# [OPE-L:4074] Dancing with Wolves

andrew kliman (Andrew_Kliman@msn.com)
Tue, 28 Jan 1997 06:26:53 -0800 (PST)

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This is a further comment on the depreciation/value transfer discussion,
focused specifically on John's position. See especially his ope-l 4062 and my

John's position, unless I'm mistaken, is that the value transferred to the
product is just equal to the loss of value that an element of fixed capital
suffers during the period. (It suffers this loss of value due to wear and
tear and due to moral depreciation.) Thus, the value transferred to the
product plus the value the thing has (in its partially or totally worn out
state) at the end of the period equals its initial value, at the start of the
period.

John gave an example in which this process could occur even though the thing's
price wasn't stationary. I objected that the example assumed that the machine
lost no value due to wear and tear, i.e., that the partially worn out machine
at the end of the period was worth just as much as a new machine of the same
type at the same time.

I then asked John to produce an example that conforms to his position in which
prices aren't stationary but the machine does wear out. You can forget about
trying to do so, John. I've realized it is impossible. Let me prove it.

The value of the machine at time 0 is Po. Assume it is a new machine at that
time. The value of a replica machine at time 1 is P1. The original machine
loses part of its USE-value by undergoing wear and tear between times 0 and 1.
Denote this part by Wo. Wo is any number between 0 and 1, inclusive. Thus,
at time 1, the original machine is worth the following fraction of what a
replica machine is worth: 1 - Wo. Since the replica machine is worth P1, the
original, in its partly worn state, is thus worth P1(1 - Wo).

The amount of value the original machine loses between times 0 and 1 is
therefore

LOSS = Po - P1(1 - Wo).

The value transferred between times 0 and 1 can be decomposed into 2 parts.
One is the value transferred due to wear and tear. Since Wo is the fraction
of its use-value which the original machine loses, the value transferred due
to wear and tear is Po*Wo. John maintains that there is a second component of
the value transferred --- the moral depreciation, i.e., the change in the
value of a new machine of this type between times 0 and 1. It equals Po - P1.
(Thus, if P1 &lt; Po, the there is a positive amount of value transferred due to
moral depreciation.) Hence, in John's view, the value transferred is

VALUE TRANSFERRED = Po*Wo + (Po - P1).

John also holds that

LOSS = VALUE TRANSFERRED

i.e., that

Po - P1(1 - Wo) = Po*Wo + (Po - P1)

which implies that

Po*Wo = P1*Wo.

If Wo = 0, as in John's example, his conception does not constrain Po and P1.
We have 0 = 0, and Po and P1 can be of any magnitudes. But note again that Wo
= 0 means that a partially used machine is worth just as much as a new machine
of the same type, at the same time.

If Wo > 0, however, then the above implies that

Po = P1.

John is dancing with wolves.

It is very interesting that John was able to come up with any type of example
at all in which his conception was compatible with nonsimultaneous prices,
because, as we see, he came up with the ONLY possible kind of example in which
they're compatible, an example in which Wo = 0. As long as a used machine is
worth less than a new machine (of the same type at the same type), John's
position implies a type of simultaneism.

EXAMPLE. New machine worth 100 at time 0. It undergoes wear and tear
equivalent to 5% = 0.05 of its use-value, so that at time 1, it is worth 95% =
0.95 of what a new machine of the same type is then worth. Hence, LOSS = 100
- 0.95*P1. The value transferred due to wear and tear is 0.05*100 = 5. The
value transferred due to moral depreciation is 100 - P1. VALUE TRANSFERRED =
5 + (100 - P1). If LOSS = VALUE TRANSFERRED,

100 - 0.95*P1 = 5 + (100 - P1)

so that

0.05*P1 = 5

or

P1 = 100 = Po.

It is also noteworthy that John's position implies something which is quite
counterintuitive, to me at least. The greater the moral depreciation, i.e.,
the bigger the drop in price a new item of this type undergoes, the better off
are the capitalists. The firm holds an asset having a value of Po at time 0.
At time 1, it has an asset worth P1(1 - Wo), plus money capital equal to the
value transferred. So it has a total of P1(1 - Wo) + Po*Wo + (Po - P1). The
increase in the value it holds at time 1 over the value it held at time 0 is

[P1(1 - Wo) + Po*Wo + (Po - P1)] - Po

= (Po - P1)*Wo.

Thus, the smaller P1 is relative to Po, the bigger the gain to the firm.
Capital should have nothing at all to fear from, but should welcome, moral
depreciation!

Note that I've said nothing about n periods, nothing that required cheaper
machines, and nothing that precluded better machines. The results have
nothing to do with expectations, and nothing to do with competition. The
matter is much more simple: (a) if capitalists recoup an amount of value that
offsets the amount that the machine loses during the period, prices must be
stationary unless a used machine is worth as much as a new machine. (b) if
capitalists fully recoup the moral depreciation of the machine, then the more
the moral depreciation, the better off they are.

In Little Red Riding Hood solidarity,

Andrew Kliman