Kidney Exchange with Immunosuppressants∗

Youngsub Chun† Eun Jeong Heo‡ Sunghoon Hong§

February 29, 2016

Abstract

We investigate the implications of using immunosuppressants as a part of a kidney exchange pro-

gram. Immunosuppressants allow patients to receive transplants from any donor, thereby relaxing all

immunological constraints. Given the limited availability of immunosuppressants, we identify which

patients are to be provided immunosuppressants and then determine which patients are matched

with which donors for the transplants. We ask whether there exist Pareto efficient solutions that

satisfy additional requirements of “monotonicity” and “maximal improvement”. We propose modi-

fications of the top-trading cycles (TTC) solutions to achieve these requirements. To quantify the

improvement on current transplantation practice made possible by the use of immunosuppressants,

we conduct counterfactual analyses using actual transplant data from South Korea. Our results

suggest that it is possible to reduce the use of immunosuppressants by up to 55 percent from the

current practice.

JEL classification Numbers: C71; D02; D63; I10

Keywords: immunosuppressants; kidney exchange; top-trading cycles solutions; Pareto efficiency;

monotonicity; maximal improvement

∗File name: KEI-2016-0227.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 and needs to receive a kidney transplant, four

options are available that depend on the immunological compatibility of the patient with her own

donor.1 First, if the patient is compatible with the donor, a direct transplant within this pair can be

performed. Otherwise, she has to look for other ways to receive a transplant. The second option is to

be registered on a waitlist to receive a transplant from a deceased donor. Another option, developed in

the last decade, is to participate in a kidney exchange program in which donors are swapped between

incompatible pairs. Unfortunately, the possibility of receiving transplants from deceased donors or

through exchanges is quite limited relative to the increasing number of patients in the waitlist, as

illustrated in Table 1 with data for the KONOS (Korean Network for Organ Sharing) program.

Recent developments in immunosuppressive protocols have introduced a fourth option. Immuno-

suppressants (suppressants, for short) eliminate all immunological compatibility constraints – both

tissue-type and blood-type – that patients might have against donors. If a patient uses a suppressant,

then she becomes compatible with any donor, so is able to receive a transplant even from an incom-

patible donor (what we call an incompatible kidney transplant).2 This protocol has been reported to

be quite successful. Specifically, the long-term survival rate of incompatible transplants is equivalent

to that of compatible transplants.3

In recent years, the number of patients using suppressants for incompatible transplants has increased

in many countries. In Korea, for example, the proportion of blood-type incompatible kidney transplants

in total living-donor transplants has increased from 4.7 percent in 2009 to 21.2 percent in 2014, as shown

in Table 2. This sharp increase is partly because the National Health Insurance Service (NHIS) of Korea

has covered a large fraction of the total cost of suppressants since 2009. Patients pay a small share of

1Immunological compatibility is mostly determined by biological characteristics of patients and donors, such as ABOblood types, Human Leukocyte Antigen (HLA) types, and the crossmatch.

The ABO blood type is determined by the inherited antigenic substances on the surface of red blood cells. For example,if a patient’s antigen is type A, then her blood type is type A and her antibody is type B; if a patient’s antigen is type AB,then blood type is AB and she has no antibody. A patient with antibody of type X cannot receive a transplant from adonor with type X antigen. For example, if a person’s blood type is A, then her antibody is type B, and thus, she cannotreceive a transplant from any donor having a type B antigen, namely, blood types B and AB. Similarly, if a person’s bloodtype is O, then her antibodies are types A and B; therefore, she cannot receive a transplant from any donor of blood typesA, B, or AB.

Tissue-type compatibility is determined by patient’s and donor’s HLAs, which are proteins on the surface of cellsthat are responsible for immunological responses. If a patient and a donor have the same HLA, then they are calledan identical match, which is very rare because of the number of combinations of HLA proteins that are possible (formore information, see the Genetics Home Reference website, provided by the U.S. National Library of Medicine, athttp://ghr.nlm.nih.gov/geneFamily/hla).

A patient’s antibodies and a donor’s HLAs also determine the “crossmatch”: if the crossmatch is positive, then thenthe patient’s antibodies react to the donor’s HLAs, thereby making a transplant unsuccessful.

2A variety of immunosuppressive protocols have been reported in the medical literature. Most of these protocols includethe use of rituximab (which is an immunosuppressant medicine that inactivates white blood cells) and plasmapheresis(which is a procedure used to remove harmful antibodies from the blood of patients). Rituximab is usually taken a monthbefore transplant surgery and plasmapheresis is carried out every second day for two weeks before surgery.

3According to the KONOS Annual Report in 2014, the five-year survival rate of ABO-incompatible living-donorkidney transplants is 95.2 percent and that of ABO-compatible living-donor kidney transplants is 96.4 percent. For otherstudies on the long-term outcomes of incompatible kidney transplants, see Takahashi et al. (2004), Tyden et al. (2007),Montgomery et al. (2012), and Kong et al. (2013).

1

YearPatients

in waitlistsTotal

transplantsTransplants fromdeceased donors

Transplants fromliving-donors

2009 4,769 1,238 488 750

2010 5,857 1,287 491 796

2011 7,426 1,639 680 959

2012 9,245 1,788 768 1,020

2013 11,381 1,761 750 1,011

2014 13,612 1,808 808 1,000

Table 1. Kidney transplantations in Korea

YearTransplants from

living-donors

Direct transplantsof ABO-compatible

pairs

Direct transplantsof ABO-incompatible

pairs

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 transplants in Korea: percentage of cases to total shown in parentheses

the total cost, as low as 20 percent depending on their medical conditions. In contrast, the proportion

of transplants made using kidney exchanges has decreased from 5.3 percent to nearly 0 percent during

the same period, as shown in Table 2. As can be seen, suppressants have largely replaced the kidney

exchange program in South Korea.

As the number of patients using suppressants increases, the expenditure of the NHIS to subsidize

these incompatible transplants also increases. Because the NHIS has a limited budget each year, it

cannot subsidize the use of suppressant for all incompatible pairs. Therefore, we should ask if there is

room for improvement on the current practice of using suppressants.

In this paper, we investigate how to optimally use suppressants in order to facilitate kidney trans-

plants as a part of a kidney exchange program. As an illustration of how the suppressants can be used

optimally, consider the following example. Assume that the compatibility is determined only by ABO

blood type, rather than by both blood type and tissue type. A pair consists of a patient and a donor,

which we represent type X-Y, where the patient’s blood type is X and the donor’s type is Y. Suppose

that there are three pairs: A-B, B-AB, and O-AB. Due to the immunological constraints, a patient of

type A can receive a transplant only from a donor of type A or O, a patient of type B can receive a

transplant only from a donor of type B or O, and a patient of type O can receive a transplant only

from a donor of type O.

Because each patient is incompatible with her own donor, in the absence of suppressants, each of

them can only receive a transplant from someone else. In a kidney exchange program, on the other

2

hand, compatible pairs are formed by swapping donors across pairs. Unfortunately, in this example,

no such exchanges are possible, because the patient in pair A-B and the patient in pair O-AB are

incompatible with all three donors.

Now suppose that the health service can provide suppressants to at most two patients. One way

to deal with this is to provide the patients in two incompatible pairs with suppressants and have them

receive transplants directly from their own donors. In our example, the patients in A-B and O-AB

pairs can be provided suppressants and receive direct transplants from their donors. The patient in

the remaining pair, however, cannot receive a transplant. This procedure for allocating suppressants is

currently used in South Korea. The number of direct transplants within ABO incompatible pairs for

South Korea in the years 2009 to 2014 is shown in Table 2.

However, there is a better way to use the limited supply of suppressants in our example, one which

enables all patients to receive transplants. Note that the two pairs A-B and B-AB form a “chain”,

(A-B)–(B-AB), in which the donor in pair A-B is compatible with the patient in pair B-AB, while the

remaining patient of type A and the remaining donor of type AB are not compatible. Such a chain can

be viewed as a trading cycle with a “missing link” in the following sense: If the donor in pair B-AB and

the patient in pair A-B were compatible, these pairs would have formed a cycle in a kidney exchange.

Providing a suppressant to the patient of type B fills in this missing link and transforms the chain into

a trading cycle, (A-B)–(B-AB)–(A-B). In this cycle, the patient of type B receives a transplant from

the donor of type B and the patient of type A receives a transplant from the donor of type AB, with the

latter transplant made possible due to the use of a suppressant. The other available unit of suppressant

can be provided to the remaining patient, thereby enabling a direct transplant within the pair O-AB.

A key feature of our proposal is that participation in a kidney exchange program is voluntary for

patient-donor pairs who become compatible through the use of suppressants. Note that in the example,

when the patient in pair A-B uses a suppressant, she can receive a transplant directly from her own

donor, and so does not need to participate in the exchange program. Nevertheless, by participating in

the program, as described above, the pair A-B can also benefit the pair B-AB. Provided that it does

not matter from whom a patient receives a transplant when using a suppressant, the pair A-B would

be willing to participate in the pool. Such “altruistic” compatible pairs have also been proposed and

studied in the standard kidney exchange context (Sonmez and Unver, 2014; Roth et al., 2005; Gentry

et al., 2007).4

Because the participation is voluntary, our proposal avoids a possible negative externality on a

kidney exchange program that can arise with tissue-type suppressants. As Sonmez and Unver (2013)

have observed, the use of tissue-type suppressants may crowd out patient-donor pairs whose donors are

in short supply in the exchange pool (usually, blood-type O donors).5 In our proposal, howerver, all

4We follow Sonmez and Unver (2014) to use the “altruistic” to refer to compatible pairs who participate in a kidneyexchange program even though a direct transplant between themselves is possible.

5Simply put, the reason is as follows. Note that donors of blood type O are always on the short side of a kidneyexchange program because any patient is blood-type compatible with them. Type O donors enter the exchange pool onlywhen they are tissue-type incompatible with the other member of their pair. If type O donors do not participate in the

3

donors in incompatible pairs still remain in the exchange pool even after their patients use tissue-type

suppressants. Therefore, the shortage of blood-type O donors does not get worse.

In what follows, we formalize the idea of using suppressants to facilitate kidney transplants as part of

kidney exchange program and investigate the implications of some fairness and efficiency requirements

for such a program. As a benchmark, we begin with the standard kidney exchange model without

suppressants. We define two Top-Trading Cycles (TTC) rules associated with a priority ordering over

patients.6 Both rules are defined by an algorithm in which patient-donor pairs form trading cycles in

each step, as in the standard TTC algorithm. In our setting, however, there can be multiple overlapping

cycles because preferences are very coarse: they are either compatible or incompatible. We propose

two different ways of choosing trading cycles from among them and they result in two versions of TTC

rule.

We next extend the model by introducing suppressants. To accommodate the limited availability

of suppressants, we assume that at most k patients can use suppressants. For each kidney exchange

problem, we first determine who are to receive the suppressants. Because these recipients are then

compatible with any other donors, we next update the compatibility profile. Based on this new profile,

we choose which patients are matched with which donors.

We regard Pareto efficiency as being a minimal requirement on matchings. A matching is Pareto

efficient if there is no other matching that makes all patients weakly better off and at least one patient

strictly better off. Note that this is an efficiency requirement on matchings for each given compatibility

profile.

In addition to requiring matchings to be Pareto efficient, we also require that suppressants be

allocated fairly and efficiently. We consider two variants of our fairness condition. Both of them ensure

that no patient gets worse off after suppressants are used to supplement the kidney transplantation

program. Strong monotonicity requires all patients to be weakly better off after suppressants are

introduced, no matter who receives them. In contrast, monotonicity requires that there be some way

of allocating the suppressants that makes all patients weakly better off.

We also consider two variants of our efficiency requirement for the allocation of suppressants. Both

require that they be allocated so as to maximize transplants in the some sense. Consider two groups

of potential recipients. Each group results in different sets of patients receiving transplants. Maximal

improvement requires that if the first group results in more transplants than the other in terms of

set inclusion, then the latter group should not be allocated the suppressants. Cardinally maximal im-

provement modifies this requirement by comparing the total number of transplants that they facilitate,

rather than comparing them in terms of set inclusion.

pool when tissue-type suppressants are available, the remaining patients suffer from the shortage of type O donors. Evenif a tissue-type incompatible pair whose donor is type O uses a suppressant, in our proposal, this pair remains in the pooland participates in the kidney exchanges.

6Shapley and Scarf (1974) propose the TTC rule for a general model with indivisible goods. Roth et al. (2004) developa TTC rule for kidney exchange problems by considering chains formed with patients on the waitlist. As in Roth et al.(2004), we impose no constraint on the size of cycles in kidney exchanges. For more discussion on the size of exchanges,see Roth et al. (2007) and Saidman et al. (2006).

4

Based on these requirements, we present two impossibility results. The first shows that no rule

jointly satisfies Pareto efficiency and strong monotonicity. The second shows that even if strong mono-

tonicity is weakened to monotonicity, we cannot achieve it together with Pareto efficiency and cardinally

maximal improvement.

In view of these impossibilities, we then weaken cardinally maximal improvement to maximal im-

provement and ask if it is compatible with Pareto efficiency and monotonicity. We show that it is: We

introduce two modifications of the TTC rules to show this. Each of these solutions operates in four

steps. First, assuming that no one uses a suppressant, each of the TTC rules is applies in order to

identify the set of patients who receive no transplant at the stage. Second, the initial priority ordering

is modified so that the patients identified in the previous stage have lower priorities than any other

patients. In the third stage, the available suppressants are allocated. This is done by selecting cycles

and chains according to the modified priority ordering and assigning suppressants to the patients at

the heads of these chains. Finally, the compatibility profile is updated based for the patients identified

in the third stage and then the TTC rule associated with the modified priority ordering is applied to

this profile.

We next conduct counterfactual analyses to see how much suppressants can be reduced relative to

the current practice. The use of suppressants for incompatible transplants is relatively new phenomenon

and, as yet, there is little data available about how many transplants have been facilitated by the use of

suppressants. However, good data is available in South Korea and so we use it for a quantitative analysis.

The KONOS transplant data include all pairs who received living-donor kidney transplants, either

compatible or incompatible, during the four year period (2011-2014). This data set also includes the

blood-types of all these pairs. Using this data, we determine the minimal quantity of suppressant that

is needed to ensure that all patients, who actually received kidney transplants by using suppressants,

still receive transplants.

For our analysis, we assume that compatibility is only determined by the blood type. There are

then seven types of incompatible pairs: A-B, B-A, A-AB, B-AB, O-A, O-B, and O-AB. During this

period in South Korea, there were almost no living-donor transplants made through a kidney exchange.

Therefore, a patient in any pair of one of these types in the data must have used a suppressant to

receive a transplant. We propose an algorithm that minimizes the use of suppressants while ensuring

that all of these patients receive transplants. This algorithm runs as follows. First, all trading cycles

are identified and patients and donors are matched along these cycles. These pairs are then excluded

from the exchange pool. Next, all 3-chains (a “k-chain” is a chain consisting of k pairs) are identified.

In this setting, these are the longest chains. After assigning suppressants to the heads of these chains,

patients and donors along these chains are matched and then excluded from the pool. This process is

then repeated for 2-chains and then for 1-chains.

We compare the outcomes obtained using this algorithm with the actual KONOS data. Using our

algorithm, only 340 pairs need to use suppressants, whereas 757 patients actually used suppressants

during the period studied. This is a reduction in usage of 55.1 percent.

5

As a robustness check, we redo our analysis for different specifications of the relevant patient-donor

pairs, and obtain the similar results. The KONOS data only include pairs (either compatible or in-

compatible) whose patients have already received transplants. Some incompatible pairs do not perform

transplants, so the KONOS data does not capture all of the individuals who would have benefited

from the use of suppressants. To estimate the benefit of our proposal when all incompatible pairs are

accounted for, we construct a hypothetical population of 1,001 pairs using the average distribution of

ABO blood types in South Korea and apply our algorithm to this population. In this scenario, we

show that it is possible to reduce the use of suppressants by 55.4 percent, while still obtaining the same

benefit as above.

The rest of this paper is organized as follows. Section 2 introduces the standard kidney exchange

model without suppressants and defines two versions of a top-trading cycles rule. Section 3 extends

the model to include suppressants and establish our two impossibility results. Section 4 considers two

modifications of a top-trading cycles solution and determine which of the properties described above

are satisfied. Section 5 presents our counterfactual analyses based on Korean experience. We provide

some concluding remarks in Section 6.

2 Standard Kidney Exchange Model without Immunosuppressants

Let N ≡ {1, . . . , n} be a set of patient-donor pairs. Pair i ∈ N consists of patient i and donor i. A

patient is either compatible or incompatible with each other donor. If a patient is compatible with a

donor, then she can receive a transplant from the donor. For each i ∈ N , let N+i ≡ {j ∈ N : patient i

is compatible with donor j} and N−i ≡ N \N+i . Each patient 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 . If a patient is incompatible with all donors, then she prefers ∅ to all

other donors. Note that compatibility of a donor and a patient is determined by their immunological

characteristics, independently of other pairs’ compatibility. Let Pi and Ii be the asymmetric and the

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 all problems.7

We introduce a graph whose nodes are pairs in N . A directed arc from one node to another, say i

to j, is denoted by i → j. We allow j = i, in which case the arc is self-directed, i → i. Let g be the

set of all directed arcs over N . For each R ∈ RN and each i, j ∈ N , let i → j if and only if donor i

is compatible with patient j at R. Let g(R) ⊆ g be the set of directed arcs defined as above at R. A

problem R is uniquely represented by g(R).

An ordered list of pairs (i1, . . . , ik, i1) forms a cycle in g(R) if i1, . . . , ik are distinct pairs and

i1 → i2 → · · · → ik → i1. As long as i1, . . . , ik keep the order, it does not matter who comes first in

7Bogomolnaia and Moulin (2004) study dichotomous preferences in a general matching context. They examine ran-domized matchings to achieve efficiency, fairness, and strategic requirements when only two-way exchanges are allowed.

6

the list (i1, . . . , ik). Note that a cycle can be a self-cycle, i → i. Let C(R) be the set of all cycles in

g(R). Two cycles c, c′ ∈ C(R) are jointly feasible if no pair is in c and c′. Let C(R) ⊆ 2C(R) be the

collection of all sets of jointly feasible cycles in C(R). For each C ∈ C(R), let N(C) be the set of pairs

in a cycle in C. Let |C| be the number of the cycles in C.

An ordered list of pairs (i1, . . . , ik) forms a chain if i1, . . . , ik are distinct pairs and i1 → i2 → · · · →ik and ik 9 i1. Unlike a cycle we defined above, it does matter who comes first in the list (i1, . . . , ik).

We call i1 the head of this chain and ik the tail of the chain. Let L(R) be the set of all chains in g(R).

Two chains l, l′ ∈ L(R) are jointly feasible if no pair is in l and l′. Let L(R) ⊆ 2L(R) be the collection

of all sets of jointly feasible chains in L(R). For each L ∈ L(R), let N(L) be the set of pairs in a chain

in L. Let |L| be the number of the chains in L.

A pair of c ∈ C(R) and l ∈ L(R) are jointly feasible if no pair is in c and l. LetH(R) ≡ C(R)∪L(R)

be the set of all cycles and chains in g(R). Let H(R) ⊆ 2H(R) be the collection of all sets of jointly

feasible cycles and chains in H(R). For each H ∈ H(R), let N(H) be the set of pairs in a cycle or a

chain in H.

A matching µ specifies which patient is matched to which donor. For each i, j ∈ N , µ(i) = j implies

that patient i is matched to donor j. 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 transplants, namely,

N(µ) = {i ∈ N : µ(i) ∈ N+i }.

We also introduce a priority ordering over pairs.8 Denote a linear ordering over pairs 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 jointly feasible. Since pair 2 is in c and c′, the two cycles c and c′ are not jointly feasible. The

collection of jointly 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 jointly feasible. Since pair 1 is in l and l′′, the two chains l and l′′ are

not jointly feasible. The collection of jointly 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′}}.8For more detailed discussion on these priorities, see Roth et al. (2005).

7

A matching rule is a mapping ϕ : RN ⇒ ∪R∈RNM(R) such that for each R ∈ RN , ϕ(R) ⊆M(R).

A matching rule 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′).9

For each R ∈ RN , 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). A matching rule 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) rules adapted to this model.10 Without loss of generality, let

1 � 2 � · · · � n. For simplicity, a priority ordering � can be written as �: 1, 2, . . . , n.

Top-trading cycles rule associated with � (simply, TTC�): For each R ∈ RN , let C0 ≡ C(R).

Step 1. Find a set of jointly 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 jointly 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 jointly feasible cycles.

Step n. Obtain Cn. For each C ∈ Cn, each patient in N(C) receives a transplant from another patient’s

donor along the directed arc of cycles in C. All other patients are assigned their own donors. Collect

the resulting matchings for all C’s in Cn.

Note that this rule can be viewed as a “sequential priority” rule since we first identify all matchings

at which patient 1 receives a transplant, and then identify all matchings at which patient 2 does, and

so on. If patients are restricted to make two-way exchanges, such a sequential priority rule is Pareto

efficient and it also maximizes the number of transplants (Roth et al., 2005). If we allow more than

two-way exchanges, however, Pareto efficient matchings do not necessarily maximize the number of

transplants. Therefore, we can formulate another version of top-trading cycles rules.

Top-trading cycles rule maximizing the number of transplants (simply, TTC�): For each

R ∈ RN , let C0 ≡ argmaxC∈C(R) |N(C)|.

Step 1. Find a set of jointly 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 jointly 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 jointly feasible cycles.

Step n. Obtain Cn. For each C ∈ Cn, each patient in N(C) receives a kidney from another patient’s

donor along the directed arc of cycles in C. All other patients stay with their own donors. Collect the

resulting matchings for all C’s in Cn.

9Suppose, 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 pairs in N+

i are strictly preferred to those in N−i , we have µ(i) Pi µ′(i), contradicting that ϕ is

essentially single-valued.10The TTC rule is proposed by Shapley and Scarf (1974) in a general model and applied to kidney exchange problems

by Roth et al. (2004).

8

Remark. It is straightforward from the 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).

Note that all patients who receive transplants at µ also receive transplants at µ′. Furthermore, there is

at least one patient who receives a transplant at µ′, but not at µ. Among such patients, let patient i be

the one with the highest priority under �. According to the definition of TTC�, this matching should

be chosen so that patient i receives a transplant, 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). Note that all patients who receive transplants at µ also receive transplants at µ′ and at least one

patient receives a transplant at µ′, but not at µ. Thus, the number of transplants at µ′ is greater than

that at µ. Hence, 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 becomes

compatible with all donors, including her own. We denote by R∗i the preference of patient i after

receiving a suppressant. For each S ⊆ N , let 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 jointly 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)}.

9

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 N composed of at

most k patients. When k = 0, the extended model coincides with the standard model.

For each problem, we first decide at most k recipients of suppressants. We then update the preference

profile and choose matchings between patients and donors. Let σ : RN → 2N (k) be a recipient choice

rule. Let ϕ : RN ⇒⋃

R∈RNM(R) be a matching rule such that for each R ∈ RN , ϕ(R) ⊆ M(R).

We represent a solution Φ as a pair (σ, ϕ). With a slight abuse of notation, for each R ∈ RN , let

Φ(R) ≡ ϕ(R(σ(R))).

The following two requirements are defined for a matching rule, ϕ.

Pareto efficiency: For each R ∈ RN , ϕ(R) is Pareto efficient at R.

Note that Pareto efficiency does not say anything about how we assign suppressants. It simply

requires Pareto efficiency of matchings at each given preference profile. A solution (σ, ϕ) is Pareto

efficient if ϕ is Pareto efficient.

Our next requirement says, no matter what a choice rule is, a matching rule should make all patients

weakly better off after suppressants are introduced to the transplantation program.11

Strong monotonicity: For each k ∈ Z+, each σ : RN → 2N (k), and each R ∈ RN , there are µ ∈ ϕ(R)

and µ′ ∈ ϕ(R(σ(R)) such that for each i ∈ N , µ′(i) Ri µ(i).

This requirement is convincing especially when we cannot choose a particular recipient set, for

example, when patients individually decide whether they use suppressants or not. Unfortunately,

TTC� and TTC� violate this requirement.

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

R, R ∈ RN be given as follows:

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

11The two monotonicity requirements can be compared with “welfare-dominance under preference replacement” inallocation problems (for a complete survey about this requirement, see Thomson (1999)). It requires when a personchanges her preferences, all the remaining people be affected in the same direction. Since the use of suppressants by agroup of patients change their preferences, our requirements share the same sprit as welfare-domination under preferencereplacement, but there is no direct logical relation between them. In our requirements, first, at most k patients maychange their preference at the same time. Second, when patient i’s preference changes, it always changes to R∗i . Third,when preferences change, all patients, including those who use suppressants, should be affected in the same direction.Lastly, all patients should be affected in a particular direction – they become weakly better off.

10

• •

••

1 2

34

• •

••

1 2

34

Since there is only one cycle at g(R) and g(R), respectively, 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 a choice rule 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, patient 1 is better off, and patient 2 remains indifferent.

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

This observation generalizes to our first impossibility result. We cannot achieve strong monotonicity

together with Pareto efficiency.

Proposition 2. No matching rule ϕ jointly 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, 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) = 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

11

better off at the new profile than at R′, we should 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.

Remark. Since a matching rule ϕ is not single-valued, there is another way to define 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). It is easy to check that this implies strong monotonicity

that we define above. Therefore, Proposition 2 still holds with this formulation.

The implication of Proposition 2 is that, as long as we impose Pareto efficiency as a minimal

requirement on matchings, there is no way to make all patients weakly better off for all possible

recipient choice rules. In what follows, we instead ask if it is possible to make all patients weakly better

off for some recipient choice rules.

Before we proceed, let us take a closer look at Example 3 above. At R, suppose that patient 1,

instead of patient 2, receives a suppressant. Independent of the cycle (2, 4, 3, 2), patient 1 forms a

self-cycle. This results in all patients being weakly better off. Similarly, at R, suppose that patient 4,

instead of patient 2, receives a suppressant. Independent of the cycle (1, 2, 1), patients 3 and 4 form a

cycle (3, 4, 3). This again results in all patient being weakly better off. In this example, it is possible to

make all patients weakly better off if we choose a particular recipient for each problem. We formulate

this possibility to the following requirement. Consider a solution (σ, ϕ).

Monotonicity: For each k ∈ Z+, each R ∈ RN , each µ ∈ ϕ(R), each µ′ ∈ ϕ(R(σ(R))), and each

i ∈ N , µ′(i) Ri µ(i).

We note that this requirement can be trivially satisfied by many solutions. Consider any essentially

single-valued matching rule ϕ. Suppose that σ(R) = ∅ for all R ∈ RN . The solution (σ, ϕ) trivially

satisfies monotonicity simply by wasting all available suppressants. To avoid such ineffective use, we

formulate another efficiency requirement on the assignment of suppressants. The following requirement

says that the recipients of suppressants should be chosen to maximize the set of patients receiving

transplants 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 any Pareto efficient solution

(σ, ϕ). 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

12

• •

••

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, it selects {µ = (µ(1) = 1, µ(2) = 2, µ(3) = 4, µ(4) =

3)} and N(µ) = {3, 4}. Next, suppose that patient 1 receives a suppressant. By Pareto efficiency, the

solution selects {µ} and N(µ) = {1, 3, 4}. Now, 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 N(µ′) =

{1, 2, 3, 4}. Because all patients can receive transplants when patient 2 is provided a suppressant,

maximal improvement requires that σ(R) 6= ∅ and σ(R) 6= {1}.

As discussed in Section 2, we can also focus on maximizing the number of transplants. Then, the

aforementioned requirement is reformulated in terms of the number of transplants.

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 solution jointly satisfies Pareto efficiency, monotonicity, and cardinally maximal

improvement.

Proof. The proof is by means of an example with six patients. Let N ≡ {1, 2, 3, 4, 5, 6} and (σ, ϕ) be a

solution satisfying the three requirements. 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)}. Next, suppose that k = 1. To satisfy CMI, it is easy to check that we should

13

have σ(R) = {2} and all patients except for patient 1 receive transplants at the resulting preference

profile. Since patient 1 is made 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 rules defined in Section 2, but we add a major modification to

assign suppressants.

The extended top-trading cycles solution (simply, eTTC�):

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

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 as follows. All patients in N \ N have higher priories

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

priority ordering by �∗ (R).

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

H0 ≡ {H ∈ H(R) : N(H) ⊇ N \ N and the number of chains in H does not exceed k},12 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 heads of all chains in H. Denote the set of

these heads 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 and find the set of patients N who

do not receive transplants at the stage. Since TTC� is essentially single-valued, the sets of patients

receiving transplants remain the same across all matchings selected by TTC�. In the second stage, we

modify the initial priority ordering � to �∗ (R) so that all patients N have lower priorities than the

rest of patients. In the third stage, we choose the sets of jointly feasible cycles and at most k chains

including all patients in N \ N and some patients in N chosen according to their priorities. We choose

any one of these sets and assign suppressants to the heads of the chains in the set. In the last stage, we

update the preference profile into R(σ∗(R)) and apply to it TTC associated with the modified priority

ordering.

12Such 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).

14

Note that to satisfy monotonicity, all patients in N \N should receive transplants after suppressants

are introduced. For this, we sort them out and give them higher priorities than all patients in N when

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

or chains including these patients will be selected ahead of the rest, guaranteeing that they still receive

transplants at the new profile. To satisfy maximal improvement, on the other hand, we identify at most

k chains including a “largest” set of patients and then assign suppressants to the heads of these chains.

By doing this, we can improve all patients’ welfare in a “maximal” way. It is straightforward from the

definition that eTTC� is essentially single-valued.

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}

• • •

•

•

• •

•

•

1 2 3 58

9 6 4 7

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

cycles is not jointly feasible, TTC� chooses the cycle c including the patient with the highest priority,

which is patient 1. Therefore, TTC�(R) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4, . . . , µ(9) = 9)} and

N = {3, 4, . . . , 9}.(1) 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 two chains l = (5, 3, 2, 1, 8, 9, 6) and l′ = (7, 3, 2, 1, 8, 9, 6)

are in the sets of jointly feasible cycles and chains including patients 1, 2, and 3. In this collection, we

identify the sets including the patient with the next highest priority in N , which 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 head of l, i.e., σ∗(R) = {5}. We also modify � to �∗ (R), which

happens to be � again. By definition, eTTC�(R) = TTC�∗(R)(R(σ∗(R))). The resulting 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).

(2) Suppose that at most two patients can receive suppressants, i.e., k = 2. As above, we identify

the set of cycles and at most two chains including patients 1, 2, 3 and 4. We obtain two sets H =

{(5, 3, 2, 1, 8, 9, 6), (7, 4)} and H ′ = {(7, 4, 2, 1, 8, 9, 6), (5, 3)}. Thus, σ∗(R) = {5, 7} for both sets. We

also modify � to �∗ (R) as above. The two resulting matchings of eTTC�(R) are µ and µ′ such that

15

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 find µ and µ′ indifferent, confirming 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 when k = 2,

the resulting matchings may be different from eTTC�(R). When suppressants are assigned one by one

sequentially, the only resulting matching is µ. All patients, however, still find both cases indifferent.

The extended top-trading cycles solution maximizing the number of transplants (simply,

eTTC�):

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

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 as follows. 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 jointly 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 heads of all chains in H. Denote the set of

these heads 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� except that we use TTC�,

instead of TTC�, in Step 1. Note that, because TTC� is essentially single-valued, the sets of patients

receiving transplants remain the same across all matchings selected by TTC� for each problem. It is

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

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

• • •

•

•

• •

•

•

1 2 3 58

9 6 4 7

16

Recall that C(R) = {c = (1, 2, 1), c′ = (2, 7, 4, 2), c′′ = (2, 7, 3, 2)}. Since any pair of these cycles is

not jointly feasible, TTC� chooses a cycle of the largest size; if there are many, it chooses a cycle

including patients with the highest priorities in order. Thus, c′′ is chosen and the resulting matching is

TTC�(R) = {(µ(1) = 1, µ(2) = 3, µ(3) = 7, µ(7) = 2, µ(4) = 4, . . . , µ(9) = 9)} and N = {1, 4, 5, 6, 8, 9}.(1) 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 jointly feasible cycles and a single chain, we

find the sets including patients 1, 2, 3, and 7. Then, we obtain H = {(1, 8, 9, 6), (2, 7, 3, 2)} and

H ′ = {(7, 3, 2, 1, 8, 9, 6)} at the last step of the eTTC� algorithm. If we choose H ′, 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 H, then σ∗(R) = {1}. The resulting matching is different from µ,

but all patient find it indifferent to µ.

(2) Now suppose that at most two patients can receive suppressants, i.e., k = 2. Similarly as

above, we identify the collection of sets of jointly feasible cycles and at most two chains including

patients 2, 3, and 7 as well as patients 1 and 4. Then, we obtain H = {(5, 3, 2, 1, 8, 9, 6), (7, 4)}and H ′ = {(7, 4, 2, 1, 8, 9, 6), (5, 3)} at the last step of the eTTC� algorithm. If we choose H ′, 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 H, then

σ∗(R) = {5, 7}. The resulting matching is different from µ, but all patients find it indifferent to µ.

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 : Let R ∈ RN . Recall that TTC� and eTTC� are essentially single-valued. Note

that at each matching, any patient who receives a transplant cannot be made strictly better off and any

patient who receives no transplant cannot be made strictly worse off. Therefore, it is enough to show

that all agents who used to receive transplants at TTC�(R) still receive transplants at eTTC�(R).

Denote the set of those patients by N \ N . Note that they are the patients included in jointly feasible

trading cycles in the TTC� algorithm. Since suppressants make patients compatible with more donors,

the jointly feasible trading cycles including N \ N still remain jointly feasible in g(R(σ∗(R))). Now,

consider the algorithm TTC�∗(R) for R(σ∗(R)). Since these patients have higher priorities than the

remaining patients at �∗ (R), these patients are always included in the sets of jointly feasible trading

cycles chosen in each step. Therefore, we end up with eTTC�(R) in which all of these patients receive

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

17

Maximal improvement : Suppose, by contradiction, that there are R ∈ RN , S ⊆ N with S ∈2N (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 transplant at TTC�(R). Let H ∈ H(R) be the set of cycles and chains chosen in the last

stage of the eTTC� algorithm for R, resulting in µ. 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.

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 N(µ) ( N(µ′) implies that the set of cycles and chains could have been chosen

to enable an additional patient i∗ to receive a transplant. This is a contradiction to the definition of

eTTC�. The same argument shows that eTTC� satisfies maximal improvement.

5 Counterfactual Analysis

To quantify how much suppressants can be reduced relative to the current practice, we next conduct

counterfactual analysis. We assume that immunological compatibility is only determined by ABO blood

types.13 We represent a pair of patient and donor by 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. We draw a directed arc from one

pair to another if the donor of the former is compatible with the patient of the latter. A k-cycle is a

cycle consisting of k pairs and a k-chain is a chain consisting of k pairs.

5.1 Counterfactual analysis based on the KONOS data

We use the KONOS data on living-donor kidney transplants during the four year period (2011-2014).

The data set includes all patient-donor pairs – either compatible or incompatible – who received trans-

plants during this period. It also includes their ABO blood types. Because there are almost no trans-

plants made through exchanges, as shown in Table 2, these patients should have received transplants

from their own donors.

In 2012, for example, there were 1,020 living-donor kidney transplants in total. Table 3 summarizes

the blood-type profile of these pairs. For instance, there were 210 compatible pairs of type A-A whose

patients received transplants directly from their donors; there were 35 incompatible pairs of type A-B

whose patients received transplants from their donors by using suppressants.

In this blood-type restricted setting, there are 16 types of pairs in total. Among them, 7 types of

pairs are incompatible: A-B, B-A, A-AB, B-AB, O-A, O-B, and O-AB. We collect the profile of these

193 incompatible pairs in Table 4. Since they received transplants directly from their own donors,

13Some analytical results in Roth et al. (2007) are also based on this assumption, which they call “upper bound”. Thisassumption is plausible when the exchange market is large enough and the tissue-type restriction does not bind.

18

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 transplants in 2012: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

# of pairs 35 28 24 24 38 39 5 193

Table 4. Blood types of incompatible pairs in 2012

the total number of these patients coincides with the number of patients who used suppressants for

transplants in 2012.

We now propose an algorithm that determines the minimal quantity of suppressant that is needed

to ensure that all patients, who actually received kidney transplants by using suppressants, still receive

transplants.

The min 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 all pairs in these 2-cycles from the pool.

Step 2. Find all 3-chains, which are the longest chains in the setting. If #(A-B) > #(B-A) in Step 1,

these 3-chains must be of the form O-A→A-B→B-AB. If #(A-B) < #(B-A) in Step 1, the 3-chains

must be of the form O-B→B-A→A-AB. If #(A-B) = #(B-A) in Step 1, there is no remaining pairs of

type (A-B) and (B-A) and no 3-chain can be formed. Remove all pairs in these 3-chains from the pool.

Step 3. Find all 2-chains. These chains must be one 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 all

pairs in these 2-chains from the pool.

Step 4. Find all 1-chains.

In Step 1, pairs A-B and B-A form trading cycles. Any patient included in these cycles can receive a

transplant even without suppressants. After all these jointly feasible 2-cycles are formed and removed,

at most one type of A-B and B-A pairs exists in the pool.

In Step 2, if A-B pairs exist in the pool but B-A pairs do not, the longest possible chain is a 3-chain,

O-A→A-B→B-AB (similarly, if B-A pairs exist in the pool but A-B pairs do not, the longest possible

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.

19

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. The min algorithm for 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. Results of the min algorithm during 2011-2014

In Step 3, if there are no remaining pairs A-B and B-A in the pool, the longest chain is a 2-chain

of the forms O-A→A-AB and O-B→B-AB. If A-B pairs exist in the pool but B-A pairs do not, then

at most one of the two pairs O-A and B-AB may exist in the pool. This is because, if both O-A and

B-AB pairs exist in the pool, they form a 3-chain O-A→A-B→B-AB, which should have been removed

in Step 2. Therefore, 2-chains are either all O-A→A-B or all A-B→B-AB. Similarly, if B-A pairs exist

in the pool, but A-B pairs do not, then at most one of the two pairs O-B and A-AB may exist in the

pool. Therefore, 2-chains are either all O-B→B-A or all B-A→A-AB.

After all jointly feasible 2-chains are formed, each remaining pair in the pool forms a 1-chain in

Step 4. Collect all chains from Step 1 to Step 4 and let the heads of these chains use suppressants. We

show that this algorithm minimizes the use of suppressants.

Theorem 2. The min algorithm minimizes the use of suppressants ensuring that all incompatible pairs

receive transplants.14

The proof of Theorem 2 is deferred to Appendix 1.

Table 5 summarizes these steps of the min algorithm applied to the data 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 represent the number of the corresponding pairs who still remain in the

14In all cases, the number of required suppressants is at least as large as the number of pairs with blood type O patients.This is because blood type O patients can receive transplants only from the type O donors, but there is no pair with typeO donor in the set. Therefore, for all patients to receive transplants, each pair of types O-A, O-B, and O-AB should forma 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.

20

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’ 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 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. The min algorithm 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. The result of the min algorithm on the hypothetical population

pool after cycles or chains are formed and removed. By incorporating suppressants to the kidney

exchange program as we propose, the use of suppressants could have been decreased from 193 to 82,

still ensuring that all incompatible pairs still receive transplants.

We have similar outcomes in other years, as summarized in Table 6. During 2011-2014 in total, 757

recipients of suppressants could have been reduced to 340, ensuring that all incompatible pairs of the

data still receive transplants. This is a reduction of 55.1% from the current practice.

5.2 Counterfactual analysis based on the hypothetical population

As a robustness check, we redo our analysis for a slightly different specification of patient-donor pairs in

this section. The KONOS data only include pairs (either compatible or incompatible) whose patients

have already received transplants. Some incompatible pairs do not perform transplants, so the KONOS

data may underrepresent incompatible pairs.

To estimate the benefit of our proposal when all incompatible pairs are accounted for, we construct

a hypothetical population of 1,001 pairs using the average distribution of ABO blood types in South

21

# 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. Results of the min algorithm when pairs of type AB-O participate.

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.15

From this distribution, we identify all incompatible pairs and apply the min algorithm. Because

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 in Step 2. Tables 8 and 9 show the result. By applying our algorithm, 453 recipients of

suppressants could have been reduced to 202. This is a reduction of 55.4% from the current practice.

5.3 Counterfactual analysis based on the KONOS data with the participation of

“altruistic” compatible pairs

We so far collected incompatible pairs when applying the min algorithm, excluding all compatible pairs

from the population. The use of suppressants can be reduced further, however, when some compatible

pairs participate in the exchange pool, facilitating transplants even more. Such compatible pairs are

called “altruistic” pairs (Sonmez and Unver, 2014). For example, if AB-O pairs participate in the

pool, then 3-chains can be transformed into 4-cycles AB-O→O-A→A-B→B-AB, and 2-chains can be

transofrmed into 3-cycles AB-O→O-A→A-AB or AB-O→O-B→B-AB. Table 10 summarizes the results

of the min algorithm when compatible pairs of type AB-O participate in the exchange pool.

More generally, we may add all compatible pairs to the pool and see how much they facilitate

transplants. It then becomes quite complicated to conduct counterfactual analyses, but it will bring in

even larger reduction of suppressants than what we identified above.

6 Conclusion

In this paper, we took a first step to investigate implications of introducing suppressants as a part

of kidney exchange program. We impose several requirements of fairness and efficiency in assigning

suppressants and matching patients with donors. We propose two extended TTC solutions and show

15The number of pairs of each type is rounded to the nearest integer, with the total equal to 1,001.

22

that they satisfy all attainable requirements we propose. By using actual transplant data from South

Korea, we also show that the use of suppressants in the current practice can be reduced significantly.

There still remain several interesting questions. First, it is possible to introduce monetary transfers

to the model. We currently assume that all expenses of suppressive treatments are fully subsidized

by the health service. This makes patients indifferent between compatible donors and incompatible

donors as long as they can receive transplants. In practice, however, patients often have to pay a

certain fraction of the cost to use suppressants. Therefore, incompatible transplants usually cost more

than compatible transplants and patients no longer find them indifferent. For instance, patients may

decline to use suppressants due to a higher cost and opt out to wait for transplants from deceased

donors. This makes the exchange pool shrink, and as a result, we may not obtain a significant gain

by introducing suppressants. To avoid such cases, it is reasonable to introduce a cost sharing process

or a compensating process for pairs who participate in the pool even after using suppressants. This is

because the use of suppressants incurs costs to these pairs, but they benefit all other patients in the

chains for free. We may consider, for instance, the equal division of the cost 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 the ABO blood types determine dichotomous

compatibility profiles, there are different levels of compatibility when tissue types (or HLA types) are

also counted in. The suppressive treatments should also be set to different levels of compatibility. 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 deferred acceptance (DA) solution to our setting. Since

there is a single priority ordering over patients, the DA solution can be defined as follows: among all sets

of jointly 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 heads of chains to be recipients of suppressants and let patients

receive kidneys from donors along the directed arcs in the cycles and chains. By doing this, we will

obtain a “stable” assignment: if a patient does not receive a transplant, then either (i) all patients with

lower priorities do not receive transplants, 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, which can be easily verified with the example in the proof of Proposition 3 after

switching labels of patients 1 and 6 except for the priority ordering.

Appendix 1. The min algorithm

We prove the following theorem in Section 5.

Theorem 2. The min algorithm minimizes the use of suppressants, ensuring that all incompatible

pairs receive transplants.

23

Proof of Theorem 2. Among all sets of jointly feasible chains and cycles including all incompatible

pairs, let H be the collection of all such sets with the smallest number of chains. Choose any H ∈ Hand let h∗ be the number of chains in H.

Claim 1. It is possible to keep the same number of chains h∗ by forming an additional 2-cycle with

pairs from the chains in H, if any.

Proof of Claim 1. Note that 2-cycles can be formed only by pairs A-B and B-A in this setting. Choose

any A-B pair and any B-A pair from the chains in H, if any. We show that the number of chains

remains h∗ when these two pairs form a cycle, while the remaining pairs in the chains are reorganized.

We consider four types of chains that may include this A-B pair in H: A-B, O-A→A-B, A-B→B-AB,

and O-A→A-B→B-AB (similarly, there are four types of chains that may include this B-A pair in H:

B-A, O-B→B-A, B-A→A-AB, and O-B→B-A→A-AB).

Case 1. A-B: We show that the B-A pair should be in O-B→B-A→A-AB at H. Suppose otherwise.

If B-A is a 1-chain in H, then the A-B and B-A pairs could have formed a cycle from H, reducing the

number of chains by 2. This is a contradiction to H ∈ H. If the B-A pair is in O-B→B-A, then the

A-B and B-A pairs could have formed a cycle and the O-B pair could have formed a 1-chain from H,

reducing the number of chains by 1. This is again a contradiction. If the B-A pair is in B-A→A-AB,

then the same argument applies. Summarizing, if this A-B pair is a 1-chain in H, then the B-A pair

should be in the 3-chain in H. Let the A-B and B-A pairs form a cycle and the remaining pairs in the

3-chain, O-B and A-AB, form two 1-chains. Keep all other cycles and chains as before. The number of

chains does not change, while an additional cycle is formed.

Case 2. O-A→A-B: We show that the B-A pair should be either in O-B→B-A or in O-B→B-A→A-

AB. If the B-A pair is in B-A→A-AB, then the A-B and B-A pairs could have formed a cycle and the

remaining pairs, O-A and A-AB, could have formed a 2-chain from H, reducing the number of chains

by 1. This is a contradiction to H ∈ H. A similar argument applies to obtain a contradiction if the

B-A pair is a 1-chain in H. Now, suppose that the B-A pair is in O-B→B-A. Let the A-B and B-A

pairs form a cycle and the remaining pairs, O-A and O-B, form two 1-chains. Keep all other cycles and

chains as before. The number of chains does not change, while an additional cycle is formed. Suppose

that the B-A pair is in the 3-chain. Let the A-B and B-A pairs form a cycle and the remaining pairs in

the chains, O-A, A-AB, and O-A, form a 2-chain O-A→A-AB and a 1-chain O-B. Keep all other cycles

and chains as before. The number of chains does not change either.

Case 3. A-B→B-AB: We show that B-A should be either in B-A→A-AB or in O-B→B-A→A-AB.

Otherwise, we can reduce the number of chains by 1 as above, a contradiction. The same argument of

Case 2 applies to show that the number of chains does not change when the A-B and B-A pairs form

a cycle and the remaining pairs are reorganized.

Case 4. O-A→A-B→B-AB: In this case, the B-A pair may be in any type of chain. Suppose that

the B-A pair is in O-B→B-A→A-AB. Let the A-B and B-A pairs form a cycle and the remaining pairs

in the chains form two 2-chains, O-A→A-AB and O-B→B-AB. Then the number of chains does not

24

change, while an additional cycle is formed. The similar argument applies to the other types of chains

that may include the B-A pair, so we omit them. �

Proof of Theorem 2. (continued) By Claim 1, we can find another set of cycles and chains in H by

forming 2-cycles as much as possible, without changing the total number of chains. Denote by H ′ a

resulting set of cycles and chains. It is straightforward that at least one type of A-B and B-A pairs

does not exist in all chains in H ′.

Claim 2. It is possible to keep the same number of chains h∗ by forming an additional 3-chain with

pairs from the 2-chains or the 1-chains in H ′, if any.

Proof of Claim 2. We consider the following three cases.

Case 1. Both A-B and B-A pairs do not exist in all chains at H ′. Then, each chain in H ′ is formed

by some of O-A, O-B, A-AB, B-AB, and O-AB pairs. However, no additional 3-chain can be formed

by these pairs and therefore we are done.

Case 2. B-A pairs do not exist in all chains at H ′. Then, each chain in H ′ is formed by some

of O-A, O-B, A-B, A-AB, B-AB, and O-AB pairs. An additional 3-chain among these pairs will be

O-A→A-B→B-AB, if any. Note that the O-AB pairs always remain as 1-chains, so we set them aside.

We focus on the remaining five types of 2-chains and 1-chains in H ′: O-A→A-B, A-B→B-AB, O-A,

A-B, and B-AB. Let the number of each type of chain in H ′ be

n1 ≡ #(O-A→A-B) n2 ≡ #(A-B→B-AB),

n3 ≡ #(O-A) n4 ≡ #(A-B) n5 ≡ #(B-AB).

Since the number of chains cannot be further reduced from H ′, we have the following observations.

(i) At least one of n3 and n4 should be zero. Otherwise, the two 1-chains, O-A and A-B, could have

changed to a 2-chain, reducing the number of chains by 1 from H ′. This is a contradiction.

(ii) At least one of n4 and n5 should be zero, by the same argument in (i).

(iii) At least one of n2 and n3 should be zero. Otherwise, a 2-chain A-B→B-AB and a 1-chain O-A

could have changed to a 3-chain, reducing the number of chains by 1 from H ′. This is again a

contradiction.

(iv) At least one of n1 and n5 should be zero, by the same argument in (iii).

Given these observations, we consider the following subcases, depending on the types of 1-chains in H ′.

Subcase 1. n4 > 0. By (i) and (ii), we have n3 = n5 = 0. Suppose that either n1 or n2 is zero.

Then no additional 3-chain can be formed. Suppose that n1, n2 > 0. Choose one O-A→A-B chain and

one A-B→B-AB chain. Let the O-A, A-B, and B-AB pairs form a 3-chain. Let the remaining pair A-B

form a 1-chain. Then, the number of chains does not change, while an additional 3-chain is formed.

Subcase 2. n4 = 0, n3 > 0, n5 > 0 : Then, by (iii) and (iv), we have n2 = 0 and n1 = 0. No

additional 3-chain can be formed in this case.

25

Subcase 3. n4 = 0, n3 > 0, n5 = 0 : Then, by (iii), we have n2 = 0. No additional 3-chain can be

formed in this case.

Subcase 4. n4 = 0, n3 = 0, n5 > 0 : Then, by (iv), we have n1 = 0. No additional 3-chain can be

formed in this case.

Subcase 5. n4 = n3 = n5 = 0 : Suppose that either n1 or n2 is zero. Then no additional 3-chain

can be formed. Suppose that n1, n2 > 0. The symmetric arguments in Subcase 1 apply here to show

that the number of chains does not change, while an additional 3-chain is formed.

Case 3. A-B pairs do not exist in all chains at H ′. The symmetric arguments in Case 2 apply here

to complete the proof. We omit it. �

Proof of Theorem 2. (continued) From Claims 1 and 2, we can find another set of cycles and chains

in H by first forming all possible 2-cycles and then forming all possible 3-chains, without changing the

total number of chains. Denote by H ′′ a resulting set of cycles and chains. There are no additional

2-chains that can be formed with pairs from the 1-chains in H ′′. If there is any, then the number of

chain could have been reduced further, a contradiction to H ′′ ∈ H.

Summarizing, all cycles and chains in H ′′ can be identified in four steps. First, indentify 2-cycles

among the pairs as much as possible and set them aside. Second, among the remaining pairs, identify

3-chains as much as possible and set these 3-chains aside. Third, among the remaining pairs, identify

2-chains as much as possible. Lastly, identify all remaining 1-chains. These cycles and chains are those

in H ′′ and the number of chains is exactly h∗. This process is exactly how the min algorithm operates.�

Appendix 2. Counterfactuals Based on the KONOS Data

We herein present the counterfactual 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 transplants in 2011

26

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. The min algorithm for incompatible pairs in 2011

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 transplants in 2013

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. The min algorithm for incompatible pairs in 2013

27

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 transplants in 2014

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. The min algorithm for incompatible pairs in 2014

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