Question

An industrial park is being planned for a tract of land near the river. To prevent flood damage to the industrial buildings that will be built on this low-lying land, an earthen embankment can be constructed. The height of the embankment will be determined by an economic analysis of the costs and benefits. The following data have been gathered: Embankment Height Above Roadway (m) Initial Cost 2.0 $100,000 2.5 165,000 3.0 300,000 3.5 400,000 4.0 550,000 Flood Level Above Roadway (m) Average Frequency That Flood Level Will Exceed Height in Col. 1 2.0 Once in 3 years 2.5 Once in 8 years 3.0 Once in 25 years 3.5 Once in 50 years 4.0 Once in 100 years The embankment can be expected to last 50 years and will require no maintenance. Whenever the flood water flows over the embankment, $300,000 of damage occurs. Determine which of the five heights above the roadway should be selected. The interest rate is 12%. (50 points)

Answer 1

Answer:

The best height will be of 3.5 as it provides the best expected present worth.

Explanation:

2.0 heights Cost $100,000 now and it is expected to have losses of 300,000 every three years:

Present Value of Annuity  

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

C 300,000

time 16.67

(50 years of useful life / 3 years expected flood)

rate 0.404928

(we capitalize the 12% annual into a 3-year rate)

$300000 \times \displaystyle \frac{1-(1+0.404928)^{-16.67} }{0.404928} = PV$  

PV $738,308.8983  

Present Worth: 100,000 + 738,308.90 = 838,308.90

2.5 height: cost $165,000, and we expected damage every eight year:

Present Value of Annuity  

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

C 300,000

time 6.25 (50 years useful life / 8 years)  

rate 1.475963176  (we capitalize the 12% annual into a 8-year rate)

$300000 \times \displaystyle \frac{1-(1+1.475963176)^{-6.25}}{1.475963176} = PV$  

PV 203,257.0478  

Present worth: 203,257.05 + 165,000 = 368,257.05

3.0 cost $300,000, and we expect a flood every 25 years

$300000 \times \displaystyle \frac{1-(1+16)^{-2} }{16} = PV$  

PV $18,685.0464  

Present worth: 300,000 + $18,685.0464   = 318,685.05

3.5 cost $400,000, and we expect a floor every 50 years:

PRESENT VALUE OF LUMP SUM  

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

Maturity  300,000.00

time   50.00  

rate  0.12

$\frac{300000}{(1 + 0.12)^{50} } = PV$  

PV   1,038.05  

Cost: 400,000 + 1,038.05 = 401,038.05