Week 11 Heavy Equipment Costs PDF

Construction Equipment 1  Ownership costs Time value of money analysis Economic useful life is estimated by hours of use  Ownership Cost Example Ex 1. A crawler tractor costs $155,000 and its annual usage will be about 2,000 hours per year....

Construction Equipment 1  Ownership costs Time value of money analysis Economic useful life is estimated by hours of use  Ownership Cost Example Ex 1. A crawler tractor costs $155,000 and its annual usage will be about 2,000 hours per year. Salvage value is 12% of the purchase price. – Interest 9%, Taxes 2%, insurance 2%, Storage 2% What is the estimated annual ownership cost of the tractor if it is operated under average conditions?  Ownership Cost Example From Table 4.1, tractor’s usage is 2,000 hours per year, the period of ownership is 12,000hr/2,000hr = 6 years. - Minimum Acceptable Rate of Return: MARR=9%+2%+2%+1%=15% - Salvage value (Present Worth) = 0.12*$155,000*(P/F,15%,6 = 0.4323) = $8,041 Net Purchase Cost = $155,000-$8,041=$146,959 For annual cost: A= $146,959*(A/P, 15%, 6 = 0.2642) = $38,827 Hourly ownership cost =($38,827,/yr)/(2,000hr/yr)=$19.41/hr  Operating Costs Operating costs are influenced by – Age of the equipment, repair, and operating conditions Maintenance and repair costs- use a percentage of the annual straight-line (SL) depreciation Separate replacement costs: Tire, cutting edges, bucket teeth, and ripper. Operating Costs: Repair Hourly repair cost = ( RF )(Useful Life in hr.)( HDR) 10,000hr – RF: Repair Factor from Table 4.2 – HDR: Hourly Depreciation Rate  Operating Costs: Tires Use separate cash flow analysis Tire maintenance and repair costs are estimated at 15% of the hourly Straight- Line depreciation amount.  Operating Costs: Fuel Hourly Fuel Cost= (Horsepower)* (Fuel Factor)*(Fuel Cost) Operating Costs: Service, downtime, labor cost Servicing (Filter, oil and grease) costs are estimated as a percentage of the hourly fuel cost. Downtime – Use an operating factor – Operating factor=actual working minutes/hour Labor cost – hourly wage rate + fringe benefits – Fringe benefits are about 20% to 30% of the actual wage rate.  Operating Cost Example A $114,000 wheeled loader will be used 2,000 hours per year for 6 years useful life with $35,000 salvage value. Tires for the loader cost $4,000 and the horsepower rating of the loader’s diesel engine is 105 HP. Diesel fuels cost $1.2 /gallon. Operator cost is $34/hr with fringe benefits. At a MARR of 12%, what are the hourly operating costs for the loader?  Operating Cost Example Repair - Purchase cost less tire = $114,000- $4,000=$110,000 – Hourly depreciation cost D = ($110,000 - $35,000)/ 12,000 hr = $6.25 /hr - Find repair factor in Table 4.2: 0.55 - Repair Cost = (0.55)(12,000hr)($6.25/hr)/10,000hr = $4.13/hr  Operating Cost Example Tire - Find tire life from Table 4.3: 3,200 hrs - Useful life of tires = (3200hr) / (2000hr/yr) = 1.6 yrs (In the H.W. and exam, round it up to 2 years.) - Hourly cost = ($4,000) (A/P, 12%, 1.6=0.803) / (2,000 hr) = $1.61/hr - Tire repair cost D = $4000 / 3200hr = $1.25/hr Repair cost = 15% * D = 0.15 * $1.25/hr = $0.19/hr  Operating Cost Example Fuel – Find the fuel factor in Table 4.4: 0.036 – Hourly fuel cost = (105 HP) (0.036 gal/HP-hr)($1.20/gal) = $4.54/hr Service cost (filter, oil, and grease) – Find the fuel factor in Table 4.5: 0.33 – Hourly service cost = (0.33) ($4.54/hr) = $1.50/hr  Operating Cost Example Total hourly operating cost – Repair : $4.13/hr – Tire (including tire repair): $1.61+$0.19 = $1.80 - Fuel: $4.54/hr - Service: $1.50/hr - Operator cost: $34.00/hr :: Total hourly operating cost: $45.97/hr  Summary Ownership cost – Depreciation: consider time value of money – MARR = interest + insurance + storage + license Operating Cost – Repair: straight-line depreciation X repair factor – Tire: Ownership + repair – Fuel: Horsepower X fuel factor X useful life/10000 – Service: Service factor X Fuel cost Unit Cost = Hourly cost / Hourly production rate