**From:** Paul Cockshott (*wpc@DCS.GLA.AC.UK*)

**Date:** Wed Sep 21 2005 - 17:39:35 EDT

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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. Paul 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 component. If you do this you then have to ask if the non basics in question enter into the necessary wage bundle. If they do, then they are basics and the problem is solved. If not, then they still require labour inputs, and hence require necessary 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, -Ian. Paul 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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