434 VALUATION, DEPRECIATION AND THE RATE-BASE 
Then the expectancy to be determined will be found from 
ne’ 
£i= (32) 
Example. — What is the expectancy of an article 8 years old, 
in fair condition, whose probable life new was 12 years? Here 
n = 12 and m = 8. 
8 
m = el 6.67. 
12 
For the age 6.67 years in the ten-year subdivision of Table 31 
(this being in the seventh year) the expectancy 5.17 years is 
found by interpolation. The required expectancy is, therefore, 
— 517 X= =6 
e = 5.17 X — = 06.2 years, 
which would be found in a twelve-year probable life table for the 
end of the eighth year or for the beginning of the ninth year. 
For general use the following formula are recommended which 
will be found to agree fairly well with the results presented in 
Table 31. They are applicable to any probable life term and 
any age to the limit beyond which the assumption is justified that 
no article will continue in service. 
When the age of an article is less than its probable life term 
new, that is, when m < n 
m2 
¢ =~ 003m + o30-". (33) 
When the age of an article is greater than its probable life 
term new, that is, when m > #» 
e =0.72% — 0.35 M. (34) 
Example. — What is the expectancy of an article whose 
probable term of usefulness when new was 12 years, which is 8 
years old and apparently in good condition? 
Here n = 12 and m = 8.