Question

Based on an annual cost comparison, using an interest rate of 15%, determine the most economical method of replacing a wearing surface on a road. Method A has a first cost of $20000 a life of 10 years and an annual maintenance cost of $1000. Method B has a first cost of $30000 a life of 14 years and an annual maintenance cost of $800.

Answer 1

Method A has a first cost of $20000 a life of 10 years and an annual maintenance cost of $1000.

Method B has a first cost of $30000 a life of 14 years and an annual maintenance cost of $800.

Commom time period is 70 years (LCM).

PW of method A = 20000 + 1000(P/A, 15%, 10) = 20000 + 1000*5.01877 = 25018.77

Repeat the cycle for 7 periods = 25018.77(1 + (P/F, 15%, 10) + (P/F, 15%, 20) + (P/F, 15%, 30) + (P/F, 15%, 40) + (P/F, 15%, 50) + (P/F, 15%, 60))

= 25018.77*(1 + (1 + 15%)^-10 + (1 + 15%)^-20 + (1 + 15%)^-30 + (1 + 15%)^-40 + (1 + 15%)^-50 + (1 + 15%)^-60))

= 25018.77*1.32827

= 33231.68

Find the annual equivalent cost = 33231.68(A/P, 15%, 70) = 6621.48

Now find PW of B = 30000 + 800(P/A, 15%, 14) = 30000 + 800*5.72448 = 34579.58

Repeat the cycle for 5 periods = 34579.58(1 + (P/F, 15%, 14) + (P/F, 15%, 28) + (P/F, 15%, 42) + (P/F, 15%, 56))

= 34579.58*1.16452

= 40268.6125

Find the AEC of B = 40268.6125(A/P, 15%, 70) = 6040.63

Since AEC of B is less, select B.