Business, 12.02.2020 05:51 lindseylewis313

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?

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Answer from: narnar5664

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:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $40,000.00

time10

rate0.07

$40000 \div \frac{1-(1+0.07)^{-10} }{0.07} = C\\$

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

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $402,000.0000

time 20.00

rate 0.07

$\frac{402000}{(1 + 0.07)^{20} } = PV$

PV 103,884.44

Then, maintencance cost:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $32,000.00

time10

rate0.07

$32000 \div \frac{1-(1+0.07)^{-10} }{0.07} = C\\$

C $ 2,316.080

$33,086.86 for the first one

the second pool will start withdrawals in 20 years so:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $33,086.8600

time 20.00

rate 0.07

$\frac{33086.86}{(1 + 0.07)^{20} } = PV$

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:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $28,571.4300

time 20.00

rate 0.07

$\frac{28571.43}{(1 + 0.07)^{20} } = PV$

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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