Question

Water supply for an irrigation system can be obtained from a stream in some nearby mountains. Two alternatives are being considered, both of which have essentially infinite lives, provided proper maintenance is performed. The first is a concrete reservioir with a steel pipe system and the second is an earthen dam with a wooden aqueduct. Below are the costs associated with each. Compare the present worths of the two alternatives, using an interest rate of 9%. Which alternative should be chosen {Perform all calculations using 5 significant figures and round any monetary answers to the nearest cent}?

	Concrete Reservoir	Earthen Dam
First Cost [dollars]	460,000	180,000
Annual Maintenance Costs [dollars]	1,950	11,000
Replacing the wood portion of the aqueduct each 15 years	N/A	105,000

The cost Concrete Reservoir is (in PW dollars): ?

The cost of the Earthen Dam plus Aqueduct is (in PW dollars): ?

Which alternative should be chosen ?

Answer 1

Interestrate = 9%

Concrete Reservoir:

First cost =460,000

Annual maintenance cost = 1,950

No replacement cost

Present value of maintenance cost of first year: [1,950 / (1 + 0.09)^¹]

Present value of maintenance cost of second year: [1,950 / (1 + 0.09)^²]

Present value of maintenance cost of third year: [1,950 / (1 + 0.09)^³]

and so on...

Sum of present value of annual maintenance cost: [1,950 / (1 + 0.09)^¹] + [1,950 / (1 + 0.09)^²] + [1,950 / (1 + 0.09)^³] + ...... whose sum can be calculated as [a / (1 - r)] as it forms a G.P.

a = [1,950 / (1 + 0.09)^¹]

r (ratio of two consecutive terms) = (1 / 1.09) = 0.9174

Sum of G.P. = [1,950 / (1 + 0.09)^¹] / (1 - 0.9174) = 21,666.7

Present value of this alternative = First cost + Present value of annual cost = 460,000 + 21,666.7 = 481,666.7

Earthen Dam:

First cost = 180,000

Present value of maintenance cost of first year: [11,000 / (1 + 0.09)^¹]

Present value of maintenance cost of second year: [11,000 / (1 + 0.09)^²]

Present value of maintenance cost of third year: [11,000 / (1 + 0.09)^³]

and so on...

Sum of present value of annual maintenance cost: [11,000 / (1 + 0.09)^¹] + [11,000 / (1 + 0.09)^²] + [11,000 / (1 + 0.09)^³] + ...... whose sum can be calculated as [a / (1 - r)] as it forms a G.P.

a = [11,000 / (1 + 0.09)^¹]

r (ratio of two consecutive terms) = (1 / 1.09) = 0.9174

Sum of G.P. = [11,000 / (1 + 0.09)^¹] / (1 - 0.9174) = 122,222.22

Cost after every year = 105,000

Present value of cost after 15 year: [105,000 / (1 + 0.09)^¹⁵]

Present value of cost after 30 year: [105,000 / (1 + 0.09)^³⁰]

Present value of cost after 45 year: [105,000 / (1 + 0.09)^⁴⁵]

and so on...

Sum of present value of annual maintenance cost: [105,000 / (1 + 0.09)^¹⁵] + [105,000  / (1 + 0.09)^³⁰] + [105,000 / (1 + 0.09)^⁴⁵] + ...... whose sum can be calculated as [a / (1 - r)] as it forms a G.P.

a = [105,000 / (1 + 0.09)^¹⁵]

r (ratio of two consecutive terms) = (1 / 1.09^¹⁵) = 0.2745

Sum of G.P. = [105,000 / (1 + 0.09)^¹⁵] / (1 - 0.2745) = 39,765.63

Present value of this alternative = First cost + Present value of annual cost = 180,000 + 122,222.2 + 39,765.63 = 341,987.83

Concrete Reservoir should be chosen due to its lowest present value of cost.