[OPE-L:3338] Re: Re: Measurement of value, computer depreciation

From: Paul Cockshott (wpc@dcs.gla.ac.uk)
Date: Thu May 25 2000 - 05:33:39 EDT


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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
>business and
>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
a competivive advantage.



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