Optimal Account Balancing II

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.

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Tags: Accounting, IOU Graphs, Linear Programming, Operations Research

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4 Responses to “Optimal Account Balancing II”

phorgyphynance Says:
July 28, 2009 at 20:37 | Reply
It would be fun to reformulate this via discrete calculus. Your weight is actually a discrete 1-form. The discrete exterior derivative maps “node vectors” (or discrete 0-forms) to “edge vectors” (or discrete 1-forms).

$\mathrm{d}\, \mathrm{e}_i = \displaystyle \sum_j (\mathrm{e}_{ji} - \mathrm{e}_{ij})$

which can be written as the commutator

$\mathrm{d}\, \mathrm{e}_i = \displaystyle [ A, \mathrm{e}_i ]$

where $A$ is the adjacency matrix.
- Rod Carvalho Says:
  July 28, 2009 at 22:10 | Reply
  Hey Eric,
  
  Is there any problem that you can’t solve using Discrete Calculus? ;-)
  
  Unfortunately, I know nothing of Differential Topology / Geometry, so I am afraid I can’t quite understand your idea. One day I hope to be able to read that long paper you wrote with Urs Schreiber, Discrete Differential Geometry on causal graphs. It looks very enticing…
  - phorgyphynance Says:
    July 29, 2009 at 08:22 | Reply
    That paper is not easy to read, but a much more pedestrian paper that gets the basic idea across is:
    
    Financial Modelling Using Discrete Stochastic Calculus
    
    Who knows? Maybe we can work out some stuff together.
Ashwinkumar Says:
August 11, 2009 at 08:09 | Reply
I propose a simple combinatorial algorithm. Tell me if there are any mistakes. First note that in both the example problems you tried min-money-flow = sum of the absolute values of divergences of nodes. I will try to show that this is the case in all the graphs.

Let $V$ be the sum of the absolute values of the divergences of the nodes. Note that $V$ is a lower bound on the min-money-flow, as in the solution to min-money-flow each node should have either the incoming or outgoing flow greater than or equal to the absolute value of the divergence.

Also note that the sum of divergences is zero. This is because if some one has to gain money, someone else has to pay for it.

Now I will propose an algorithm whose flow value is $V$ and hence gives the optimal flow.

- Divide the set into two sets. $S_1 = \{ i, \text{div}_i (G)\}$ . Delete all nodes with $\text{div}_i (G) = 0$ .

- Let $\text{val}(i)$ = absolute value of divergence of node $i$ .

- Iteratively consider nodes in $S_2$ and do the following two steps.

- Let the node in $S_2$ be $v_2$ . Consider node $v_1$ in $S_1$ . Add an edge from $v_2$ to $v_1$ with $\text{flow} = \min ( \text{val}(v_1), \text{val}(v_2))$ . Do

$\text{val}(v_1)= \text{val}(v_1) - \text{flow}$ and $\text{val}(v_2) = \text{val}(v_2) - \text{flow}$ .

- delete any vertex in $S_1$ or $S_2$ with $\text{val}=0$ .

The properties which I state follow quite obviously. I think you can further modify the algorithm to greatly improve the running time. This algorithm is quite similar to what we usually do when we are in groups. Whoever owes money at the end throw it on the table and from that whoever owes money takes it.

Rod Carvalho

Optimal Account Balancing II

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