a). If annual amount = A, interest rate = i and n = infinity then capitalized cost is
A*(P/A, i, n)
Note: For n = infinity, (P/A, i, n) will be reduced to 1/i
b). Repair cost/year = Aroutine; major repair every 12 years = Amajor
Convert Amajor to an annulized cost first: Amajor*(A/F, i, 12)
Convert this annualized cost into the PW: Amajor*(A/F, i, 12)*(P/A, i, n)
Capitalized cost = Aroutine*(P/A, i, n) + Amajor*(A/F, i, 12)*(P/A, i, n)
c). Repair cost/year = Aroutine; major repair every 10 years = Amajor; cyclical cost for every 6 years = Acyclical increasing by 1,000 per year
Step 1). Calculate the PW of Acyclical arithmetic gradient at the beginning of year 1 (or beginning of year 7, etc.)
Gradient (G) = 1,000 (amount by which Acyclical is increasing every year)
PW of Acyclical (PW1) = Acyclical*((P/A, i, 6) + G*(P/G, i, 6)
Step 2). Now, this PW1 will be repeating every 5 years (beginning of year 6), so its annualized cost over a 5 year period will be PW1*(A/F, i, 5)
Step 3). Covert this to the capitalized cost as PW1*(A/F, i, 5)*(P/A, i, n)
Step 4). PW of Amajor = Amajor*(A/F, i, 120)*(P/A, i, n)
Step 5). PW of Aroutine = Aroutine*(P/A, i, n)
Step 6). Total capitalized cost = step 5 + step 4 + step 3
= Aroutine*(P/A, i, n) + Amajor*(A/F, i, 120)*(P/A, i, n) + PW1*(A/F, i, 5)*(P/A, i, n) OR
Aroutine*(P/A, i, n) + Amajor*(A/F, i, 120)*(P/A, i, n) + [Acyclical*((P/A, i, 6) + G*(P/G, i, 6)]*(A/F, i, 5)*(P/A, i, n)
d). Perpetuity amount = Annual withdrawal/annual interest rate = 16,000/30% = 53,333.33
2. Using such quantities as P, i , A , (A I P i% n), G, ( Al G i%n), (AIF î% n) algebraic express...