Actually, Patrick, I'm a bit confused. How do you get 1/c from s/c, or
is it l/c (if so, what "l")? Also, aren't you confirming what I was
writing? Paul
On Wed, 28 Aug 1996, Patrick Mason wrote:
> Paul Z:
>
> You are of course correct in your reply to my comment on your post.
> Specifically, I should not have stated that the maximal profit rate occurs
> when s/v = 0, but when v = 0, i.e., workers work for free. Sorry about the typo.
> Aside from this bad editing, I think the rest of my post holds.
>
> peace, patrick l mason
>
> At 08:36 AM 8/28/96 -0700, you wrote:
> >On Mon, 26 Aug 1996, Patrick Mason wrote:
> >
> >> In response to Duncan's post, Paul Z. writes:
> >> > ...
> >> >I always find it useful to write the rate of profit r from s/(c+v) to
> >> >s/v divided by c/v+1 and rewriting the divisor to
> >> >
> >> > c v + s c
> >> >------- ------- + 1 = ----- [1 + s/v] + 1
> >> > v + s v v + s
> >> >
> >> >
> >> >Thus, with s/v fixed, the movement in the rate of profit depends upon
> >> >movements in c/(v+s), the technical value composition of capital, the
> >> >ratio of labor time in fixed capital to the living labor time working with
> >> >it (rising implying falling r).
> >>
> >> The necessity for holding s/v fixed confuses me. Consider the standard
> >> formulation for the average rate of profit:
> >>
> >> r = s/(c + v).
> >>
> >> Even we assume a maximal rate of exploitation, i.e., s/v = 0, it is still
> >> the case that "the movement in the rate of profit depends upon movements in
> >> c/(v+s), the technical value composition of capital, the
> >> ratio of labor time in fixed capital to the living labor time working with
> >> it (rising implying falling r)." Under maximal exploitation, we would have:
> >>
> >> r (max) = l/c.
> >>
> >> Clearly, as the technical composition of capital increases the maximal rate
> >> of profit will fall. Since this is in fact the maximum potential rate of
> >> profit the argument is independent of any assumption regarding the rate of
> >> exploitation.
> >
> >Pat, I don't understand you. Clearly under maximal r, s/v is not 0 but
> >rather infinity as v goes to 0 (using your conception).
> >
> >Paul Z.
> >
> >
>
>