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

From: Philip Dunn (pscumnud@DIRCON.CO.UK)
Date: Thu Sep 22 2005 - 06:04:14 EDT

```Quoting Gerald_A_Levy@MSN.COM:

> Hi Phil:
>
> > In the example I gave no value was destroyed -- aggregate value added
> > was positive.
>
> You're example was a bit vague on details.  You previously wrote:
>
> > Suppose we have an all agricultural economy and there are bad harvests,
> > so that the physical net product is all negative.
>
> The following numerical example is a special case of your example.
>
> Let's stay at the aggregate level.    If I recall correctly, you like
> numerical examples.   OK ...
>
> Previously -- in the last harvest before the 'bad harvest' -- the value of
> the total product was \$150  where C=50, V=50 & S=50.
>
> Assume that the entire S is then productively consumed.  So,  'before the
> harvest'  the total capital invested equals \$150 -- \$75 V and \$75 C.
>
> Assume further for now that C consists entirely of constant circulating
> capital (but see 'PS' below).
>
> Now assume -- as a result of bad harvests -- that the value of the total
> product now equals \$50.   In this case, there is a decline in the value
> of the total product from \$150 to \$50.
>
> Hasn't  a value equivalent to \$100 been destroyed in this case?
>
> In solidarity, Jerry
>

Hi Jerry

Comparisons of sums of money paid at different times are only valid if the value
of money has not changed.

Suppose aggregate labour is 25 hours. Then aggregate value added is 25 hours.
Let tha value if money initially be 1. Then C is 75 hours.  Let the value of
money at the end be v hours/\$. The value of revenue, C', is 50v hours.

C' - C = 50v - 75 = 25

v=2 hours/\$, c' = 100 hours

Philip Dunn
```

This archive was generated by hypermail 2.1.5 : Fri Sep 23 2005 - 00:00:01 EDT