P is its price
C is the accumulated capital-value advanced for this means
of production, i.e., the running total value advanced in
acquiring it
d is the fraction of the means of production that is
destroyed (i.e., turns over), per period, due to wear and
tear. d ranges between 0 and 1, inclusive. If the means of
production are non-depreciating, d = 0; if we have
circulating capital only, d = 1.
Physical Investment
===================
Between periods t and t+1, *net* investment in physical
terms is M[t+1] - M[t].
In addition, some investment *replaces* the worn-out means
of production. The amount of the means of production that
wears out between periods t and t+1 is dM[t]. dM[t] is
likewise the *replacement* investment in physical terms.
*Gross* investment in physical terms is the sum of net and
replacement investment:
(1) {M[t+1] - M[t]} + dM[t].
Example: assume that the growth rate of the means of
production is 10%, and that M[t] = 10. Hence, M[t+1] = 11.
Assume also that d = .2. Thus, of the 10 units that existed
at the start of period t, 2 units wore out during the
period, leaving only 8 remaining at the start of period t+1.
Thus, gross investment in physical terms is
{M[t+1] - M[t]} + dM[t]
= {11 - 10} + .2*10
= 1 + 2
= 3
In other words, beginning with 8 units remaining from period
t, the firm, in order to wind up with 11 units in period
t+1, must acquire 3 additional units. 2 of these replace the
worn-out means of production; 1 of these is a net additional
to the stock of the means of production.
Determination of C
==================
What is the magnitude of the total capital-value advanced
for this means of production? Let's examine the magnitude at
the start of period t+1. It is the sum of two parts:
(a) the value of the capital *remaining* at the end of
period t. This is
(2) C[t] - P[t]dM[t],
where P[t] is the start-of-period (input) price. Since dM[t]
is the used-up portion of the means of production, P[t]dM[t]
is the value transferred to the product. This using-up of
means of production reduces the value of the capital; the
reduction is equal to the current (pre-production) price of
the means of production used-up.
(b) the value of gross investment. Since both the net
investment and the replacement investment are made at the
start of period t+1, the value of gross investment is the
physical investment times the price of the means of
production at the start of period t+1, i.e.,
(3) P[t+1]*({M[t+1] - M[t]} + dM[t]).
Hence, the total capital-value advanced at the start of
period t+1 is
(4) C[t+1] = C[t] - P[t]dM[t] + P[t+1]*({M[t+1] - M[t]} +
dM[t]).
Note that this implies that the *change* in the
capital-value advanced is
(5) C[t+1] - C[t] = P[t+1]{M[t+1] - M[t]} + (P[t+1] -
P[t])dM[t].
In other words, the change consists of 2 parts. First, the
cost, at time t+1, of the net investment; and second, the
*change* in the cost of the replacement investment.
Special Cases
=============
When the capital does not depreciate physically, d = 0, and
(5) becomes
(6) C[t+1] - C[t] = P[t+1]{M[t+1] - M[t]}.
In the pure circulating capital case, d = 1, and (5) becomes
(7) C[t+1] - C[t] = P[t+1]M[t+1] - P[t]M[t].
Both results are exactly what we would expect.
Rate of Depreciation and Change in C
====================================
How does the rate of depreciation, d, affect the change in
the value of capital advanced? One way of assessing this is
to measure the difference between the changes in the
capital-value advanced at two different rates of
depreciation.
More notation: Let D(+1,d) stand for the change (D) in the
value of capital advanced between t and t+1 (+1) when the
rate of depreciation is d. Hence, (6) is D(+1,0) and (7) is
D(+1,1).
The difference between these two changes is (6) minus (7),
or
(8) D(+1,0) - D(+1,1) = (P[t] - P[t+1])M[t].
Likewise, in the general case, when d is greater than 0 but
less than 1, the difference between the *actual* change and
the change that would occur in the pure circulating capital
case is (5) minus (7), or
(9) D(+1,d) - D(+1,1) = P[t+1]{M[t+1] - M[t]} + (P[t+1] -
P[t])dM[t] - (P[t+1]M[t+1] - P[t]M[t]) = (1-d)(P[t] -
P[t+1])M[t].
Now notice that (9) divided by (8) is
D(+1,d) - D(+1,1)
(10) ----------------- = 1 - d.
D(+1,0) - D(+1,1)
This is an important result. It means that the difference
between the *actual* and *minimum* changes (numerator) is
equal to the fraction 1 - d of the difference between the
*maximum* and *minimum* changes (denominator). In other
words, it lets us assess how "close" a particular case is to
the pure circulating capital case, and how "close" it is to
the non-depreciation case.
Assuming that the age-structure of the means of production
is fairly evenly distributed, d is approximately equal to
1/N, the reciprocal of the average number of periods the
means of production lasts. If the means of production last
5 periods, N = 5, which is obviously much closer to N = 1
than it is to N = infinity. However, as Table 1 (below)
shows, the difference between the *actual* and *minimum*
changes in the capital-value advanced in the 5-period case
is 80% of the maximum difference. So the 5-period case is
much closer to the non-depreciating case. Moreover, the
measly 2-PERIOD case is the midway point between the
1-period and the infinite-period case.
===========================
TABLE 1
N (approx) d (10)
========== === ====
1 1 0
2 .5 .5
3 .33 .67
4 .25 .75
5 .2 .8
10 .1 .9
20 .05 .95
infinity 0 1
===========================
This makes perfect sense. In the 1-period case,
100% of the capital turns over each period; in the
infinite-period case, 0% turns over. Midway between these is
the case in which 50% turns over, the 2-period case.
Another Measure of Closeness
============================
Another way of assessing the relation between the rate of
depreciation, and the change in the value of capital
advanced is to express D(+1,d) in terms of D(+1,1) and
D(+1,0). From (10), we can derive
(11) D(+1,d) = d*D(+1,1) + (1-d)*D(+1,0).
This is another important result. (11) shows that the actual
change is a WEIGHTED AVERAGE of the minimum and maximum
changes, with d, the depreciation rate, serving as the
weight. It again indicates that, for instance, the change in
capital-value in the 2-period case is halfway between the
minimum and maximum changes, and the change in the 5-period
case lies 80% of the way between the minimum and maximum
changes.
Table 2 provides an example of the relationship between d
and D(+1,d), assuming that D(+1,1) = 10 and D(+1,0) = 30.
==============================
TABLE 2
N (approx) d D(+1,d)
========== === =======
1 1 10
2 .5 20
3 .33 23.3
4 .25 25
5 .2 26
10 .1 28
20 .05 29
infinity 0 30
==============================
Note also that (11) implies that the actual change in the
capital-value advanced is more than (1-d) times the
maximum change. Thus, in the 5-period case, no matter what
the minimum and maximum changes may be, the actual change
must be more than 80% of the maximum change.
Rate of Depreciation and Value of C
===================================
The results above pertain to changes between one particular
period and the next. Yet they are completely general, so
they pertain to changes between *all* periods, beginning
with the change between periods 0 and 1.
By summing the various terms, we obtain the *total* change
in C between periods 0 and the current period. Since our
results concerning the relation between d and the
inter-period changes in C apply to each and every period,
the results likewise apply to the total change. Thus, for
instance, the actual total change is again the weighted
average of the minimum and maximum total changes, with d
again serving as the weight.
The total capital-value advanced through the current period
is simply this *total* change in C plus the initial C, the
capital-value advanced at the start of period 0. If we
assume that the capital-value advanced continually grows,
the initial C becomes a vanishing share of the total C.
Hence, as time approaches infinity, the total change in C
and the total capital-value advanced converge. And, hence,
as time approaches infinity, the relationships we found
between d and the interperiod change in C apply not only to
the total change, but also to the total capital-value
advanced.
For instance, as time approaches infinity, we can say that
the capital-value advanced in the 2-period case is halfway
between the capital-value advanced in the 1-period and
infinite-period cases. And so on.
Thus, if at some "large" time t, the capital-value
advanced would be 10 in the 1-period case and 30 in the
infinite-period case, the capital-value advanced in the
5-period case is 26, *much closer* to the latter than to the
former.
Depreciation and the Profit Rate
================================
The algebraic example in my paper ("A Value-Theoretic
Critique of the Okishio Theorem," in _Marx and
Non-Equilibrium Economics_") can be translated readily into
the notation laid out above. I called the nondepreciating
"fixed capital" F, and the "circulating capital" A. But F[t]
= (1-d)M[t] and A[t] = dM[t]. Using the lowercase letters
"f" and "a" to indicate the initial levels of F and A per
unit of output, I obtained the following limiting value (as
t approaches infinity) to the profit rate, r:
1 - a(b/c)
(12) lim r[t] = -----------------------
f[(b-1)/(c-1)] + a(b/c)
where b was the growth factor for means of production and
output, and c was the growth factor for living labor.
Using the alternative notation, and letting m be the initial
level of M per unit of output, we have
1 - dm(b/c)
(12') lim r[t] = -----------------------------
(1-d)m[(b-1)/(c-1)] + dm(b/c)
(12') again illustrates the result obtained in (11), since
the denominator would be m[(b-1)/(c-1)] in the
infinite-period case, and m(b/c) in the 1-period case. In
other words, in the limit, the value of capital advanced is
a weighted average of the minimum and maximum denominators,
with d serving as the weight.
What is the effect of depreciation on the profit rate? Not
much, once we move beyond the 1-period, and perhaps the
2-period, cases.
Letting initial net output per unit of gross output equal y,
and noting that net output plus depreciation equals gross
output, we have y + dm = 1. Substituting the left-hand side
of this equality for 1 in (12'), we have
y + dm - dm(b/c)
(12'') lim r[t] = -----------------------------
(1-d)m[(b-1)/(c-1)] + dm(b/c)
In my numerical example, I had m = 2.4, d = 1/6, y = .6, b =
1.06, and c = 1.02. The profit rate, initially 20%, fell to
9.11%.
Had I assumed d = 0, it would not have fallen much more; it
would have fallen only to 8.33%. Put in reverse, a moderate
rate of depreciation raises the limit of the profit rate
only slightly over the rate that would obtain were constant
capital fully nondepreciating.
Table 3 shows the limit of the profit rate in this example,
for these and other rates of depreciation.
==============================
TABLE 3
N (approx) d lim r
========== === =======
1 1 20.28%
2 .5 11.41
3 .33 10.10
4 .25 9.57
5 .2 9.29
6 .17 9.11
10 .1 8.78
20 .05 8.55
infinity 0 8.33
==============================
Declining Extraction of Living Labor
====================================
In my chapter in M&NE, I showed that, if extraction of
living labor declines absolutely, c < 1, the profit rate
falls asymptotically to zero. However, it is known that, if
there's only circulating capital, the profit rate remains
positive even if c < 1.
Yet, the d = 1 case is the ONLY case in which the profit
rate remains positive. Assuming an initial static
equilibrium, my unit
price equation (5'' of my chapter in M&NE), and eq. (5) of
the present post, the limit of the capital-value advanced,
when c < 1, turns out to be
nMo(b-c)
(13) lim C[t] = -----------*(1-d)
(1-ab)(1-c)
where n is the initial living labor per unit of output. As I
discussed in M&NE, 1 - ab > 0, and in the present case b-c
and 1-c are also positive. Hence, if d is less than 1, the
constant capital-value advanced remains positive, while (as
I showed in M&NE) the numerator falls to zero. Hence the
profit rate falls to zero.
Only when d is EXACTLY 1, i.e., ALL constant capital turns
over each period, does the capital-value fall to zero and
the profit rate therefore remain positive.
Conclusion
==========
(to follow)
Andrew Kliman