# [OPE-L:3342] Re: Re: Re: Measurement of value, computerdepreciation

From: Michael Perelman (michael@ecst.csuchico.edu)
Date: Thu May 25 2000 - 10:44:38 EDT

Paul, if you still prior knowledge of the rate of future technological change as
well as the market conditions that would do with it determines the timing of the
replacement of the computers, then you have little problem in determining value. I
often think of this sort of calculation as rational expectations Marxism.

Paul Cockshott wrote:

> At 16:18 24/05/00 -0700, you wrote:
> >Think of the simple formula c + v + s. The first element represents the value
> >that is transferred from the capital goods to the final product. It is an
> >analog
> >to depreciation in conventional theory. So suppose that you start a
> >you're using a new computer. If the computer will be obsolete in a year, the
> >amount of value that would be transferred to the final good -- assuming
> >that the
> >final good is using the socially necessary labor required for production
> >-- would
> >be much higher than if the computer would be expected to last for 20
> >years. How
> >you know in advance how long the computer will be in years.
>
> We know that there is a long term growth rate in the computer industry that
> involves a doubling of productivity in terms of logic gates* operations per
> second
> every 18 months or so.
>
> If we know the long term exponential rate of growth of material
> productivity, I dont
> see why it is difficult to compute what the depreciation period should be.
>
> Suppose that we take 1996 as a starting date and that machines of the period
> required 100 hours or 360,000 seconds of abstract labour to produce.
> These machines could perform 10^8 operations per second.
> Assume that machine power grows by a factor of 1.5 per year
> we get the following table:
>
> computer speedyearcyles per year
> 1.00E+0819963.15E+15
> 1.50E+0819974.73E+15
> 2.25E+0819987.10E+15
> 3.38E+0819991.06E+16
> 5.06E+0820001.60E+16
>
> Now let us compare depreciation periods of 1,2,4 years and get what the
> cost in terms of labour time of cycles will be in different years:
>
> depreciation period
> year124
> 19961.1E-095.71E-102.85E-10
> 19977.6E-105.71E-102.85E-10
> 19985.1E-102.54E-102.85E-10
> 19993.4E-102.54E-102.85E-10
>
> costs in terms of seconds of labour per machine cycle
>
> consider the firm that depreciates over 1 year.
> Each year it will have ha hdigher cost than the firm that depreciates over
> 2 years.
> Consider the firm that depreciates over 2 years compared with the firm that
> depreciates over 4 years, its buys its cycles in 1998 and 1999 at a cheaper
> price than the firm that bought computers in 1996 and depreciated over 4 years.
> These cheaper machine cycles enable it to replace more living labour in
> 1998 and 1999 than the machine that depreciated over 4 years, giving it
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