Question

An airline is considering two types of engine systems for use in its planes. Each has the same life and the same maintenance and repair record.

System A costs ​$90,000 and uses 39,000 gallons per​ 1,000 hours of operation at the average load encountered in passenger service.

System B costs ​$220,000 and uses 26,000 gallons per​ 1,000 hours of operation at the same level.

Both engine systems have​ three-year lives before any major overhaul is required. On the basis of the initial​ investment, the systems have 18​% salvage values. If jet fuel costs ​$2.18 a gallon (year 1) and fuel consumption is expected to increase at the rate of 5​% per year because of degrading engine​ efficiency, which engine system should the firm​ install? Assume 2,000 hours of operation per year and a MARR of 8​%. Use the AE criterion. What is the equivalent operating cost per hour for each​ engine? Assume an​ end-of-year convention for the fuel cost.

The equivalent annual costs for system A are ​$______. ​(Round to the nearest​ dollar.)

The equivalent annual costs for system B are ​$______. ​(Round to the nearest​ dollar.)

The equivalent operating cost per hour for process A is $________. (Round to the nearest cent.)

The equivalent operating cost per hour for process B is $________. (Round to the nearest cent.)

Which engine system should be selected? Choose the correct answer below.

System B or System A

Answer 1

MARR=i=8%

Useful life=n=3 years

Calculate the Present value of annuity of $1 with the help of given information

System A

Fuel consumption in Year 1=Q1=39000/1000*2000=78000 gallons

Fuel Cost in year 1=FC1=2.18*78000=$170040

Fuel consumption in Year 2=Q2=78000*1.05 =81900 gallons

Fuel Cost in year 2=FC2=2.18*81900=$178542.00

Fuel consumption in Year 3=Q3=81900*1.05 =85995 gallons

Fuel Cost in year 2=FC3=2.18*85995=$187469.10

PW of Fuel Cost=PWf=FC1/(1+i)+FC2/(1+i)^2+FC3/(1+i)^3

PWf=170040/(1+8%)+178542.00/(1+8%)^2+187469.10/(1+8%)^3=$459334.45

Cost of System=Co=$90,000

Salvage value=S=18%*90000=$16200

Present Worth of Salvage value=PWs=S/(1+i)^3=16200/(1+8%)^3=$12860.08

Present Worth of Costs of system=PW =-Co-PWf+PFs=-90000-459334.45+12860.08=-$536474.37

(negative sign indicates the cash outflow)

Equivalent Annual cost=PW/PWFa=536474.37/2.577097=$208170.03

Equivalent operating cost per hour=Equivalent annual cost/2000=208170.03/2000=$104.09

System B

Fuel consumption in Year 1=Q1=26000/1000*2000=52000 gallons

Fuel Cost in year 1=FC1=2.18*52000=$ 113360.00

Fuel consumption in Year 2=Q2=52000*1.05 = 54600 gallons

Fuel Cost in year 2=FC2=2.18* 54600=$ 119028.00

Fuel consumption in Year 3=Q3=54600*1.05 =57330 gallons

Fuel Cost in year 2=FC3=2.18*57330=$124979.40

PW of Fuel Cost=PWf=FC1/(1+i)+FC2/(1+i)^2+FC3/(1+i)^3

PWf=113360.00/(1+8%)+119028.00/(1+8%)^2+124979.40/(1+8%)^3=$306222.97

Cost of System=Co=$220,000

Salvage value=S=18%*220000=$39600

Present Worth of Salvage value=PWs=S/(1+i)^3=39600/(1+8%)^3=$31435.76

Present Worth of Costs of system=PW =-Co-PWf+PFs=-220000-306222.97+31435.76=-$494787.21

(negative sign indicates the cash outflow)

Equivalent Annual cost=PW/PWFa=494787.21/2.577097=$191994.02

Equivalent operating cost per hour=Equivalent annual cost/2000 =191994.02/2000=$96.00

Equivalent operating cost for system B is lower. It must be selected.