In: Economics
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 40,000 gallons per 1,000 hours of operation at the average load encountered in passenger service. System B costs $220,000 and uses 25,000 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 10% salvage values. If jet fuel costs $2.24 a gallon (year 1) and fuel consumption is expected to increase at the rate of 5% per year because of degrading engine efficiency, which engine system should the firm install? Assume 5,000 hours of operation per year and a MARR of 9%. Use the AE criterion. What is the equivalent operating cost per hour for each engine? Assume an end-of-year convention for the fuel cost.
The equivalent annual costs for system A are:
The equivalent annual costs for system B are:
Given
Slno |
System A |
Amount ($) |
System B |
Amount ($) |
1 |
Initial cost |
90000 |
220000 |
|
2 |
Estimated life |
3 years |
Estimated life |
3 years |
3 |
Salvage value |
9000 |
Salvage value |
22000 |
4 |
Fuel cost Year 01 (40000*5*2.24) |
448000 |
Fuel cost Year 01 (25000*5*2.24) |
280000 |
5 |
Fuel cost Year 02 (42000*5*2.24) |
470400 |
Fuel cost Year 02 (26250*5*2.24) |
294000 |
6 |
Fuel cost Year 02 (44000*5*2.24) |
492800 |
Fuel cost Year 02 (27500*5*2.24) |
308000 |
Solution
System A
Annual worth = 90000(A/P,9,3) - 9000(A/F,9,3) + 448000(P/F,9,1) (A/P,9,3) + 470400(P/F,9,2) (A/P,9,3) +492800(P/F,9,3) (A/P,9,3)
Using DCIF Tables
Annual worth = 90000(0.3951) - 9000(0.3051) + 448000(0.9174) (0.3951) + 470400(0.8417) (0.3951) +492800(0.7722) (0.3951)
The equivalent annual worth for system A are = $501982.9
System B
Annual worth = 220000(A/P,9,3) - 22000(A/F,9,3) + 280000(P/F,9,1) (A/P,9,3) + 294000(P/F,9,2) (A/P,9,3) +308000(P/F,9,3) (A/P,9,3)
Using DCIF Tables
Annual worth = 220000(0.3951) - 22000(0.3051) + 280000(0.9174) (0.3951) + 294000(0.8417) (0.3951) +308000(0.7722) (0.3951)
The equivalent annual worth for system B are = $373440.9