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Because I have white hair and that so sparse as to resemble the Soviet harvest of 1963, I am asked for advice. Just recently I was asked about `hot’ research topics in the sharing economy. `You mean a pure exchange economy?, said I in reply.  Because I have white hair etc, I sometimes forget to bite my tongue.

Returning to topic, the Economist piece I linked to above, gets it about right. With a fall in certain transaction costs, trades that were otherwise infeasible, are realized. At a high level there is nothing more to be said beyond what we know already about exchange economies.

A closer looks suggests something of interest in the role of the mediator (eBay, Uber) responsible for the reduction in transaction costs. They are not indifferent Walrasian auctioneers but self interested ones. eBay and Uber provide an interesting contrast in `intrusiveness’. The first reduces the costs with respect to search, alleviates the lemons problem and moral hazard by providing information and managing payments. It does not, however, set prices. These are left to participants to decide. In sum, eBay it appears,  tries to eliminate the textbook obstacles to a perfectly competitive market. Uber, also does these things but more. It chooses prices and the supplier who will meet the reported demand. One might think eBay does not because of the multitude of products it would have to track. The same is true for Uber. A product on Uber is a triple of origin, destination and time of day.    The rider and driver may not be thinking about things in this way, but Uber certainly must in deciding prices and which supplier will be chosen to meet the demand. Why doesn’t Uber allow riders to post bids and drivers to post asks?

Fox News and CNN have been in knots recently about how to allocate scarce debate slots to the many Republican pretenders to the oval office. Should Poll numbers be the criteria? What about gender, race, state office etc. etc. There is a simple and transparent way to allocate the slots: an auction. Neither CNN or Fox are charities. If the debates are a public good, there is no reason why Fox and CNN alone should forgo profits to host them. They should, instead auction off the rights to participate in the debates.

What design questions would have to be answered? First, how much time to allocate to the debate, i.e., what is the supply of the resource to be allocated? Second, should one auction slots or time. For example, should it be 10 slots of equal time lengths in a 2 hour time period? Or, should one allow the bidders to bid on the actual amount of time they would get in the 2 hour period? Are candidates the only ones allowed to bid, or can anyone submit bids on their behalf? Can one bid to deny time to the candidates? Actually, as these are not debates at all, but parallel news conference, perhaps one should auction off the right to answer questions. In real time. That would be entertainment!

According to the NY Times,  some Californians

would have to cut their water consumption by 35 percent under the terms of a preliminary plan issued by state officials on Tuesday to meet a 25 percent mandatory statewide reduction in urban water use.

There is an obvious way to achieve this, raise the price of water. If its obvious why isn’t it the first thing California did some years back when it was clear that water was scarce? In some cases the hands of the state are tied by water rights allocated for commercial purposes, so lets focus on household consumption.

We know that the first tool the state reaches for is regulation. See, for example, the following memo from the California State water board. Interestingly, it begins by noting that the state is in the 4th year of a drought! Eventually, it is clear that regulation is insufficient and then, price increases are considered. In fact, the water reduction targets quoted from the NY Times above come from a report by the Governor of the state that also urges the use of

rate structures and other pricing mechanisms

to achieve reductions. Again, why is price last rather first? Is this because the state must maintain a credible reputation for not exploiting its monopoly power with water?

If one is going reduce consumption by raising prices, should it be an across the board price increase? Note that consumption is metered so the exact amount that is purchased by a household is known to the provider. The state also has access to other information: location, home size, family size and income. In principle, the state could price discriminate. Here is an example from the Irvine Ranch Water district. Each household is given an initial `allotment’ that depends on household size and area that is landscaped. Exceed the allotment and the price of water rises. For more details and the impact on consumption see the following paper. Is it obvious that this is the `correct’ mechanism?



Penn state runs auctions to license its intellectual property. For each license on the block there is a brief description of what the relevant technology is and an opening bid which I interpret as a reserve price. It also notes whether the license is exclusive or not. Thus, the license is sold for a single upfront fee. No royalties or other form of contingent payment. As far as I can tell the design is an open ascending auction.

In an earlier pair of posts I discussed a class of combinatorial auctions when agents have binary quadratic valuations. To formulate the problem of finding a welfare maximizing allocation let {x^k_j = 1} if object {j \in M} is given to agent {k \in N} and zero otherwise. Denote the utility of agent {k \in N} from consuming bundle {S \subseteq M} by

\displaystyle u_k(S) = \sum_{i \in S}u^k_i + \sum_{i, j \in S}w^k_{ij}.

The problem of maximizing total welfare is

\displaystyle \max \sum_{k \in N}\sum_{i \in M}u^k_ix^k_i + \sum_{k \in N}\sum_{i \neq j}w^k_{ij}x^k_ix^k_j

subject to

\displaystyle \sum_{k \in N}x^k_i \leq 1\,\, \forall i \in M

\displaystyle x^k_i \in \{0,1\}\,\, \forall i \in M, k \in N

I remarked that Candogan, Ozdaglar and Parrilo (2013) identified a solvable instance of the welfare maximization problem. They impose two conditions. The first is called sign consistency. For each {i,j \in M}, the sign of {w^k_{ij} } and {w^r_{ij}} for any {k, r \in N} is the same. Furthermore, this applies to all pairs {i, j \in M}.

Let {G^w} be a graph with vertex set {M} and for any {i,j \in M} such that {w^k_{ij} \neq 0} introduce an edge {(i,j)}. Because of the sign consistency condition we can label the edges of {G^w} as being positive or negative depending on the sign of {w^k_{ij}}. Let {E^+ = \{(i,j): w^k_{ij} \geq 0\}} and {E^- = \{(i,j): w^k_{ij} < 0\}}. The second condition is that {G^w} be a tree.

The following is the relaxation that they consider:

\displaystyle \max \sum_{k \in N}\sum_{i \in M}u^k_ix^k_i + \sum_{k \in N}\sum_{(i,j) \in E^+ \cup E^-}w^k_{ij}z^k_{ij}

subject to

\displaystyle \sum_{k \in N}x^k_i \leq 1\,\, \forall i \in M

\displaystyle z^k_{ij} \leq x^k_i, x^k_j\,\, \forall k \in N, (i,j) \in E^+

\displaystyle z^k_{ij} \geq x^k_i + x^k_j - 1\,\, \forall k \in N, (i,j) \in E^-

\displaystyle x^k_i, z^k_{ij} \in \{0,1\}\,\, \forall i \in M, k \in N

Denote by {P} the polyhedron of feasible solutions to the last program. I give a new proof of the fact that the extreme points of {P} are integral. My thanks to Ozan Candogan for (1) patiently going through a number of failed proofs and (2) being kind enough not to say :“why the bleep don’t you just read the proof we have.”

Let {{\cal C}} be the maximal connected components of {G^w} after deletion of the edges in {E^-} (call this {G^+}). The proof will be by induction on {|{\cal C}|}. The case {|{\cal C}| = 1} follows from total unimodularity. I prove this later.

Suppose {|{\cal C}| > 1}. Let {({\bar z}, {\bar x})} be an optimal solution to our linear program. We can choose {({\bar z}, {\bar x})} to be an extreme point of {P}. As {G^w} is a tree, there must exist a {C \in {\cal C}} incident to exactly one negative edge, say {(p,q)}. Denote by {P'} the polyhedron {P} restricted to just the vertices of {C} and by {Q} the polyhedron {P} restricted to just the vertices in the complement of {C}. By the induction hypothesis, both {P'} and {Q} are integral polyhedrons. Each extreme point of {P'} ({Q}) assigns a vertex of {C} (the complement of {C}) to a particular agent. Let {X_1, X_2, \ldots, X_a} be the set of extreme points of {P'}. If in extreme point {X_r}, vertex {p} is assigned to agent {k} we write {X_{rk} = 1} and zero otherwise. Similarly with the extreme points {Y_1, Y_2, \ldots, Y_b} of {Q}. Thus, {Y_{rk} = 1} is {Y_r} assigns vertex {q} to agent {k}. Let {v(X_{r})} be the objective function value of the assignment {X_r}, similarly with {v(Y_{rk})}.

Now {({\bar z}, {\bar x})} restricted to {P'} can be expressed as {\sum_r\lambda_{r}X_{r}}. Similarly, {({\bar z}, {\bar x})} restricted to {Q} can be expressed as {\sum_r\mu_{r}Y_{r}}. We can now reformulate our linear program as follows:

\displaystyle \max \sum_r\lambda_{r}v(X_{r}) + \sum_r\mu_{r}v(Y_{r}) -\sum_{k \in N} |w^k_{pq}|y^k_{pq}

subject to

\displaystyle -\sum_r\lambda_{rk} = -1

\displaystyle -\sum_r\mu_{rk} = -1

\displaystyle \sum_{r: X_{rk} = 1}\lambda_{r}X_{r} + \sum_{r: Y_{rk} = 1}\mu_{r}Y_{r} \leq y^k_{pq} + 1\,\, \forall k \in N

\displaystyle \lambda_{r}, \mu_{r}, y^k_{pq} \geq 0\,\, \forall r, k

The constraint matrix of this last program is totally unimodular. This follows from the fact that each variable appears in at most two constraints with coefficients of opposite sign and absolute value 1 (this is because {X_{rk}} and {X_{rk'}} cannot both be 1, similarly with the {Y}‘s). Total unimodularity implies that the last program has integral optimal solution and we are done. In fact, I believe the argument can be easily modified to to the case where in {G^w} every cycle must contain a positive even number of negative edges.

Return to the case {|{\cal C}| = 1}. Consider the polyhedron {P} restricted to just one {C \in {\cal C}}. It will have the form:

\displaystyle \sum_{k \in N}x^k_i \leq 1\,\, \forall i \in C

\displaystyle z^k_{ij} \leq x^k_i, x^k_j\,\, \forall k \in N, (i,j) \in E^+ \cap C

\displaystyle x^k_i, z^k_{ij} \in \{0,1\}\,\, \forall i \in C, k \in N

Notice the absence of negative edges. To establish total unimodularity we use the Ghouila-Houri (GH) theorem. Fix any subset, {S}, of rows/constraints. The goal is to partition them into two sets {L} and {R} so that column by column the difference in the sum of the non-zero entries in {L} and and the sum of the nonzero entries in {R} differ by at most one.

Observe that the rows associated with constraints {\sum_{k \in N}x^k_i \leq 1} are disjoint, so we are free to partition them in any way we like. Fix a partition of these rows. We must show to partition the remaining rows to satisfy the GH theorem. If {y^k_{ij} - x^k_i \leq 0} is present in {S} but {y^k_{ij} -x^k_j \leq 0} is absent (or vice-versa), we are free to assign the row associated with {y^k_{ij} - x^k_i \leq 0} in any way to satisfy the GH theorem. The difficulty will arise when both {y^k_{ij}}, {x^k_i} and {x^k_j} are present in {S}. To ensure that the GH theorem is satisfied we may have to ensure that the rows associated with {y^k_{ij} - x^k_i \leq 0} and {y^k_{ij} -x^k_j \leq 0} be separated.

When {S} is the set of all constraints we show how to find a partition that satisfies the GH theorem. We build this partition by sequentially assigning rows to {L} and {R} making sure that after each assignment the conditions of the GH theorem are satisfied for the rows that have been assigned. It will be clear that this procedure can also be applied when only a subset of constraints are present (indeed, satisfying the GH theorem will be easier in this case).

Fix an agent {k \in N}. The following procedure will be repeated for each agent in turn. Pick an arbitrary vertex in {C} (which is a tree) to be a root and direct all edges `away’ from the root (when {S} is a subset of the constraints we delete from {C} any edge {(i,j)} in which at most one from the pair {y^k_{ij} - x^k_i \leq 0} and {y^k_{ij} -x^k_j \leq 0} appears in {S}) . Label the root {L}. Label all its neighbors {R}, label the neighbors of the neighbors {L} and so on. If vertex {i \in C} was labeled {L} assign the row {\sum_{k \in N}x^k_i \leq 1} to the set {L} otherwise to the row {R}. This produces a partition of the constraints of the form {\sum_{k \in N}x^k_i \leq 1} satisfying GH.

Initially, all leaves and edges of {C} are unmarked. Trace out a path from the root to one of the leaves of {C} and mark that leaf. Each unmarked directed edge {(i,j)} on this path corresponds to the pair {y^k_{ij} - x^k_i \leq 0} and {y^k_{ij} -x^k_j \leq 0}. Assign {y^k_{ij} - x^k_i \leq 0} to the same set that is the label of {i}. Assign {y^k_{ij} - x^k_j \leq 0} to the same set that is the label of vertex {j}. Notice that in making this assignment the conditions of the GH theorem continues to satisfied. Mark the edge {(i,j)}. If we repeat this procedure again with another path from the root to an unmarked leaf, we will violate the GH theorem. To see why suppose the tree contains edge {(i,j)} as well as {(i,t)}. Suppose {i} was labeled {L} on the first iteration and {(i,j)} was marked. This means {y^k_{ij} - x^k_{i} \leq 0} was assigned to {L}. Subsequently {y^k_{it} - x^k_i \leq 0} will also be assigned to {L} which will produce a partition that violates the GH theorem. We can avoid this problem by flipping the labels on all the vertices before repeating the path tracing procedure.

What is the institutional detail that makes electricity special? Its in the physics that I will summarize with a model of DC current in a resistive network. Note that other sources, like Wikipedia give other reasons, for why electricity is special:

Electricity is by its nature difficult to store and has to be available on demand. Consequently, unlike other products, it is not possible, under normal operating conditions, to keep it in stock, ration it or have customers queue for it. Furthermore, demand and supply vary continuously. There is therefore a physical requirement for a controlling agency, the transmission system operator, to coordinate the dispatch of generating units to meet the expected demand of the system across the transmission grid.

I’m skeptical. To see why, replace electricity by air travel.

Let {V} be the set of vertices and {E^*} the set of edges a the network. It will be convenient in what follows to assign (arbitrarily) an orientation to each edge in {E^*}. Let {E} be the set of directed arcs that result. Hence, {(i,j) \in E} mens that the edge {(i,j)} is directed from {i} to {j}. Notice, if {(i,j) \in E}, then {(i,j) \not \in E}.

Associated with each {(i,j) \in E} is a number {x_{ij}} that we interpret as a flow of electricity. If {x_{ij} > 0} we interpret this to be a flow from {i} to {j}. If {x_{ij} < 0} we interpret this as a flow from {j} to {i}.

  1. Let {\rho_{ij}} is the resistance on link {(i,j)}.
  2. {c_i} unit cost of injecting current into node {i}.
  3. {v_i} marginal value of current consumed at node {i}.
  4. {d_i} amount of current consumed at node {i}.
  5. {s_i} amount of current injected at node {i}.
  6. {K_{ij}} capacity of link {(i,j)}.

Current must satisfy two conditions. The first is conservation of flow at each node:

\displaystyle s_i + \sum_{k: (k,i) \in E}x_{ji} = \sum_{j: (i,j) \in E}x_{ij} + d_i\,\, \forall i \in V

The second is Ohm’s law. There exist node potentials {\{\phi_i\}_{i \in V}} such that

\displaystyle \rho_{ij}x_{ij} = \phi_i - \phi_j\,\, \forall (i,j) \in E.

Using this systems equations one can derive the school boy rules for computing the resistance of a network (add them in series, add the reciprocals in parallel). At the end of this post is a digression that shows how to formulate the problem of finding a flow that satisfies Ohm’s law as an optimization problem. Its not relevant for the economics, but charming nonetheless.

At each node {i \in V} there is a power supplier with constant marginal cost of production of {c_i} upto {S_i} units. At each {i \in V} there is a consumer with constant marginal value of {v_i} upto {D_i} units. A natural optimization problem to consider is

\displaystyle \max \sum_{i \in V}[v_id_i - c_is_i]

subject to

\displaystyle \sum_{j: (i,j) \in E}x_{ij} -\sum_{j: (j,i) \in E}x_{ji} - s_i + d_i= 0\,\, \forall i \in V

\displaystyle \rho_{ij}x_{ij} = \mu_i - \mu_j\,\, \forall (i,j) \in E

\displaystyle -K_{ij} \leq x_{ij} \leq K_{ij}\,\, \forall (i,j) \in E

\displaystyle 0 \leq s_i \leq S_i\,\, \forall i \in V

\displaystyle 0 \leq d_i \leq D_i\,\, \forall i \in V

This is the problem of finding a flow that maximizes surplus.

Let {{\cal C}} be the set of cycles in {(V, E^*)}. Observe that each {C \in {\cal C}} corresponds to a cycle in {(V, E)} if we ignore the orientation of the edges. For each cycle {C \in {\cal C}}, let {C^+} denote the edges in {E} that are traversed in accordance with their orientation. Let {C^-} be the set of edges in {C} that are traversed in the opposing orientation.

We can project out the {\phi} variables and reformulate as

\displaystyle \max \sum_{i \in V}[v_id_i - c_is_i]

subject to

\displaystyle \sum_{j: (i,j) \in E}x_{ij} -\sum_{j: (j,i) \in E}x_{ji} - s_i + d_i= 0\,\, \forall i \in V

\displaystyle \sum_{(i,j) \in C^+}\rho_{ij}x_{ij} - \sum_{(i,j) \in C^-}\rho_{ij}x_{ij} = 0\,\, \forall \,\, C \in {\cal C}

\displaystyle -K_{ij} \leq x_{ij} \leq K_{ij}\,\, \forall (i,j) \in E

\displaystyle 0 \leq s_i \leq S_i\,\, \forall i \in V

\displaystyle 0 \leq d_i \leq D_i\,\, \forall i \in V

Recall the scenario we ended with in part 1. Let {V = \{1, 2, 3\}}, {E = \{(1,3), (1,2), (2,3)\}} and in addition suppose {\rho_{ij} =1} for all {(i,j)}. Only {(1,3)} has a capacity constraint of 600. Let {D_1 = D_2 = 0} and {S_3 = 0}. Also {c_1 = 20} and {c_2 = 40} and each have unlimited capacity. At node 3, the marginal value is {V > 40} upto 1500 units and zero thereafter. The optimization problem is

\displaystyle \max Vd_3 - 20s_1 - 40 s_2

subject to

\displaystyle x_{12} + x_{13} - s_1 = 0

\displaystyle x_{23} - s_2 - x_{12} = 0

\displaystyle d_3 - x_{13} - x_{23} = 0

\displaystyle x_{13} - x_{23} - x_{12} = 0

\displaystyle -600 \leq x_{13} \leq 600

\displaystyle 0 \leq d_3 \leq 1500

Notice, for every unit of flow sent along {(1,3)}, half a unit of flow must be sent along {(1,2)} and {(2,3)} as well to satisfy the cycle flow constraint.

The solution to this problem is {x_{13} = 600}, {x_{12} = -300}, {x_{23} = 900}, {s_1 = 300}, {s_2 = 1200} and {d_3 = 1500}. What is remarkable about this not all of customer 3’s demand is met by the lowest cost producer even though that producer has unlimited capacity. Why is this? The intuitive solution would have been send 600 units along {(1,3)} and 900 units along {(1,2) \rightarrow (2,3)}. This flow violates the cycle constraint.

In this example, when generator 1 injects electricity into the network to serve customer 3’s demand, a positive amount of that electricity must flow along every path from 1 to 3 in specific proportions. The same is true for generator 2. Thus, generator 1 is unable to supply all of customer 3’s demands. However, to accommodate generator 2, it must actually reduce its flow! Hence, customer 3 cannot contract with generators 1 and 2 independently to supply power. The shared infrastructure requires that they co-ordinate what they inject into the system. This need for coordination is the argument for a clearing house not just to manage the network but to match supply with demand. This is the argument for why electricity markets must be designed.

The externalities caused by electricity flows is not a proof that a clearing house is needed. After all, we know that if we price the externalities properly we should be able to implement the efficient outcome. Let us examine what prices might be needed by looking at the dual to the surplus maximization problem.

Let {y_i} be the dual variable associated with the flow balance constraint. Let {\lambda_C} be associated with the cycle constraints. Let {\nu_i} and {\theta_i} be associated with link capacity constraints. Let {\mu_i} and {\sigma_i} be associated with the remaining tow constraints. These can be interpreted as the profit of supplier {i} and the surplus of customer {i} respectively. For completeness the dual would be:

\displaystyle \min \sum_{(i,j) \in E}[\nu_{ij} + \theta_{ij}]K_{ij} + \sum_{i \in V}[S_i \mu_i + D_i \sigma_i]

subject to

\displaystyle -\theta_{ij} + \nu_{ij} + \rho_{ij}\sum_{C^+ \ni (i,j)}\lambda_C - \rho_{ij}\sum_{C^- \ni (i,j)}\lambda_C + y_i - y_j = 0\,\, \forall (i,j) \in E

\displaystyle \mu_i - y_i \geq -c_i\,\, \forall i \in V

\displaystyle \sigma_i + y_i \geq v_i\,\, \forall i \in V

\displaystyle \nu_{ij}, \theta_{ij}, \mu_i, \sigma_i \geq 0\,\, \forall i \in V,\,\,\forall (i,j) \in E

Now {y_i} has a natural interpretation as a price to be paid for consumption at node {i} for supply injected at node {i}. {\mu_i} and {\nu_i} can be interpreted as the price of capacity. However, {\lambda_C} is trickier, price for flow around a cycle? It would seem that one would have to assign ownership of each link as well as ownership of cycles in order to have a market to generate these prices.

In part two, as promised, I turn to the welfare maximization problem. To formulate the problem of finding a welfare maximizing allocation let {x^k_j = 1} if object {j \in M} is given to agent {k \in N} and zero otherwise. Denote the utility of agent {k \in N} from consuming bundle {S \subseteq M} by

\displaystyle u_k(S) = \sum_{i \in S}u^k_i + \sum_{i, j \in S}w^k_{ij}.

The problem of maximizing total welfare is

\displaystyle \max \sum_{k \in N}\sum_{i \in M}u^k_ix^k_i + \sum_{k \in N}\sum_{i \neq j}w^k_{ij}x^k_ix^k_j

subject to

\displaystyle \sum_{k \in N}x^k_i \leq 1\,\, \forall i \in M

\displaystyle x^k_i \in \{0,1\}\,\, \forall i \in M, k \in N

Welfare maximization with BQP preferences is in general NP-hard. One proof relies on a reduction to the multi-way cut problem. Given a graph {G=(V,E)} with edge weight {c_{ij}} for each {(i,j) \in E}, and a set of terminal vertices {T = \{v_1, v_2, \ldots, v_n\}}, a multiway cut is a set of edges whose removal disconnects every pair of terminal vertices. The problem of finding the multiway cut of minimum total weight is called the multiway cut problem. When {T} consists of only two terminals ({k = 2}) the problem reduces to the well known minimum cut problem. For {k \geq 3}, it is known that the problem is {NP-}hard even on planar graphs.

We can obtain the multiway cut problem by setting {u^k_i = 0} for all {k \in N} and {i \in M} and {w^k_{ij} = c_{ij}} for all {k \in N} and {(i,j) \in E}. Any pair {(i,j)} such that {x^k_ix^r_j = 1} for {k \neq r} will be part of a multi-way cut. This reduction implies that welfare maximization when {w^k_{ij} \geq 0} for all {k \in N} and {i,j \in M} is NP-hard. This is in contrast to the case of surplus maximization.

,Candogan, Ozdaglar and Parrilo (2013), the paper that prompted this post, identify a solvable instance of the welfare maximization problem. They impose two conditions. The first is called sign consistency. For each {i,j \in M} the sign of {w^k_{ij} } and {w^r_{ij}} for any {k, r \in N} is the same. Furthermore, this applies to all pairs {i, j \in M}. Sign consistency by itself is not sufficient to obtain a solvable instance. Another condition is needed. Let {G^w} be a graph with vertex set {M} and for any {i,j \in M} such that {w^k_{ij} \neq 0} introduce an edge {(i,j)}. The second condition is that {G^w} be a tree. Interestingly, Erdos and Sz\'{e}kely (1995) show that a generalization of the multiway cut problem which corresponds to welfare maximization under sign consistency and {u^k_i = 0} for all {k \in N} and {i \in M}, is polynomially solvable when the underlying graph {G} is a tree. The Candogan, Ozdaglar and Parrilo (COP) proof is based on a dynamic programming argument similar to the one used in Erdos and Sz\'{e}kely (1994).

The key result in COP is the following natural linearization of the welfare maximization problem admits an integral optimal solution.

\displaystyle \max \sum_{k \in N}\sum_{i \in M}u^k_ix^k_i + \sum_{k \in N}\sum_{i \neq j}w^k_{ij}y^k_{ij}

subject to

\displaystyle \sum_{k \in N}x^k_i \leq 1\,\, \forall i \in M

\displaystyle y^k_{ij} \leq \{x^k_i, x^k_j\}^-\,\, \forall w^k_{ij} > 0

\displaystyle y^k_{ij} \geq x^k_i + x^k_j -1\,\, \forall w^k_{ij} < 0

\displaystyle x^k_i \in \{0,1\}\,\, \forall i \in M, k \in N

There is a connection between the welfare maximization problem and packing subtrees into trees that I want to highlight. It suggests a possible avenue by which one might enlarge the class of preferences COP consider.

Because of the sign consistency condition we can label the edges of {G^w} as being positive or negative depending on the sign of {w^k_{ij}}. Let {E^+ = \{(i,j): w^k_{ij} \geq 0\}} and {E^- = \{(i,j): w^k_{ij} < 0\}}. Let {{\cal C}} be the maximal connected components of {G^w} after deletion of the edges in {E^-} (call this {G^+}). For any {C \in {\cal C}} and {S \subseteq C} let

\displaystyle v_k(S) = \sum_{i \in S}u^k_i + \sum_{i,j \in S}w^k_{ij}.

By the way, for each {C \in {\cal C}} and {k \in N}, {v_k} is supermodular over the subsets of {C}. Let {z_k(S) = 1} if {S \subseteq C} is assigned to agent {k} and zero otherwise. The problem of finding a welfare maximizing allocation can be expressed as follows:

\displaystyle \max \sum_{k \in N}\sum_{C \in {\cal C}}\sum_{S \subseteq C}v_k(S)z_k(S) - \sum_{k \in N}\sum_{(i,j) \in E^-}|w^k_{ij}|y^k_{ij}

subject to

\displaystyle \sum_{S \ni i}\sum_{k \in N}z_k(S) \leq 1\,\, \forall i \in M,\,\, S \subseteq C \in {\cal C}

\displaystyle y^k_{ij} \geq \sum_{S \ni i}z_k(S) + \sum_{S \ni j}z_k(S) - 1\,\, \forall k \in N, (i,j) \in E^-

\displaystyle z_k(S) \in \{0,1\}\,\, \forall k \in N, S \subseteq C \in {\cal C}

\displaystyle y^k_{ij} \in \{0,1\}\,\, \forall k \in N, (i,j) \in E^-

In the above program, we can, without loss, restrict attention to subsets {S} that a subtrees (connected subgraphs) of {E^+}. To see why, suppose in that in some optimal solution to the above integer program, {z_k(S) = 1} where {S} is not a subtree. Then, we can write {S = A \cup B} where both {A} and {B} are in the same component of {C} as {S} is. Furthermore, it must be the the case that there is no edge {(i,j) \in E^+} such that {i \in A} and {j \in B}. Therefore, {v_k(S) = v_k(A) + v_k(B)}. That means we can construct a new optimal solution by setting {z_k(S) = 0} and raising {z_k(A)} and {z_k(B)} to 1. Note, in the original solution {z_k(A), z_k(B) =0} by virtue of the first constraint. As long as {S} is not a subtree we can repeat this argument.

Hence, if we let {{\cal T}} be the set of subtrees of {G^+}, the welfare maximization problem can be expressed as follows:

\displaystyle \max \sum_{k \in N}\sum_{T \in {\cal T}}v_k(T)z_k(T) - \sum_{k \in N}\sum_{(i,j) \in E^-}|w^k_{ij}|y^k_{ij}

subject to

\displaystyle \sum_{T \in {\cal T}}\sum_{T \ni i}\sum_{k \in N}z_k(T) \leq 1\,\, \forall i \in M,

\displaystyle y^k_{ij} \geq \sum_{T \ni i}z_k(T) + \sum_{S \ni j}z_k(S) - 1\,\, \forall k \in N, (i,j) \in E^-

\displaystyle z_k(S) \in \{0,1\}\,\, \forall k \in N, T \in {\cal T}

\displaystyle y^k_{ij} \in \{0,1\}\,\, \forall k \in N, (i,j) \in E^-

Its important to emphasize that no {T \in {\cal T}} contains a negative edge.

Were it not for the second set of the constraints, integrality would follow from Barany, Edmonds and Wolsey (1986). I’m not aware of this variation of the tree packing problem having been considered. A follow up paper by Aghezzaf and Wolsey (1994) comes close in the sense of allowing for a piecewise linear concave objective function.

A paper by Azevdo, Weyl and White in a recent issue of Theoretical Economics caught my eye. It establishes existence of Walrasian prices in an economy with indivisible goods, a continuum of agents and quasilinear utility. The proof uses Kakutani’s theorem. Here is an argument based on an observation about extreme points of linear programs. It shows that there is a way to scale up the number of agents and goods, so that in the scaled up economy a Walrasian equilibrium exists.
First, the observation. Consider {\max \{cx: Ax = b, x \geq 0\}}. The matrix {A} and the RHS vector {b} are all rational. Let {x^*} be an optimal extreme point solution and {\Delta} the absolute value of the determinant of the optimal basis. Then, {\Delta x^*} must be an integral vector. Equivalently, if in our original linear program we scale the constraints by {\Delta}, the new linear program has an optimal solution that is integral.

Now, apply this to the existence question. Let {N} be a set of agents, {G} a set of distinct goods and {u_i(S)}the utility that agent {i} enjoys from consuming the bundle {S \subseteq G}. Note, no restrictions on {u} beyond non-negativity and quasi-linearity.

As utilities are quasi-linear we can formulate the problem of finding the efficient allocation of goods to agents as an integer program. Let {x_i(S) = 1} if the bundle {S} is assigned to agent {i} and 0 otherwise. The program is

\displaystyle \max \sum_{i \in N}\sum_{S \subseteq G}u_i(S)x_i(S)

subject to
\displaystyle  \sum_{S \subseteq G}x_i(S) \leq 1\,\, \forall i \in N

\displaystyle \sum_{i \in N}\sum_{S \ni g} x_i(S) \leq 1 \forall g \in G

\displaystyle x_i(S) \in \{0,1\}\,\, \forall i \in N, S \subseteq G

If we drop the integer constraints we have an LP. Let {x^*} be an optimal solution to that LP. Complementary slackness allows us to interpret the dual variables associated with the second constraint as Walrasian prices for the goods. Also, any bundle {S} such that {x_i^*(S) > 0} must be in agent {i}‘s demand correspondence.
Let {\Delta} be the absolute value of the determinant of the optimal basis. We can write {x_i^*(S) = \frac{z_i^*(S)}{\Delta}} for all {i \in N} and {S \subseteq G} where {z_i^*(S)} is an integer. Now construct an enlarged economy as follows.

Scale up the supply of each {g \in G} by a factor of {\Delta}. Replace each agent {i \in N} by {N_i = \sum_{S \subseteq G}z_i^*(S)} clones. It should be clear now where this is going, but lets dot the i’s. To formulate the problem of finding an efficient allocation in this enlarged economy let {y_{ij}(S) = 1} if bundle {S} is allocated the {j^{th}} clone of agent {i} and zero otherwise. Let {u_{ij}(S)} be the utility function of the {j^{th}} clone of agent {i}. Here is the corresponding integer program.

\displaystyle \max \sum_{i \in N}\sum_{j \in N_i}\sum_{S \subseteq G}u_{ij}(S)y_{ij}(S)

subject to
\displaystyle  \sum_{S \subseteq G}y_{ij}(S) \leq 1\,\, \forall i \in N, j \in N_i

\displaystyle \sum_{i \in N}\sum_{j \in N_i}\sum_{S \ni g} y_{ij}(S) \leq \Delta \forall g \in G

\displaystyle y_{ij}(S) \in \{0,1\}\,\, \forall i \in N, j \in N_i, S \subseteq G

Its easy to see a feasible solution is to give for each {i} and {S} such that {z_i^*(S) > 0}, the {z_i^*(S)} clones in {N_i} a bundle {S}. The optimal dual variables from the relaxation of the first program complements this solution which verifies optimality. Thus, Walrasian prices that support the efficient allocation in the augmented economy exist.

In this post I describe an alternative proof of a nifty result that appears in a forthcoming paper by Goeree, Kushnir, Moldovanu and Xi. They show (under private values) given any Bayesian incentive compatible mechanism, M, there is a dominant strategy mechanism that gives each agent the same expected surplus as M provides.

For economy of exposition only, suppose 2 agents and a finite set of possible outcomes, {A}. Suppose, also the same type space, {\{1, \ldots, m\}} for both. Let {f(\cdot)} be the density function over types. To avoid clutter, assume the uniform distribution, i.e., {f(\cdot) = \frac{1}{m}}. Nothing in the subsequent analysis relies on this.

When agent 1 reports type {s} and agent 2 reports type {t}, denote by {z_r(s,t) } the probability that outcome {r \in A} is selected. The {z}‘s must be non-negative and satisfy {\sum_{r \in A}z_r(s,t) = 1.}

Associated with each agent {i} is a vector {\{a_{ir}\}_{r \in A}} that determines, along with her type, the utility she enjoys from a given allocation. In particular, given the allocation rule {z}, the utility that agent {1} of type {t} enjoys when the other agent reports type {s} is {\sum_{r \in A}ta_{1r}z_r(t,s).} A similar expression holds agent 2.

Let {q_i(s,t) = \sum_{r \in A}a_{ir}z_r(s,t).} Interpret the {q}‘s as the `quantity’ of goods that each agent receives. Dominant strategy implies that {q_1(s,t)} should be monotone increasing in {s} for each fixed {t} and {q_2(s,t)} should be monotone increasing in {t} for each fixed {s}. The interim `quantities will be: {mQ_i(s) = \sum_t\sum_{r \in A}a_{ir}z_r(s,t).} Bayesian incentive compatibility (BIC) implies that the {Q_i}‘s should be monotone. To determine if given BIC interim `quantities’ {Q_i}‘s can be implemented via dominant strategies, we need to know if the following system is feasible

\displaystyle \sum_{r \in A}a_{1r}[z_r(i,j) - z_r(i-1,j)] \geq 0\,\, \forall \,\, i = 2, \ldots, m \ \ \ \ \ (1)

\displaystyle \sum_{r \in A}a_{2r}[z_r(i,j) - z_r(i,j-1)] \geq 0\,\, \forall \,\, j = 2, \ldots, m \ \ \ \ \ (2)

\displaystyle \sum_t\sum_{r \in A}a_{1r}z_r(s,t) = mQ_1(s) \ \ \ \ \ (3)

\displaystyle \sum_s\sum_{r \in A}a_{2r}z_r(s,t) = mQ_2(t) \ \ \ \ \ (4)

\displaystyle \sum_{r \in A}z_r(s,t) = 1\,\, \forall s,t \ \ \ \ \ (5)

\displaystyle z_r(i,j) \geq 0\,\, \forall i,j,r \ \ \ \ \ (6)

System (1-6) is feasible iff the optimal objective function value of the following program is zero:

\displaystyle \min \sum_{i,j}w_{ij} + \sum_{i,j}h_{ij}

subject to
\displaystyle \sum_{r \in A}a_{1r}[z_r(i,j) - z_r(i-1,j)] + w_{ij} - u_{ij} = 0\,\, \forall \,\, i = 2, \ldots, m\,\, (\mu_{ij}) \ \ \ \ \ (7)

\displaystyle \sum_{r \in A}a_{2r}[z_r(i,j) - z_r(i,j-1)] + h_{ij} - v_{ij} = 0\,\, \forall \,\, j = 2, \ldots, m\,\, (\nu_{ij}) \ \ \ \ \ (8)

\displaystyle \sum_j\sum_{r \in A}a_{1r}z_r(i,j) = mQ_1(i)\,\, (\alpha_i) \ \ \ \ \ (9)

\displaystyle \sum_i\sum_{r \in A}a_{2r}z_r(i,j) = mQ_2(j)\,\, (\beta_j) \ \ \ \ \ (10)

\displaystyle \sum_{r \in A}z_r(i,j) = 1\,\, \forall i,j\,\, (\lambda(i,j)) \ \ \ \ \ (11)

\displaystyle w_{ij},\,\, h_{ij}\,\,z_r(i,j) \geq 0\,\, \forall i,j,r \ \ \ \ \ (12)

Let {(z^*, w^*, h^*, u^*, v^*)} be an optimal solution to this program.
Suppose, for a contradiction there is a pair {(p,q)} such that {w^*_{pq} > 0}. I will argue that there must exist an {s} such that {u^*_{ps} > 0}. Suppose not, then for each {j \neq q}, either {w^*_{pj} > 0} or {w^*_{pj} = 0} and {u^*_{pj} = 0} (at optimality, it cannot be that {w^*_{pq}} and {u^*_{pq}} are both non-zero). In this case

\displaystyle m[Q_1(p)-Q_1(p-1)] = \sum_j\sum_{r \in A}a_{1r}z^*_r(p,j) - \sum_j\sum_{r \in A}a_{1r}z^*_r(p-1,j)

= \sum_{j: w^*_{pj} > 0}a_{1r}[z^*_r(p,j) - z^*_r(p-1,j)] = -\sum_j w^*_{pj}. This last term is negative, a contradiction. Therefore, there is a s such that w^*_{ps} = 0 but u^*_{ps} > 0.

Let {Z_1(i,j) = \sum_{r \in A}a_{1r}z^*_r(i,j)}, {Z_2(i,j) = \sum_{r \in A}a_{2r}z^*_r(i,j)} and denote by {Z(i,j)} the point {(Z_1(i,j), Z_2(i,j))}. Observe that {Z(i,j)} is in the convex hull, {K} of {\{(a_{1r}, a_{2r}\}_{r \in A}} for all {i,j}. Thus choosing {\{z_r(i,j)\}_{r \in A}}‘s amounts to choosing a point {Z(i,j) \in K}. Equivalently, choosing a point {Z(i,j) \in K} gives rise to a set of {\{z_r(i,j)\}_{r \in A}}‘s. For convenience assume that {Z(i,j)} is in the strict interior of {K} for all {(i,j)} and that K is full dimensional. This avoids having to deal with secondary arguments that obscure the main idea.

Recall, {w^*_{pq} > 0} implies {Z_1(p,q)  0} implies that {Z_1(p,s) > Z_1(p,s)}. Take all points {\{Z(i,q)\}_{i \geq p}} and shift them horizontally to the right by {\delta}. Call these new points {\{Z'(i,q)\}_{i \geq p}}. Observe that {Z'(i,q) \in K} for all {i \geq p}. Next, take all points {\{Z(i,s)\}_{i \geq p}} and shift them horizontally to the left by {\delta} to form new points {\{Z'(i,s)\}_{i \geq p}}. These points are also in {K}. Leave all other points {\{Z(p,j)\}_{j \neq, q,s}} unchanged.

Because the vertical coordinates of all points were left unchanged, (8) and (10) are satisfied by this choice of points. Because {\{Z(i,q)\}_{i \geq p}} and {\{Z(i,s)\}_{i \geq p}} were shifted in opposite directions along the horizontal, (9) is still true. Finally, because all points in {\{Z(i,q)\}_{i \geq p}} and {\{Z(i,s)\}_{i \geq p}} were shifted by the same amount, (7) continues to hold.

The shift leftwards of {Z(p,s)} reduces {u_{ps}} while the rightward shift of {Z(p,q)} reduces {w_{pq}}. Thus, we get a new solution with higher objective function value, a contradiction.

If {Z(p,q)} and {Z(p,s)} are not the interior of {K} but on the boundary, then horizontal shifts alone may place them outside of {K}. In the case of {Z(p,q)} this can only happen if {Z_2(p,q) > Z_2(p-1,q)}. In this case, shift {Z(p,q)} across and to the right by {\delta} as well and then downwards by the same amount. This would have to be matched by a corresponding upward shift by some point {Z_2(h,q)}. Similarly with {Z(p,s)}.

Thanks to Alexey Kushnir and Ahmad Peivandi for comments.

This session was devoted to three different perspectives on Gale and Shapley’s stable matching problem. The first, is traditional wherein nothing more than a command of the English language coupled with a patience for moderately long chains of reasoning is needed. The second, is Adachi’s (from his OR letters paper) lattice formulation of the question and an example to illustrate how it can be generalized to other settings. The third, based on Vandevate’s characterization of the convex hull of stable matchings. I also stuck in a new result that fills a much needed gap in the literature that I hope is correct (if incorrect perhaps it will stimulate others).

1. Stable Matchings

The stable matching problem was introduced by Gale and Shapley as a model of how to assign students to colleges. One can view it as the ordinal version of the SS model. The problem has a set~{M} of men and a set~{W} of women. Each {m \in M} has a strict preference ordering over the elements of {W} and each {w \in W} has a strict preference ordering over the men. The preference ordering of agent~{i} is denoted {\succ_i} and {x \succ_i y} means agent~{i} ranks {x} above {y}. One can accommodate the possibility of an agent choosing to remain single by including for each man (woman) a dummy woman (man) in the set {W } ({M}) that corresponds to being single (or matched with oneself). With this construction we can assume {|M| = |W|}.

A matching is called \index{matching!unstable}unstable if there are two men {m, m'} and two women {w}, {w'} such that

  1. {m} is matched to {w},
  2. {m'} is matched to {w'},
  3. and {w' \succ_m w} and {m \succ_{w'} m'}

The pair {(m,w')} is called a blocking pair. A matching that has no blocking pairs is called stable.

Given the preferences of the men and women, its always possible to find a stable matching using the deferred acceptance algorithm.

Deferred Acceptance Algorithm, male-propose version

  1. First, each man proposes to his top ranked choice.
  2. Next, each woman who has received at least two proposals keeps (tentatively) her top ranked proposal and rejects the rest.
  3. Then, each man who has been rejected, proposes to his top ranked choice amongst the women who have not rejected him.
  4. Again each woman who has at least two proposals (including ones from previous rounds), keeps her top ranked proposal and rejects the rest.
  5. Process repeats until no man has a woman to propose to or each woman has at most one proposal.

Theorem 1 The male propose algorithm terminates in a stable matching.

Proof: When it terminates, it clearly does so in a matching. Suppose the matching is not stable. Then, there exists a blocking pair {(m_1, w_1)} with {m_1} matched to {w_2}, say, and {w_1} matched to {m_2}. Since {(m_1, w_1)} is blocking and {w_1 \succ_{m_1} w_2}, in the proposal algorithm, {m_1} would have proposed to {w_1} before {w_2}. Since {m_1} was not matched with {w_1} by the algorithm it must be because {w_1} received a proposal from a man that she ranked higher than {m_1}. Since the algorithm matches her to {m_2} it follows that {m_2 \succ_{w_1} m_1}. This contradicts the fact that {(m_1, w_1)} is a blocking pair. {\Box}

One could describe an algorithm where the females propose and the outcome would also be a stable matching and possibly different from the one returned using the male propose version. Thus, not only is a stable matching guaranteed to exist but there can be more than one. If there can be more than one stable matching, is there a reason to prefer one to another?

Denote a matching by {\mu}. The woman matched to man {m} in the matching {\mu} is denoted {\mu(m)}. Similarly, {\mu(w)} is the man matched to woman {w}. A matching {\mu} is male-optimal if there is no stable matching {\nu} such that {\nu(m) \succ_m \mu(m)} or {\nu(m) = \mu(m)} for all {m} with {\nu(j) \succ_j \mu(j)} for at least one {j \in M}. Similarly define female-optimal.

Theorem 2 The stable matching produced by the (male-proposal) Deferred Acceptance Algorithm is male-optimal.

Proof: Let {\mu} be the matching returned by the male-propose algorithm. Suppose {\mu} is not male optimal. Then, there is a stable matching {\nu} such that {\nu(m) \succ_m \mu(m)} or {\nu(m) = \mu(m)} for all {m} with {\nu(j) \succ_j \mu(j)} for at least one {j \in M}. Therefore, in the application of the proposal algorithm there must be an iteration where some man {j} proposes to {\nu(j)} before {\mu(j)} since {\nu(j) \succ_j \mu(j)} and is rejected by woman {\nu(j)}. Consider the first such iteration. Since woman {\nu(j)} rejects {j} she must have received a proposal from a man {i} she prefers to man {j}. Since this is the first iteration at which a male is rejected by his partner under {\nu} it follows that man {i} ranks woman {\nu(j)} higher than {\nu(i)}. Summarizing, {i \succ_{\nu(j)} j} and {\nu(j) \succ_i \nu(i)} implying that {\nu} is not stable, a contradiction. {\Box}

You can replace the word “male” by the word “female” in the statement of the theorem. There is no stable matching that is simultaneously optimal with respect to males and females.

A stable matching is immune to a pair of agents opting out of the matching (the blocking pair). We could ask that no subset of agents should have an incentive to opt out of the matching. Formally, a matching {\mu'} dominates a matching {\mu} if there is a set {S \subset M \cup W} that satisfies the following two conditions:

  1. {\mu'(m), \mu'(w) \in S} for all {m,w \in S}
  2. {\mu'(m) \succ_m \mu(m)} and {\mu'(w) \succ_w \mu(w)} for all {m,w \in S}.

Stability is a special case of this dominance condition when we restrict attention to sets~{S} consisting of a single couple. The set of undominated matchings is called the core of the matching game.

Theorem 3 The core of the matching game is the set of all stable matchings.

Now assume the preference orderings of the agents are private information to the agent.

Theorem 4 The direct mechanism associated with the male propose algorithm is strategy proof for the males.

Proof: If not, there is a preference profile {\pi =(\succ_{m_1}, \succ_{m_2}, \ldots, \succ_{m_n})} for the men, such that man {m_1}, say, can misreport his preferences and obtain a better match. Formally, let {\mu} be the stable matching obtained by applying the male proposal algorithm to the profile {\pi}. Suppose {m_1} reports the preference ordering {\succ_* } instead. Let {\nu} be the matching that results when the male-proposal algorithm is applied to the profile {\pi^1 = (\succ_*, \succ_{m_2}, \ldots, \succ_{m_n})}. For a contradiction suppose {\nu(m_1) \succ_{m_1} \mu(m_1)}. For convenience write {a \succeq _{m} b} to mean that {a \succ_m b} or {a=b}.

Let {R = \{m: \nu(m) \succ_m \mu(m)\}} be the set of men better off under matching {\nu} than matching {\mu}. We show that for any {m \in R} and {w = \nu(m)} that {m' = \mu(w) \in R}. In other words, {S = \{w: \nu(w) \in R\} = \{w: \mu(w) \in R\}}. For a contradiction suppose {m' \not \in R}. In particular {m' \neq m_1}. Now, {w \succ_m \mu(m)} because {m \in R}, and stability of {\mu} wrt {\pi} implies {m' \succ_w m}. Stability of {\nu} wrt {\pi^1} requires that {\nu(m') \succ_{m'} w}, otherwise {(m',w)} would block {\nu}. Therefore, {m' \in R}.

Now, {\mu(w) \succ_w \nu(w)} for any {w \in S}, because during execution of the male-propose algorithm on {\pi}, each {w \in S} rejects {\nu(w) \in R} at some iteration. Let {m} be the last man in {R} to make a proposal during execution of male-propose algorithm. This proposal is made to {w = \mu(m) \in S} who, by choice of {m}, must have rejected {\nu(w)} at some earlier iteration. This means that when {m} proposes to {w}, she must reject a pending proposal from {m' \not \in R} such that {m' \succ_w \nu(w)}. As {m' \not \in R}, we have {w \succ_{m'} \mu(m') \succeq_{m'} \nu(m')}. Therefore, {(m', w)} form a blocking pair for {\nu} wrt {\pi^1} (since {m' \neq m_1}). {\Box}

The mechanism associated with the male propose algorithm is not strategy proof for the females.

2. A Lattice Formulation

We describe a proof (due to Adachi; Fleiner also independently but later) of the existence of stable matchings using Tarski’s fixed point theorem.

Call an assignment of women to men such that each man is assigned to at most one woman (but a woman may be assigned to more than one man) a male semi-matching. The analogous object for women will be called a female semi-matching. For example, assigning each man his first choice would be a male semi-matching. Assigning each woman her third choice would be an example of a female semi-matching.

A pair of male and female semi-matchings will be called a semi-matching denoted {\mu}, {\nu}, etc. An example of a semi-matching would consist of each man being assigned his first choice and each woman being assigned her last choice.

The woman assigned to man {m} under the semi-matching {\mu} will be denoted {\mu(m)}. If man {m} is assigned to no woman under {\mu}, then {\mu(m) = m}. Similarly for {\mu(w)}. Next, define a partial order over the set of semi-matchings. Write {\mu \succeq \nu} if

  1. {\mu(m) \succ_m \nu(m)} or {\mu(m) = \mu(m)} for all {m \in M}, and,
  2. {\mu(w) \prec_w \nu(w)} or {\mu(w) = \nu(w)} for all {w \in W}.

Therefore {\mu \succeq \nu} if all the men are better off under {\mu} than in {\nu} and all the women are worse off under {\mu} than in {\nu}.

Next, define the meet and join operations. Given two semi-matchings {\mu} and {\nu} define {\lambda = \mu \vee \nu} as follows:

  1. {\lambda(m) = \mu(m)} if {\mu(m) \succ_m \nu(m)} otherwise {\lambda (m) = \nu (m)},
  2. {\lambda(w) = \mu(w)} if {\mu(w) \prec_w \nu(w)} otherwise {\lambda (w) = \nu (w).}

Define {\lambda' = \mu \wedge \nu} as follows:

  1. {\lambda'(m) = \mu(m)} if {\mu(m) \prec_m \nu(m)} otherwise {\lambda (m) = \nu (m)},
  2. {\lambda(w) = \mu(w)} if {\mu(w) \succ_w \nu(w)} otherwise {\lambda (w) = \nu (w).}

With these definitions it is easy to check that the set of semi-matchings form a compact lattice.

Now define a function {f} on the set of semi-matchings that is non-decreasing. Given a semi-matching {\mu} define {f(\mu)} to be the following semi-matching:

  1. {f(\mu)(m)} is man {m}‘s most preferred woman from the set {\{w: m \succeq_w \mu(w) \} }. If this set is empty, set {f(\mu)(m) = m}.
  2. {f(\mu)(w)} is woman {w}‘s most preferred man from the set {\{m: w \succeq_m \mu(m)\}}. If this set is empty set {f(\mu)(w) = w}.

It is clear that {f} maps semi-matchings into semi-matchings.

Theorem 5 There is a semi-matching {\mu} such that {f(\mu) = \mu} and that {\mu} is a stable matching.

Proof: We use Tarski’s theorem. To apply it we need to check that {f} is non-decreasing. Suppose {\mu\succeq \nu}. Pick any {m \in M}. From the definition of {\succeq}, the women are worse off under {\mu} than in {\nu}. Thus

\displaystyle \{w: m \succ_w \nu(w)\} \subseteq \{w: m \succ_w \mu(w)\}

and so {f(\mu)(m) \succ_m f(\nu)(m)} or {f(\mu)(m) = f(\nu)(m).} When considering the women, for each {w \in W} we must have (because the women are worse off under {\mu} than under {\nu}):

\displaystyle \{m: w \succ_m \mu(m)\} \subseteq \{m: w \succ_m \nu(m)\}.

Therefore, {f(\nu)(w) \succ_w f(\mu)(w) }.

As the conditions of Tarski’s theorem hold, there is a semi-matching {\mu} such that {f(\mu) = \mu}. We show that the semi-matching is a stable matching.

By the definition of a semi-matching we have for every {m \in M}, {\mu(m)} single valued as is {\mu(w)} for all {w \in W}. To show that {\mu} is a matching, suppose not. Then there is a pair {m_1, m_2 \in M}, say, such that {\mu(m_1) = \mu(m_2) = w^*}. Since {f(\mu) = \mu} it follows that {w^*} is {m_1}‘s top ranked choice in {\{w: m_1 \succeq_w \mu(w) \}} and {m_2}‘s top ranked choice in {\{w: m_2 \succeq_w \mu(w) \}}. From this it follows that if {\mu(w^*) = m_3}, then, {m_1, m_2 \succeq _{w^*} m_3}. As {m_1 \neq m_2}, we can assume, WLOG that {m_2 \succ_{w^*} m_3}. However, {m_3 =\mu(w^*) = f(\mu^*)(w^*)} which is woman {w^*}‘s top ranked choice in {\{m: w^* \succeq_m \mu(m) \}}. Since {m_1} belongs to this set, we get a contradiction.

To show that {\mu} is stable, suppose not. Then, there is a blocking pair {(m^*,w^*)}. Let {w' = \mu(m^*)} and {m' = \mu(w^*)}, {m' \neq m^*} and {w^* \neq w'}. As {(m^*,w^*)} blocks {\mu}, {m^* \succ_{w^*} m'} and {w^* \succ_{m^*} w'}. Now {w' = \mu(m^*) = f(\mu)(m^*)} which is man {m^*}‘s top ranked choice from {\{w: m^* \succ_w \mu(w), m^* = \mu(w) \} }. But this set contains {w^*} which is ranked higher by man {m^*} than {w'}, a contradiction. {\Box}

2.1. Splittable Stable Marriage

To show how versatile Adachi’s lattice approach is, we examine an extension due to Sethuraman and Teo called the splittable stable marriage problem.

We have two disjoint agents, say,`sellers” (indexed by {i} or {m}) and “buyers” (indexed by {j} or {w}). Imagine a “market” for a product that is infinitely divisible. Seller {i} has {s_i \geq 0} units to sell; buyer {j} wishes to buy {t_j \geq 0} and the quantity of sale between {i} and {j} can be at most {u_{ij}} (“capacity” constraint). Each agent has a strict preference ordering over agents on the opposite side. For convenience assume each agent ranks themselves last. Assume that the number of buyers is equal to the number of sellers. In the sequel the set of sellers will be {\{m_1, m_2, \ldots, m_n\}}, and the set of buyers will be {\{w_1, w_2, \ldots, w_n\}}.

A splittable marriage is defined as an {(n+1) \times (n+1)} matrix {Z = (z_{ij})} such that:

\displaystyle  \sum_{i=1}^{n+1} z_{ij} = t_j\,\, \forall j

\displaystyle \sum_{j=1}^{n+1} z_{ij} = s_i, \,\, \forall i

\displaystyle z_{ij} \leq u_{ij}, \,\, \forall i, j

\displaystyle z_{ij} \geq 0, \,\, \forall i, j

\displaystyle z_{n+1, n+1} = 0

Here {z_{i,j}} is the quantity of the product sold by seller {i} to buyer {j}, if neither {i} nor {j} is {n+1}; if {i = n+1} and {j < n+1}, then {z_{ij}} is the additional quantity that buyer {j} would like to buy but cannot; and if {i < n+1} and {j = n+1}, then {z_{ij}} is the additional quantity that seller {i} would like to sell but cannot.

A blocking pair for a splittable marriage {Z} is a pair of agents {m_i} and {w_j} such that

  1. {z_{ij} < u_{ij}};
  2. {\sum_{k: k \geq_i j} z_{ik} < s_i}; and
  3. {\sum_{k: k \geq_j i} z_{kj} < t_j}.

If {(i,j)} blocks {Z}, the agents {i} and {j} can do better by trading more with each other, and giving up some of the quantities they transact with less-preferred agents. A splittable stable marriage is a splittable marriage that is not blocked.

Next we define the appropriate generalization of semi-matching. A demand-list for seller {i} is a vector {x^i \equiv (x_{i,1}, x_{i,2}, \ldots, x_{i,n}, x_{i,(n+1)})} that satisfies (1) and (3). Similarly, a demand-list for buyer {j} is a vector {y^j \equiv (y_{1,j}, y_{2,j}, \ldots, y_{n,j}, y_{(n+1),j})} that satisfies (2) and (3).

The demand-lists of sellers and buyers can be encoded by matrices {X} and {Y}. Assume matrices {X} and {Y} have their rows indexed by the sellers and the columns indexed by the buyers. Specifically, {X} has {n} rows and {(n+1)} columns, whereas {Y} has {(n+1)} rows and {n} columns.

The seller demand-list matrix {X} is a valid “matching” of the sellers, but not a valid matching of the sellers to the buyers. Similarly, for the buyer demand-list matrix {Y}.

Let {L_s} and {L_b} be a collection of seller and buyer demand-lists respectively. Define partial orders {\geq_s} and {\geq_b} on the sets {L_s} and {L_b} respectively. For any two elements {X, \hat{X} \in L_s},

\displaystyle X \geq_s \hat{X} \;\;\;\; {\rm if} \;\;\; \forall i, j,\;\; 		\sum_{k:k \geq_i j} x_{ik} \geq 		\sum_{k:k \geq_i j} \hat{x}_{ik}.

Similarly, for any two elements {Y, \hat{Y} \in L_b},

\displaystyle Y \geq_b \hat{Y} \;\;\;\; {\rm if} \;\;\; \forall i, j,\;\; 		\sum_{l:l \geq_j i} x_{lj} \leq 		\sum_{l:l \geq_j i} \hat{x}_{lj}.

(Notice the change in the direction of the inequality in the definition of {\geq_b}.)

Given any {X, \hat{X} \in L_s}, let {M \equiv X \wedge \hat{X}} be defined as follows:

\displaystyle M_{ij} \; = \; \min \{ 		\sum_{k:k \geq_i j} x_{ik}, 		\sum_{k:k \geq_i j} \hat{x}_{ik} \} - 		\sum_{k:k >_i j} M_{ik}.

Similarly, let {J \equiv X \vee \hat{X}} be defined as follows:

\displaystyle J_{ij} \; = \; \max \{ 		\sum_{k:k \geq_i j} x_{ik}, 		\sum_{k:k \geq_i j} \hat{x}_{ik} \} - 		\sum_{k:k >_i j} J_{ik}.

The matrices {M} and {J} can be computed as follows: for each seller, compute the entries corresponding to his row, starting with the most preferred buyer. Clearly, {M} and {J} are in {L_s}, and are, respectively, the greatest lower bound and the least upper bound of {X} and {\hat{X}}. Again, the meet and join of any two elements of {L_b} can be defined in a similar way. Thus {(L_s, \vee, \wedge)} and {(L_b, \vee, \wedge)} are lattices. By replacing “min” and “max” with “inf” and “sup” respectively, we can infer that meets and joins exist for arbitrary subsets of elements of {L_s} or {L_b}; hence {(L_s, \vee, \wedge)} and {(L_b, \vee, \wedge)} are complete lattices. The lattice of interest is the product lattice {L \equiv (L_s, L_b)} with the partial order {\geq} defined as follows:

\displaystyle (X, Y) \geq (\hat{X}, \hat{Y}) \;\;\; {\rm if} 	\;\; X \geq_s \hat{X} \; {\rm and} \; 	Y \geq_b \hat{Y}.

From standard results from lattice theory we see that {L} is a complete lattice.

Call demand-lists {X} and {Y} compatible if {X_{ij} = Y_{ij}} for all { 1 \leq i, j \leq n}. Any splittable stable marriage {Z} can be viewed as a pair of compatible demand-lists: for any seller {i} and buyer {j}, set {X_{ij}} and {Y_{ij}} to be {Z_{ij}}; allocate the remaining supply and unfulfilled demand to the last entries in each row and column of {X} and {Y} respectively. The converse is false.

Next we need an appropriate monotone function on the lattice {L}. As intuition, suppose the buyers form a coalition and “agree” to hand over their demand-list {Y} to the sellers. How should the sellers respond? Clearly, the sellers would “update” their demand-lists as follows:

\displaystyle   X_{ij} = \min \{ u_{ij}, \max (t_j - \sum_{l: l >_j i} Y_{lj}, 0),\; \max (s_i - \sum_{k: k >_i j} X_{ik}, 0) \}. \ \ \ \ \ (1)

Notice that {X_{.,.}} appears on both sides of (1). One way to perform this update is as follows: for each seller, update the entries corresponding to his row, starting with the most preferred buyer. In this manner, the “old” demand-list of {X} is never used in the update. We call this the canonical computation.

Similar considerations suggest:

\displaystyle   Y_{ij} = \min \{ u_{ij}, \max (t_j - \sum_{l: l >_j i} Y_{lj}, 0),\; \max (s_i - \sum_{k: k >_i j} X_{ik}, 0) \}. \ \ \ \ \ (2)

Again, Eq.(2) can be computed as follows: for each buyer, update the entries corresponding to his row, starting with the most preferred seller.

Equations (1) and (2) have identical right hand sides. Unsurprisingly, compatible demand-lists arising from solutions of Equations (1) and (2) correspond to splittable stable marriages.

Lemma 6 If {X} and {Y} are solutions to Equations (1) and (2), then {Z \equiv (X_{ij})} for {1 \leq i, j \leq n} defines a splittable stable marriage.

Proof: By the hypothesis of the lemma, {X} and {Y} are identical. Thus,

\displaystyle  X_{ij} = \min \bigg\{ u_{ij}, \; \max\bigg(t_j - \sum_{l: l >_j i} X_{lj}, 0\bigg),\; \max\bigg(s_i - \sum_{k: k >_i j} X_{ik}, 0\bigg) \bigg\}.

So if {X_{ij} < u_{ij}}, either {\sum_{l: l \geq_j i} X_{lj} \; = \; t_j} or {\sum_{k: k \geq_i j} X_{ik} \; = \; s_i}. {\Box}

We next show that the computations of Eqs.(1) and (2) define monotone maps.

Theorem 7 For {Y \in L_b}, let {f(Y)} be the seller demand-list obtained from Eq.~(1). Then,

\displaystyle Y \geq_b \hat{Y} \;\; \Rightarrow \;\; f(Y) \geq_s f(\hat{Y}).

Proof: Consider seller {i} and buyer {j}. Since {Y \geq_b \hat{Y}}, by definition,

\displaystyle  \sum_{l:l \geq_j i} y_{lj} \; \leq \; 		\sum_{l:l \geq_j i} \hat{y}_{lj}.


\displaystyle  t_j - \sum_{l:l \geq_j i} y_{lj} \; \geq \; 		t_j - \sum_{l:l \geq_j i} \hat{y}_{lj}.

If {j} is {i}‘s most preferred buyer, {(f(Y))_{ij} \geq_i (f(\hat{Y}))_{ij}} follows from the definition. Inductively, we can argue along these lines that {f(Y) \geq_s f(\hat{Y})}. {\Box}


Theorem 8 For {X \in L_s}, let {g(X)} be the buyer demand-list obtained from Eq.~(2). Then,

\displaystyle X \geq_s \hat{X} \;\; \Rightarrow \;\; g(X) \geq_b g(\hat{X}).

Consider the function {h} defined on {L} as follows:

\displaystyle h(X,Y) := (g(Y), f(X)).

We claim that {h} is monotone. Why? Let {(X, Y) \geq (\hat{X}, \hat{Y})}. Then {Y \geq_b \hat{Y}}, and so {g(Y) \geq_s g(\hat{Y})}; similarly, {X \geq_s \hat{X}}, and so {f(X) \geq_b f(\hat{X})}. Thus {h(X,Y) \geq h(\hat{X}, \hat{Y})}. By Tarski’s theorem, {h} has a fixed point. This fixed point is clearly a splittable stable marriage.

One can generalize even further. Let {f} and {g} be two monotone integer valued submodular functions. Define a splittable matching to be any {z \in P(f) \cap P(g)}. Preferences would defined similarly. Each buyer seeks to get at much as possible from their most preferred seller subject to feasibility etc.

3. An LP Formulation

Vandevate identified a collection of linear inequalities that describe the convex hull of stable matchings. For each man {m} and woman {w} let {x_{mw} = 1} if man {m} is matched with woman {w} and zero otherwise.

Then, every stable matching must satisfy the following.

\displaystyle  \sum_{w \in W}x_{mw} = 1\,\, \forall m \in M

\displaystyle \sum_{m \in M}x_{mw} =1\,\, \forall w \in W

\displaystyle \sum_{j \prec_m w}x_{mj} + \sum_{i \prec_w m}x_{iw} + x_{mw} \leq 1\,\, \forall m \in M, w \in W

\displaystyle x_{mw} \geq 0 \,\, \forall m \in M, w \in W

Let {P} be the polyhedron defined by these inequalities.

The first two constraints of {P} ensure that each agent is matched with exactly one other agent of the opposite sex. The third constraint ensures stability by ruling out blocking pairs (call it the blocking constraint). To see why, suppose {\sum_{j \prec_m w}x_{mj} = 1} and {\sum_{i \prec_w m} x_{iw} = 1}. Then man {m} is matched to a woman, {j} that he ranks below {w}. Similarly, woman {w} is matched to a man she ranks below {m}. This would make the pair {(m,w)} a blocking pair.

Lemma 9 Suppose {P \neq \emptyset}. Let {x \in P}. Then, {x_{ij} > 0} implies that

\displaystyle \sum_{j \prec_m w}x_{mj} + \sum_{i \prec_w m}x_{iw} + x_{mw} = 1.

Proof: Consider {\min \{\sum_i \sum_j x_{ij}: x \in P\}}.The dual to this program is

\displaystyle  \max \sum_{i \in M}\alpha_i + \sum_{j \in W}\beta_j - \sum_{i \in M}\sum_{j \in W} \nu_{ij}

subject to

\displaystyle  \alpha_i + \beta_j - \sum_{k \in W: k \succeq_{i} j}\nu_{ik} - \sum_{k \in M: k \succ_{j} i}\nu_{kj} \leq 1\,\, \forall\,\, i \in M \,\,,j \in W

\displaystyle \nu_{ij} \geq 0\,\, \forall\,\, i \in M \,\,,j \in W

Set {\alpha_i = \sum_{j \in W}\nu_{ij}} and {\beta_j = \sum_{i \in M}\nu_{ij}}. Substituting this into the dual constraints yields:

\displaystyle  \sum_{k \in W:k \prec_i j}\nu_{ik} + \sum_{k \in M:k \prec_j i}\nu_{kj} + \nu_{ij} \leq 1\,\, \forall\,\, i \in M \,\,,j \in W.

Choose any {x^* \in P} and set {\nu_{ij} = x^*_{ij}}. This choice of {\nu} is clearly dual feasible. It has an objective function value of {\sum_{i \in M}\sum_{j \in W}x^*_{ij}} and so, is dual optimal. The lemma now follows by complementary slackness.{\Box}

The proof below is due to Sethuraman and Teo.

Theorem 10 Suppose {P \neq \emptyset}. Then, {P} is the convex hull of stable matchings.

Proof: Choose any weight vector {\{c_{ij}\}_{i \in M, j \in W}} and let

\displaystyle x^* \in \arg \max \{\sum_i \sum_j c_{ij}x_{ij}: x \in P\}.

With each member of {M \cup W} we associate an interval {(0,1]}. For each {i \in M}, partition the associated interval {(0,1]_i} into subintervals of length {x^*_{ij}} for all {j \in W}. Arrange these subintervals left to right by by man {i}‘s decreasing preference over {W}. For each woman {j \in W}, partition the associated interval {(0,1]_j} into subintervals of length {x^*_{ij}} for all {i \in M}. Arrange these subintervals left to right by by woman {j}‘s increasing preference over {M}.

Lemma 6 means the subinterval spanned by {x^*_{ij}} in {(0,1]_i} and {(0,1]_j} coincides. Pick a random number {U} uniformly in (0, 1] and construct a semi-matching in the following way:

  1. Match {i \in M} to {k \in W} if {U} lies in the subinterval of {(0,1]_i} spanned by {x^*_{ik}}.
  2. Match {j \in W} to {i \in M} if {U} lies in the subinterval of {(0,1]_j} spanned by {x^*_{ij}}.

By Lemma 6, {i \in M} is matched to {j \in W} iff {j \in W} is matched to {i \in M}. Furthermore, no two men can be matched to the same woman, and similarly no two women can be matched to the same man. So the above semi-matching gives rise to a perfect matching. To show this matching is stable, consider man {i} who is matched to {j \in W} but prefers {k \in W}. Then, in {(0,1]_i}, the subinterval corresponding to {x^*_{ik}} is to the left of the subinterval corresponding to {x^*_{ij}}. Because {i \in M} is matched to {k \in W}, it means that {U} is to the right of the subinterval corresponding to {x^*_{ik}} in {(0,1]_i}. Therefore, {U} is to the left of the subinterval corresponding to {x^*_{ik}} in {(0,1]_j}. In other words, {j \in W} is matched to someone she prefers to {i \in M}. Set {X^U_{ij} = 1} iff man {i} is matched to woman {j} by the randomized scheme above. Then

\displaystyle E(\sum_{i \in M}\sum_{j \in W}c_{ij}X^U_{ij}) = \sum_{i \in M}\sum_{j \in W}c_{ij}E(X^U_{ij})) = \sum_{i \in M}\sum_{j \in W}c_{ij}x^*_{ij}.

Thus, we have exhibited a probability distribution over stable matchings whose expected objective function value coincides with {c \cdot x^*}. It follows then, that there is a stable matching with objective function value {c \cdot x^*}.{\Box}

3.1. A Subgradient Algorithm

The LP approach requires that we know that {P \neq \emptyset}. It would be nice to verify this independently of the proposal algorithm. Here I will outline (not warranted correct) how one can show directly from the LP that {P \neq \emptyset}. If woman { j} is man { i}‘s first choice set { r_{ij} = n}. If woman { j} is man { i}‘s second choice set { r_{ij} = n-1} and so on. Consider:

\displaystyle \max \{\sum_{m \in M}\sum_{w \in W}r_{mw}x_{mw}: x \in P\}.

Let {\nu} be the multiplier associated with the blocking constraints and {p} the multiplier associated with the constraint {\sum_{w \in W}x_{mw} = 1} for all {m}. Consider the following Lagrangean relaxation of the optimization problem:

\displaystyle  \max \sum_{m \in M}\sum_{w \in W} [r_{mw} - p_w - \nu_{mw} - \sum_{k \in W:k \succ_m w } \nu_{mk} - \sum_{k \in M: k \succ_w m} \nu_{km}]x_{mw}

subject to

\displaystyle \sum_{w \in W}x_{mw} = 1\,\, \forall m \in M

\displaystyle  x_{mw} \geq 0\,\, \forall (m,w)

This is easy to solve. For each man {m \in M} choose the woman {w \in W} that maximizes

\displaystyle r_{mw} - p_w - \nu_{mw} - \sum_{k \in W:k \succ_m w } \nu_{mk} - \sum_{k \in M: k \succ_w m} \nu_{km}.

In case of a tie, chooses the top ranked woman.

Initially set {(p, \nu) = 0}. Let {(p^t, \nu^t)} be the value of the multipliers at start of iteration {t}. Denote by {x^t} the optimal choice of {x} given the multipliers {(p^t, \nu^t)}. If {x^t_{mw} = 1} we will say that man {m} proposed to woman {w}. Given {x^t} we adjust {(p^t, \nu^t)} by {(p, \nu)} as follows:

  1. Set {p_j = 0}
  2. Set {\nu_{ij} = x^t_{ij}} for all {i,j}.
  3. Set {(p^{t+1}, \nu^{t+1}) = (p^t, \nu^t) + (p, \nu)}
  4. Set {r^{t+1}_{mw} = r^t_{mw} - x^t_{mw} - \sum_{k \in W:k \succ_m w } x^t_{mk} - \sum_{k \in M: k \succ_w m} x^t_{km}.}

Stop once {x^t} is an assignment. By complementary slackness this must be an optimal stable assignment. Call {r^t} the dual adjusted rank.

Suppose at iteration {t}, {x^t_{mw} = 1}.

Case 1 If among all {i \in M} such that {x^t_{iw} = 1}, man {m} is woman {w}‘s top ranked choice, {r^{t+1}_{mw} = r^t_{mw} - 1}.
Because {i} is woman {w}‘s top ranked choice it follows that {\sum_{k \in M: k \succ_w m} x^t_{km} = 0}. Because {x^t_{mw} = 1} it follows that {\sum_{k \in W:k \succ_m w } x^t_{mk} = 0}.

Case 2 If among all {i \in M} such that {x^t_{iw} = 1}, man {m} is not woman {w}‘s top ranked choice, {r^{t+1}_{mw} \leq r^t_{mw} - 2}.
Because {i} is not woman {w}‘s top ranked choice it follows that {\sum_{k \in M: k \succ_w m} x^t_{km} \geq 1}. Because {x^t_{mw} = 1} it follows that {\sum_{k \in W:k \succ_m w } x^t_{mk} = 0}.

Case 3 If {w \succ_m j} and {x^t_{kj} = 0} for all {k \in M} such that {k \succ_j m}, then {r^{t+1}_{mj} = r^t_{mj}-1}.
Clearly {\sum_{k \in M: k \succ_w m} x^t_{km} = 0} and {x^t_{mj} = 0}. Also, {\sum_{k \in W:k \succ_m w } x^t_{mk} = 1}.

Case 4 If {w \succ_m j} and {x^t_{kj} = 1} for at least one {k \in M} such that {k \succ_j m}, then {r^{t+1}_{mj} \leq r^t_{mj} - 2}.
Clearly {\sum_{k \in M: k \succ_w m} x^t_{km} \geq 1} and {x^t_{mj} = 0}. Also, {\sum_{k \in W:k \succ_m w } x^t_{mk} = 1}.

Case 5 If {j \succ_m w} then, then {r^{t+1}_{mj} = r^t_{mj} - \sum_{k \in M: k \succ_j m} x^t_{km}}.
Clearly, {x^t_{mj} = 0}. Also, {\sum_{k \in W:k \succ_m w } x^t_{mk} = 0}.

Let {P^t = \{w \in W: \sum_{m \in M}x_{mw} \geq 1\}}. I show first that {P^t \subseteq P^{t+1}}. Consider first the case {t=0}. Suppose man {m} proposed to woman {w} and man {m} is woman {w}‘s top choice among those who have proposed. Then the dual adjusted rank of woman {w} by man {m} declines by exactly 1 (see case 1). The dual adjusted rank of all other women decline by at least 1 (case 2 to 4). Note that case 5 does not apply in this iteration. Therefore, in the next iteration man {m} will continue to propose to woman {w}. More generally, any woman who receives a proposal in iteration {0} continues to receive a proposal in iteration {1}. Suppose this holds until iteration {t}. Suppose man {m} proposes to woman {w} in iteration {t} and man {m} is woman {w}‘s top choice among those who have proposed at iteration {t}. By cases (1-4) the dual adjusted rank for all woman ranked below {w} declines by at least 1. Consider a woman {j \succ_m w}. By case 5

\displaystyle r^{t+1}_{mj} = r^t_{mj} - \sum_{k \in M: k \succ_j m} x^t_{km}.

However, {\sum_{k \in M: k \succ_j m} x^t_{km} \geq 1} because any woman who received a proposal before iteration {t} continues to receive a proposal at iteration {t}. Thus the dual adjusted rank of all women ranked above {w} decline by at least 1. Because man {m} did not proposes to these women in iteration {t} their dual adjusted rank must be at least 2 smaller than {r^t_{mw}}. Thus, woman {w} continues to have the highest dual adjusted rank for man {m} and at iteration {t+1}, man {m} will continue to propose to woman {w}.

Next, I show there must be some {t} such that {P^t = W}. If not, then, it must be the case that {P^t = P^{t+1} = \ldots \subset W}. Because {P^t \subset W} there must be a woman {w} who is overdemanded. Consider a man {m} who has proposed to her at iteration {t} who is not her top choice among those proposing to her. Notice the following:

  1. By case 2, {r^{t+1}_{mw} \leq r^t_{mw} - 2.}
  2. By case 5 for all {j \in P^t} such that {w \succ_m j}, {r^{t+1}_{mj} \leq r^t_{mj} -2}.
  3. By case 5, for all {j \in P^t} such that {j \succ_m w}, {r^{t+1}_{mj} \leq r^t_{mj} -1}.
  4. By case 3, for all {j \not \in P^t}, {r^{t+1}_{mj} = r^t_{mj} - 1}. Note, for all such {j} it must be that {w \succ_m j}.

To summarize, the dual adjusted rank of all women {j \in P^t} such that {w \succeq_m j}, goes down by at least 2. That means in all subsequent iterations, man {m} must propose to a woman ranked at or above {w} in {P^t}. Eventually, there is a woman, {w' \in P^t}, say, that man {m} keeps proposing to. This means that the dual adjusted rank of all women in {P^t} declines by at least 2 while the dual adjusted rank of all women outside of {P^t} declines by exactly 1. Eventually, some woman outside of {P^t} must have a dual adjusted rank that is largest and man {m} will propose to her, contradiction.


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