Re: [OPE-L] basics vs. non-basics

From: Paul Cockshott (wpc@DCS.GLA.AC.UK)
Date: Wed Sep 21 2005 - 17:39:35 EDT


        Please explain why this is the case.

Take a look at, for example, pages 70-73 of Kurz's & Salvadori's "Theory
of Production", under the section heading "non-basic commodities
reconsidered". They give a simple two sector example of the problem of
indeterminacy. I am not satisifed with the general presentation of this
paradox in the literature, as I think the incompleteness has not been
revealed in its full glory.

The problem arises due to the distinction between basics and non-basics.
According to the neo-Ricardian approach to self-replacing equilibrium,
the rate of profit and rate of growth is determined only in the basic
sector. But what happens if the maximum rate of growth of the non-basic
sector is less than that which obtains in the basic sector? This is the
problem of self-reproducing non-basics, or "beans", which require
themselves for their production.


I think that to solve this one needs to follow Sraffa's suggestion that
one divides the

wage into two parts, that necessary for reproduction, and a surplus

If you do this you then have to ask if the non basics in question enter

the necessary wage bundle. If they do, then they are basics and the

is solved.

If not, then they still require labour inputs, and hence require

wage goods as input, and are no longer self reproducing, so the

problem goes away again.


It appears that all the Sraffian price solutions that we see in the
literature should be qualified with the assumption that the maximum
eigenvalue of the i/o matrix A lies on the principal diagonal of the
submatrix that refers to basic commodities. For the infinite set of
other, logically possible, technical conditions of production, the
Sraffian price equation is indeterminate. This is not a healthy
state-of-affairs, to say the least.

Best wishes,



I don't understand your remark about the maximum eigenvalue lying on the
principle diagonal.

The diagonal is a vector not a value. Did you mean the eigenvector
corresponding to the

main eigenvalue lies on the diagonal?

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