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# Von Neumann Growth Model (and a Generalization)¶

Co-author: Balint Szoke

This notebook uses the class Neumann to calculate key objects of a linear growth model of John von Neumann [vN37] that was generalized by Kemeny, Morgenstern and Thompson [KMT56].

Objects of interest are the maximal expansion rate ($\alpha$), the interest factor ($β$), and the optimal intensities ($x$) and prices ($p$).

In addition to watching how the towering mind of John von Neumann formulated an equilibrium model of price and quantity vectors in balanced growth, this notebook shows how fruitfully to employ the following important tools:

• a zero-sum two-player game
• linear programming
• the Perron-Frobenius theorem

We’ll begin with some imports:

In [1]:
import numpy as np
import matplotlib.pyplot as plt
from scipy.linalg import solve
from scipy.optimize import fsolve, linprog
from textwrap import dedent
%matplotlib inline

np.set_printoptions(precision=2)


The code below provides the Neumann class

In [2]:
class Neumann(object):

"""
This class describes the Generalized von Neumann growth model as it was
discussed in Kemeny et al. (1956, ECTA) and Gale (1960, Chapter 9.5):

Let:
n ... number of goods
m ... number of activities
A ... input matrix is m-by-n
a_{i,j} - amount of good j consumed by activity i
B ... output matrix is m-by-n
b_{i,j} - amount of good j produced by activity i

x ... intensity vector (m-vector) with non-negative entries
x'B - the vector of goods produced
x'A - the vector of goods consumed
p ... price vector (n-vector) with non-negative entries
Bp - the revenue vector for every activity
Ap - the cost of each activity

Both A and B have non-negative entries. Moreover, we assume that
(1) Assumption I (every good which is consumed is also produced):
for all j, b_{.,j} > 0, i.e. at least one entry is strictly positive
(2) Assumption II (no free lunch):
for all i, a_{i,.} > 0, i.e. at least one entry is strictly positive

Parameters
----------
A : array_like or scalar(float)
Part of the state transition equation.  It should be n x n
B : array_like or scalar(float)
Part of the state transition equation.  It should be n x k
"""

def __init__(self, A, B):

self.A, self.B = list(map(self.convert, (A, B)))
self.m, self.n = self.A.shape

# Check if (A, B) satisfy the basic assumptions
assert self.A.shape == self.B.shape, 'The input and output matrices \
must have the same dimensions!'
assert (self.A >= 0).all() and (self.B >= 0).all(), 'The input and \
output matrices must have only non-negative entries!'

# (1) Check whether Assumption I is satisfied:
if (np.sum(B, 0) <= 0).any():
self.AI = False
else:
self.AI = True

# (2) Check whether Assumption II is satisfied:
if (np.sum(A, 1) <= 0).any():
self.AII = False
else:
self.AII = True

def __repr__(self):
return self.__str__()

def __str__(self):

me = """
Generalized von Neumann expanding model:
- number of goods          : {n}
- number of activities     : {m}

Assumptions:
- AI:  every column of B has a positive entry    : {AI}
- AII: every row of A has a positive entry       : {AII}

"""
# Irreducible                                       : {irr}
return dedent(me.format(n=self.n, m=self.m,
AI=self.AI, AII=self.AII))

def convert(self, x):
"""
Convert array_like objects (lists of lists, floats, etc.) into
well-formed 2D NumPy arrays
"""
return np.atleast_2d(np.asarray(x))

def bounds(self):
"""
Calculate the trivial upper and lower bounds for alpha (expansion rate)
and beta (interest factor). See the proof of Theorem 9.8 in Gale (1960)
"""

n, m = self.n, self.m
A, B = self.A, self.B

f = lambda α: ((B - α * A) @ np.ones((n, 1))).max()
g = lambda β: (np.ones((1, m)) @ (B - β * A)).min()

UB = np.asscalar(fsolve(f, 1))  # Upper bound for α, β
LB = np.asscalar(fsolve(g, 2))  # Lower bound for α, β

return LB, UB

def zerosum(self, γ, dual=False):
"""
Given gamma, calculate the value and optimal strategies of a
two-player zero-sum game given by the matrix

M(gamma) = B - gamma * A

Row player maximizing, column player minimizing

Zero-sum game as an LP (primal --> α)

max (0', 1) @ (x', v)
subject to
[-M', ones(n, 1)] @ (x', v)' <= 0
(x', v) @ (ones(m, 1), 0) = 1
(x', v) >= (0', -inf)

Zero-sum game as an LP (dual --> beta)

min (0', 1) @ (p', u)
subject to
[M, -ones(m, 1)] @ (p', u)' <= 0
(p', u) @ (ones(n, 1), 0) = 1
(p', u) >= (0', -inf)

Outputs:
--------
value: scalar
value of the zero-sum game

strategy: vector
if dual = False, it is the intensity vector,
if dual = True, it is the price vector
"""

A, B, n, m = self.A, self.B, self.n, self.m
M = B - γ * A

if dual == False:
# Solve the primal LP (for details see the description)
# (1) Define the problem for v as a maximization (linprog minimizes)
c = np.hstack([np.zeros(m), -1])

# ... non-negativity constraints
bounds = tuple(m * [(0, None)] + [(None, None)])
# ... inequality constraints
A_iq = np.hstack([-M.T, np.ones((n, 1))])
b_iq = np.zeros((n, 1))
# ... normalization
A_eq = np.hstack([np.ones(m), 0]).reshape(1, m + 1)
b_eq = 1

res = linprog(c, A_ub=A_iq, b_ub=b_iq, A_eq=A_eq, b_eq=b_eq,
bounds=bounds, options=dict(bland=True, tol=1e-7))

else:
# Solve the dual LP (for details see the description)
# (1) Define the problem for v as a maximization (linprog minimizes)
c = np.hstack([np.zeros(n), 1])

# ... non-negativity constraints
bounds = tuple(n * [(0, None)] + [(None, None)])
# ... inequality constraints
A_iq = np.hstack([M, -np.ones((m, 1))])
b_iq = np.zeros((m, 1))
# ... normalization
A_eq = np.hstack([np.ones(n), 0]).reshape(1, n + 1)
b_eq = 1

res = linprog(c, A_ub=A_iq, b_ub=b_iq, A_eq=A_eq, b_eq=b_eq,
bounds=bounds, options=dict(bland=True, tol=1e-7))

if res.status != 0:
print(res.message)

# Pull out the required quantities
value = res.x[-1]
strategy = res.x[:-1]

return value, strategy

def expansion(self, tol=1e-8, maxit=1000):
"""
The algorithm used here is described in Hamburger-Thompson-Weil
(1967, ECTA). It is based on a simple bisection argument and utilizes
the idea that for a given γ (= α or β), the matrix "M = B - γ * A"
defines a two-player zero-sum game, where the optimal strategies are
the (normalized) intensity and price vector.

Outputs:
--------
alpha: scalar
optimal expansion rate
"""

LB, UB = self.bounds()

for iter in range(maxit):

γ = (LB + UB) / 2
ZS = self.zerosum(γ=γ)
V = ZS[0]     # value of the game with γ

if V >= 0:
LB = γ
else:
UB = γ

if abs(UB - LB) < tol:
γ = (UB + LB) / 2
x = self.zerosum(γ=γ)[1]
p = self.zerosum(γ=γ, dual=True)[1]
break

return γ, x, p

def interest(self, tol=1e-8, maxit=1000):
"""
The algorithm used here is described in Hamburger-Thompson-Weil
(1967, ECTA). It is based on a simple bisection argument and utilizes
the idea that for a given gamma (= alpha or beta),
the matrix "M = B - γ * A" defines a two-player zero-sum game,
where the optimal strategies are the (normalized) intensity and price
vector

Outputs:
--------
beta: scalar
optimal interest rate
"""

LB, UB = self.bounds()

for iter in range(maxit):
γ = (LB + UB) / 2
ZS = self.zerosum(γ=γ, dual=True)
V = ZS[0]

if V > 0:
LB = γ
else:
UB = γ

if abs(UB - LB) < tol:
γ = (UB + LB) / 2
p = self.zerosum(γ=γ, dual=True)[1]
x = self.zerosum(γ=γ)[1]
break

return γ, x, p


## Notation¶

We use the following notation.

$\mathbf{0}$ denotes a vector of zeros. We call an $n$-vector - positive or $x\gg \mathbf{0}$ if $x_i>0$ for all $i=1,2,\dots,n$

• non-negative or $x\geq \mathbf{0}$ if $x_i\geq 0$ for all $i=1,2,\dots,n$ - semi-positive or $x > \mathbf{0}$ if $x\geq \mathbf{0}$ and $x\neq \mathbf{0}$.

For two conformable vectors $x$ and $y$, $x\gg y$, $x\geq y$ and $x> y$ mean $x-y\gg \mathbf{0}$, $x-y \geq \mathbf{0}$, and $x-y > \mathbf{0}$.

By default, all vectors are column vectors, $x^{T}$ denotes the transpose of $x$ (i.e. a row vector).

Let $\iota_n$ denote a column vector composed of $n$ ones, i.e. $\iota_n = (1,1,\dots,1)^T$.

Let $e^i$ denote the vector (of arbitrary size) containing zeros except for the $i$ th position where it is one.

We denote matrices by capital letters. For an arbitrary matrix $A$, $a_{i,j}$ represents the entry in its $i$ th row and $j$ th column.

$a_{\cdot j}$ and $a_{i\cdot}$ denote the $j$ th column and $i$ th row of $A$, respectively.

## Model Ingredients and Assumptions¶

A pair $(A,B)$ of $m\times n$ non-negative matrices defines an economy.

• $m$ is the number of activities (or sectors)
• $n$ is the number of goods (produced and/or used in the economy)
• $A$ is called the input matrix; $a_{i,j}$ denotes the amount of good $j$ consumed by activity $i$
• $B$ is called the output matrix; $b_{i,j}$ represents the amount of good $j$ produced by activity $i$

Two key assumptions restrict economy $(A,B)$:

• Assumption I: (every good which is consumed is also produced)
$$b_{.,j} > \mathbf{0}\hspace{5mm}\forall j=1,2,\dots,n$$
• Assumption II: (no free lunch)
$$a_{i,.} > \mathbf{0}\hspace{5mm}\forall i=1,2,\dots,m$$

A semi-positive $m$-vector:math:x denotes the levels at which activities are operated (intensity vector).

Therefore,

• vector $x^TA$ gives the total amount of goods used in production
• vector $x^TB$ gives total outputs

An economy $(A,B)$ is said to be productive, if there exists a non-negative intensity vector $x \geq 0$ such that $x^T B > x^TA$.

The semi-positive $n$-vector $p$ contains prices assigned to the $n$ goods.

The $p$ vector implies cost and revenue vectors

• the vector $Ap$ tells costs of the vector of activities
• the vector $Bp$ tells revenues from the vector of activities

A property of an input-output pair $(A,B)$ called irreducibility (or indecomposability) determines whether an economy can be decomposed into multiple ‘’sub-economies’’.

Definition: Given an economy $(A,B)$, the set of goods $S\subset \{1,2,\dots,n\}$ is called an independent subset if it is possible to produce every good in $S$ without consuming any good outside $S$. Formally, the set $S$ is independent if $\exists T\subset \{1,2,\dots,m\}$ (subset of activities) such that $a_{i,j}=0$, $\forall i\in T$ and $j\in S^c$ and for all $j\in S$, $\exists i\in T$, s.t. $b_{i,j}>0$. The economy is irreducible if there are no proper independent subsets.

We study two examples, both coming from Chapter 9.6 of Gale [Gal89]

In [3]:
# (1) Irreducible (A, B) example: α_0 = β_0
A1 = np.array([[0, 1, 0, 0],
[1, 0, 0, 1],
[0, 0, 1, 0]])

B1 = np.array([[1, 0, 0, 0],
[0, 0, 2, 0],
[0, 1, 0, 1]])

# (2) Reducible (A, B) example: β_0 < α_0
A2 = np.array([[0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 1, 0, 0, 1],
[0, 0, 0, 0, 1, 0]])

B2 = np.array([[1, 0, 0, 1, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 2, 0],
[0, 0, 0, 1, 0, 1]])


The following code sets up our first Neumann economy or Neumann instance

In [4]:
n1 = Neumann(A1, B1)
n1

Out[4]:
Generalized von Neumann expanding model:
- number of goods          : 4
- number of activities     : 3

Assumptions:
- AI:  every column of B has a positive entry    : True
- AII: every row of A has a positive entry       : True

In [5]:
n2 = Neumann(A2, B2)
n2

Out[5]:
Generalized von Neumann expanding model:
- number of goods          : 6
- number of activities     : 5

Assumptions:
- AI:  every column of B has a positive entry    : True
- AII: every row of A has a positive entry       : True


## Dynamic Interpretation¶

Attach a time index $t$ to the preceding objects, regard an economy as a dynamic system, and study sequences

$$\{(A_t,B_t)\}_{t\geq 0}, \hspace{1cm}\{x_t\}_{t\geq 0},\hspace{1cm} \{p_t\}_{t\geq 0}$$

An interesting special case holds the technology process constant and investigates the dynamics of quantities and prices only.

Accordingly, in the rest of this notebook, we assume that $(A_t,B_t)=(A,B)$ for all $t\geq 0$.

A crucial element of the dynamic interpretation involves the timing of production.

We assume that production (consumption of inputs) takes place in period $t$, while the associated output materializes in period $t+1$, i.e. consumption of $x_{t}^TA$ in period $t$ results in $x^T_{t}B$ amounts of output in period $t+1$.

These timing conventions imply the following feasibility condition:

\begin{aligned} x^T_{t}B \geq x^T_{t+1} A \hspace{1cm}\forall t\geq 1 \end{aligned}

which asserts that no more goods can be used today than were produced yesterday.

Accordingly, $Ap_t$ tells the costs of production in period $t$ and $Bp_t$ tells revenues in period $t+1$.

### Balanced Growth¶

We follow John von Neumann in studying “balanced growth”.

Let $./$ denote an elementwise division of one vector by another and let $\alpha >0$ be a scalar.

Then balanced growth is a situation in which

$$x_{t+1}./x_t = \alpha , \quad \forall t \geq 0$$

With balanced growth, the law of motion of $x$ is evidently $x_{t+1}=\alpha x_t$ and so we can rewrite the feasibility constraint as

$$x^T_{t}B \geq \alpha x^T_t A \hspace{1cm}\forall t$$

In the same spirit, define $\beta\in\mathbb{R}$ as the interest factor per unit of time.

We assume that it is always possible to earn a gross return equal to the constant interest factor $\beta$ by investing “outside the model”.

Under this assumption about outside investment opportunities, a no-arbitrage condition gives rise to the following (no profit) restriction on the price sequence:

$$\beta Ap_{t} \geq B p_{t} \hspace{1cm}\forall t$$

This says that production cannot yield a return greater than that offered by the investment opportunity (note that we compare values in period $t+1$).

The balanced growth assumption allows us to drop time subscripts and conduct an analysis purely in terms of a time-invariant growth rate $\alpha$ and interest factor $\beta$.

## Duality¶

The following two problems are connected by a remarkable dual relationship between the technological and valuation characteristics of the economy:

Definition: The technological expansion problem (TEP) for the economy $(A,B)$ is to find a semi-positive $m$-vector $x>0$ and a number $\alpha\in\mathbb{R}$, s.t.

\begin{aligned} &\max_{\alpha} \hspace{2mm} \alpha\\ &\text{s.t. }\hspace{2mm}x^T B \geq \alpha x^T A \end{aligned}

Theorem 9.3 of David Gale’s book [Gal89] asserts that if Assumptions I and II are both satisfied, then a maximum value of $\alpha$ exists and it is positive.

It is called the technological expansion rate and is denoted by $\alpha_0$. The associated intensity vector $x_0$ is the optimal intensity vector.

Definition: The economical expansion problem (EEP) for $(A,B)$ is to find a semi-positive $n$-vector $p>0$ and a number $\beta\in\mathbb{R}$, such that

\begin{aligned} &\min_{\beta} \hspace{2mm} \beta\\ &\text{s.t. }\hspace{2mm}Bp \leq \beta Ap \end{aligned}

Assumptions I and II imply existence of a minimum value $\beta_0>0$ called the economic expansion rate.

The corresponding price vector $p_0$ is the optimal price vector.

Evidently, the criterion functions in technological expansion problem and the economical expansion problem are both linearly homogeneous, so the optimality of $x_0$ and $p_0$ are defined only up to a positive scale factor.

For simplicity (and to emphasize a close connection to zero-sum games), in the following, we normalize both vectors $x_0$ and $p_0$ to have unit length.

A standard duality argument (see Lemma 9.4. in (Gale, 1960) [Gal89]) implies that under Assumptions I and II, $\beta_0\leq \alpha_0$.

But in the other direction, that is $\beta_0\geq \alpha_0$, Assumptions I and II are not sufficient.

Nevertheless, von Neumann [vN37] proved the following remarkable “duality-type” result connecting TEP and EEP.

Theorem 1 (von Neumann): If the economy $(A,B)$ satisfies Assumptions I and II, then there exists a set $\left(\gamma^{*}, x_0, p_0\right)$, where $\gamma^{*}\in[\beta_0, \alpha_0]\subset\mathbb{R}$, $x_0>0$ is an $m$-vector, $p_0>0$ is an $n$-vector and the following holds true

\begin{aligned} x_0^T B &\geq \gamma^{* } x_0^T A \\ Bp_0 &\leq \gamma^{* } Ap_0 \\ x_0^T\left(B-\gamma^{* } A\right)p_0 &= 0 \end{aligned}

Proof (Sketch): Assumption I and II imply that there exist $(\alpha_0, x_0)$ and $(\beta_0, p_0)$ solving the TEP and EEP, respectively. If $\gamma^*>\alpha_0$, then by definition of $\alpha_0$, there cannot exist a semi-positive $x$ that satisfies $x^T B \geq \gamma^{* } x^T A$. Similarly, if $\gamma^*<\beta_0$, there is no semi-positive $p$ so that $Bp \leq \gamma^{* } Ap$. Let $\gamma^{* }\in[\beta_0, \alpha_0]$, then $x_0^T B \geq \alpha_0 x_0^T A \geq \gamma^{* } x_0^T A$. Moreover, $Bp_0\leq \beta_0 A p_0\leq \gamma^* A p_0$. These two inequalities imply $x_0\left(B - \gamma^{* } A\right)p_0 = 0$.

Here the constant $\gamma^{*}$ is both expansion and interest factor (not necessarily optimal).

We have already encountered and discussed the first two inequalities that represent feasibility and no-profit conditions.

Moreover, the equality compactly captures the requirements that if any good grows at a rate larger than $\gamma^{*}$ (i.e., if it is oversupplied), then its price must be zero; and that if any activity provides negative profit, it must be unused.

Therefore, these expressions encode all equilibrium conditions and Theorem I essentially states that under Assumptions I and II there always exists an equilibrium $\left(\gamma^{*}, x_0, p_0\right)$ with balanced growth.

Note that Theorem I is silent about uniqueness of the equilibrium. In fact, it does not rule out (trivial) cases with $x_0^TBp_0 = 0$ so that nothing of value is produced.

To exclude such uninteresting cases, Kemeny, Morgenstern and Thomspson [KMT56] add an extra requirement

$$x^T_0 B p_0 > 0$$

and call the resulting equilibria economic solutions.

They show that this extra condition does not affect the existence result, while it significantly reduces the number of (relevant) solutions.

## Interpretation as a Game Theoretic Problem (Two-player Zero-sum Game)¶

To compute the equilibrium $(\gamma^{*}, x_0, p_0)$, we follow the algorithm proposed by Hamburger, Thompson and Weil (1967), building on the key insight that the equilibrium (with balanced growth) can be considered as a solution of a particular two-player zero-sum game. First, we introduce some notations.

Consider the $m\times n$ matrix $C$ as a payoff matrix, with the entries representing payoffs from the minimizing column player to the maximizing row player and assume that the players can use mixed strategies: - row player chooses the $m$-vector $x > \mathbf{0}$, s.t. $\iota_m^T x = 1$ - column player chooses the $n$-vector $p > \mathbf{0}$, s.t. $\iota_n^T p = 1$.

Definition: The $m\times n$ matrix game $C$ has the solution $(x^*, p^*, V(C))$ in mixed strategies, if

\begin{aligned} (x^* )^T C e^j \geq V(C)\quad \forall j\in\{1, \dots, n\}\quad \quad \text{and}\quad\quad (e^i)^T C p^* \leq V(C)\quad \forall i\in\{1, \dots, m\} \end{aligned}

The number $V(C)$ is called the value of the game.

From the above definition, it is clear that the value $V(C)$ has two alternative interpretations:

• by playing the appropriate mixed stategy, the maximizing player can assure himself at least $V(C)$ (no matter what the column player chooses)
• by playing the appropriate mixed stategy, the minimizing player can make sure that the maximizing player will not get more than $V(C)$ (irrespective of what is the maximizing player’s choice)

From the famous theorem of Nash (1951), it follows that there always exists a mixed strategy Nash equilibrium for any finite two-player zero-sum game.

Moreover, von Neumann’s Minmax Theorem [vN28] implies that

$$V(C) = \max_x \min_p \hspace{2mm} x^T C p = \min_p \max_x \hspace{2mm} x^T C p = (x^*)^T C p^*$$

### Connection with Linear Programming (LP)¶

Finding Nash equilibria of a finite two-player zero-sum game can be formulated as a linear programming problem.

To see this, we introduce the following notation - For a fixed $x$, let $v$ be the value of the minimization problem: $v \equiv \min_p x^T C p = \min_j x^T C e^j$ - For a fixed $p$, let $u$ be the value of the maximization problem: $u \equiv \max_x x^T C p = \max_i (e^i)^T C p$.

Then the max-min problem (the game from the maximizing player’s point of view) can be written as the primal LP

\begin{aligned} V(C) = & \max \hspace{2mm} v \\ \text{s.t. } \hspace{2mm} v \iota_n^T &\leq x^T C \\ x &\geq \mathbf{0} \\ \iota_n^T x & = 1 \end{aligned}

while the min-max problem (the game from the minimizing player’s point of view) is the dual LP

\begin{aligned} V(C) = &\min \hspace{2mm} u \\ \text{s.t. } \hspace{2mm}u \iota_m &\geq Cp \\ p &\geq \mathbf{0} \\ \iota_m^T p & = 1 \end{aligned}

Hamburger, Thompson and Weil [HTW67] view the input-output pair of the economy as payoff matrices of two-player zero-sum games. Using this interpretation, they restate Assumption I and II as follows

$$V(-A) < 0\quad\quad \text{and}\quad\quad V(B)>0$$

Proof (Sketch): $\Rightarrow$ $V(B)>0$ implies $x_0^T B \gg \mathbf{0}$, where $x_0$ is a maximizing vector. Since $B$ is non-negative, this requires that each column of $B$ has at least one positive entry, which is Assumption I. $\Leftarrow$ From Assumption I and the fact that $p>\mathbf{0}$, it follows that $Bp > \mathbf{0}$. This implies that the maximizing player can always choose $x$ so that $x^TBp>0$, that is it must be the case that $V(B)>0$.

In order to (re)state Theorem I in terms of a particular two-player zero-sum game, we define the matrix for $\gamma\in\mathbb{R}$

$$M(\gamma) \equiv B - \gamma A$$

For fixed $\gamma$, treating $M(\gamma)$ as a matrix game, we can calculate the solution of the game

• If $\gamma > \alpha_0$, then for all $x>0$, there $\exists j\in\{1, \dots, n\}$, s.t. $[x^T M(\gamma)]_j < 0$ implying that $V(M(\gamma)) < 0$.
• If $\gamma < \beta_0$, then for all $p>0$, there $\exists i\in\{1, \dots, m\}$, s.t. $[M(\gamma)p]_i > 0$ implying that $V(M(\gamma)) > 0$.
• If $\gamma \in \{\beta_0, \alpha_0\}$, then (by Theorem I) the optimal intensity and price vectors $x_0$ and $p_0$ satisfy
\begin{aligned} x_0^T M(\gamma) \geq \mathbf{0}^T \quad \quad \text{and}\quad\quad M(\gamma) p_0 \leq \mathbf{0} \end{aligned}

That is, $(x_0, p_0, 0)$ is a solution of the game $M(\gamma)$ so that $V\left(M(\beta_0)\right) = V\left(M(\alpha_0)\right) = 0$.

• If $\beta_0 < \alpha_0$ and $\gamma \in (\beta_0, \alpha_0)$, then $V(M(\gamma)) = 0$.

Moreover, if $x'$ is optimal for the maximizing player in $M(\gamma')$ for $\gamma'\in(\beta_0, \alpha_0)$ and $p''$ is optimal for the minimizing player in $M(\gamma'')$ where $\gamma''\in(\beta_0, \gamma')$, then $(x', p'', 0)$ is a solution for $M(\gamma)$, $\forall \gamma\in (\gamma'', \gamma')$.

Proof (Sketch): If $x'$ is optimal for a maximizing player in game $M(\gamma')$, then $(x')^T M(\gamma')\geq \mathbf{0}^T$ and so for all $\gamma<\gamma'$.

$$(x')^T M(\gamma) = (x')^T M(\gamma') + (x')^T(\gamma' - \gamma)A \geq \mathbf{0}^T$$

hence $V(M(\gamma))\geq 0$. If $p''$ is optimal for a minimizing player in game $M(\gamma'')$, then $M(\gamma)p \leq \mathbf{0}$ and so for all $\gamma''<\gamma$

$$M(\gamma)p'' = M(\gamma'') + (\gamma'' - \gamma)Ap'' \leq \mathbf{0}$$

hence $V(M(\gamma))\leq 0$.

It is clear from the above argument that $\beta_0$, $\alpha_0$ are the minimal and maximal $\gamma$ for which $V(M(\gamma))=0$.

Moreover, Hamburger et al. [HTW67] show that the function $\gamma \mapsto V(M(\gamma))$ is continuous and nonincreasing in $\gamma$.

This suggests an algorithm to compute $(\alpha_0, x_0)$ and $(\beta_0, p_0)$ for a given input-output pair $(A, B)$.

### Algorithm¶

Hamburger, Thompson and Weil [HTW67] propose a simple bisection algorithm to find the minimal and maximal roots (i.e. $\beta_0$ and $\alpha_0$) of the function $\gamma \mapsto V(M(\gamma))$.

#### Step 1¶

First, notice that we can easily find trivial upper and lower bounds for $\alpha_0$ and $\beta_0$.

• TEP requires that $x^T(B-\alpha A)\geq \mathbf{0}^T$ and $x > \mathbf{0}$, so if $\alpha$ is so large that $\max_i\{[(B-\alpha A)\iota_n]_i\} < 0$, then TEP ceases to have a solution.

Accordingly, let UB be the $\alpha^{*}$ that solves $\max_i\{[(B-\alpha^{*} A)\iota_n]_i\} = 0$.

• Similar to the upper bound, if $\beta$ is so low that $\min_j\{[\iota^T_m(B-\beta A)]_j\}>0$, then the EEP has no solution and so we can define LB as the $\beta^{*}$ that solves $\min_j\{[\iota^T_m(B-\beta^{*} A)]_j\}=0$.

The bounds method calculates these trivial bounds for us

In [6]:
n1.bounds()

Out[6]:
(1.0, 2.0)

#### Step 2¶

Compute $\alpha_0$ and $\beta_0$

• Finding $\alpha_0$
1. Fix $\gamma = \frac{UB + LB}{2}$ and compute the solution of the two-player zero-sum game associated with $M(\gamma)$. We can use either the primal or the dual LP problem.
2. If $V(M(\gamma)) \geq 0$, then set $LB = \gamma$, otherwise let $UB = \gamma$.
3. Iterate on 1. and 2. until $|UB - LB| < \epsilon$.
• Finding $\beta_0$
1. Fix $\gamma = \frac{UB + LB}{2}$ and compute the solution of the two-player zero-sum game associated. with $M(\gamma)$. We can use either the primal or the dual LP problem.
2. If $V(M(\gamma)) > 0$, then set $LB = \gamma$, otherwise let $UB = \gamma$.
3. Iterate on 1. and 2. until $|UB - LB| < \epsilon$.
Existence: Since $V(M(LB))>0$ and $V(M(UB))<0$ and $V(M(\cdot))$ is a continuous, nonincreasing function, there is at least one $\gamma\in[LB, UB]$, s.t. $V(M(\gamma))=0$.

The zerosum method calculates the value and optimal strategies associated with a given $\gamma$.

In [7]:
γ = 2

print(f'Value of the game with γ = {γ}')
print(n1.zerosum(γ=γ)[0])
print('Intensity vector (from the primal)')
print(n1.zerosum(γ=γ)[1])
print('Price vector (from the dual)')
print(n1.zerosum(γ=γ, dual=True)[1])

Value of the game with γ = 2
-0.24000000097850305
Intensity vector (from the primal)
[0.32 0.28 0.4 ]
Price vector (from the dual)
[4.00e-01 3.20e-01 2.80e-01 2.54e-10]

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland
/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:176: OptimizeWarning: Unknown solver options: bland

In [8]:
numb_grid = 100
γ_grid = np.linspace(0.4, 2.1, numb_grid)

value_ex1_grid = np.asarray([n1.zerosum(γ=γ_grid[i])[0]
for i in range(numb_grid)])
value_ex2_grid = np.asarray([n2.zerosum(γ=γ_grid[i])[0]
for i in range(numb_grid)])

fig, axes = plt.subplots(1, 2, figsize=(14, 5), sharey=True)
fig.suptitle(r'The function $V(M(\gamma))$', fontsize=16)

for ax, grid, N, i in zip(axes, (value_ex1_grid, value_ex2_grid),
(n1, n2), (1, 2)):
ax.plot(γ_grid, grid)
ax.set(title=f'Example {i}', xlabel='$\gamma$')
ax.axhline(0, c='k', lw=1)
ax.axvline(N.bounds()[0], c='r', ls='--', label='lower bound')
ax.axvline(N.bounds()[1], c='g', ls='--', label='upper bound')

plt.show()

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland


The expansion method implements the bisection algorithm for $\alpha_0$ (and uses the primal LP problem for $x_0$)

In [9]:
α_0, x, p = n1.expansion()
print(f'α_0 = {α_0}')
print(f'x_0 = {x}')
print(f'The corresponding p from the dual = {p}')

α_0 = 1.2599210478365421
x_0 = [0.33 0.26 0.41]
The corresponding p from the dual = [4.13e-01 3.27e-01 2.60e-01 1.82e-10]

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland
/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:176: OptimizeWarning: Unknown solver options: bland


The interest method implements the bisection algorithm for $\beta_0$ (and uses the dual LP problem for $p_0$)

In [10]:
β_0, x, p = n1.interest()
print(f'β_0 = {β_0}')
print(f'p_0 = {p}')
print(f'The corresponding x from the primal = {x}')

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:176: OptimizeWarning: Unknown solver options: bland

β_0 = 1.2599210478365421
p_0 = [4.13e-01 3.27e-01 2.60e-01 1.82e-10]
The corresponding x from the primal = [0.33 0.26 0.41]

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland


Of course, when $\gamma^*$ is unique, it is irrelevant which one of the two methods we use.

In particular, as will be shown below, in case of an irreducible $(A,B)$ (like in Example 1), the maximal and minimal roots of $V(M(\gamma))$ necessarily coincide implying a ‘‘full duality’’ result, i.e. $\alpha_0 = \beta_0 = \gamma^*$, and that the expansion (and interest) rate $\gamma^*$ is unique.

### Uniqueness and Irreducibility¶

As an illustration, compute first the maximal and minimal roots of $V(M(\cdot))$ for Example 2, which displays a reducible input-output pair $(A, B)$

In [11]:
α_0, x, p = n2.expansion()
print(f'α_0 = {α_0}')
print(f'x_0 = {x}')
print(f'The corresponding p from the dual = {p}')

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland

α_0 = 1.1343231229111552
x_0 = [1.67e-11 1.83e-11 3.24e-01 2.61e-01 4.15e-01]
The corresponding p from the dual = [5.04e-01 4.96e-01 2.96e-12 2.24e-12 3.08e-12 3.56e-12]

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:176: OptimizeWarning: Unknown solver options: bland

In [12]:
β_0, x, p = n2.interest()
print(f'β_0 = {β_0}')
print(f'p_0 = {p}')
print(f'The corresponding x from the primal = {x}')

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:176: OptimizeWarning: Unknown solver options: bland

β_0 = 1.2579826870933175
p_0 = [5.11e-01 4.89e-01 2.73e-08 2.17e-08 1.88e-08 2.66e-09]
The corresponding x from the primal = [1.61e-09 1.65e-09 3.27e-01 2.60e-01 4.12e-01]

/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:158: OptimizeWarning: Unknown solver options: bland


As we can see, with a reducible $(A,B)$, the roots found by the bisection algorithms might differ, so there might be multiple $\gamma^*$ that make the value of the game with $M(\gamma^*)$ zero. (see the figure above).

Indeed, although the von Neumann theorem assures existence of the equilibrium, Assumptions I and II are not sufficient for uniqueness. Nonetheless, Kemeny et al. (1967) show that there are at most finitely many economic solutions, meaning that there are only finitely many $\gamma^*$ that satisfy $V(M(\gamma^*)) = 0$ and $x_0^TBp_0 > 0$ and that for each such $\gamma^*_i$, there is a self-sufficient part of the economy (a sub-economy) that in equilibrium can expand independently with the expansion coefficient $\gamma^*_i$.

The following theorem (see Theorem 9.10. in Gale [Gal89]) asserts that imposing irreducibility is sufficient for uniqueness of $(\gamma^*, x_0, p_0)$.

Theorem II: Consider the conditions of Theorem 1. If the economy $(A,B)$ is irreducible, then $\gamma^*=\alpha_0=\beta_0$.

### A Special Case¶

There is a special $(A,B)$ that allows us to simplify the solution method significantly by invoking the powerful Perron-Frobenius theorem for non-negative matrices.

Definition: We call an economy simple if it satisfies 1. $n=m$ 2. Each activity produces exactly one good 3. Each good is produced by one and only one activity.

These assumptions imply that $B=I_n$, i.e., that $B$ can be written as an identity matrix (possibly after reshuffling its rows and columns).

The simple model has the following special property (Theorem 9.11. in Gale [Gal89]): if $x_0$ and $\alpha_0>0$ solve the TEP with $(A,I_n)$, then

$$x_0^T = \alpha_0 x_0^T A\hspace{1cm}\Leftrightarrow\hspace{1cm}x_0^T A=\left(\frac{1}{\alpha_0}\right)x_0^T$$

The latter shows that $1/\alpha_0$ is a positive eigenvalue of $A$ and $x_0$ is the corresponding non-negative left eigenvector.

The classical result of Perron and Frobenius implies that a non-negative matrix always has a non-negative eigenvalue-eigenvector pair.

Moreover, if $A$ is irreducible, then the optimal intensity vector $x_0$ is positive and unique up to multiplication by a positive scalar.

Suppose that $A$ is reducible with $k$ irreducible subsets $S_1,\dots,S_k$. Let $A_i$ be the submatrix corresponding to $S_i$ and let $\alpha_i$ and $\beta_i$ be the associated expansion and interest factors, respectively. Then we have

$$\alpha_0 = \max_i \{\alpha_i\}\hspace{1cm}\text{and}\hspace{1cm}\beta_0 = \min_i \{\beta_i\}$$
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