Question

1) The equivalent annual costs for system A are?

2) The equivalent annual costs for system B are?

3) the equiavlent operating cost per hour for process A is?

4) the equiavlent operating cost per hour for process B is?

5) which engine should you choose?

An airt e condenng wo types of engine systems for use in its planes Each has the same ife and the same maintenance and repair record. System System B co $28 nd uses 43.000 gallons per 1,000 hours of operation at the average load encourtered in passenger servicee ,000 and uses 30,000 galions 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 inital investment, the systems have 18% salvage values If jet fuel costs $2.15 a gallon year 1) and fuel consumption is expected to increase at the rate of 5% per year because of degrading engine eficiency, which engine system should the frm install? Assume 4,000 hours of operation per year and a MARR of 13 % Use the AE oriterion Whet is he eguvelent operating cost per hous for each engine Assume an end-of-year convention for the fuel cost Cick the icon to view the interest factors for discrete compounding when MARR 13 % per year The equivalent annual costs for system A are S (Round to the nearest dollar)y

Answer 1

System A

Initial Cost=Co=-$90000

Fuel Consumption in first year=F=43000*(4000/1000)*2.15=-$369800

Growth rate of fuel consumption=g=5%

Salvage =S=90000*18%=$16200

MARR=i=13%

Useful life=n=3 years

Let us calculate annualized costs for Initial Cost, Fuel and Salvage.

EUAC(I)=-90000*(A/P,13%,3)

EUAC(S)=16200*(A/F,13%,3)

EUAC(F)=-PW(F)*(A/P,13%,3)

Now we estimate the interest factors and PW(F)

A/P, i, n) 1 1 (1+1

0.13 A/P, 0.13,3 1- 1+0.13)T 0.423522 1

(A/F, i, n) 1+ i )n -1

0.13 A/F,0.13,3)=(1+0.13)3-1 0.293522

(1 g)" (1" F 1 PV g= g

(10.05)3 369800 1 PV F 0.13- 0.05 (1+0.13)3 913904.33

EUAC(I)=-90000*(A/P,13%,3)=-90000*0.423522=-$38116.98

EUAC(S)=16200*(A/F,13%,3)=16200*0.293522=$4755.06

EUAC(F)=-PW(F)*(A/P,13%,3)=-913904.33*0.423522=-$387058.59

EUAC for system A=-38116.98+4755.06-387058.59=-$420420.51 or say -$420421

EUOC for system A=-387058.59/4000=-$96.76 per hour

System B

Initial Cost=Co=-$280000

Fuel Consumption in first year=F=30000*(4000/1000)*2.15=-$258000

Growth rate of fuel consumption=g=5%

Salvage =S=280000*18%=$50400

MARR=i=13%

Useful life=n=3 years

Let us calculate annualized costs for Initial Cost, Fuel and Salvage.

EUAC(I)=-280000*(A/P,13%,3)

EUAC(S)=50400*(A/F,13%,3)

EUAC(F)=-PW(F)*(A/P,13%,3)

Now we estimate the PW(F)

(1 g)" (1" F 1 PV g= g

(10.05)3 (1+0.13)3) 258000 637607.67 1 PV F 0.13- 0.05

EUAC(I)=-280000*(A/P,13%,3)=-280000*0.423522=-$118586.16

EUAC(S)=50400*(A/F,13%,3)=50400*0.293522=$14793.51

EUAC(F)=-PW(F)*(A/P,13%,3)=-637607.67*0.423522=-$270040.88

EUAC for system B=-118586.16+14793.51-270040.88=-$373833.53 or say -$373834

EUOC for system B=-270040.88/4000=-$67.51 per hour

EUAC for system A=-38116.98+4755.06-387058.59=-$420420.51 or say -$420421

EUOC for system A=-387058.59/4000=-$96.76 per hour

EUAC for system B=-118586.16+14793.51-270040.88=-$373833.53 or say -$373834

EUOC for system B=-270040.88/4000=-$67.51 per hour

Absolute annual cost of system B is lower. It should be selected.