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Bargaining for assembly

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Abstract

We study a multilateral bargaining problem, where the buyer intends to purchase a subset of available items, each owned by a seller. The subset purchased must satisfy a notion of contiguity, which is modeled using graphs. The graph theoretic approach allows us to study different degrees of complementarity and substitutability between items. It also allows us to examine how degrees of complementarity and substitutability affect the share of surplus obtained by the buyer in the equilibrium of the bargaining game. We characterize necessary and sufficient conditions on the graphs for the buyer to extract full surplus in subgame perfect equilibrium. When the conditions are not met, we provide upper bounds on the equilibrium surplus share of the buyer.

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Notes

  1. A mechanism is “successful” in this sense if it is ex-post efficient, interim incentive compatible, interim individually rational and ex post budget balanced.

  2. This can be relaxed to include any special graph of a fixed size. Rights of passage directly motivates the desire to purchase a path in our case.

  3. Note that the buyer is assumed to be located outside the graph.

  4. Allowing sellers to make redundant offers makes more sense when there are multiple buyers.

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Correspondence to Soumendu Sarkar.

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This paper builds on a chapter from Dhritiman’s PhD thesis at Indian Statistical Institute.We are grateful to Professors Prabal Roy Chowdhury and Arunava Sen for their guidance and encouragement. We are also grateful to Albin Erlansson, Steven Williams, Jun Xiao, Huseyin Yildirim, participants of the Delhi Economic Theory Workshop, Conference of Economic Design, York, 2017, Asia Pacific Industrial Organization Conference, Auckland, 2017, and Economics Seminar, Shiv Nadar University, Delhi, 2017. Special thanks to an anonymous referee of this journal for many useful suggestions. Travel grants from Royal Economic Society and TERI School of Advanced Studies are gracefully acknowledged.

Appendices

A. Proof of Theorem 1

Propositions 13 constitute the if part of the argument. They cover the three mutually exclusive and exhaustive classes of graphs where no critical seller exists. In all these cases valuations of sellers are identical, and therefore, all feasible paths have the same sum of seller valuations. In each of these cases we provide equilibrium strategies such that the buyer extracts full surplus within two periods. The arguments of these claims are generalizable to graphs with two or more feasible paths corresponding to the minimum sum of valuations. Propositions 4 and 5 constitute the only if part. Proposition 4 covers the classes of graphs where critical sellers exist and Proposition 5 covers the classes of graphs where only one feasible path has the minimum sum of seller valuations. In both these cases we show that there cannot be an equilibrium where the buyer extracts full surplus.


Proposition 1Consider an assembly problem \(\langle \Gamma ^{\triangle }, k, v, \delta \rangle\) such that \(v_{1}=\cdots =v_{n}, v_{0}>kv_{1}\). The buyer extracting full surplus is an equilibrium outcome.

Proof

We will prove the case of a cycle of length \(k+1\). It follows immediately that such an equilibrium can be obtained for any graph containing a cycle of length \(k+1\) as a subgraph.

Consider the following strategy of the buyer. She picks a feasible path. Consider any continuation game where \(m< k\) plots have already been acquired. If the buyer is making offers, she offers \(k-m\) sellers zero surplus and makes negative offers to the remaining seller. If sellers are making offers, buyer accepts the lowest \(k-m\) claims provided her share of the surplus is non-negative and rejects all other claims. If more than \(k-m\) sellers are making identical lowest offers, she accepts \(k-m\) offers with equal probability.

We claim that given the above strategy, all active sellers claiming zero surplus at any continuation game they are required to make an offer is a best response. Let \(x_{i}\) be the surplus claim of active seller i. No seller can gain by deviating for one stage when \(x_{i}=x_{j}=0, i\ne j\). Hence it is an equilibrium. If \(x_{i}=x_{j}>0, i\ne j\), either seller i or j can reduce his claim by a small amount and make a gain. If \(x_{i}>x_{j}\ge 0, i\ne j\), then seller j can increase his claim by a small amount and make a gain. Hence these are not equilibrium claims.

At any continuation where the buyer is making an offer and m plots have already been acquired, the active seller who is made a negative offer rejects it. Simultaneously, \(k-m\) sellers immediately accept corresponding zero offers. If any of these sellers rejects such offers, they reach a continuation game where the maximum they can gain by rejecting buyer’s offers is zero. Hence this is an equilibrium. Trade takes place in the first period itself when \(m=0\), with k sellers who are made zero surplus offers. \(\square\)

Since any complete graph of order \(n>k\) contains a cycle of length \(k+1\), we get the following Corollary.

Corollary 3

Consider an assembly problem \(\langle n, k, v, \delta \rangle\) such that \(v_{1}=\cdots =v_{n}, v_{0}>kv_{1}\). The buyer extracting full surplus is an equilibrium outcome.

Remark 3

Note that the equilibrium outcome does not change whether the buyer moves first, or the sellers.

Remark 4

Notice that in this case the existence of equilibrium with full surplus extraction is not dependent on the magnitude of the discount factor \(\delta\).

Remark 5

When \(k=2\), then the above result is also true for any graph containing a cycle of length more than 3. But it is not true for \(k>2\). For instance, consider the cycle of length 5 when \(k=3\) (see Figure 6). Suppose the buyer wants to make offers that are acceptable to sellers 1,2 and 3 in the first period itself. Sellers 1 and 3 will accept a zero surplus offer since if they reject, they have to compete with sellers 5 or 4. Seller 2, on the other hand, will not accept a surplus of less than \(\delta v_1\), since if he rejects an offer, he has to compete with sellers 4 and 5 together. Therefore, the buyer has two ways to complete the transaction in the first period: either (i) she makes zero surplus offers to 4 sellers on the graph and makes a negative offer to the remaining seller, or (ii) she makes zero surplus offers to sellers 1 and 3, make a surplus offer of \(\delta v_1\) to seller 2, and negative offers to the remaining sellers.

Fig. 6
figure 6

A cycle of length 5; \(\Gamma ^{SO}\) in blue


Proposition 2 Consider an assembly problem \(\langle \Gamma ^{D}, k, v, \delta \rangle\) such that \(v_{1}=\cdots =v_{n}, v_{0}>kv_{1}\). (a) If the sellers move first, the buyer extracts full surplus in the first period. (b) If the buyer moves first, there exists \({\overline{\delta }}\) such that \(\forall \delta >{\overline{\delta }}\) there is an equilibrium where the buyer extracts full surplus in the second period

Proof

Consider the following strategy of the buyer: In any continuation game where the buyer makes the offers, she makes negative offers to all sellers. In any continuation game where sellers have the first move, the buyer accepts the claims of sellers on a path with the lowest sum of claims provided her share of surplus is non-negative and rejects all other claims. If the sum of claims on the two feasible paths are same, she accepts claims from one of the paths with equal probability.

We claim that given the above strategy, sellers in the two disjoint feasible paths claiming zero surplus at any continuation game they are required to make an offer is a best response. Let \({\mathcal {P}}_{1}\) and \({\mathcal {P}}_{2}\) be the two feasible paths in \(\Gamma ^{D}\). Let \(x_{i}\) be the surplus claim of active seller i. No seller can gain by deviating for one stage when \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}=\sum _{i\in {\mathcal {P}}_{2}}x_{i}=0\). Hence it is an equilibrium. If \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}=\sum _{i\in {\mathcal {P}}_{2}}x_{i}>0\), a seller on one of the paths can reduce his claim by a small amount and make a gain. If \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}>\sum _{i\in {\mathcal {P}}_{2}}x_{i}\), then any seller on \({\mathcal {P}}_{2}\) can increase his claim by a small amount and make a gain. Hence these are not equilibrium claims.

Part (a) of the claim follows immediately. For part (b), note that buyer can make zero surplus offers to sellers on both paths, and negative surplus offers to all other sellers; sellers on both paths would accept these offers. To ensure that this deviation in the first stage is not profitable for the buyer, we require \(\delta >\frac{v_{0}-2kv_1}{v_{0}-kv_1}\). The buyer can also make acceptable offers of surplus shares, \(\delta (k-1)v_1\), to each seller on one path and negative offers to all other sellers, provided \(v_{0}-kv_1-\delta k(k-1)v_1>0\). This is because, by rejecting a first period offer from the buyer, a seller on the chosen path competes with sellers on the other path; the highest surplus he can claim in a continuation game where he and the other sellers are making offers is \((k-1)v_1\).To ensure that this deviation in the first stage is not profitable for the buyer, we require \(\delta >\frac{v_{0}-kv_1-\delta k(k-1)v_1}{v_{0}-kv_1}\). Thus, provided \(\delta >\max \{ \frac{v_{0}-2kv_1}{v_{0}-kv_1},\frac{v_{0}-kv_1-\delta k(k-1)v_1}{v_{0}-kv_1}\}\), the buyer extracting full surplus in the second period is an equilibrium outcome in the strategies described above. \(\square\)

Remark 6

If \(\delta \le \max \{ \frac{v_{0}-2kv_1}{v_{0}-kv_1},\frac{v_{0}-kv_1-\delta k(k-1)v_1}{v_{0}-kv_1}\}\) then full surplus extraction is not possible in the equilibrium. Either the buyer purchases items from all sellers on a single path by paying positive surplus shares; or, she purchases 2k items by offering zero surplus shares to all sellers on two disjoint paths.


Proposition 3 Consider an assembly problem \(\langle \Gamma ^{O}, k, v, \delta \rangle\) such that \(v_{1}=\cdots =v_{n}, v_{0}>kv_{1}\). (a) If the sellers move first, the buyer extracts full surplus in the first period. (b) If the buyer moves first, there exists \({\overline{\delta }}\) such that \(\forall \delta >{\overline{\delta }}\) there is an equilibrium where the buyer extracts full surplus in the second period.

Proof

We introduce some notation in the next two paragraphs that would be useful in proving the next result.

We note that each graph \(\Gamma ^{O}\) has a subgraph \(\Gamma ^{SO}\) such that (i) it contains a feasible path \({\mathcal {P}}\), (ii) for each node \(x\in {\mathcal {P}}\) there exists a node \(y\in \Gamma ^{O}-\Gamma ^{SO}\) and an edge \(e(y, z), z\in \Gamma^{SO},\) such that \(\Gamma ^{SO}-x+z\) contains a feasible path of length k. For instance, in Fig. 6, the path \(\{1234\}\) qualifies as \(\Gamma ^{SO}\). Figure 5 shows one more example. Observe that the order of any \(\Gamma ^{SO}\) would vary from k to \(n-1\). For any given \(\Gamma ^{O}\), let \(\Gamma ^{SO*}\) be the smallest of all \(\Gamma ^{SO}\subset \Gamma ^{O}\) with order \(m^{*}\).

Further, pick any feasible path \({\mathcal {P}}\) of length k on \(\Gamma ^{O}\). For each x on \({\mathcal {P}}\), let s(x) be the order of the smallest subgraph \(\Gamma _{S}\) of \(\Gamma ^{O}\) such that \(({\mathcal {P}}-x)\cup \Gamma ^{S}\) is a feasible path of length k. For example, in Fig. 5, \(s(1)=s(5)=1\) and \(s(2)=2\).

Consider the following strategy of the buyer: In any continuation game where the buyer has the first move , the buyer makes negative offers to all sellers. In any continuation game where sellers have the first move, the buyer accepts the claims of sellers on a path with the lowest sum of claims provided her share of surplus is non-negative and rejects all other claims. In case the sum of claims on the two feasible paths are same, she accepts claims from one of the paths chosen with equal probability.

We claim that given the above strategy, sellers claiming zero surplus at any continuation game they are required to make an offer is a best response. Let \({\mathcal {P}}_{1}, \ldots {\mathcal {P}}_{m}\) be the feasible paths in \(\Gamma ^{O}\). Let \(x_{i}\) be the surplus claim of active seller i. No seller can gain by deviating for one stage when \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}=\cdots =\sum _{i\in {\mathcal {P}}_{m}}x_{i}=0\). This is because, for each \(x_{i}\), there is always a feasible path in \(\Gamma ^{O}\) that does not contain \(x_{i}\). Hence it is an equilibrium. If \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}=\cdots =\sum _{i\in {\mathcal {P}}_{m}}x_{i}>0\), a seller on either path can reduce his claim by a small amount and make a gain. If \(\sum _{i\in {\mathcal {P}}_{1}}x_{i}>\sum _{i\in {\mathcal {P}}_{2}}x_{i}\), then any seller on \({\mathcal {P}}_{2}\) can increase his claim by a small amount and make a gain. Hence these are not equilibrium claims.

Part (a) of the claim follows immediately. For part (b), note that buyer can make zero surplus offers to all sellers on \(\Gamma ^{SO*}\), and negative surplus offers to all other sellers; sellers on \(\Gamma ^{SO*}\) would accept these offers. To ensure that this deviation in the first stage is not profitable for the buyer, we require \(\delta >\frac{v_{0}-m^{*}v_1}{v_{0}-kv_1}\). The buyer can also make acceptable offers of surplus shares to sellers on a path and negative offers to all other sellers. If \({\mathcal {P}}\) is the picked path and \(x_{i}\) is the node corresponding to seller i, he accepts any surplus share at least equal to \(\delta (s(x_{i})-1)v_1\). This is possible when \(v_{0}-kv_1-\delta \sum _{i\in {\mathcal {P}}}(s(x_{i})-1)v_1>0\). This is because, by rejecting a first period offer from the buyer, a seller on the chosen path competes with sellers on the other path; the highest surplus he can claim in a continuation game where he and the other sellers are making offers is \((s(x_{i})-1)v_1\).To ensure that this deviation in the first stage is not profitable for the buyer, we require \(\delta >\frac{v_{0}-kv_1-\delta \sum _{i\in {\mathcal {P}}}(s(x_{i})-1)v_1}{v_{0}-kv_1}\). Thus, provided \(\delta >\max \{ \frac{v_{0}-m^{*}v_1}{v_{0}-kv_1},\frac{v_{0}-kv_1-\delta \sum _{i\in {\mathcal {P}}}(s(x_{i})-1)v_1}{v_{0}-kv_1}\}\), the buyer extracting full surplus in the second period is an equilibrium outcome in the strategies described above. \(\square\)

Remark 7

If \(\delta \leq\frac{v_{0}-kv_1-\delta \sum _{i\in {\mathcal {P}}}(s(x_{i})-1)v_1}{v_{0}-kv_1}\), full surplus extraction is not possible in the equilibrium: either buyer purchases items from all sellers on a single path by paying positive surplus shares; or, she purchases more than k items by offering zero surplus shares to corresponding sellers.

Remark 8

Observe that in Propositions 1-3 the underlying graph does not contain a critical seller. If the graph has only one component, then the corresponding bargaining game has an equilibrium where the buyer extracts full surplus in at most two periods. Suppose the graph has multiple components. When seller valuations are identical, there exists an equilibrium where the buyer extracts full surplus in the first period itself if and only if the graph contains a \(k+1\)-cycle. Otherwise, (a) competing paths lie in different components or (b) form an oddball graph. In these cases, there exists an equilibrium where the buyer extracts full surplus in the second period if she is making the first set of offers, or in the first period itself if sellers are making the first set of offers.

Now consider assembly problems where the underlying graph contains at least one critical seller. The following result is obtained without any assumption on valuations.


Proposition 4Suppose \(\Gamma =\Gamma ^*\). The buyer cannot extract full surplus in an equilibrium.

Proof

Suppose there is an equilibrium where the buyer obtains full surplus. We will show that in any such equilibrium a critical seller has profitable deviation.

If \(k=1\), then there is a unique equilibrium by Rubinstein (1982) where buyer gets \(\frac{1}{1+\delta }\) if offering first and \(\frac{\delta }{1+\delta }\) if the seller is offering first. Suppose \(k\ge 2\). Suppose, if possible, that the buyer obtains full surplus in an equilibrium at period t. This implies that all sellers on an efficient path are selling their items at period t or some period before t. Consider a critical seller whom trades at period \({\hat{t}}\leq t\) at zero surplus share. Now by rejecting buyers zero offer or by quoting an offer unacceptable to the buyer, a critical seller moves to a continuation game in period \(t+1\) where he is the only active seller. By the result due to Rubinstein (1982) (see Sect. 3.3) the critical seller obtains a positive surplus share in the continuation game. This constitutes a profitable deviation for the critical seller.\(\square\)

Remark 9

The proof above implies that in any equilibrium of \(\Gamma =\Gamma ^*\) at least one critical seller earns a positive surplus share. The non-critical sellers may or may not get positive shares.

Now consider the case where seller valuations are not equal. In this case, the sum of seller valuations may differ over paths. The path corresponding to the least sum of seller valuations is efficient in the sense that it corresponds to highest potential surplus. It follows that if possible, the buyer would prefer to purchase the efficient path.

Let \({\mathcal {P}}_{i}\) denote the path corresponding to the i-th smallest sum of valuations on a path in \(\Gamma\). We will refer to a set of assembly problems as rich if there does not exist two distinct paths \({\mathcal {P}}_{1}\) and \({\mathcal {P}}_{2}\) such that \({\mathcal {S}}_{1}={\mathcal {S}}_{2}\). Suppose the richness condition is not satisfied. Consider the following strategy for the buyer: she offers negative surplus shares to all sellers whenever she is supposed to make an offer, and agrees to trade with sellers on the path with lowest sum of valuations provided it leads to a positive surplus. If the buyer is proposing first, all sellers reject buyer’s offers. In the next period, sellers on \({\mathcal {P}}_{1}\) and \({\mathcal {P}}_{2}\) cannot claim any surplus: the buyer extracts full surplus in the second period. If the sellers are making offers first, sellers on these two paths cannot claim any surplus share.


Proposition 5 Consider the rich class of assembly problems \(\langle \Gamma , k, v, \delta \rangle\) such that \(v_{1}\le \cdots \le v_{n}\) with at least one strict inequality. There does not exist any equilibrium where the buyer extracts full surplus.

Proof

Case 1 (\(\Gamma =\Gamma ^{\Delta }\)): Suppose the buyer obtains full surplus in an equilibrium at period t. This implies that all sellers on the efficient path sell their items at t or prior to t. Let us pick a seller i on the efficient path and suppose he is last active at period \({\hat{t}}\leqslant t\). Since buyer extracts full surplus in the proposed equilibrium either i proposes zero surplus share at \({\hat{t}}\) or accepts a zero surplus share offer at \({\hat{t}}\). Now pick the seller j who is on \({\mathcal {P}}_{2}\) but not on \({\mathcal {P}}_{1}\). Since by assumption \({\mathcal {S}}_{1}<{\mathcal {S}}_{2}\), \(v_i<v_j\). Consider the following deviation for seller i at \({\hat{t}}\): i makes a surplus offer of \(v_j-v_i-\epsilon\) and accepts offer greater than \(\delta (v_j-v_i-\epsilon )\). Here \(\epsilon\) is a small positive quantity. If the buyer keeps rejecting i’s offer and keeps offering less than the claim of i then we reach a continuation game where i is the only active seller on the efficient path. Note that for small \(\epsilon\), buyer would never agree to trade with seller j. In this continuation game i can ensure a positive surplus.

Case 2 (\(\Gamma =\Gamma ^D\)): Suppose the buyer obtains full surplus in an equilibrium at period t. This implies that all sellers on the efficient path sell their items at t or prior to t. Let us pick a seller i on the efficient path and suppose he is last active at period \({\hat{t}}\leqslant t\). Since buyer extracts full surplus in the proposed equilibrium either i proposes zero surplus share at \({\hat{t}}\) or accepts a zero surplus share offer at \({\hat{t}}\). Now consider the deviation strategy of i where he makes an offer of \({\mathcal {S}}_{2}-v_i-\epsilon\) and accepts offers of at least \(\delta ({\mathcal {S}}_{2}-v_i-\epsilon )\). Then there exists a continuation game at period \(t+1\) where i is the only remaining active seller on the efficient path and can guarantee himself a positive payoff.

Case 3 (\(\Gamma =\Gamma ^O\)): Suppose the buyer obtains full surplus in an equilibrium at period t. This implies that all sellers on the efficient path sell their items at t or prior to t. Let us pick a seller i at the intersection of \({\mathcal {P}}_1\) and \({\mathcal {P}}_2\) and suppose he is last active at period \({\hat{t}}\leqslant t\). Since buyer extracts full surplus in the proposed equilibrium either i proposes zero surplus share at \({\hat{t}}\) or accepts a zero surplus share offer at \({\hat{t}}\). Now consider a deviation strategy for seller i: suppose the cheapest path on the subgraph excluding i is \({\mathcal {P}}_R\) and the corresponding sum of valuations is \({\mathcal {S}}_R\); seller i claims \({\mathcal {S}}_R-{\mathcal {S}}_1-\epsilon\) and accepts no less than \(\delta ({\mathcal {S}}_R-{\mathcal {S}}_1-\epsilon )\). In this case there is a continuation game at \(t+1\) where i is the only active seller on the efficient path and can guarantee himself a positive surplus.\(\square\)

B. Proof of Theorem 2

We prove by contradiction. Suppose, if possible, the buyer realizes the highest possible surplus \(X>v_{0}-{\mathcal {S}}_{2}\) in an equilibrium, completing trading at period t. It follows that the buyer purchases the efficient path in this equilibrium, since purchasing any other path would imply that the total surplus realized can be at most \(v_{0}-{\mathcal {S}}_{2}\). This implies that there exists at least one seller i in \({\mathcal {P}}_{1}\) who is claiming a surplus share x such that \(v_{i}+x\) is less than the valuation \(v_j\) of some seller j in \({\mathcal {P}}_{2}\).

Suppose, if possible, this seller i trades at t. Consider the following deviation strategy for player i at t, which we call an \(\epsilon\)-deviation. He can claim an increment \(\epsilon <v_{j}-v_{i}\) on his surplus share in the continuation game beginning period t. By accepting this offer, the buyer would realize \(X-\epsilon\), while by rejecting, she would get at most \(\delta X\). For \(\epsilon <(1-\delta )X\), the buyer would accept the claim of i. If the buyer is making an offer of surplus share x such that \(v_{i}+x<v_{j}\) at t, seller i can reject this offer and claim a surplus share \(\frac{x}{\delta }+\epsilon\) at \(t+1\). If the buyer rejects this offer, her share can be at most \(\delta X\) in the continuation game, whereas by accepting, she gets \(X+x-\frac{x}{\delta }-\epsilon\). For \(\epsilon <(1-\delta )(X-\frac{x}{\delta })\), the buyer must accept this offer. Notice that the buyer cannot preserve her surplus in response to such deviations by changing the offers made to other sellers: if it were possible in the continuation game, it is also possible to increase buyer’s surplus at t when the seller does not deviate, hence it cannot be an equilibrium. Thus no seller trading at period t can be claiming a price less than the valuation of some seller j in \({\mathcal {P}}_{2}\).

Now suppose, if possible, that this seller i is making offers at \(t-1\). Suppose the seller applies the \(\epsilon\) deviation. Then either the buyer agrees to trade at the new claim, or the seller moves to period t, where he cannot be getting a price less than the valuation of some seller j in \({\mathcal {P}}_{2}\). The buyer would agree to trade at the new claim as long as \(X-\epsilon >\max (\delta X, X+x-(v_{j}-v_{i}))\). If the buyer is making an offer of surplus share x such that \(v_{i}+x<b_{j}\) at \(t-1\), seller i can reject this offer and claim \(\frac{x}{\delta }+\epsilon\) at t. If the buyer rejects this offer, her share can be at most \(\delta X\) in the continuation game, whereas by accepting, she gets \(X+x-\frac{x}{\delta }-\epsilon\). For \(\epsilon <(1-\delta )(X-\frac{x}{\delta })\), the buyer must accept this offer.

The argument is then completed by applying an induction on t. Suppose the \(\epsilon\)-deviation is profitable for seller i at periods \(t, t-1,\ldots , t-l\). Applying the argument of previous paragraph again, we can show that the \(\epsilon\)-deviation is profitable for seller i at periods \(t-l-1\). This completes the proof.

C. Proof of Theorem 3

We prove by contradiction. If possible, suppose there exists an equilibrium where the buyer gets a surplus share strictly higher than \(\frac{1}{1+\delta }\). Since the buyer realizes a strictly positive surplus, the game terminates at some finite period t. This implies that the critical seller gets a surplus share strictly less than \(\frac{\delta }{1+\delta }\). We show a profitable deviation strategy for the critical seller. Note that no other seller changes his strategy. Consequently, the buyer continues to trade with the same sellers in the continuation game at the same prices as agreed upon in the equilibrium proposed. It is not possible for the buyer to trade with other sellers or with the same sellers at lower prices and increase her payoff in the continuation game: if it were possible in the continuation game, then it is also possible in the original configuration, which therefore is no longer an equilibrium.

Suppose the critical seller agrees to trade at t. If \(k=1\), by Rubinstein (1982), in the continuation game beginning at t, he gets a surplus share equal to \(\frac{\delta }{1+\delta }\) if the buyer is making an offer and \(\frac{1}{1+\delta }\) if himself making an offer. Consequently, buyer’s surplus share cannot exceed \(\frac{1}{1+\delta }\). Suppose \(k\ge 2\) and if possible, buyer’s share is the highest \(X>\frac{1}{1+\delta }\). Then the critical seller is getting a share \(x<\frac{\delta }{1+\delta }\). We show that this cannot be an equilibrium. Suppose the critical seller trades at t: if buyer is making an offer \(x<\frac{\delta }{1+\delta }\) at t, the seller can reject and claim \(\frac{x}{\delta }+\epsilon\) at \(t+1\). If the buyer rejects this offer, her share can be at most \(\delta X\) in the continuation game, whereas she gets \(X+x-\frac{x}{\delta }-\epsilon\) at \(t+1\) by accepting this offer. For \(\epsilon <(1-\delta )\left( X-\frac{x}{\delta }\right)\), the buyer must accept this offer. If the seller is making an offer \(x<\frac{\delta }{1+\delta }\), she can claim an increment of \(\epsilon\) on this offer: if the buyer accepts, she realizes \(X-\epsilon\) at t, whereas by rejecting she can get at most \(\delta X\) in the continuation game beginning at \(t+1\). For \(\epsilon <(1-\delta )X\), she will accept this offer.

Suppose the critical seller trades at \(t-1\), : if buyer is making an offer \(x<\frac{\delta }{1+\delta }\) at \(t-1\), the seller can reject and claim \(\frac{x}{\delta }+\epsilon\) at t. If the buyer rejects this offer, her maximum possible share is \(\delta X\) in the continuation game, whereas she gets \(X+x-\frac{x}{\delta }-\epsilon\) at t by accepting this offer. For \(\epsilon <(1-\delta )\left( X-\frac{x}{\delta }\right)\), the buyer must accept this offer. If the seller is making an offer \(x<\frac{\delta }{1+\delta }\), she can claim an increment of \(\epsilon\) on this offer: if the buyer accepts, she realizes \(X-\epsilon\), whereas by rejecting she can get at most \(\max \left( \frac{1}{1+\delta }, \delta X\right)\) in the continuation game beginning at t. For \(\epsilon <X-\max \left( \frac{1}{1+\delta }, \delta X\right)\), she will accept this offer.

Suppose the following claim is true: if the critical seller trades at \(t-s\), \(s\in \{ 1, 2\ldots , \}\), buyer’s surplus share cannot be more than \(\frac{1}{1+\delta }\). We argue that it is true for \(t-(s+1)\) also : if buyer is making an offer \(x<\frac{\delta }{1+\delta }\) at \(t-s-1\), the seller can reject and claim \(\frac{x}{\delta }+\epsilon\) at \(t-s\). If the buyer rejects this offer, her share is \(\max \left( \frac{1}{1+\delta }, \delta X\right)\) in the continuation game, whereas she gets \(X+x-\frac{x}{\delta }-\epsilon\) at \(t-s\) by accepting this offer. For \(\epsilon <X-\max \left( \frac{1}{1+\delta }, \delta X\right) -(1-\delta )\frac{x}{\delta }\), the buyer must accept this offer. If the seller is making an offer \(x<\frac{\delta }{1+\delta }\), she can claim an increment of \(\epsilon\) on this offer: if the buyer accepts, she realizes \(X-\epsilon\) at \(t-s-1\), whereas by rejecting she can get at most \(\max \left( \frac{1}{1+\delta }, \delta X\right)\) in the continuation game beginning at \(t-s\). For \(\epsilon <X-\max \left( \frac{1}{1+\delta }, \delta X\right)\), she will accept this offer. Since this is true for \(s=1\), this completes the proof by induction.

D. Proof of Theorem 4

We show a profitable deviation strategy for at least one critical seller. As noted in the proof of Theorem 3 above, the buyer continues to trade with the same sellers in the continuation game at the same prices as agreed upon in the equilibrium proposed.

Consider an equilibrium in an assembly problem with \(k\ge m\ge 2\) critical sellers. If \(k=m\), then by Roy Chowdhury and Sengupta (2012) (see Sect. 3.3), buyer’s surplus share cannot exceed \(\frac{1-\delta }{1+\delta }\). Suppose \(k>m\). Suppose the final agreement takes place at period t.

Step 1: Suppose trade takes place with at least one critical seller at t and buyer is getting her highest surplus share \(X>\frac{1-\delta }{1+\delta }\). We argue that critical sellers trading at t cannot be getting \(x<\frac{\delta }{1+\delta }\). If buyer is making an offer at t, a critical seller can reject buyer’s offer and successfully claim \(\frac{x}{\delta }+\epsilon\) in the next period for a small \(\epsilon <(1-\delta )\left( X-\frac{x}{\delta }\right)\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(X-\epsilon >\delta X\). It also implies that if two or more critical sellers trade at t, buyer’s surplus cannot be more than \(\frac{1-\delta }{1+\delta }\).

Step 2: Now suppose no critical seller trades at t while at least one critical seller trades at \(t-1\) and buyer is getting her highest surplus share \(X>\frac{1-\delta }{1+\delta }\). We argue that critical sellers trading at \(t-1\) cannot be getting \(x<\frac{\delta }{1+\delta }\). If buyer is making an offer at \(t-1\), this critical seller can reject buyer’s offer and claim \(\frac{x}{\delta }+\epsilon\) at t. Buyer accepts this claim for a small \(\epsilon <(1-\delta )\left( X-\frac{x}{\delta }\right)\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(x+\epsilon <\frac{\delta }{1+\delta }\). It also implies that if two or more critical sellers trade at \(t-1\) but none at t, buyer’s surplus cannot be more than \(\frac{1-\delta }{1+\delta }\).

Step 3: Now suppose no critical seller trades at t, at least one critical seller trades at \(t-1\), some critical sellers trade at some period \(t-s-1\), \(s\in \{1, 2, \ldots \}\), and buyer is getting her highest surplus share \(X>\frac{1-\delta }{1+\delta }\). We know by Step 2 that critical sellers trading at \(t-1\) cannot be getting \(x<\frac{\delta }{1+\delta }\). So if two or more critical sellers trade at \(t-1\), buyer cannot realize more than \(\frac{1-\delta }{1+\delta }\). So suppose exactly one critical seller trades at \(t-1\), and rest trade at \(t-2\). If buyer is getting X, then at least one critical seller is trading at \(x<\frac{\delta }{1+\delta }\) at \(t-2\). If buyer is making an offer at \(t-2\), this critical seller can reject buyer’s offer and claim \(\frac{x}{\delta }+\epsilon\) at \(t-1\). Buyer accepts this claim for a small \(\epsilon <X-\max \left( \frac{1-\delta }{1+\delta }, \delta X\right) +\frac{(1-\delta )x}{\delta }\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(X-\epsilon <\max \left( \frac{1-\delta }{1+\delta }, \delta X\right)\). Hence this cannot be an equilibrium. Thus the claim is true for \(s=1\). Suppose the claim is true for \(s=1, \ldots , l\), then it must be true for \(s=l+1\): if buyer is realizing X, then at least one seller getting \(x<\frac{\delta }{1+\delta }\) has a similar profitable deviation as shown above.

Step 4: Now suppose some critical sellers trade at t while others trade at \(t-1\) and buyer is getting her highest surplus share \(X>\frac{1-\delta }{1+\delta }\). We know by Step 1 that critical sellers trading at t must be getting at least \(\frac{\delta }{1+\delta }\). We now argue that at least one critical seller trading at \(t-1\) cannot be getting \(x<\frac{\delta }{1+\delta }\). If buyer is making an offer at \(t-1\), such a critical seller can reject buyer’s offer and claim \(\frac{x}{\delta }+\epsilon\) at t. Buyer accepts this claim for a small \(\epsilon <(1-\delta )\left( X-\frac{x}{\delta }\right)\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(X-\epsilon <\max \left( \frac{1-\delta }{1+\delta }, \delta X\right)\).

Step 5: Now suppose some critical sellers trade at t while others trade at \(t-s\), \(s\ge 1\). We claim that buyer cannot be getting a surplus share greater than \(\frac{1-\delta }{1+\delta }\). We know by Step 3 that the claim is true for \(s=1\). We now use an induction argument. Suppose the claim is true for \(s=1, 2, \ldots , l\). We show that the claim must be true for \(s=l+1\). By Step 1, no critical seller trading at t must be getting less than \(\frac{\delta }{1+\delta }\). So, if there are at least two critical sellers trading at t, then the claim is true. So suppose only one critical seller is trading at t and getting at least \(\frac{\delta }{1+\delta }\) while others trade at \(t-l-1\). If the claim is false, then a critical seller is trading at \(x<\frac{\delta }{1+\delta }\) at \(t-l-1\). If buyer is making an offer at \(t-l\), such a critical seller can reject buyer’s offer and claim \(\frac{x}{\delta }+\epsilon\) at \(t-l\). Buyer accepts this claim for a small \(\epsilon <X-\max\left( \frac{1-\delta }{1+\delta }, \delta X\right) +\frac{(1-\delta )x}{\delta }\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(X-\epsilon <\max \left( \frac{1-\delta }{1+\delta }, \delta X\right)\). Hence this cannot be an equilibrium.

Step 6: Suppose critical sellers trade for the last time at \(t-s\), \(s\in \{1, 2, \ldots \}\), while others trade at \(s'<t-s\), and so on. We claim that buyer’s equilibrium surplus share cannot be more than \(\frac{1-\delta }{1+\delta }\). We know that the claim is true for \(s=1\) by Step 3. Suppose the claim is true for \(s=1, \ldots l\). We claim that it must be true for \(s=l+1\). Suppose buyer is realizing her highest surplus share \(X>\frac{1-\delta }{1+\delta }\). First notice that critical sellers trading at \(t-l-1\) cannot be getting \(x<\frac{\delta }{1+\delta }\). If buyer is making an offer at \(t-l-1\), such a critical seller can reject buyer’s offer and claim \(\frac{x}{\delta }+\epsilon\) at \(t-l\). Buyer accepts this claim for a small \(\epsilon <X-\max\left( \frac{1-\delta }{1+\delta }, \delta X\right) +\frac{(1-\delta )x}{\delta }\). If the sellers are making an offer, then if this critical seller claims an increment \(\epsilon\) over x, then the buyer accepts as long as \(X-\epsilon <\max \left( \frac{1-\delta }{1+\delta }, \delta X\right)\). Consequently if there are two or more critical sellers trading at \(t-l-1\), then buyer’s surplus share cannot be more than \(\frac{1-\delta }{1+\delta }\). So suppose that one critical seller trades at \(t-l-1\) and at least one critical seller trades at some \(t-l-l'\), \(l'\in \{2, 3, \ldots \}\). If \(l'=2\), then the critical seller trading at \(x<\frac{\delta }{1+\delta }\) has a profitable deviation similar to the one discussed above. An induction on \(l'\) then completes the argument.

Steps 1, 4 and 5 prove the claim of the Theorem when trade last occurs with critical sellers at t, while Steps 2, 3 and 6 prove the claim for the case when trade last occurs with critical sellers at some period prior to t.

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Sarkar, S., Gupta, D. Bargaining for assembly. Theory Decis 95, 229–254 (2023). https://doi.org/10.1007/s11238-022-09917-5

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