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A city has developed a plan to provide for future municipal water needs. The plan proposes an aqueduct that passes through 500 feet of tunnel in a nearby mountain. Two alternatives are being considered. The first proposes to build a full-capacity tunnel now for $556,000. The second proposes to build a half capacity tunnel now (cost = $402,000), which should be adequate for 20 years, and then to build a second parallel half-capacity tunnel. The maintenance cost of the tunnel lining for the full-capacity tunnel is $40,000 every 10 years, and for each half-capacity tunnel it is $32.000 every 10.

The Friction losses in the half-capacity tunnels will be greater than in the full-capacity tunnel. The estimated additional pumping costs in each half-capacity tunnel will be $2000 per year. Based on capitalized cost and a 7% interest rate, which alternative should be selected?

Respuesta :

TomShelby

Answer:

The second alternative capitalized cost is lower thus, it is convient to go for the half-capacity tanks.

Explanation:

Full capacity net worth:

556,000 tunnel investment

+ present value of maintenance cost at perpetuity:

[tex]PV \div \frac{1-(1+r)^{-time} }{rate} = C\\[/tex]  

PV  $40,000.00  

time 10

rate 0.07

[tex]40000 \div \frac{1-(1+0.07)^{-10} }{0.07} = C\\[/tex]  

C  $ 2,895.100  

This will be the annual cost for the maintenance as it is every 10 year we calcualte the perpetuity which gneerates this amount:

$2,895.1 / 0.07 = $41,358.57  

Total cost: 556,000 + 41,358.57 = 597,358.57‬

Now, we solve for the cost of the half-capacity. As we already know the value of the first alternative our analisys should stop if we surpass it.

402,000

+ PV of the second

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]  

Maturity  $402,000.0000  

time   20.00  

rate  0.07

[tex]\frac{402000}{(1 + 0.07)^{20} } = PV[/tex]  

PV   103,884.44  

Then, maintencance cost:

[tex]PV \div \frac{1-(1+r)^{-time} }{rate} = C\\[/tex]  

PV  $32,000.00  

time 10

rate 0.07

[tex]32000 \div \frac{1-(1+0.07)^{-10} }{0.07} = C\\[/tex]  

C  $ 2,316.080  

$33,086.86   for the first one

the second pool will start withdrawals in 20 years so:

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]  

Maturity  $33,086.8600  

time   20.00  

rate  0.07

[tex]\frac{33086.86}{(1 + 0.07)^{20} } = PV[/tex]  

PV   8,550.27  

Then, we have a perpetuity of 2,000 dollar for additional pumping cost

on each one:

2,000/0.07 = 28,571.43

the second again is discounted for 20 years:

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]  

Maturity  $28,571.4300  

time   20.00  

rate  0.07

[tex]\frac{28571.43}{(1 + 0.07)^{20} } = PV[/tex]  

PV   7,383.40  

capitalized cost for the second alternative:

402,000 + 103,884.44

+ 33,086.86 + 8,550.27

+ 28,571.45 + 7,383.4

Total capitalized cost: 583,476.42‬

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