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 $93,000 and uses 36,000 gallons of fuel per 1,900 hours of operation at the average load encountered in passenger service.
SYSTEM B costs $186,000 and uses 32,000 gallons of fuel per 1,900 hours of operation at the same level.
Both engine systems have three-year lives. Each system's salvage value is 11.8% of its initial investment. If jet fuel currently costs $2.8 a gallon and fuel consumption is expected to increase at the rate of 7.5% per year because of degrading engine efficiency, which engine system should the firm install? Assume 2,900 hours of operation per year and a MARR of 10.2%. Use the annual equivalent cost criterion. What is the annual equivalent cost of the preferred engine?"

Answer 1

Working note:

(1)

For System A, Annual Fuel consumption, year 1 = (32,000 / 2,000) x 2,400 = 38,400 gallons

For System B, Annual Fuel consumption, year 1 = (23,000 / 2,000) x 2,400 = 27,600 gallons

(2)

Annual fuel cost = Annual fuel consumption x $2.66

(3)

For System A, Salvage value ($) = 98,000 x 10% = 9,800

For System B, Salvage value ($) = 196,000 x 10% = 19,600

(4)

In year 3, Annual cost for each System decreases by the amount of its salvage value.

(5)

PV Factor in year N = (1.142)-N

(6)

First, we compute Present Worth (PW) of Costs for both systems as follows.

SYSTEM - A                  
Year   First Cost ($)   Fuel Consumption   Fuel Cost ($)   Total Cost ($)   PV Factor @14.2%   Discounted Cost ($)
(A)   (B)   (C)=(B) x $2.66   (D) = (A) + (C)   (E)   (D) x (E)
0   98,000       0   98,000   1.0000   98,000
1       38,400   1,02,144   1,02,144   0.8757   89,443
2       40,512   1,07,762   1,07,762   0.7668   82,629
3       42,740   1,13,689   1,03,889   0.6714   69,754
PW ($) =   3,39,826
SYSTEM - B                  
Year   First Cost ($)   Fuel Consumption   Fuel Cost ($)   Total Cost ($)   PV Factor @14.2%   Discounted Cost ($)
(A)   (B)   (C)=(B) x $2.66   (D) = (A) + (C)   (E)   (D) x (E)
0   1,96,000       0   1,96,000   1.0000   1,96,000
1       27,600   73,416   73,416   0.8757   64,287
2       29,118   77,454   77,454   0.7668   59,390
3       30,719   81,714   62,114   0.6714   41,705
PW ($) =   3,61,382
(6) Annual equivalence cost (EUAC) = PW / P/A(14.2%, 3) = PW / 2.3139**

EUAC, System A ($) = 339,826 / 2.3139 = 146,863

EUAC, System B ($) = 361,382 / 2.3139 = 156,179

Since System A has lower EUAC of costs, this is preferred with an EUAC of $146,863.

**P/A(r%, N) = [1 - (1 + r)-N] / r

P/A(14.2%, 3) = [1 - (1.142)-3] / 0.142 = (1 - 0.6714) / 0.142 = 0.3286 / 0.142 = 2.3139