# [OPE-L:4077] Another Sheep?

john erns (ernst@pipeline.com)
Tue, 28 Jan 1997 23:53:26 -0800 (PST)

[ show plain text ]

Andrew,

Here's a response to your OPE 4074.

1. To some extent you caught me as I held the prices of machines
constant regardless of age. Yet, I still feel we are caught in
a discussion limited to cheaper machines. Witness your demonstration
of that p1=p0 in which you show that the price of a new machine, p0,
at t0 is equal to the of new machine at t1. The comparison is meaningful
as we assume that the two machine are the same save for their ages.

2. It is not altogether clear that there are not instances where
w0=0. Consider two identical pentium computers of today. Does it
really matter which is older? Is there really wear and tear? Would
not one using it in profitable endeavors not expect to discard long
before it wears out?

3. I'm still not sure how "moral depreciation" fits into calculations
you would make concerning the rate of profit. Are you subtracting
the losses due to moral depreciation from profits prior figuring
s/(c+v).

4. Let's try a different way of looking at matters. Ignoring all
costs save those of fixed capital, let's assume a capitalist
buys a machine for 2732.053, expects to use it for 4 periods and
also expects that the prices for the total output produced in
periods 1,2,3,and 4 are 1000,900,800, and 700 respectively. At
the end of the 4th period, it has no scrap value. At the end
of the first period, a new machine becomes available that sells
for 2005.259 that can produce the same output as the other
machine. The output prices expected in its 4 periods of life
are 900,800,700,600.

Obviously, my assumptions are not accidental. Using these
machines in each period a rate of profit of .1 can be obtained.
All investment is recovered for each at the end of 4th period.
Here I assume the rate of profit is calculated for the 1st machine
by breaking down the original investment in the following fashion.

2732.053 = 1000/(1+.1) + 900/(1+.1)^2 + 800/(1+.1)^3 + 700/((1+.1)^4