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\begin{document}


\title{An economic system of subsistence by Sraffa}

\maketitle


\textbf{Sophie} Today I would like to talk about some economics,
particularly
a simple model of the economic system. Let us turn to Chapter 1
of Pierro Sraffa's
book Production of Commodities by means of
commodities.
I like this book, because it is very compact, less than 100 pages
long, and Chapter 1 Production of subsistence is just two and a
half page long. Here it is. He describes a simple society that
sustains itself with just two products wheat and iron. Each year,
this society produces \( 400qr \) of
wheat and \( 20t \) of iron. For the production of the \( 400qr
\) of wheat,
\( 280qr \) of wheat and \( 12t \) of iron are required, and for
the production
of \( 20t \) of iron, \( 120qr \) of wheat and \( 8t \) of iron
are required.
Therefore, this society consumes each year exactly the same
amount of wheat
and iron that it produces. For this society to continue to exist,
Of the \( 400qr \)
of wheat produced by the wheat industry, \( 120qr \) of wheat is
exchanged
for the \( 12t \) of iron of the \( 20t \) of iron produced in
the iron industry.
Now, if they use money for the trade, then what are the prices of
what and iron?

\textbf{Gena} Let the price of \( 1qr \) of wheat be \( \$p_{1}
\)
and the
price of \( 1t \) of iron be \( \$p_{2} \). The fact is that \(
120qr \)
of wheat is exchanged for \( 12t \) of iron. Therefore, wheat
industry get
\( \$120p_{1} \) by selling \( 120qr \) of wheat to iron
industry. But the
wheat industry must in turn pay \( \$12p_{2} \), since it buys \(
12t \) of
iron from the iron industry. Therefore, in order that for the
both industries
to have no deficit or surplus, it follows that \(
120p_{1}=12p_{2} \). 

\textbf{Sophie} That's right. There is no hard trade problem in
this society,
since only relative prices of the two products matter. We can
formulate the
above situation in the following equations. And that is one use
of mathematics. 

{\centering \( 280\times p_{1}+12\times p_{2}=400\times p_{1} \)
Wheat industry,\par}

{\centering \( 120\times p_{1}+8\times p_{2}=20\times p_{2} \)
Iron industry,\par}

{\raggedright where \( p_{2}/p_{1}=10 \). Here, the total prices
of consumed
commodities are on the left side of each equation and the total
prices of the
produced commodities are on the right side of each equation. Now,
if the society
consists of three industries, those of wheat, iron, and pigs,
then do prices
exist for these three commodities so that the three industries
can smoothly
exchange commodities among them? If so, how can we determine the
unit price
of these three commodities?. For example in the second case that
Sraffa described,
a mathematical question arises as to whether there exist prices
\( p_{1} \),
\( p_{2} \), \( p_{3} \) that satisfy the following equations.
\par}

{\centering \( 240\times p_{1}+12\times p_{2}+18\times
p_{3}=450\times p_{1} \)
Wheat industry, ~~~~(1)\par}

{\centering \( 90\times p_{1}+6\times p_{2}+12\times
p_{3}=21\times p_{2} \)
Iron industry, ~~~~(2)\par}

{\centering \( 120\times p_{1}+3\times p_{2}+30\times
p_{3}=60\times p_{3} \)
Pigs industry, ~~~~(3)\par}

\textbf{Luis} Why do you make things harder? Can this society not
do away with
money, forget prices, and just transport commodities? The wheat
industry can
send \( 90qr \) of wheat to the iron industry and \( 120qr \) of
wheat to
Pigs industry. The iron industry can send \( 12t \) of iron to
the wheat industry
and \( 3t \) of iron to the pigs industry. The pigs industry can
send 18 pigs
to the wheat industry and 12 pigs to the iron industry. That
solves all the
trade issues.

\textbf{Sophie} A utopian may think like you. But although this
economic model
is highly simplified, we assume that the utopian idea does not
work. 

\textbf{Luis} OK! let me first check the method Gena showed in
the
first
case. If there
is no trade deficit or surplus between the wheat industry and the
iron industry,
we have 

{\centering \( p_{2}/p_{1}=90/12 \). ~~~~(4) \par}

{\raggedright If there is no trade deficit or surplus between the
wheat industry
and the pigs industry, then \par}

{\centering \( p_{3}/p_{1}=120/18 \) . ~~~~(5) \par}

{\raggedright These determine the relative prices of the three
commodities.
But wait a minute! What happens between the iron industry and the
pigs industry.
If there is no deficit or surplus between these industries, \par}

{\centering \( p_{2}/p_{3}=12/3=4 \). \par}

{\raggedright But from (4) and (5), \par}

{\centering \( p_{2}/p_{3}=(p_{2}/p_{1})\times
(p_{1}/p_{3})=(90/12)\times (18/120)=9/8 \)
.\par}

{\raggedright That is a contradiction. In fact, since the iron
industry get
\( \$3p_{2} \) from the pigs industry and pay \( \$12p_{3} \) to
the pigs
industry, if the relative prices determined by (4) and (5) are
applied between
them, then the iron industry has a deficit of \par}

{\centering \( 12p_{3}-3p_{2}=12p_{3}-3\times (9/8)p_{3} \) \par}

{\raggedright with the pigs industry. More than that, this
deficit is the deficit
with the wheat and pigs industry combined. This means the
equation (2) does
not hold. Similarly, equation (3) does not hold. That is, if the
relative prices
determined by (4) and (5) are applied , only equation (1) holds.
\par}

\textbf{Sophie} That's right. You also showed that how impossible
it is for
a country to achieve a bilateral trade balance with every foreign
country. 

\textbf{Gena} I see. Then, let's forget the bilateral trade
balance and look
at (1), (2), (3) as a system of simultaneous equations of unknown
variables
\( p_{1} \), \( p_{2} \), \( p_{3} \). Since relative prices
matter, let
me assume \( p_{1} \) = 1, and solve the simultaneous equations. 

{[}After a few minutes later.{]} I obtained \( p_{1}=1 \), \(
p_{2}=10 \),
\( p_{3}=5 \). And these values satisfy all of (1), (2), and (3).

\textbf{Sophie} Good!. Now, In a general case, we must further go
to more advanced
math. I want to determine whether the following system of
equations has always
a solution such that \( p_{i}\geq 0 \) for every \( i \), and \(
p_{i}>0 \)
for some \( i \). 

{\centering \( A_{11}p_{1}+A_{12}p_{2}+A_{13}p_{3}=A_{1}p_{1} \)
~~~~(6) \par}

{\centering \( A_{21}p_{1}+A_{22}p_{2}+A_{23}p_{3}=A_{2}p_{2} \)
~~~~(7) \par}

{\centering \( A_{31}p_{1}+A_{32}p_{2}+A_{33}p_{3}=A_{3}p_{3} \)
~~~~(8) \par}

{\raggedright where \( A_{ij}\geq 0 \) for every \( i \), \( j \)
and \( \sum _{i}A_{ij}=A_{j}>0 \)
for every \( j \). To this problem, Sraffa gives no satisfactory
answer. He
merely says, for general \( k \) industries, that there are \( k
\) - 1 independent
equalities, and since the price of one commodity can be made 1,
the equations
can be solved. This is a sloppy erroneous mathematical reasoning.
Even junior
high school students know that a system of linear simultaneous
equations has
sometimes no solution. Moreover, in this case, we must prove the
equation has
a \emph{non-negative} solution \( p_{1} \), \( p_{2} \), \( p_{3}
\) and at least one of them is positive. Let me fill
the gap. \par}

First, in order to normalize the knowns and unknown variables,
let \( x_{i}=A_{i}p_{i} \).
Then the equations become 

{\centering \( a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}=x_{1} \) 
\par}

{\centering \( a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}=x_{2} \) 
\par}

{\centering \( a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}=x_{3} \) 
\par}

{\raggedright where \( a_{ij}=A_{ij}/A_{i}\geq 0 \) for every \(
i \), \( j \)
and \( \sum _{i}a_{ij}=1. \) See these are simpler than (6), (7),
(8). To put
the unknow vector \( x=(x_{1},x_{2},x_{3}) \) on the equal
footing with the
known vectors \( a^{(j)}=((a_{1j},a_{2j},a_{3j}) \), we can
assume \( \sum _{j}x_{j}=1 \),
since only relative prices matter. Of course, \( x_{j}\geq 0 \)
for every \( j \),
Then we obtain the expression, further simpler \par}

{\centering \( x_{1}a^{(1)}+x_{2}a^{(2)}+x_{3}a^{(3)}=x \)
~~~~(9) \par}

{\raggedright Now consider the set \par}

{\centering \( C=\{x=(x_{1},x_{2},x_{3})\, |\, \sum _{j}x_{j}=1\,
\mathrm{and}\, x_{j}\geq 0\, \mathrm{for}\, \mathrm{every}\, i\}
\) \par}

{\raggedright Now, \( a^{(1)} \), \( a^{(2)} \), \( a^{(3)} \)
are members
of \( C \), and if \( x \) is a member of \( C \), then so is \(
x_{1}a^{(1)}+x_{2}a^{(2)}+x_{3}a^{(3)} \).
Therefore, I can talk about a function \( f \): \( C\rightarrow C
\) defined
by \par}

{\centering \( fx=x_{1}a^{(1)}+x_{2}a^{(2)}+x_{3}a^{(3)} \).
\par}

{\raggedright What we have to prove is that there is a fixed
point, i.e. a point
\( x \) such that \( fx=x \), for this function \( f \). Since \(
f \) is
clearly a continuous function, and \( C \) is a compact convex
set, \( f \)
has a fixed point by Brouwer's well-known fixed-point theorem.
\par}

\end{document}
