Question

An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at i = 8% per year on the basis of an annual worth evaluation.

	Land	Incineration	Contract
First Cost	$-140,000	$-700,000	0
AOC per Year	$-85,000	$-40,000	$-130,000
Salvage Value	$20,000	$250,000	0
Life	4 years	6 years	2 years

The environmental engineer selects a ---------------

a. private disposal contract
b. fluidized-bed incineration  
c. land application  .

Answer 1

# Annual worth by Private Disposal Contract:

The annual worth will be the Annual Operating Costs as there is no other expense involved under this contract.

Therefore, Annual worth of Private Disposal Contract = $130,000

# Annual worth by Fluidized-bed Incineration method:

Annual depreciation = (First Cost - Salvage Value) / Life

= (700,000 - 250,000) / 6

= $75,000

Interest per annum = First Cost * Interest rate

= 700,000 * 8%

= $56,000

Total Annual worth = Annual depreciation + Interest per annum + AOC per Year

= 75,000 + 56,000 + 40,000

= $171,000

Therefore, Annual worth of Incineration = $171,000

# Annual worth by Land Application method:

Annual depreciation = (First Cost - Salvage Value) / Life

= (140,000 - 20,000) / 4

= $30,000

Interest per annum = First Cost * Interest rate

= 140,000 * 8%

= $11,200

Total Annual worth = Annual depreciation + Interest per annum + AOC per Year

= 30,000 + 11,200 + 85,000

= $126,200

Therefore, Annual worth of Land Application = $126,200.

It can be seen from the calculation of annual worth of each contract that the the Land application will be the most economical to use since it has the least cost among the three options.

Therefore the correct answer is option C. Land application.