[OPE-L:5290] Re: x+a = revenue

andrew kliman (Andrew_Kliman@msn.com)
Wed, 18 Jun 1997 05:42:52 -0700 (PDT)

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A reply to Ajit's ope-l 5289:

Ajit: "The quotation from Marx, which I interpreted, does not support
your interpretation. There Marx is not talking about revenue being larger
than the surplus value."

Of course Marx is talking about revenue being larger than surplus-value. x =
surplus-value, and he notes that, if it costs an additional sum a > 0 for
capitalists to attain the same standard of living, and they do so, then they
spend x+a as revenue. Therefore revenue is larger than surplus-value.
Algebraic symbols don't lie.

Ajit: "I doubt that this is what Fred's position is. How can one deny that a
system could be run down to the ground."

Fred doesn't deny that. He reasons this way, more or less:

Marx states that the part of surplus-value spent on capitalist consumption is
Therefore, Marx defines the part of surplus-value spent on capitalist
consumption as "revenue."
Therefore, Marx defines "revenue" as the part of surplus-value spent on
capitalist consumption.
Therefore, "revenue" is a part of surplus-value.
No part can be larger than the whole of which it is part.
Surplus-value is the whole of which "revenue" is part.
Therefore, "revenue" cannot be larger than surplus-value.

Ajit: "As far as your 'difference equation' is concerned. I must admit, I
don't understand what is going on there."

Why is "difference equation" in scare quotes? Do you deny that I wrote down a
difference equation?

The essence of the matter is very simple. The equation gives cyclical
behavior for total value. In the expansionary phase of the cycle,
surplus-value exceeds revenue, in the downturns the opposite occurs. The
point is that this is *long run* behavior yet, because the system is not a
*static equilibrium* system, the existence of revenue > surplus-value even in
the long run does not mean that the system runs down into the ground, as you
alleged that it did.

Your mistake was to assume wrongly that, in the long run, stationary (growth)
equilibria and breakdown are the only possibilities. You've made that mistake
before. For instance, you confuse average prices with stationary equilibrium
prices, though the two will be equal in general only if the system in question
is a static equilibrium one. You also seem to argue at times that the mere
persistence of a system in the long-run means that its behavior can be
approximated by means of static equilibrium equations.

Ajit: "But then I'm not much interested in carrying on this strand of the
debate. I have to start grading the exams like a mad man from tomorrow."

Yes, and you still have to meet the challenge of producing a set of prices
that acquits your price theory of internal inconsistency.

Andrew Kliman