Q1 A new highway is being evaluated at an interest rate of i
Solution
FOr A
Initial cost = 18000 k
Annual maintenance = 342k
i = 5.5%
Periodic maintenance = 1600k
n = 12
Capitalized cost = -18000k - 1600 k (P/F,i%,n) - 342k/i
Capitalized cost = -18000k - 1600 k (P/F,5.5%,n) - 342k/5.5%
(P/F,i,n) = 1 / (1+i)n
(P/F,5.5%,12) = 1 / (1 + 0.055)12 = 0.526
Capitalized cost = -18000k - 1600 k x0.526 - (342k/0.055)
Capitalized cost = -25059k
For B
Initial cost = 28000 k
Annual maintenance = 238k
i = 5.5%
Periodic maintenance = 1200k
n = 18
Capitalized cost = -28000k - 1200 k (P/F,i%,n) - 238k/i
Capitalized cost = -28000k - 1200 k (P/F,5.5%,n) - 238k/5.5%
(P/F,i,n) = 1 / (1+i)n
(P/F,5.5%,18) = 1 / (1 + 0.055)18 = 0.382
Capitalized cost = -28000k - 1200 k x0.382 - (238k/0.055)
Capitalized cost = -32785.67
As the value for A is greater than B we select A
