An existing electrical power line needs to have its capacity increased, and this can be done in either of two ways. The first method is to add a second conductor to each phase wire using the same poles, insulators and fittings, at a construction cost of $15,000. The second method for increasing capacity is to build a second line parallel to the existing line, using new poles, insulators and fittings, at a construction cost of $23,000. At some time in the future, the line will require another increase in capacity, with the first alternative now requiring a second line at a cost of $32,500, and the second alternative requiring added conductors at a cost of $23,000. If interest rate is 6%, how many years between the initial expenditure and the future expenditure will make the two methods economically equal

3 years

Explanation:

First method

The PV of the investment can be written as:

PV1 = $15,000 + $32,500/(1+0.06)^n

Second method

The PV of the investment can be written as:

PV2 = $23,000 + $23,000/(1+0.06)^n

After n years both projects will be economically equal. Hence their present value must be equal PV1 = PV2

$15,000 + $32,500/(1.06)^n = $23,000 + $23,000/(1.06)^n

$23,000 - $15,000 = $32,500/(1.06)^n - $23,000/(1.06)^n

$8,000 = $9,500/(1.06)^n

(1.06)^n = $9,500/$8,000

(1.06)^n = 1.1875

Taking log on both sides we get:

nlog1.06 = log 1.1875

n = log 1.1875/log 1.06

n = 0.07463361829/0.02530586526

n = 2.94926166417121

n = 3

So, the answer is 3 years

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