Question

Which of the following is closest to the capitalized cost of an initial investment of $200,000, refurbishment costs of $50,000 every five years, and annual operating costs of $25,000 continuing forever? Assume an interest rate of 6% per year.

$350,000

$620,000

$760,000

$860,000

Answer 1

Capitalization cost = Initial investment + [ Annual operating cost / Interest rate] + [ Refurbished costs ( A/F, Interest rate(i), year(n) /Interest rate

Capitalization cost = $ 200,000 + [$ 25,000/ 0.06] + [ $ 50,000 X (0.1774) / 0.06

Capitalization cost = $ 200,000 + $ 416,667 + $ 147,833 = $ 764,500

So, it will be closest to $ 760,000.

Note : From A/F table you can get the value of A/F:

(A/F, i, n) = (A/F 0.06 , 5) = 0.1774

Mathematical formula for A/F( Manually you can do without table) :

A/F ( interest :0.06, time: 5 years) = i/ (1+i)n- 1

A/ F = 0.06/ ( 1+ 0.06)5-1

A/F = 0.06/ 1.3382255776 - 1

A/F = 0.06/ 0.3382255776

A/F = 0.17739640043

A/F = 0.1774 ( rounded up)