Question

Turbine	United Hydro Services	Kiser Hydro, LLC
Initial Cost	$200,000	$1,500,000
Maintenance Cost	$200 per year, increasing by 4% per year	$1,000 per year for the first ten years, then $2,000 per year thereafter
Overhaul Cost	$50,000 at year ten	$750,000 at year thirty
Income	$20,000 per year	$25,000 per year
Salvage Value	$15,000	$50,000
Life	20 Years	Infinite

. Memphis Light, Gas and Water Division is exploring the possibility of installing hydrokinetic turbines in the Mississippi River to generate power for the Memphis and Shelby County area. Consulting engineers have identified two potential types of hydrokinetic turbines to meet the needs of MLG&W. The costs and expected income for the two types of turbines are shown in the table below. Using a perpetual annual worth analysis and a MARR of 6% per year, determine which type of hydrokinetic turbine, if any, should be selected. Complete all calculations to the nearest dollar.

Answer 1

The formula for computation of Annual worth of a machine = Initial Cost of the machine(A/P, i,n) + salvage Value(A/F, i, n) + Total Income- other cost, where n= machine total life (in years), i = MARR rate

For United Hydro Services turbine,

Overhaul Cost at 10 years = $50,000 i.e., annual overhaul cost = $10,000

Maintenance Cost = $200, increasing at 4% per annum, total income = $20,000

Iniitial Cost of the turbine 1 = $200,000, salvage value = $15,000, MARR = 6% per year, machine life = 20 years

Therefore,

AW¹ = $200,000(A/P, 6, 20) + $15,000(A/F, 6,20) + $20,000- $10,000 -( $200 * 4/100 +$200)

        =1200000+ 0.0000000000011 + 10,000 -204 = $1209796.00 (Approximate)

Similarly, for Kiser Hydro, LLC turbine,

Initial Cost = $1,500,000

Turbine life = infinite but from the MCA table, we get useful life of Turbine = 40 years i.e., n = 40

Maintenance Cost = $1,000 annually for first 10 years then $2,000 per year

Therefore Average Annual maintenance cost for 40 years = (10 *$1,000 + 30 *$2,000)/40 = $1,750

Overhaul cost = $750,000 at thirty years

Therefore, Annual overhaul cost = 750,000/30 = $2,500

Income = $25,000 per year

Salvage Value = $50,000

Therefore,

Annual Worth for second turbine, AW² = $1,500,000(A/P, 6, 40) + $50,000(A/F, 6, 40) +$25,000-$1,750-$2,500

                                                      = 900000 +0.000000000000000000000000000047+25,000-4,250 = $9020750 (Approximately)

It is clear that AW² is greater than AW¹ Therefore, Memphis Light, Gas and Water Division should select Kiser Hydro, LLC turbine