Problems in the Mathematical Theory of Investment

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How much can one afford to expend on a piece of equipment which must be replaced at a cost of flOO every three years for an unlimited time, in order to get better equip- ment which will need replacement only every five years ?
Assume money steadily worth 4%.
Let X = the cost of the better article.
Then we have to compare $100 every three years in x every five years in perpetuity. perpetuity.
To compare two plans we must find the value of each at the same time, in this case preferably at the pre
...sent. Since here we must add the cost of first installation to the present value of the perpetuity, this is equivalent to finding the capitalized cost of each article. These plans will be equivalent in the long run if their capitalized costs are equal. This gives the equation of value 100 1 a: 1 . . ^ ■ ■:o4-^=:o4-^'^*^%' whence x = 100 — a-. = 160.42.
One can therefore afford to expend x — 100 = 60.42 for the improvement of the piece of equipment in order to make it last five years instead of three.


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