−i . 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, c′ ∈ C(R) with c 6= c′. We say that

c and c′ are feasible if no patient appears both in c and c′. 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, l′ ∈ L(R) with l 6= l′. We say that l and l′ are feasible if no patient appears both in l

and l′. 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) /∈ N+i , 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 N+1 = {2}, N+

2 = {1, 3}, and N+3 = {2, 3}. The

corresponding R and g(R) are given as follows:

R1 R2 R3

2 1, 3 2, 3

1, 3 2 1

• • •1 2 3

The set of cycles in g(R) is C(R) = {c ≡ (1, 2, 1), c′ ≡ (2, 3, 2), c′′ ≡ (3, 3)}. Note that c and c′′ are

feasible. Since patient 2 appears both in c and c′, c and c′ are not feasible. The collection of feasible

cycles is C(R) = {{c}, {c′}, {c′′}, {c, c′′}}. Let C ≡ {c, c′′} ∈ C(R). Then, N(C) = {1, 2, 3}. The set of

chains in g(R) is L(R) = {l ≡ (1), l′ ≡ (2), l′′ ≡ (1, 2, 3), l′′′ ≡ (3, 2, 1)}. Note that l and l′ are feasible.

Since patient 1 appears both in l and l′′, l and l′′ are not feasible. The collection of feasible chains

is L(R) = {{l}, {l′}, {l′′}, {l′′′}, {l, l′}}. Let L ≡ {l, l′} ∈ L(R). Then, N(L) = {1, 2}. Also, H(R) =

{c, c′, c′′, l, l′, l′′, l′′′} and H(R) = {{c}, {c′}, {c′′}, {l}, . . . , {l′′}, {l, l′′}, {c′, l}, {c′′, l}, {c′′, l′}, {c′′, l, l′}}.

A solution is a mapping ϕ : RN ⇒ ∪R∈RNM(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, µ′ ∈ ϕ(R), and each i ∈ N , µ(i) Ii µ′(i).

If ϕ is essentially single-valued, then for each R ∈ RN and each µ, µ′ ∈ ϕ(R), N(µ) = N(µ′).7

For each R ∈ RN , we say that µ ∈ M(R) is Pareto efficient at R if there is no other µ′ ∈ M(R)

such that for each i ∈ N , µ′(i) Ri µ(i) and for some i ∈ N , µ′(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.

6For more detailed discussion on these priorities, see Roth et al. (2005).7Suppose, by contradiction, that there are R ∈ RN , i ∈ N , and a pair µ, µ′ ∈ ϕ(R) such that µ(i) ∈ N+

i and µ′(i) /∈ N+i .

Since all patients in N+i are strictly preferred to those in N−i , we have µ(i) Pi µ

′(i), contradicting that ϕ is essentiallysingle-valued.

8The TTC solution is proposed by Shapley and Scarf (1974) in a general model and applied to kidney exchangeproblems by Roth et al. (2004).

6

Top-trading cycles solution associated with � (simply TTC�): 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 sequen-

tial 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 TTC�): 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 TTC� and TTC� are essentially single-valued.

Proposition 1. For every priority ordering �, TTC� and TTC� are Pareto efficient.

Proof. Suppose, by contradiction, that TTC� is not Pareto efficient. Then, there are R ∈ RN , µ ∈TTC�(R), and µ′ ∈ M(R) such that for each i ∈ N , µ′(i) Ri µ(i) and for some i ∈ N , µ′(i) Pi

µ(i). This implies that all patients who receive transplantations at µ also receive transplantations at

µ′. Furthermore, there is at least one patient, say i who additionally receives a transplantation at µ′.

Among such patient i’s, choose the one with the highest priority under �. According to the definition

of TTC�, the matchings should be chosen so that patient i receives a transplantation, a contradiction.

Next, suppose, by contradiction, that TTC� is not Pareto efficient. Then, there are R ∈ RN , µ ∈TTC�(R), and µ′ ∈M(R) such that for each i ∈ N , µ′(i) Ri µ(i) and for some i ∈ N , µ′(i) Pi µ(i). This

again implies that all patients who receive transplantations at µ also receive transplantations at µ′ and

7

at least one patient additionally receives a transplantation at µ′. Thus, the number of transplantations

at µ′ is greater than that at µ. Hence, it must be that µ /∈ TTC�(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 com-

patible with all donors, including her own. We denote such preference by R∗i . If a set of patients

S receive suppressants, we denote their preferences by R∗S ≡ (R∗i )i∈S . For each R ∈ RN and each

S ⊆ N , let R(S) ≡ (R∗S , 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 R2 R3

2 1, 3 2, 3

1, 3 2 1

• • •1 2 3

Now suppose that S = {2}. Then, R(S) = (R1, R∗2, R3) and g(R(S)) are represented as follows:

R1 R∗2 R3

2 1, 2, 3 2, 3

1, 3 1

• • •1 2 3

The set of cycles in g(R∗2, R−2) is C(R∗2, R−2) = {c ≡ (1, 2, 1), c′ ≡ (2, 3, 2), c′′ ≡ (3, 3), c′′′ ≡ (2, 2)}.Thus, the collection of feasible cycles is C(R∗2, R−2) = {{c}, {c′}, {c′′}, {c′′′}, {c, c′′}, {c′′, c′′′}}. The set

of chains in g(R∗2, R−2) is L(R∗2, 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

recipient-selection rule. Let ϕ : RN ⇒⋃

R∈RNM(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

µ′ ∈ ϕ(R(σ(R))), and each i ∈ N , µ′(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 R′2 R3

2 1 3 2

1, 3 2, 3 1, 2 1, 3

Suppose that k = 0. Let R ≡ (R1, R2, R3) and R′ ≡ (R1, R′2, R3). By Pareto efficiency, we have

ϕ(R) = {µ = (µ(1) = 2, µ(2) = 1, µ(3) = 3)} and ϕ(R′) = {µ′ = (µ′(1) = 1, µ′(2) = 3, µ′(3) = 2)}.Now suppose that k = 1. Let σ : RN → 2N (k) be such that σ(R) = {2} and σ(R′) = {2}. The

new preference profile resulting from σ is (R1, R∗2, R3) for both problems. To make everyone weakly

better off at the new profile than at R, we should have µ ∈ ϕ(R1, R∗2, R3). Again, to make everyone

weakly better off at the new profile than at R′, we should also have µ′ ∈ ϕ(R1, R∗2, R3). Altogether,

{µ, µ′} ⊆ ϕ(R1, R∗2, R3). However, patient 1 is worse off at µ′ than at µ and patient 3 is worse off at µ

than at µ′, 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 µ′ chosen by a solution for a problem. Let T be the

set of patients receiving transplantations at µ. Let T ′ be the set of patients receiving transplantations

at µ′. Suppose that a set of patients receive suppressants. If the solution is essentially single-valued,

T = T ′ and strong monotonicity requires that all patients in T = T ′ receive transplantations at the

new preference profile. If the solution is not essentially single-valued, it may be that T 6= T ′, and strong

monotonicity requires that all patients in T ∪ T ′ 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 TTC� and TTC� as we show below.

Example 3. (TTC� and TTC� violating strong monotonicity) Let N ≡ {1, 2, 3, 4}, �: 1, 2, 3, 4 and

R, R ∈ RN be given as follows:

9

R1 R2 R3 R4

2 3 4 2

1, 3, 4 1, 2, 4 1, 2, 3 1, 3, 4

R1 R2 R3 R4

2 1 4 2

1, 3, 4 2, 3, 4 1, 2, 3 1, 3, 4

• •

••

1 2

34

• •

••

1 2

34

It is easy to check that TTC�(R) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)} and TTC�(R) =

{(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(R) = {2}. Patient 2’s preference changes to R∗2 as illustrated in the following graph:

• •

••

1 2

34

• •

••

1 2

34

Let R′ ≡ (R1, R∗2, R3, R4) and R′ ≡ (R1, R

∗2, R3, R4). Then,

TTC�(R′) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4)}TTC�(R′) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)}.

Under TTC�, patients 3 and 4 are worse off while patient 1 is better off and patient 2 remains indifferent.

Under TTC�, 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 TTC� and TTC�, 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 R, 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 µ′ ∈ ϕ(R(σ(R))), and each i ∈ N , µ′(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 ′ ∈ 2N (k), if there are

µ ∈ ϕ(R(T )) and µ′ ∈ ϕ(R(T ′)) such that N(µ′) ( N(µ), then T ′ 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

• •

••

1 2

34

S = ∅

• •

••

1 2

34

S′ = {1}

• •

••

1 2

34

S′′ = {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 {µ′ = (µ′(1) = 2, µ′(2) = 1, µ′(3) = 4, µ′(4) =

3)} and then N(µ′) = {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

9Note that any essentially single-valued solutions trivially satisfy monotonicity by setting σ(R) = ∅ for all R ∈ RN . Toavoid such triviality, we impose monotonicity together with an “efficiency” requirement of maximal improvement definedlater.

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 ′ ∈ 2N (k),

if there are µ ∈ ϕ(R(T )) and µ′ ∈ ϕ(R(T ′)) such that |N(µ′)| < |N(µ)|, then T ′ 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:

R1 R2 R3 R4 R5 R6

5 ∅ 5 1, 2 4 3

N \ {5} N N \ {5} N \ {1, 2} N \ {4} N \ {3}

•

•

•

•

•1

2

3

4

5 6• •

•

•

•

•1

2

3

4

5 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, eTTC�:

Stage 1. For each R ∈ RN , apply TTC�. Denote by N the set of patients who receive no trans-

plantations at any matching selected by TTC�. Without loss of generality, let N ≡ {j1, . . . , jn} and

j1 � · · · � jn.

Stage 2. Modify � into a new priority ordering. All patients in N \ N have higher priories than those

in N , while the relative priorities within N and N \ N 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 \ N and the number of chains in H does not exceed k},10 and

for each t ∈ {1, . . . , n},

Ht ≡

{H ∈ Ht−1 : jt ∈ N(H)}, if {H ∈ Ht−1 : jt ∈ N(H)} 6= ∅,

Ht−1, otherwise.

Choose any H ∈ Hn 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 eTTC�(R) ≡ TTC�∗(R)(R(σ∗(R))).

In the first stage, we apply TTC� to the initial preference profile of the standard model and find

the set of patients N who do not receive transplantations. Since TTC� is essentially single-valued, the

sets of patients receiving transplantations remain the same across all matchings selected by TTC� for

each problem. In the second stage, we replace an initial priority ordering � with the modified priority

ordering �∗ (R). Now all patients in N 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 \ N and some

selected patients in N 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 TTC� associated with the modified priority ordering.

Note that to satisfy monotonicity, all patients in N \ N should receive transplantations after sup-

pressants are introduced. For this, we sort them out and give them higher priorities than those in N

when we apply TTC� for the new preference profile. As long as they have higher priorities than N ,

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

eTTC� is essentially single-valued, which is straightforward from the definition.

Example 5. (eTTC�) 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}10Such H0 is always non-empty. This is because at g(R), (i) patients outside N form cycles among themselves and

(ii) patients in N do not form cycles, but only chains, among themselves (if there were any cycle among patients in N ,such a cycle should have been chosen in Stage 1 of the algorithm, making these patients not belong to N). Therefore, wecan choose the cycles from (i) and at most k chains from (ii).

13

• • •

•

•

• •

•

•

1 2 3 58

9 6 4 7

The set of cycles is C(R) = {c = (1, 2, 1), c′ = (2, 4, 7, 2), c′′ = (2, 3, 7, 2)}. Since all these cycles are

not feasible with each other, TTC� chooses the cycle involving the patient with the highest priority.

Therefore, TTC�(R) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4, . . . , µ(9) = 9)} and N = {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 N . 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 l′ = (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 N , 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, eTTC�(R) = TTC�∗(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 N in order of their priorities. We end up with two sets of chains, L = {(6, 9, 8, 1, 2, 3, 5), (4, 7)} and

L′ = {(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 eTTC�(R) are µ and µ′ 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)

µ′ = (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).

Note that all patients receive transplantations under µ and µ′, showing that eTTC� 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 eTTC�. Example 5 shows that the result-

ing matchings may be different from eTTC�(R): the only resulting matching is µ in this case, while

eTTC�(R) includes both of µ and µ′.

The extended top-trading cycles solution maximizing the number of transplantations,

eTTC�:

Stage 1. For each R ∈ RN , apply TTC�. Denote by N the set of patients who receive no trans-

plantations at any matching selected by TTC�. Without loss of generality, let N ≡ {j1, . . . , jn} and

j1 � · · · � jn.

14

Stage 2. Modify � into a new priority ordering. All patients in N \ N have higher priories than those

in N , while the relative priorities within N and N \ N remain the same, respectively. Denote this new

priority ordering by �∗(R).

Stage 3. Find a set of feasible cycles and chains. Let

H0 ≡ {H ∈ H(R) : N(H) ⊇ N \ N and the number of chains in H does not exceed k}, and

for each t ∈ {1, . . . , n},

Ht ≡

{H ∈ Ht−1 : jt ∈ N(H)}, if {H ∈ Ht−1 : jt ∈ N(H)} 6= ∅,

Ht−1, otherwise.

Choose any H ∈ Hn 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

eTTC�(R) ≡ TTC�∗(R)(R(σ∗(R))).

Basically we make the same modifications of TTC� as we did for TTC� with an exception of Stage

1 where we use TTC� instead of TTC�. Note that, because TTC� is essentially single-valued, the sets

of patients receiving transplantations remain the same across all matchings selected by TTC� for each

problem. Also, eTTC� is essentially single-valued, which is straightforward from the definition.

Example 6. (eTTC�) Consider the problem in Example 5.

• • •

•

•

• •

•

•

1 2 3 58

9 6 4 7

Recall that C(R) = {c = (1, 2, 1), c′ = (2, 4, 7, 2), c′′ = (2, 3, 7, 2)}. Since all these cycles are not feasible

with each other, TTC� chooses a cycle of the largest size; if there are many, it chooses a cycle including

patients with higher priorities in order. Thus, c′′ is chosen and the matching is TTC�(R) = {(µ(1) =

1, µ(2) = 3, µ(3) = 7, µ(7) = 2, µ(4) = 4, . . . , µ(9) = 9)} with N = {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 N . 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 L′ ≡ {(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

the eTTC� algorithm. Suppose that we choose L′. Then, σ∗(R) = {7}. We modify � to �∗(R) :

2, 3, 7, 1, 4, 5, 6, 8, 9. By definition, eTTC�(R) = TTC�∗(R)(R(σ∗(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 σ∗(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 eTTC� algorithm, we

obtain L = {(6, 9, 8, 1, 2, 3, 5), (4, 7)} and L′ = {(6, 9, 8, 1, 2, 4, 7), (3, 5)} that include the largest set of

patients. Suppose that we choose L′. Then, σ∗(R) = {5, 7}. We modify � to �∗(R) as above. The

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 σ∗(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. eTTC� and eTTC� satisfy Pareto efficiency, monotonicity, and maximal improvement.

Proof. (Pareto efficiency) For each R ∈ RN , note that R(σ∗(R)), R(σ∗(R)) ∈ RN . By definition,

eTTC�(R) ≡ TTC�∗(R)(R(σ∗(R)) and eTTC�(R) ≡ TTC�∗(R)(R(σ∗(R)). By Proposition 1, TTC�

and TTC� are Pareto efficient for every �. Therefore, for each R ∈ RN , eTTC�(R) and eTTC�(R)

are Pareto efficient at R(σ∗(R)) and R(σ∗(R)), respectively.

(Monotonicity) We show that eTTC� is monotonic. Let R ∈ RN . Recall that TTC� and

eTTC� 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

TTC�(R) still receive transplantations at eTTC�(R). Denote the set of those patients by N \ N . Note

that they are the only patients involved in feasible trading cycles that generate matchings in TTC�(R)

and the preferences of these patients are not affected by the use of suppressants under the eTTC�

algorithm. Thus, the feasible trading cycles involving N \N still remain feasible in g(R(σ∗(R))). Given

ordering �∗ (R), these patients have higher priorities than those in N . Consider the algorithm of

TTC�∗(R) applied to R(σ∗(R)). The feasible trading cycles involving N \ N at the TTC� 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 \ N . Among these sets, we continue to sort out the set of feasible cycles in

order of priorities among N . This guarantees that there are trading cycles involving all patients in

N \ N . Therefore, we end up with eTTC�(R) in which all of these patients receive transplantations,

completing the proof. The same argument shows that eTTC� is monotonic.

(Maximal improvement) We show that eTTC� satisfies maximal improvement. Suppose, by

contradiction, that there are R ∈ RN , S ⊆ N with |S| ≤ k, µ ∈ eTTC�(R), and µ′ ∈ M(R(S)) such

that N(µ) ( N(µ′). Let N be the set of patients who receive no transplantations at TTC�(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 eTTC� algorithm for R. Let H ′ ∈ H(R) be the set of cycles and chains that results

in µ′ when the patients in S use suppressants. According to the definition of eTTC�, H should have

been chosen to include all patients in N \ N as well as patients in N in the order of priorities over N ,

16

as long as the number of chains does not exceed k. Since N(µ) ( N(µ′), there is i∗ ∈ N(µ′) \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 eTTC� algorithm. Since i∗ /∈ N(H), this is a contradiction to

the definition of eTTC�. 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 ′.

Since i∗ /∈ N(H), again, this is a contradiction to the definition of eTTC�. The same argument shows

that eTTC� 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 transplan-

tations 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 210 35 80 24 349

B 28 172 77 24 301

O 38 39 164 5 246

AB 34 37 20 33 124

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 28 (7) 28 (0) 56

3-chain 7 (0) 7 (17) 7 (31) 21

2-chain 24 (0) 24 (7) 48

2-chain 17 (0) 17 (22) 34

1-chain 7 (0) 7

1-chain 22 (0) 22

1-chain 5 (0) 5

Table 5. Simulation algorithm for the set of incompatible pairs in 2012

# pairs 2-cycle 3-chain 2-chain 1-chain # Suppressants

2011 131 17 2 30 31 63

2012 193 28 7 41 34 82

2013 216 29 10 38 52 100

2014 217 34 3 48 44 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.

11In 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 withtype O donor in the set. Therefore, for all patients to receive transplantations, each pair of types O-A, O-B, and O-ABshould form a chain, and any two pairs of such types cannot form a chain together. Thus, the total number of pairs oftypes O-A, O-B, and O-AB is a lower bound for the number of chains.

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Type A B O AB Total

A 116 92 95 37 340

B 92 73 76 30 271

O 95 76 78 31 280

AB 37 30 31 12 110

Total 340 271 280 110 1001

Table 7. The profile of blood types of a hypothetical population: patients’ types are in leftmost column; donors’ typesare 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 92 (0) 92 (0) 184

3-chain 0

2-chain 37 (0) 37 (58) 74

2-chain 30 (0) 30 (46) 60

1-chain 58 (0) 58

1-chain 46 (0) 46

1-chain 31 (0) 31

Table 8. Simulation for incompatible pairs in a hypothetical population

# pairs 2-cycle 3-chain 2-chain 1-chain # Suppressants

Hypotheticalpopulation 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

12The number of pairs of each type is rounded to the nearest integer, with the total equal to 1,001.

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# pairs AB-O 2-cycle 4-cycle 3-cycle 2-chain 1-chain # Suppressants

2011 131 20 17 2 18 12 31 43

2012 193 20 28 7 13 28 34 62

2013 216 17 29 10 7 31 52 83

2014 217 11 34 3 8 40 44 84

Hypotheticalpopulation 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 incom-

patible 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 (Sonmez and Unver, 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.

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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←B-

A←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 O-

B←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.

13If they exist together, these two 1-chains can form a 2-chain, reducing the total number of chains, a contradiction toH ∈ 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) 34

3-chain 2 (0) 2 (17) 2 (31) 6

2-chain 13 (0) 13 (18) 26

2-chain 17 (0) 17 (4) 34

1-chain 18 (0) 18

1-chain 4 (0) 4

1-chain 9 (0) 9

Table A2. Simulation for incompatible pairs in 2011

25

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) 58

3-chain 10 (0) 10 (14) 10 (41) 30

2-chain 24 (0) 24 (17) 48

2-chain 14 (0) 14 (30) 28

1-chain 17 (0) 17

1-chain 30 (0) 30

1-chain 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) 68

3-chain 3 (0) 3 (22) 3 (43) 9

2-chain 26 (0) 26 (17) 52

2-chain 22 (0) 22 (18) 44

1-chain 17 (0) 17

1-chain 18 (0) 18

1-chain 9 (0) 9

Table A6. Simulation for incompatible pairs in 2014

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