John E. and Paul C. are sceptical re. my IRR calculation
relating to John's example of two capitals. Two points:
(1) If you assume that capitalist 2 sets aside $100 per year
as depreciation allowance, and is able -- as Paul says -- to
deposit this at interest, then we can recalculate the IRR
for capital 2 as the r that solves:
2000 = sum(t = 1 to 9)[300/(1 + r)^t] + 1200/[(1 + r)^10]
(We've shifted $1000 of the final amount back in time,
adding $100 of it to each year's profit.) This gives r =
11.9 percent.
(2) However, I wonder if that's right. Shouldn't the
depreciation amount be endogenous to the problem? If the
capitalist "invests" his annual depreciation allowance at
the same rate of return as he's making on the basic project,
the thing solves out at an IRR of .10 and an annual
depreciation amount of $62.75 (i.e., if he sets aside $62.75
per year, which compounds at 10 percent, he'll have the
$1000 to replace the machine at the end of 10 years). To
put it differently, the $100 per year depreciation
assumption means that the capitalist is actually getting in
depreciation allowance more than the present value of the
replacement machine, which is what raises his profit rate
above 10 percent in the first calculation above.
Allin Cottrell