Accounting | Rod Carvalho

Archive for the ‘Accounting’ Category

Optimal Account Balancing II

July 28, 2009

Consider a network of $n = 10$ agents, each owing money to and being owed money by other agents. The debts are represented pictorially by the weighted directed graph

which I called an IOU graph on my previous post on this topic. The IOU graph representing this network is an ordered triple $\mathcal{G} = (\mathcal{N}, \mathcal{E}, w)$ , where $\mathcal{N} = \{1, 2, \ldots, 10\}$ is the node set (each node of the IOU graph represents a different agent), $\mathcal{E} \subset \mathcal{N} \times \mathcal{N}$ is the edge set (containing $|\mathcal{E}| = 20$ elements), and $w: \mathcal{E} \rightarrow \mathbb{R}^{+}$ is a weight function that assigns positive weights to each edge in $\mathcal{E}$ . Each directed edge of graph $\mathcal{G}$ is an ordered pair $(i,j)$ , where $i$ is the edge’s tail and $j$ is the edge’s head. If $(i,j) \in \mathcal{E}$ then node $i$ owes money to node $j$ . We do not allow nodes to owe money to themselves and, therefore, for each $i \in \mathcal{N}$ we must have $(i,i) \notin \mathcal{E}$ . The weight of edge $(i, j) \in \mathcal{E}$ is given by $w\left( (i, j) \right) > 0$ , and it tells us the amount of money that node $i$ owes to node $j$ .

Alternatively, the information contained in the IOU graph can be encoded in the following nonnegative matrix

$\left[\begin{array}{cccccccccc} 0 & 10 & 0 & 20 & 20 & 5 & 15 & 0 & 0 & 0\\ 0 & 0 & 5 & 10 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 15 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 10 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 10 & 0 & 0 & 0 & 0\\ 5 & 20 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0\\ 0 & 15 & 0 & 0 & 0 & 0 & 0 & 20 & 0 & 10\\ 0 & 5 & 15 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right]$

where entry $(i,j)$ is the amount of money node $i$ owes to node $j$ . Since we do not allow nodes to owe money to themselves, the entries on the main diagonal of this matrix are zero. We will call this matrix the IOU matrix. Note that the IOU matrix is the adjacency matrix of the IOU graph. Every nonnegative matrix with zero entries on the main diagonal is an admissible IOU matrix, and we can associate with it a corresponding IOU graph.

If the IOU matrix contains all the information about the IOU graph, then we can work with the IOU matrix instead of the IOU graph. Is there any advantage in doing so? I believe there is. In my opinion, the IOU graph is good for conceptual thinking and intuition, while the IOU matrix is a data structure that is convenient for computation. However, the matrix is not the only data structure that one can use to encode the information in an IOU graph. Another possibility would be to use adjacency lists, for example.

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Divergence and IOU Matrices

Suppose we are given an IOU graph $\mathcal{G}_0 = (\mathcal{N}, \mathcal{E}_0, w_0)$ , whose node set is $\mathcal{N} = \{1, 2, \ldots, n\}$ . Note that $(i,i) \notin \mathcal{E}_0$ for each $i \in \mathcal{N}$ , because nodes cannot owe money to themselves. We now introduce variables $y_{ij} \geq 0$ as follows

$y_{ij} = \displaystyle\left\{\begin{array}{cl} w_0 ((i,j)) & \text{if} \quad{} (i,j) \in \mathcal{E}_0\\ 0 & \text{if} \quad{} (i,j) \notin \mathcal{E}_0\\ \end{array}\right.$

and build the (nonnegative) adjacency matrix of the IOU graph $\mathcal{G}_0$

$Y = \left[\begin{array}{ccccc} 0 & y_{12} & y_{13} & \ldots & y_{1n}\\ y_{21} & 0 & y_{23} & \ldots & y_{2n}\\ y_{31} & y_{32} & 0 & \ldots & y_{3n}\\\vdots & \vdots & \vdots & \ddots & \vdots\\ y_{n1} & y_{n2} & y_{n3} & \ldots & 0\\\end{array}\right]$

which is the IOU matrix associated with the IOU graph $\mathcal{G}_0$ . The $(i,j)$ entry of matrix $Y \in \mathbb{R}^{n \times n}$ denotes the amount of money that node $i$ owes to node $j$ . The divergence of node $i \in \mathcal{N}$ is a scalar defined as

$\mbox{div}_i (\mathcal{G}_0) = \displaystyle \sum_{j \neq i} y_{ij} - \displaystyle \sum_{j \neq i} y_{ji}$

and it gives us the amount of money node $i$ owes to other nodes minus the amount of money node $i$ is owed to by other nodes. In terms of the IOU matrix $Y$ , the divergence of node $i \in \mathcal{N}$ is given by

$\mbox{div}_i (\mathcal{G}_0) = e_i^T Y 1_n - e_i^T Y^T 1_n = e_i^T Y 1_n - 1_n^T Y e_i$

where $e_i \in \mathbb{R}^{n}$ is the $i$ -th vector of the canonical basis for $\mathbb{R}^n$ , i.e., the $i$ -th column of the $n \times n$ identity matrix $I_n$ , and $1_n$ is an $n$ -dimensional vector of ones. Let $\mbox{vec}$ denote the vectorization operator, which stacks a given matrix’s columns in one large column vector. Note that if we apply operator $\mbox{vec}$ to a scalar, we obtain the same scalar. Hence, we have that

$e_i^T Y 1_n = \mbox{vec} (e_i^T Y 1_n) = (1_n^T \otimes e_i^T) \mbox{vec} (Y)$

and

$1_n^T Y e_i = \mbox{vec} (1_n^T Y e_i) = (e_i^T \otimes 1_n^T) \mbox{vec} (Y)$

where $\otimes$ denotes the Kronecker product. Therefore, the divergence of node $i \in \mathcal{N}$ can be written as

$\mbox{div}_i (\mathcal{G}_0) = \left[(1_n^T \otimes e_i^T) - (e_i^T \otimes 1_n^T)\right] \mbox{vec} (Y)$ .

We define also the divergence vector of graph $\mathcal{G}_0$ as the $n$ -dimensional vector whose $i$ -th component is $\mbox{div}_i (\mathcal{G}_0)$

$\mbox{div} (\mathcal{G}_0) = \left[\begin{array}{c} (1_n^T \otimes e_1^T) - (e_1^T \otimes 1_n^T)\\ (1_n^T \otimes e_2^T) - (e_2^T \otimes 1_n^T)\\ \vdots\\ (1_n^T \otimes e_n^T) - (e_n^T \otimes 1_n^T)\\\end{array}\right] \mbox{vec} (Y)$ .

Alternatively, we can use equation $\mbox{div}_i (\mathcal{G}_0) = e_i^T Y 1_n - e_i^T Y^T 1_n$ , and write the divergence vector as

$\mbox{div} (\mathcal{G}_0) = I_n Y 1_n - I_n Y^T 1_n = (Y - Y^T) 1_n$

and, since

$I_n Y 1_n = \mbox{vec} (I_n Y 1_n) = (1_n^T \otimes I_n) \mbox{vec} (Y)$

and

$I_n Y^T 1_n = \mbox{vec}(I_n Y^T 1_n) = (1_n^T \otimes I_n) \mbox{vec} (Y^T)$ ,

we obtain

$\mbox{div} (\mathcal{G}_0) = (1_n^T \otimes I_n) \mbox{vec} (Y) - (1_n^T \otimes I_n) \mbox{vec} (Y^T)$ .

Note that $n^2$ -dimensional vectors $\mbox{vec} (Y)$ and $\mbox{vec} (Y^T)$ contain the same entries, but in a different order. Hence, there exists a $n^2 \times n^2$ permutation matrix $P$ such that

$\mbox{vec} (Y^T) = P \mbox{vec} (Y)$

which allows us to write

$\mbox{div} (\mathcal{G}_0) = (1_n^T \otimes I_n) (I_{n^2} - P) \mbox{vec} (Y)$ .

Do note that the permutation matrix depends on $n$ only. It does not depend on the entries of the matrix to which operator $\mbox{vec}$ is applied. Henceforth, we will denote $\tilde{y} = \mbox{vec}(Y)$ . Finally, we have a rather compact equation for the divergence vector

$\mbox{div} (\mathcal{G}_0) = (1_n^T \otimes I_n) (I_{n^2} - P) \tilde{y}$ .

The divergence vector of an IOU graph $\mathcal{G}_0$ tells us how much money each node in the graph will lose after debts have been paid. We say that two IOU graphs (on the same node set) are equi-divergent if their divergence vectors are the same. This means that, although the debt relationships are different, both IOU graphs result in every node getting paid what he is owed and paying what he owes.

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Minimum Money Flow

Suppose we are given an IOU graph $\mathcal{G}_0 = (\mathcal{N}, \mathcal{E}_0, w_0)$ . We would like to find a new IOU graph $\mathcal{G} = (\mathcal{N}, \mathcal{E}, w)$ on the same node set $\mathcal{N}$ , such that the two IOU graphs are equi-divergent, i.e., $\mbox{div} (\mathcal{G}) = \mbox{div} (\mathcal{G}_0)$ . Moreover, we would like the new IOU graph to be “optimal” in some sense. On this post, we will consider the following optimality criterion:

minimum money flow: we want to minimize the total amount of money being exchanged between the nodes. This is the same as minimizing the sum of the weights of the IOU graph $\mathcal{G}$ .

Let us start with matrix $X \in \mathbb{R}^{n \times n}$ , which will be an admissible IOU matrix if we impose two constraints:

nonnegativity: the entries of matrix $X$ must be nonnegative. This is equivalent to imposing the constraint that vector $\tilde{x} = \mbox{vec}(X)$ has nonnegative entries, i.e., $\tilde{x} \geq 0_{n^2}$ , which is equivalent to $(-I_{n^2}) \tilde{x} \leq 0_{n^2}$ .

zero entries on the main diagonal: we must have $x_{ii} = 0$ for all $i \in \mathcal{N}$ . Note that $x_{ii} = e_i^T X e_i = \mbox{vec}(e_i^T X e_i)$ , and therefore $x_{ii} = 0$ is equivalent to $(e_i^T \otimes e_i^T) \tilde{x} = 0$ . Hence, imposing the constraint that the entries on the main diagonal be zero is equivalent to imposing the $n$ equality constraints

$\left[\begin{array}{c} e_1^T \otimes e_1^T\\ e_2^T \otimes e_2^T\\ \vdots\\ e_n^T \otimes e_n^T\\\end{array}\right] \tilde{x} = 0_n$ .

Once these two constraints have been imposed, matrix $X$ is the adjacency matrix of an IOU graph. Finally, we must impose the constraint that the divergence vector of the new IOU graph is the same as the divergence vector of the given IOU graph, i.e., $\mbox{div} (\mathcal{G}) = \mbox{div} (\mathcal{G}_0)$ , which yields

$(1_n^T \otimes I_n) (I_{n^2} - P) \tilde{x} = \mbox{div} (\mathcal{G}_0)$

where $\mbox{div} (\mathcal{G}_0) = (Y - Y^T) 1_n$ can be computed from the given IOU graph. Considering the minimum money flow criterion, the cost function to be minimized is the following

$\displaystyle\sum_{i \in \mathcal{N}} \sum_{j \neq i} x_{ij} = 1_n^T X 1_n = (1_n^T \otimes 1_n^T) \tilde{x} = 1_{n^2}^T \tilde{x}$

and, hence, we obtain the following linear program in $\tilde{x} \in \mathbb{R}^{n^2}$

$\begin{array}{ll} \text{minimize} & \displaystyle 1_{n^2}^T \tilde{x}\\\\ \text{subject to} & (1_n^T \otimes I_n) (I_{n^2} - P) \tilde{x} = \mbox{div} (\mathcal{G}_0)\\\\ & \left[\begin{array}{c} e_1^T \otimes e_1^T\\ e_2^T \otimes e_2^T\\ \vdots\\ e_n^T \otimes e_n^T\\\end{array}\right] \tilde{x} = 0_n\\\\ & (-I_{n^2}) \tilde{x} \leq 0_{n^2}\\\\\end{array}$

which has $2 n$ equality and $n^2$ inequality constraints. One can easily solve this linear program with MATLAB using function linprog.

However, this formulation looks a bit silly. Note that initially we have $n^2$ degrees of freedom (the $n^2$ entries of matrix $X$ ), then we impose $n$ equality constraints $x_{ii} = 0$ to ensure that the entries on the main diagonal are zero, which leaves us with $m = n^2 - n = n (n-1)$ degrees of freedom. If we already know that the entries on the main diagonal are zero, why consider them as decision variables? Wouldn’t it make much more sense to consider only the $m = n (n-1)$ entries off the main diagonal as decision variables?

Let $R$ be a $m \times n^2$ binary matrix such that $\hat{x} = R \tilde{x}$ is the reduced vector of $m$ decision variables, containing solely the entries of $X$ off the main diagonal. We now eliminate the $n$ columns of $n \times n^2$ matrix $(1_n^T \otimes I_n) (I_{n^2} - P)$ which were to be multiplied by the $x_{ii}$ variables. We do so by multiplying matrix $(1_n^T \otimes I_n) (I_{n^2} - P)$ on the right by $R^T$ , which produces a $n \times m$ matrix $(1_n^T \otimes I_n) (I_{n^2} - P) R^T$ .

The new vector of decision variables is $\hat{x} \in \mathbb{R}^m$ . Since we no longer consider the entries on the main diagonal as decision variables, we don’t need to impose the $n$ equality constraints $x_{ii} = 0$ . Hence, we obtain a lower-dimensional linear program

$\begin{array}{ll} \text{minimize} & \displaystyle 1_{m}^T \hat{x}\\\\ \text{subject to} & (1_n^T \otimes I_n) (I_{n^2} - P) R^T \hat{x} = \mbox{div} (\mathcal{G}_0)\\\\ & (-I_m) \hat{x} \leq 0_{m}\\\\\end{array}$

where we now have $n$ equality constraints (divergences at the nodes), and $m$ inequality constrains ( $x_{ij} \geq 0$ ). My intuition tells me that

$(1_n^T \otimes I_n) (I_{n^2} - P) R^T = \left[\begin{array}{c} (1_n^T \otimes e_1^T) - (e_1^T \otimes 1_n^T)\\ (1_n^T \otimes e_2^T) - (e_2^T \otimes 1_n^T)\\ \vdots\\ (1_n^T \otimes e_n^T) - (e_n^T \otimes 1_n^T)\\\end{array}\right] R^T$

is an incidence matrix of the directed complete graph on node set $\mathcal{N}$ . What do you think? Do you agree?

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A Numerical Example

Consider a network of $n = 10$ agents whose debts are represented by the following IOU graph (with $15$ edges)

The IOU matrix associated with this given IOU graph is

$Y = \left[\begin{array}{cccccccccc} 0 & 0 & 0 & 20 & 20 & 5 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 10 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 10 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 10 & 0 & 0 & 0 & 0\\ 5 & 20 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 20 & 0 & 10\\ 0 & 5 & 15 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right]$

Solving the linear program, we obtain a new IOU matrix

$X = \left[\begin{array}{cccccccccc} 0 & 6.097 & 3.924 & 10.776 & 10.776 & 8.427 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0.850 & 0.644 & 1.234 & 1.234 & 1.039 & 0 & 0 & 0 & 0\\ 0 & 1.690 & 1.185 & 2.505 & 2.505 & 2.114 & 0 & 0 & 0 & 0\\ 0 & 4.672 & 3.063 & 7.980 & 7.980 & 6.306 & 0 & 0 & 0 & 0\\ 0 & 1.690 & 1.185 & 2.505 & 2.505 & 2.114 & 0 & 0 & 0 & 0\\\end{array}\right]$

which is the adjacency matrix of a new IOU graph. The given IOU graph and the new IOU graph are equi-divergent. The given IOU graph had $15$ edges and a total of $165$ flowing in it, while the new IOU graph has $25$ edges and a total of $95$ flowing in it. While the flow of money has been reduced, the number of edges (i.e., the number of transactions) has increased dramatically!!

On my previous post on this topic, I gave two very simple examples, and in both of them the minimum money flow solution also yielded the least number of transactions. I wondered back then if that was always the case. This example clearly shows that it is not. Generally speaking, not only does the minimum money flow criterion fail to minimize the number of transactions, but it can actually result in more transactions than the ones in the initial IOU graph. I cannot think of any reason why minimizing the total amount of money flowing in the IOU graph at the cost of a larger number of transactions would be a good idea…

Therefore, the next step is to abandon the minimum money flow criterion, and focus solely on the least number of transactions. Minimizing the number of transactions is a hard combinatorial problem, but maybe there are approximation algorithms that make the problem tractable.

Tags:Accounting, IOU Graphs, Linear Programming, Operations Research
Posted in Accounting, Operations Research | 4 Comments »

Optimal Account Balancing

June 20, 2009

Suppose that Alice owes Bob $5 and is owed $10 by Bob. These debts can be represented as the weighted directed graph

which we will call an IOU graph. Alice pays Bob $5 and Bob pays Alice $10. The debts are paid and $15 have been exchanged in a total of two transactions. Let us now consider the following cash flow diagram

and, accounting for the net cash flows, we obtain

which illustrates that if Bob paid Alice $5, then the debts would still be paid, though only $5 would be exchanged and only one transaction would be made. This looks much cleverer and more efficient than exchanging $15 in a total of two transactions. Hence, the following IOU graph

is “equivalent” to the original IOU graph. In both IOU graphs, if debts are paid then Alice will be $5 up, while Bob will be $5 down.

We add a third agent, Charlie. Consider now the IOU graph

where:

Alice must pay a total of $20 and be paid a total of $30.

Bob must pay a total of $15 and be paid a total of $20.

Charlie must pay a total of $35 and be paid a total of $20.

Hence, after the debts are paid, Alice will have $30 – $20 = $10 more in her account, Bob will have $20 – $15 = $5 more in his account, and Charlie will have $20 – $35 = -$15 more (= $15 less) in his account. Note that $10 + $5 – $15 = $0, i.e., money is conserved. In total, $70 have been exchanged in six transactions. We can do better than that. For example, we could pay all debts in just three transactions, as illustrated in the following IOU graph

where a total of $20 are exchanged. Note that Alice is $10 up, Bob is $5 up, and Charlie is $15 down, just like before. We can do even better than this. If Charlie pays Bob’s debt to Alice, then all debts can be paid in just two transactions as shown in the IOU graph below

where only $15 are exchanged. This is an “optimal” debt-paying scheme because the debts cannot be paid in less than two transactions, nor exchanging less than $15. Do you agree?

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Debts and IOU Graphs

An IOU graph is a weighted directed graph given by an ordered triple $\mathcal{G} = (\mathcal{N}, \mathcal{E}, w)$ , where:

$\mathcal{N}$ is the node set. Each node of the IOU graph represents a different agent.

$\mathcal{E} \subset \mathcal{N} \times \mathcal{N}$ is the edge set. Each directed edge is an ordered pair $(i,j)$ , where $i$ is the edge’s tail and $j$ is the edge’s head. If $(i,j) \in \mathcal{E}$ then node $i$ owes money to node $j$ . We do not allow nodes to owe money to themselves and, therefore, for each $i \in \mathcal{N}$ we must have $(i,i) \notin \mathcal{E}$ .

$w: \mathcal{E} \rightarrow \mathbb{R}^{+}$ is a weight function that assigns positive weights to each edge in $\mathcal{E}$ . The weight of edge $(i, j) \in \mathcal{E}$ is given by $w\left( (i, j) \right) > 0$ , and it tells us the amount of money that node $i$ owes to node $j$ .

We define the in-neighborhood and out-neighborhood of node $i$ as follows

$\begin{array}{ll} \mathcal{N}_i^{in} &= \displaystyle\{ j \in \mathcal{N} \mid (j,i) \in \mathcal{E}\}\\\\ \mathcal{N}_i^{out} &= \displaystyle\{ j \in \mathcal{N} \mid (i,j) \in \mathcal{E}\}\end{array}$

and illustrate them below

The divergence [1] of node $i$ of graph $\mathcal{G}$ is a scalar defined as

$\textbf{div}_i (\mathcal{G}) = \displaystyle \sum_{j \in \mathcal{N}_i^{out}} x_{ij} - \displaystyle \sum_{j \in \mathcal{N}_i^{in}} x_{ji}$

and it gives us the amount of money node $i$ owes to other nodes minus the amount of money node $i$ is owed to by other nodes. The divergence of a node can be viewed as the discrete, graph-theoretic analogue of the divergence operator in vector calculus. We define also the divergence vector of graph $\mathcal{G}$ as the vector whose $i$ -th component is $\textbf{div}_i (\mathcal{G})$

$\textbf{div} (\mathcal{G}) = \left[\begin{array}{c} \vdots\\ \textbf{div}_i (\mathcal{G})\\ \vdots \\ \end{array}\right]$ .

The divergence vector of an IOU graph $\mathcal{G}$ tells us how much money each node in the graph will lose after debts have been paid. We say that two IOU graphs (on the same node set) are equi-divergent if their divergence vectors are the same. This means that, although the debt relationships are different, both IOU graphs result in every node getting paid what he is owed and paying what he owes.

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Optimal Account Balancing

$\textbf{div} (\mathcal{G}) = \textbf{div} (\mathcal{G}_0)$ .

Moreover, we would like the new IOU graph to be “optimal” in some sense. I can think of the two following optimality criteria:

minimum money flow: we want to minimize the total amount of money being exchanged between the nodes. This is the same as minimizing the sum of the (positive) weights of the edges of the IOU graph.

least number of transactions: we want to minimize the total number of transactions being made. This is the same as minimizing the number of edges in the IOU graph, i.e., maximizing the sparsity of the IOU graph.

Can you think of other interesting optimality criteria?

It would be reasonable to consider the case where the transaction fees are nonzero. If these are variable and proportional to the amount of money being transferred, then we would like to minimize the amount of money flowing in the network of agents. If the transaction fees are fixed, then we would like to minimize the total number of transactions being made. For example, in the two examples above the minimum money flow solution is the same as the least number of transactions solution. Is this always the case?

In order to find the optimal IOU graph, we start with a weighted directed graph on node set $\mathcal{N}$ whose edge set is the one of the directed complete graph on $\mathcal{N}$ . Hence, every two distinct nodes are connected by two directed edges with opposite directions. The weight of edge $(i,j)$ is denoted by $x_{ij} \geq 0$ . Since some of these weights might be zero, there might be edges in this graph which do not represent debt relationships! Therefore, although this graph is weighted and directed, it is not necessarily an IOU graph. We optimize for the $n (n-1)$ nonnegative variables $\{x_{ij}\}$ and, in the end, we remove the edges whose weight is zero so that we obtain an IOU graph. I view this approach as somewhat akin to scaffolding ;-)

For example, considering the minimum money flow criterion, we have the following minimum cost flow problem [1]

$\begin{array}{ll} \text{minimize} & \displaystyle \sum_{i \in \mathcal{N}}\sum_{j \neq i} x_{ij}\\\\ \text{subject to} & \displaystyle \sum_{j \neq i} x_{ij} - \displaystyle \sum_{j \neq i} x_{ji} = \textbf{div}_i (\mathcal{G}_0), \quad{} \forall i \in \mathcal{N}\\\\ & x_{ij} \geq 0, \quad{} \forall i \neq j \in \mathcal{N}\\\\\end{array}$

which can be solved via linear programming [2]. In words: we want to find the debts $x_{ij} \geq 0$ such that all debts are paid (each node should be paid what it is owed minus what it owes to other nodes) and the total amount of debt is minimized, which results in the total amount of money flowing to be also minimized.

The least number of transactions criterion is much harder to tackle because we want to maximize the sparsity of the graph, while ensuring that the graph and the original IOU graph equi-divergent. We want to optimize the topology of the graph and, hence, a combinatorial flavor is added to the problem. I will write more about it on future posts.

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A Numerical Example

Let us again consider the $3$ -agent example where the agents are Alice, Bob and Charlie. The IOU graph is

where the node set is $\mathcal{N} = \{\text{a}, \text{b}, \text{c}\}$ . We introduce vector

$x = (x_{ab}, x_{ac}, x_{ba}, x_{bc}, x_{ca}, x_{cb})$

where $x_{ij} \geq 0$ is the amount of money node $i$ owes to node $j$ . The divergence of the new graph at each node must be the same as the divergence of the given IOU graph at each node, which yields the equality constraint $C x = b$ , where

$C = \left[\begin{array}{rrrrrr} -1 & -1 & 1 & 0 & 1 & 0\\ 1 & 0 & -1 & -1 & 0 & 1\\ 0 & 1 & 0 & 1 & -1 & -1\\\end{array}\right]$

is the incidence matrix of the graph, and $b = [\begin{array}{ccc} 10 & 5 & -15\end{array}]^T$ is the opposite of the divergence vector of the given IOU graph above. This equality constraint guarantees that the given IOU graph and the new graph we obtain are equi-divergent.

Let $I_6$ be the $6 \times 6$ identity matrix, let $1_6$ be the $6$ -dimensional vector of ones, and let $0_6$ be the $6$ -dimensional vector of zeros. The $6$ nonnegativity constraints $x_{ij} \geq 0$ can be written in matrix form as $(-I_6) x \leq 0_6$ . Considering the minimum money flow criterion, we obtain the linear program

$\begin{array}{ll} \text{minimize} & 1_6^T x\\\\ \text{subject to} & C x = b, \quad{} (-I_6) x \leq 0_6\\\\\end{array}$

Note that the equality constraint $C x = b$ is equivalent to two inequality constraints: $C x \leq b$ and $C x \geq b$ . Thus, we can transform the linear program above into a linear program with inequality constraints only

$\begin{array}{ll} \text{minimize} & 1_6^T x\\\\ \text{subject to} & \left[\begin{array}{r} C\\ -C\\ -I_6\\\end{array}\right] x \leq \left[\begin{array}{r} b\\ -b\\ 0_6\\\end{array}\right]\\\\\end{array}$

If you have MATLAB, you can easily solve these linear programs using function linprog. The following Python 2.5 script solves the latter linear program using CVXOPT:

from cvxopt import matrix
from cvxopt import solvers

# -------------------
# auxiliary matrices
# -------------------

# defines incidence matrix
C = matrix([ [-1.0, 1.0, 0.0],
             [-1.0, 0.0, 1.0],
             [1.0, -1.0, 0.0],
             [0.0, -1.0, 1.0],
             [1.0, 0.0, -1.0],
             [0.0, 1.0, -1.0] ])

# defines desired divergence vector
d = matrix( [-10.0, -5.0, 15.0] )

# defines 6x6 identity matrix
Id = matrix([ [1, 0, 0, 0, 0, 0],
              [0, 1, 0, 0, 0, 0],
              [0, 0, 1, 0, 0, 0],
              [0, 0, 0, 1, 0, 0],
              [0, 0, 0, 0, 1, 0],
              [0, 0, 0, 0, 0, 1] ])

# ---------------
# linear program
# ---------------

# defines objective vector
c = matrix(1.0, (6,1))

# defines inequality constraint matrices
G = matrix( [C, -C, -Id] )
h = matrix( [-d, d, matrix(0.0, (6,1))] )

# solves linear program
solvers.options['show_progress'] = True
solution = solvers.lp(c, G, h)

# prints solution
print solution['x']

The solution is

$x = \left[\begin{array}{r}5.86\text{e-}008\\ -1.50\text{e-}008\\ -1.36\text{e-}008\\ -1.99\text{e-}008\\ 1.00\text{e+}001\\ 5.00\text{e+}000\\\end{array}\right]$ ,

which is “contaminated” with numerical noise. Adjusting the tolerances of the numerical solver, one can reduce the numerical noise and obtain the true solution, which is $x = (0, 0, 0, 0, 10, 5)$ . Removing the edges with zero weight, we obtain an IOU graph with 2 edges

and, hence, only two transactions are required, and only $15 need to be exchanged. This is the exact same solution we obtained previously via heuristic methods! :-)

__________

Concluding Remarks

The two optimal debt-paying schemes considered on this post rely on an “accountant” who has perfect global information about all the debts in the network of agents. This accountant solves an optimization problem to find out what transactions should be carried out. This is a centralized approach, as the decision power is centralized in the accountant. Another (and arguably more interesting) approach would be a decentralized one, where the nodes have only local information (i.e., each node only knows who owes him money and to whom he owes money) and where each node solves a local optimization problem using, for example, message-passing algorithms. I will write about it on future posts.

The only modeling of this problem I have found was on Demetri Spanos‘ PhD thesis [3]. If you know of any papers on this topic, please let me know.

Last but not least, allow me to add a personal note: I first started thinking of this problem years ago, while going to dinners with friends and trying to figure out how to split the bill in an efficient manner. The real world is a great source of inspiration when it comes to interesting problems.

__________

References

[1] Dimitri Bertsekas, Network Optimization: Continuous and Discrete Models, Athena Scientific, 1998.

[2] Stephen Boyd and Lieven Vandenberghe, Convex Optimization, Cambridge University Press, 2004.

[3] Demetri Spanos, Distributed gradient systems and dynamic coordination, Ph.D. thesis, California Institute of Technology, December 2005.

Tags:Accounting, CVXOPT, IOU Graphs, Linear Programming, Operations Research
Posted in Accounting, Operations Research, Python | 21 Comments »

Rod Carvalho

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Rod Carvalho

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