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 $92,000 and uses 28,000 gallons of fuel per 1,100 hours of operation at the average load encountered in passenger service. SYSTEM B costs $184,000 and uses 21,000 gallons of fuel per 1,100 hours of operation at the same level. Both engine systems have three-year lives. Each system's salvage value is 10.5% of its initial investment. If jet fuel currently costs $2.6 a gallon and fuel consumption is expected to increase at the rate of 6% per year because of degrading engine efficiency, which engine system should the firm install? Assume 2,600 hours of operation per year and a MARR of 8.3%. Use the annual equivalent cost criterion. What is the annual equivalent cost of the preferred engine?

Answer 1

Present worth of geometric series = A *[1 - (1+g)^n /(1+i)^n] /(i-g)

For System A

Consumption of Fuel per year = 2600 / 1100 * 28000 = 66181.818

First yr cost of Fuel = 66181.818 * 2.6 = 172072.72

Present value of fuel cost = 172072.72 * [1 - (1+0.06)^3 /(1+0.083)^3] /(0.083-0.06)

= 172072.72 * [1 - (1.06)^3 /(.083)^3] /(0.023)

= 172072.72 * 2.7116705

= 466604.52

Equivalent annual cost = 92000*(A/P,8.3%,3) + 466604.52 *(A/P,8.3%,3) - 0.105*92000*(A/F,8.3%,3)

= 92000*(0.083*((1 + 0.083)^3)/((1 + 0.083)^3-1)) + 466604.52 *(0.083*((1 + 0.083)^3)/((1 + 0.083)^3-1)) - 0.105*92000*(0.083 /((1 + 0.083)^3-1))

= 92000*(0.083*((1.083)^3)/((1.083)^3-1)) + 466604.52 *(0.083*((1.083)^3)/((1.083)^3-1)) - 0.105*92000*(0.083 /((1.083)^3-1))

= 92000*0.390136 + 466604.52 *0.390136 - 0.105*92000* 0.307136

= 214964.799 ~ 214964.80

For System B

Consumption of Fuel per year = 2600 / 1100 * 21000 = 49636.363

First yr cost of Fuel = 49636.363 * 2.6 = 129054.54

Present value of fuel cost = 129054.54 * [1 - (1+0.06)^3 /(1+0.083)^3] /(0.083-0.06)

= 129054.54 * [1 - (1.06)^3 /(.083)^3] /(0.023)

= 129054.54 * 2.7116705

= 349953.389

Equivalent annual cost = 184000*(A/P,8.3%,3) + 349953.389 *(A/P,8.3%,3) - 0.105*184000*(A/F,8.3%,3)

= 184000*(0.083*((1 + 0.083)^3)/((1 + 0.083)^3-1)) + 349953.389 *(0.083*((1 + 0.083)^3)/((1 + 0.083)^3-1)) - 0.105*184000*(0.083 /((1 + 0.083)^3-1))

= 184000*(0.083*((1.083)^3)/((1.083)^3-1)) + 349953.389 *(0.083*((1.083)^3)/((1.083)^3-1)) - 0.105*184000*(0.083 /((1.083)^3-1))

= 184000*0.390136 + 349953.389 *0.390136 - 0.105*184000* 0.307136

= 202380.572 ~ 202380.57

As annual cost of System B is less, it should be selected

EUAC of System B = 202380.57