Kidney Exchange with Immunosuppressants

Transcription

1 Kidney Exchange with Immunosuppressants Youngsub Chun Eun Jeong Heo Sunghoon Hong October 2, 2015 Abstract This paper investigates implications of introducing immunosuppressants (or suppressants) in kidney exchange problems. We begin with the standard kidney exchange model without suppressants and define two versions of the top-trading cycles (TTC) solutions satisfying Pareto efficiency. We then extend the model by introducing suppressants that make a patient compatible with any donor by relaxing immunological compatibility constraints. We examine the existence of solutions satisfying Pareto efficiency together with monotonicity and maximal improvement. Monotonicity requires that no patient should get worse off after the introduction of suppressants. Maximal improvement requires that if one set of recipients of suppressants enables more transplantations than the other (in terms of set inclusion), then the latter should not be chosen as the set of recipients. We propose two modified versions of TTC solutions satisfying these requirements. We also provide simulation analyses based on the actual transplantation data to see how much reduction we can expect on the current practice of using suppressants. In a blood-type restricted environment, we introduce an algorithm that minimizes the use of suppressants, given that all patients in the data receive transplantations. Our simulation result suggests that the use of suppressants can be reduced by 55 percent when the recipients of suppressants are chosen as we propose. JEL classification Numbers: C71; D02; D63; I10 Keywords: immunosuppressants; kidney exchange; top-trading cycles solutions; Pareto efficiency; monotonicity; maximal improvement File name: KEI tex Seoul National University. ychun@snu.ac.kr Vanderbilt University. heoeunjeong@gmail.com Korea Institute of Public Finance. sunghoonhong@kipf.re.kr

2 1 Introduction When a patient suffers from End Stage Renal Disease (ESRD) and needs to receive a kidney transplantation, three options are available depending on the immunological compatibility with her own donor. If the patient is compatible with the donor, a direct transplantation within the pair can be performed immediately. Otherwise, she has to list herself in a waitlist to get a transplantation from a deceased donor, or she has to participate in a kidney exchange program to seek other incompatible patient-donor pairs. Unfortunately, the transplantations from deceased donors or through exchanges are very limited in general to cover the increasing number of patients in the waitlists, as the KONOS (Korean Network for Organ Sharing) data show in Tables 1 and 2. Recent developments in immunosuppressive protocols brought in the fourth possibility of getting transplantations. Immunosuppressants, or simply suppressants, relax immunological compatibility constraints, which are mainly determined by ABO blood types, HLA (Human Leukocyte Antigen) typing, and crossmatching. 1 By using a suppressant, a patient becomes compatible with any donor, being able to receive incompatible kidney transplantations. It is reported that the long-term survival rates of incompatible kidney transplantations using suppressants are equivalent to those of compatible kidney transplantations. 2 This new protocol has led to the increased practice of incompatible kidney transplantations in a number of countries including Germany, Japan, Korea, Sweden, and the United States. In Korea, for example, the proportion of ABO-incompatible living donor kidney transplantations using suppressants has been increased from 4.7 percent in 2009 to 21.2 percent in 2014 as shown in Table 2. 3 Such a sharp increase is due to the National Health Insurance Service (NHIS) of Korea that covers a large fraction of the total cost of using suppressants since Patients are paying a small share, down to 20 percent, of the total cost, depending on their medical conditions. As the number of such incompatible kidney transplantations increases, the NHIS expenditure also increases to subsidize the cost. Since the NHIS has a limited budget assigned for suppressants each year, we should ask whether suppressants (and the corresponding expenditure) are allocated in an efficient manner and there is a room for further improvement. In this paper, we investigate whether it is possible to reduce the use of suppressants while maintaining the same set of patients receiving transplantations. We also investigate how to assign the unused 1 Human Leukocyte Antigens (HLAs) are proteins on the surface of cells that are responsible for immune systems. There are three subsets of HLAs, namely, HLA-A, HLA-B, and HLA-DR. Within each of these subsets, there are many different HLA proteins, say HLA-A1, HLA-A2, and so on. If the donor and the patient share the same HLAs, they are called an identical match. Two siblings are an identical match with probability of 25 percent. A parent and a child share the half of HLAs. On the other hand, if the crossmatch is positive between the patient s antibodies and the donor s HLAs, the patient s antibodies will attack the organ transplanted from the donor, and thus, transplantation is incompatible. In addition to immunological compatibility, kidney size and age may also affect transplantation outcomes. 2 There has been a variety of immunosuppressive protocols reported in the medical literature but most of these protocols include the use of rituximab (which is an immunosuppressant medicine) and plasmapheresis (which is a procedure to remove harmful antibodies from the blood of patients). For studies on long-term outcomes of incompatible kidney transplantations, see Takahashi et al. (2004), Tyden et al. (2007), Montgomery et al. (2012), and Kong et al. (2013). 3 In contrast, the proportion of living donor kidney transplantations through an exchange program had been decreased from 5.3 percent to near zero percent during the same period. 1

3 Year Patients in waitlists Deceased donor transplants Living donor transplants , , , , , , , , ,000 Table 1. Kidney transplantations in Korea Year Living donor transplants in total Transplants within compatible pairs Transplants using suppressants Transplants through exchanges (90.0%) 35 (4.7%) 40 (5.3%) (86.6%) 78 (9.8%) 29 (3.6%) (86.3%) 113 (11.8%) 18 (1.9%) , (81.1%) 193 (18.9%) 0 (0.0%) , (78.6%) 212 (21.0%) 4 (0.4%) , (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 immunological compatibility is determined only by ABO blood types. A pair of a patient and a donor is represented as type X-Y, where the patient s blood type is X and the donor s type is Y. Suppose that there are three pairs of types A-B, B-AB, and O-AB. 4 Since each patient is incompatible with her own donor, no patient can get a transplantation within the pair. In a kidney exchange program, on the other hand, patients switch their donors and get transplantations across pairs, if possible. Unfortunately, there is no such trading cycle among these pairs, since the patient of pair A-B is incompatible with the donor of pair B-AB and the patient of pair O-AB is incompatible with any other donor. Now suppose that the NHIS can afford at most two patients to use suppressants. One possibility is that two pairs, for example, A-B and O-AB, receive suppressants and they perform direct transplantations within the pairs. Note that the Transplants using suppressants in the fourth column of Table 2 represents this case. The remaining pair B-AB can not receive a transplantation. We find that, however, there exists a more efficient way of using suppressants. Note that the two pairs A-B and B-AB form a kind of chain in that the donor of A-B is compatible with the patient of pair B-AB, although the remaining patient and donor in these pairs are not compatible. Such a chain can be viewed as a 4 The blood type X specifies the types of antigen and antibody a person has. For example, if a person s blood type is A, then her antigen is type A and her antibody is type B; if a person s blood type is AB, then her antigen is type A and type B, and she has no antibody. A person with antibody X cannot get a transplantation from a donor with antigen X. For example, if a person s blood type is A, then her antibody is type B, and thus, she cannot get a transplantation from any donor having antigen B, namely, blood types B and AB. Similarly, if a person s blood type is O, then her antibody is A and B, therefore, she cannot get a transplantation from any donor of blood types A, B, and AB. 2

4 trading cycle with a missing link. We point out that a suppressant can be used to fill this missing link to form a trading cycle. That is, it is possible that the patient of pair B-AB gets a transplantation from the (compatible) donor of pair A-B, while the patient of pair A-B uses a suppressant and gets a transplantation from the (initially incompatible) donor of pair B-AB. The pair O-AB can use the remaining suppressant and perform a direct transplantation within the pair. Then, all patients receive transplantations. In what follows, we formalize this idea, together with additional considerations on fairness. Before we proceed, it is worth emphasizing that our analysis assumes the voluntary participation of patientdonor pairs who become compatible with any donor by using suppressants. In the example above, if the patient of pair A-B uses a suppressant, then she can get a transplantation directly from her own donor, without having to participate in the exchange program. What we propose, however, is that the pair A-B participates in the program to benefit the pair B-AB without bearing any welfare loss. As long as it results in transplantations that are equally successful as those within pairs, it is plausible to add such altruistic pairs to the exchange program, as proposed in the proceeding literature (Sönmez and Ünver, 2014; Roth et al., 2005; Gentry et al., 2007). We begin with the standard kidney exchange model without suppressants. We define two 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 compatibility constraints. We propose two different ways of choosing trading cycles among multiple cycles, which define two versions of TTC. We show that both are Pareto efficient for all priority orderings. We then extend the model by introducing suppressants. To reflect the limited availability of suppressants, we assume that at most k patients can use suppressants. For each problem, we have to decide (i) the set of at most k recipients of suppressants and (ii) the matching of pairs who participate in the exchange program. We first require that all patients should be weakly better off after suppressants are introduced. We regard it as a fairness requirement: since the use of suppressants is (fully or partially) subsidized by public health insurance, as in Korea, its benefit should be distributed to everyone. We refer to this requirement as monotonicity. We next require that the set of recipients of suppressants should be chosen to maximize transplantations. This is an efficiency requirement in assigning suppressants subject to the limited availability. Consider two sets of potential recipients of suppressants. They can result in two different sets of patients getting transplantations in the end. If one set of recipients enables more transplantations than the other in terms of set inclusion, then it is reasonable not to choose the latter set. We refer to this requirement as maximal improvement. We can also formulate the same idea, but in terms of 5 Shapley and Scarf (1974) propose the TTC solution in a general model with indivisible goods. Roth et al. (2004) develop the TTC solution for kidney exchange problems by considering chains formed with patients on waitlists. Note that Roth et al. (2004) impose no constraint on the size of exchanges, i.e., the length of cycles or chains, when designing their Top-Trading Cycles and Chains (TTCC) solution. For more discussion on the size of exchanges, see Roth et al. (2007) and Saidman et al. (2006). 3

5 total number of transplantations, not in terms of set inclusion. We refer to it as cardinally maximal improvement. We present two impossibility results. First, there is no solution satisfying Pareto efficiency and a strong form of monotonicity, which requires that all patients should be weakly better off after suppressants are introduced, no matter who becomes recipients of suppressants. Given this impossibility, we weaken monotonicity requirement: all patients should be weakly better off for some recipients of suppressants. In other words, it requires that, for each problem, there exists a right set of recipients who can make all patients weakly better off by using suppressants. Unfortunately, this requirement is again incompatible with Pareto efficiency and cardinally maximal improvement. We therefore turn to see if there is any solution satisfying the weak notion of monotonicity, Pareto efficiency, and maximal improvement. We introduce two TTC solutions for the extended model. Each of these extended TTC solutions runs in four stages. First, apply the two standard TTC solutions respectively, assuming that no one uses a suppressant; and identify the sets of patients who receive no transplantations under these solutions, respectively (this is the only step where the two versions of extended TTC solution diverge). Second, modify an initial priority ordering so that the patients identified in the previous step have lower priorities than the remaining patients. Third, select feasible cycles and chains according to the modified priority ordering, subject to the availability of suppressants; and assign suppressants to the patients at the end of these selected chains. Finally, obtain a new compatibility profile and apply the first standard TTC solution associated with the modified priority ordering. We prove that these extended TTC solutions satisfy the aforementioned requirements. To see how much we can improve on the current practice of using suppressants, we provide simulation results based on the actual transplantation data in South Korea. The KONOS data provides the profiles of blood types of all patient-donor pairs who received living donor kidney transplantations during the recent four years ( ). Assuming that the ABO blood-types determine immunological compatibility, we identify seven types of incompatible pairs, A-B, B-A, A-AB, B-AB, O-A, O-B, and O-AB, who used suppressants to get transplantations. We propose an algorithm that minimizes the use of suppressants given that all incompatible pairs in the data receive transplantations. This algorithm runs as follows. We first identify all trading cycles. We match patients and donors along these cycles and then exclude them from the pool. We next identify all 3-chains (a k-chain is a chain consisting of k pairs), which is the longest chain in this setting. We assign suppressants to the tails of these chains, match patients and donors along the chains, and exclude them from the pool. We repeat this process for 2-chains, and then for 1-chains. We compare what we obtain from the algorithm with the actual use of suppressants in the KONOS data. According to our algorithm, 340 pairs need to use suppressants, while 757 patients had used suppressants in the actual data during the four years. This is a reduction of 55.1 percent in total. Note that the KONOS data only includes compatible/incompatible pairs that have received transplantations. Therefore, the KONOS data can be biased to represent the whole population of patientdonor pairs (especially, it may underrepresent the population of incompatible pairs). We construct a 4

6 hypothetical population of 1,001 pairs according to the distribution of ABO blood types in Korea. We collect all incompatible pairs and run the algorithm. We again obtain a reduction of 55.4 percent. The rest of this paper is organized as follows. Section 2 introduces the standard kidney exchange model without suppressants, and defines two versions of top-trading cycles solutions. Section 3 extends the model to include suppressants, and shows two impossibility results. Section 4 provides two modifications of top-trading cycles solutions, and examines whether both solutions satisfy desirable properties. Section 5 presents simulation results with an algorithm to minimize the use of suppressants for incompatible pairs. Section 6 concludes. 2 Standard Kidney Exchange Model without Immunosuppressants Let N {1,..., n} be the set of patients. Each patient has her donor. For each i, j N, patient i is either compatible or incompatible with the donor of patient j. Note that j can be i herself. For each i N, let N i + {j N : patient i is compatible with the donor of patient j} and Ni N \N i +. Each i N has a weak preference R i over N as follows: Each patient prefers all elements in N i + to all elements in Ni. All elements in N i + are equally desirable and so are those in Ni. For patients incompatible with all donors, without loss of generality, is preferred to all donors. Note that compatibility of a donor and a patient is verifiable and determined independently of other pairs compatibility. Let P i and I i be the asymmetric and symmetric part of R i, 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 = (R i ) i N. Since we fix N, we represent a problem as R. Let R N 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 R N 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 (i 1,..., i k, i 1 ) forms a cycle in g(r) if i 1,..., i k are distinct patients and i 1 i 2 i k i 1. As long as i 1,..., i k are ordered, it does not matter who comes first in the list (i 1,..., i k ). 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 c. We say that c and c are feasible if no patient appears both in c and c. Let C(R) 2 C(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 (i 1,..., i k ) forms a chain if i 1,..., i k are distinct patients and i 1 i 2 i k and i k i 1. We call i k 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 l. We say that l and l are feasible if no patient appears both in l and l. Let L(R) 2 L(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

7 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) 2 H(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) R i 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: R 1 R 2 R 3 2 1, 3 2, 3 1, 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 : R N R R N M(R) such that for each R R N, ϕ(r) M(R). We say that ϕ is essentially single-valued if for each R R N, each µ, µ ϕ(r), and each i N, µ(i) I i µ (i). If ϕ is essentially single-valued, then for each R R N and each µ, µ ϕ(r), N(µ) = N(µ ). 7 For each R R N, 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) R i µ(i) and for some i N, µ (i) P i µ(i). We say that ϕ is Pareto efficient if for each R R N and each µ ϕ(r), µ is Pareto efficient at R. We define Top-Trading Cycles (TTC) solutions adapted to this model. 8 Without loss of generality, let 1 2 n. For simplicity, a priority ordering can be written as : 1, 2,..., n. 6 For more detailed discussion on these priorities, see Roth et al. (2005). 7 Suppose, by contradiction, that there are R R N, 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 essentially single-valued. 8 The TTC solution is proposed by Shapley and Scarf (1974) in a general model and applied to kidney exchange problems by Roth et al. (2004). 6

8 Top-trading cycles solution associated with (simply T T C ): For each R R N, let C 0 C(R). Step 1. Find a set of feasible cycles C C 0 such that 1 N(C). If there is no such C, then C 1 C 0. Otherwise, let C 1 be the collection of all such sets of feasible cycles. Step t 2. Find C C t 1 such that t N(C). If there is no such C, then C t C t 1. Otherwise, let C t be the collection of all such sets of feasible cycles. Step n. Obtain C n. For each C C n, 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 C n. Note that this solution can also be viewed as a sequential priority solution as we first figure out all matchings that patient 1 receives a transplantation, and then figure out all matchings that patient 2 does, and so on. When only two-way exchanges are allowed between patients and donors, such a sequential priority solution is Pareto efficient and it also maximizes the number of transplantations, which is another important issue in kidney exchange problems. If we allow more than two-way exchanges, however, Pareto efficient matchings do not necessarily maximize the number of transplantations. We thereby propose another version of top-trading cycles solutions. Top-trading cycles solution maximizing the number of transplantations (simply T T C ): For each R R N, let C 0 argmax C C(R) N(C). Step 1. Find a set of feasible cycles C C 0 such that 1 N(C). If there is no such C, then C 1 C 0. Otherwise, let C 1 be the collection of all such sets of feasible cycles. Step t 2. Find C C t 1 such that t N(C). If there is no such C, then C t C t 1. Otherwise, let C t be the collection of all such sets of feasible cycles. Step n. Obtain C n. For each C C n, 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 C n. Remark. It is straightforward from definitions that T T C and T T C are essentially single-valued. Proposition 1. For every priority ordering, T T C and T T C are Pareto efficient. Proof. Suppose, by contradiction, that T T C is not Pareto efficient. Then, there are R R N, µ T T C (R), and µ M(R) such that for each i N, µ (i) R i µ(i) and for some i N, µ (i) P i 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 T T C, the matchings should be chosen so that patient i receives a transplantation, a contradiction. Next, suppose, by contradiction, that T T C is not Pareto efficient. Then, there are R R N, µ T T C (R), and µ M(R) such that for each i N, µ (i) R i µ(i) and for some i N, µ (i) P i µ(i). This again implies that all patients who receive transplantations at µ also receive transplantations at µ and 7

9 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 / T T C (R), a contradiction. 3 Extended Kidney Exchange Model with Immunosuppressants We introduce suppressants to the standard model. If patient i receives a suppressant, she gets compatible with all donors, including her own. We denote such preference by Ri. If a set of patients S receive suppressants, we denote their preferences by RS (R i ) i S. For each R R N 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) R N, 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: R 1 R 2 R 3 2 1, 3 2, 3 1, Now suppose that S = {2}. Then, R(S) = (R 1, R 2, R 3) and g(r(s)) are represented as follows: R 1 R 2 R 3 2 1, 2, 3 2, 3 1, The set of cycles in g(r2, R 2) is C(R2, R 2) = {c (1, 2, 1), c (2, 3, 2), c (3, 3), c (2, 2)}. Thus, the collection of feasible cycles is C(R2, R 2) = {{c}, {c }, {c }, {c }, {c, c }, {c, c }}. The set of chains in g(r2, R 2) is L(R2, R 2) = {(1), (1, 2, 3), (3, 2, 1)}. To accommodate the limited availability of suppressants, we introduce a non-negative integer k to be an upper bound on the number of patients who can receive suppressants. Let Z + be the set of non-negative integers and for each k Z +, let 2 N (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 : R N 2 N (k) be a recipient-selection rule. Let ϕ : R N R R N M(R) such that for each R RN, ϕ(r) M(R), be a matching rule. A solution is represented as a pair (σ, ϕ). The following two requirements are defined for a matching rule, ϕ: Pareto efficiency: For each R R N, ϕ(r) is Pareto efficient at R. 8

10 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 : R N 2 N (k), each R R N, each µ ϕ(r), each µ ϕ(r(σ(r))), and each i N, µ (i) R i 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: R 1 R 2 R 2 R , 3 2, 3 1, 2 1, 3 Suppose that k = 0. Let R (R 1, R 2, R 3 ) and R (R 1, R 2, R 3). 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 : R N 2 N (k) be such that σ(r) = {2} and σ(r ) = {2}. The new preference profile resulting from σ is (R 1, R2, R 3) for both problems. To make everyone weakly better off at the new profile than at R, we should have µ ϕ(r 1, R2, R 3). Again, to make everyone weakly better off at the new profile than at R, we should also have µ ϕ(r 1, R2, R 3). Altogether, {µ, µ } ϕ(r 1, R2, R 3). 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 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 T T C and T T C as we show below. Example 3. (T T C and T T C violating strong monotonicity) Let N {1, 2, 3, 4}, : 1, 2, 3, 4 and R, R R N be given as follows: 9

11 R 1 R 2 R 3 R , 3, 4 1, 2, 4 1, 2, 3 1, 3, 4 R 1 R2 R3 R , 3, 4 2, 3, 4 1, 2, 3 1, 3, It is easy to check that T T C (R) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)} and T T C ( R) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4)}. Now suppose that k = 1 and consider σ : R N 2 N (k) such that σ(r) = σ( R) = {2}. Patient 2 s preference changes to R2 as illustrated in the following graph: Let R (R 1, R2, R 3, R 4 ) and R ( R 1, R2, R 3, R 4 ). Then, T T C (R ) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4)} T T C ( R ) = {(µ(1) = 1, µ(2) = 3, µ(3) = 4, µ(4) = 2)}. Under T T C, patients 3 and 4 are worse off while patient 1 is better off and patient 2 remains indifferent. Under T T C, patient 1 is worse off while patients 3 and 4 are better off and patient 2 remains indifferent. As shown in the above example, even T T C and T T C, which seem most natural solutions in this setting, violate strong monotonicity. Having a closer look at this example, however, we observe that it is possible to make everyone weakly better off by assigning suppressants to carefully chosen patients. Given R of the above example, consider another assignment of suppressants at which patient 1 receives a suppressant. While maintaining the cycle (2, 3, 4, 2), we find an additional self-cycle of patient 1, which allows all patients weakly better off. Similarly, at 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

12 Monotonicity: For each k Z +, there is σ : R N 2 N (k) such that for each R R N, each µ ϕ(r), each µ ϕ(r(σ(r))), and each i N, µ (i) R i 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 R N, and each T, T 2 N (k), if there are µ ϕ(r(t )) and µ ϕ(r(t )) such that N(µ ) N(µ), then T σ(r). Example 4. (Maximal improvement) Let N {1, 2, 3, 4}. associated with σ. Suppose that R R N is given as follows: Consider a Pareto efficient solution ϕ R 1 R 2 R 3 R , 3, 4 1, 2, 4 1, 2, 3 1, 2, S = 1 2 S = {1} 1 2 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) and σ(r) {1}. As discussed in Section 2, it is also considered important in kidney exchange problems to maximize the number of transplantations. We propose the following requirement to accommodate this idea. It 9 Note that any essentially single-valued solutions trivially satisfy monotonicity by setting σ(r) = for all R R N. To avoid such triviality, we impose monotonicity together with an efficiency requirement of maximal improvement defined later. 11

13 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 R N, and each T, T 2 N (k), if there are µ ϕ(r(t )) and µ ϕ(r(t )) such that N(µ ) < N(µ), then T 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 : R N 2 N (k) associated to ϕ. Consider the following preferences: R 1 R 2 R 3 R 4 R 5 R , N \ {5} N N \ {5} N \ {1, 2} N \ {4} N \ {3} Suppose first that k = 0. By Pareto efficiency, ϕ(r) = {µ = (µ(1) = 5, µ(2) = 2, µ(3) = 3, µ(4) = 1, µ(5) = 4, µ(6) = 6)}. Suppose instead that k = 1. To satisfy CMI, it is easy to check that we should have σ(r) = {2} and all patients except for patient 1 get transplantations at the resulting preference profile. Since patient 1 gets worse off, monotonicity is violated. 4 Top-Trading Cycles Solutions with Immunosuppressants We propose two solutions satisfying Pareto efficiency, monotonicity, and maximal improvement. They are based on the top-trading cycles solutions defined in Section 2, but a major modification is made in specifying the assignment of suppressants. The extended top-trading cycles solution, et T C : Stage 1. For each R R N, apply T T C. Denote by N the set of patients who receive no transplantations at any matching selected by T T C. Without loss of generality, let N {j 1,..., j n } and j 1 j n. 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

14 priority ordering by (R). Stage 3. Find a set of feasible cycles and chains. Let H 0 {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}, {H H t 1 : j t N(H)}, if {H H t 1 : j t N(H)}, H t H t 1, otherwise. Choose any H H n 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 R N, let et T C (R) T T C (R)(R(σ (R))). In the first stage, we apply T T C to the initial preference profile of the standard model and find the set of patients N who do not receive transplantations. Since T T C is essentially single-valued, the sets of patients receiving transplantations remain the same across all matchings selected by T T C for each problem. In the second stage, we replace an initial priority ordering with the modified priority ordering (R). Now all patients in 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 T T C associated with the modified priority ordering. Note that to satisfy monotonicity, all patients in N \ N should receive transplantations after suppressants are introduced. For this, we sort them out and give them higher priorities than those in N when we apply T T C 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 et T C is essentially single-valued, which is straightforward from the definition. Example 5. (et T C ) Let N {1,..., 9} and : 1,..., 9. Let R R N and the corresponding g(r) be given as follows: R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 2 1, 3, 4 5, 7 7 1, N \ {2} N \ {1, 3, 4} N \ {5, 7} N \ {7} N N \ {1, 9} N \ {2} N \ {1} N \ {8} 10 Such H 0 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, we can choose the cycles from (i) and at most k chains from (ii). 13

15 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, T T C chooses the cycle involving the patient with the highest priority. Therefore, T T C (R) = {(µ(1) = 2, µ(2) = 1, µ(3) = 3, µ(4) = 4,..., µ(9) = 9)} and 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, et T C (R) = T T C (R)(R(σ (R))). The matching is (µ(1) = 2, µ(2) = 3, µ(3) = 5, µ(4) = 4, µ(5) = 6, µ(6) = 9, µ(7) = 7, µ(8) = 1, µ(9) = 8). Next suppose that at most two patients can receive suppressants, i.e., k = 2. Similarly as above, we search for the set of cycles and at most two chains including patients 1 and 2 as well as other patients in 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 et T C (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 et T C is essentially single-valued. Remark. When multiple suppressants are available, it is also possible to assign them one by one sequentially, instead of assigning them all at once as in et T C. Example 5 shows that the resulting matchings may be different from et T C (R): the only resulting matching is µ in this case, while et T C (R) includes both of µ and µ. The extended top-trading cycles solution maximizing the number of transplantations, et T C : Stage 1. For each R R N, apply T T C. Denote by N the set of patients who receive no transplantations at any matching selected by T T C. Without loss of generality, let N {j 1,..., j n } and j 1 j n. 14

16 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 H 0 {H H(R) : N(H) N \ N and the number of chains in H does not exceed k}, and for each t {1,..., n}, {H H t 1 : j t N(H)}, if {H H t 1 : j t N(H)}, H t H t 1, otherwise. Choose any H H n 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 R N, let et T C (R) T T C (R)(R( σ (R))). Basically we make the same modifications of T T C as we did for T T C with an exception of Stage 1 where we use T T C instead of T T C. Note that, because T T C is essentially single-valued, the sets of patients receiving transplantations remain the same across all matchings selected by T T C for each problem. Also, et T C is essentially single-valued, which is straightforward from the definition. Example 6. (et T C ) Consider the problem in Example 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, T T C chooses a cycle of the largest size; if there are many, it chooses a cycle including patients with higher priorities in order. Thus, c is chosen and the matching is T T C (R) = {(µ(1) = 1, µ(2) = 3, µ(3) = 7, µ(7) = 2, µ(4) = 4,..., µ(9) = 9)} with 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 et T C 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, et T C (R) = T T C (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

17 Now suppose that at most two patients can receive suppressants, i.e., k = 2. Similarly as above, we figure out the collection of sets of feasible cycles and at most two chains including patients 2, 3, and 7 as well as other patients in order of priorities. At the last step of the et T C algorithm, we obtain L = {(6, 9, 8, 1, 2, 3, 5), (4, 7)} and 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. et T C and et T C satisfy Pareto efficiency, monotonicity, and maximal improvement. Proof. (Pareto efficiency) For each R R N, note that R(σ (R)), R( σ (R)) R N. By definition, et T C (R) T T C (R)(R(σ (R)) and et T C (R) T T C (R)(R( σ (R)). By Proposition 1, T T C and T T C are Pareto efficient for every. Therefore, for each R R N, et T C (R) and et T C (R) are Pareto efficient at R(σ (R)) and R( σ (R)), respectively. (Monotonicity) We show that et T C is monotonic. Let R R N. Recall that T T C and et T C are essentially single-valued. Note also that at each matching, any patient who receives a transplantation cannot be better off further and any patient who receives no transplantation cannot be worse off further. Therefore, it is enough to show that all agents who used to receive transplantations at T T C (R) still receive transplantations at et T C (R). Denote the set of those patients by N \ N. Note that they are the only patients involved in feasible trading cycles that generate matchings in T T C (R) and the preferences of these patients are not affected by the use of suppressants under the et T C algorithm. Thus, the feasible trading cycles involving N \ 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 T T C (R) applied to R(σ (R)). The feasible trading cycles involving N \ N at the T T C algorithm are not ruled out at the step where we figure out the sets of feasible cycles including the patient with the lowest priority in N \ 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 et T C (R) in which all of these patients receive transplantations, completing the proof. The same argument shows that et T C is monotonic. (Maximal improvement) We show that et T C satisfies maximal improvement. Suppose, by contradiction, that there are R R N, S N with S k, µ et T C (R), and µ M(R(S)) such that N(µ) N(µ ). Let N be the set of patients who receive no transplantations at T T C (R). Let H H(R) be the set of cycles and chains that results in µ, that is, a set of cycles and chains chosen in the last stage of the et T C algorithm for R. Let H 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 et T C, 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

18 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 et T C algorithm. Since i / N(H), this is a contradiction to the definition of et T C. If i has a lower priority than all patients in N(µ), then the set of cycles and chains should have been chosen to include all patients in N(µ) as well as i, which is possible as in H. Since i / N(H), again, this is a contradiction to the definition of et T C. The same argument shows that et T C satisfies maximal improvement. 5 Simulation Results We present simulation results to show that the use of suppressants can be significantly reduced from the current practice. For this analysis, we assume that immunological compatibility is determined only by ABO blood types. We denote a pair by its type X-Y where the patient s blood type is X and the donor s type is Y. Let #(X-Y) be the number of pairs of type X-Y. A k-cycle is a cycle consisting of k pairs and a k-chain is a chain consisting of k pairs. 5.1 Simulation based on the KONOS data We use the KONOS (Korean Network for Organ Sharing) data on living donor kidney transplantations in the recent four years ( ). The data set collects all patient-donor pairs who receive transplantations during this period and specifies their ABO blood types. As shown in Table 2, almost no pairs in the set receive transplantations through kidney exchanges. For instance, in 2012, no pair participated in kidney exchanges. For In other words, almost all transplantations in the data are performed within pairs: if a pair is compatible, the patient receives a kidney directly from her own donor; otherwise, the pair uses a suppressant. In 2012, there are 1,020 living donor kidney transplantations in total. Table 3 summarizes the blood type profile. For instance, there are 210 compatible pairs of type A-A, who receive the kidneys from their own donors; there are 35 incompatible pairs of type A-B, who use suppressants to get transplantations from their donors. In the blood-type restricted setting, there are seven types of incompatible pairs, A-B, B-A, A-AB, B-AB, O-A, O-B, and O-AB. It turns out that there are 193 incompatible pairs in 2012, as summarized in Table 4. Since there are 193 incompatible pairs in the data, suppressants are used 193 times for incompatible kidney transplantations. We propose an algorithm that minimizes the use of suppressants, while all these incompatible pairs still receive transplantations. Simulation Algorithm For any set of incompatible pairs and their blood type profile: Step 1. Find all 2-cycles. Since only pairs of types A-B and B-A can form cycles, the number of all 2-cycles is min {#(A-B),#(B-A)}. Remove the pairs of these 2-cycles from the pool. 17