Off-Road Equipment Management TSM 262: Spring 2016 LECTURE 25: Machinery Costing I Off-Road Equipment Engineering Dept of Agricultural and Biological Engineering achansen@illinois.edu

Homework and Lab

Class Objectives Students should be able to: Differentiate between fixed and operating costs of machines Identify the different cost components that are represented under both fixed and operating costs Determine the costs of owning and operating machines Evaluate the effect of machine size on timeliness and overall machinery costs

Machinery Costs Fixed/Ownership/Overhead Costs Independent of machine use Variable/Operating Costs Dependent only on machine use Timeliness Costs Costs incurred from any loss of crop quantity or quality due to use of set of machinery to complete desired field operation

Fixed Costs: Depreciation Reduction of machine value with time and use Age, obsolescence, other factors? (wear connected with machine use?) Often largest single cost of machine ownership =Purchase price - current or salvage value Life of machine=time when further repair cannot be justified Depreciation has a big impact on income tax Estimation methods

Depreciation Estimation Straight-line method Equal reduction in value each year machine owned Depreciation=(purchase price-salvage value)/years of use Declining balance method Percentage of remaining value depreciated each year Closer to reality than straight-line method Depreciation rate higher in earlier years, decreases as machine gets older Sinking-fund method Establishment of fund with annual payments that accumulate to an amount that will purchase a new machine

Depreciation Method

Depreciation Estimation (ASAE D497.7) Actual trade-in or “as is” value Obtained from publications Blue book Can be estimated Remaining value estimation (ASAE D497.7 see 6.1) Based on auction sales of used equipment RVn=100[C1-C2(n0.5)-C3(h0.5)]2 where RVn=remaining value as % of list price for equipment at end of n years of age and after h average hours of use per year C1, C2, and C3=coefficients from Table 4 (D497.7) Inflation effects: multiply list price by (1+i)n where i is the average annual inflation rate Depreciation=purchase price-RVn

Class Problem A farmer has recently purchased a new 36 ft chisel plow and is wanting to estimate how much it will be worth after 10 years, taking into account an annual inflation rate of 1%. With the aid of the ASAE D497.7 standard and the tables published by the University of Illinois Extension that include the list price and typical annual use, calculate: the remaining value of this implement after 10 years Depreciation relative to the list price Remaining value taking into account inflation rate http://www.farmdoc.illinois.edu/manage/index.asp

Class Problem RVn=100[C1-C2(n0.5)-C3(h0.5)]2 From Table 4 of ASAE D497.7 C1=0.738, C2=0.051, C3=0 n = 10 years h=? From Appendix Table 1 of UIUC Extension publication For 36ft Chisel Plow, ac/h=22 and ac/yr=1,296 h=1296/22=59 h/yr RVn=100[0.738-0.051(100.5)-0]2 RVn=33.3% of list price (list price=$49,500 from Appendix Table 1)) Remaining value after 10 years = $16,464 Purchase price? From Table 2 of UIUC publication, use 85% of list price Depreciation = (0.85*$49,500-$16,464)= $25,611 http://www.farmdoc.illinois.edu/manage/machinery/machinery%20field%20operations%202012.pdf

Class Problem Taking into account inflation rate of 1% Adjusted list price = $49,500*(1+ 1/100)10 ALP=$54,679 Adjusted RVn = 0.333*$54,679= $18,208

Machine Life Machine life terminated through wear or obsolescence Wear: repair costs become so high that it is no longer economical to continue making repairs Obsolescence: machine out of production and repair parts no longer available OR it can be replaced by another machine to produce greater profit Refer to Table 3 of ASAE D497.7 for estimated life of machines based on wear out Number of years of life=total hours/annual hours Economic life: length of time after which it is more economical to replace the machine because of wear/obsolescence

Machine Reliability (ASAE D497.7) Operational reliability: probability of satisfactory machine function over any given time period (=1 – probability of failure) Based on Midwestern US reports by farmers

Machine Reliability (ASAE D497.7) Average SD of total downtime/year for farms > 200 ha

Machine Reliability (ASAE D497.7) Breakdown probabilities for machine systems increase with increase in farm size

Machine Reliability (ASAE D497.7) Downtime for specific machines relative to ha covered or hours of machine use: Moldboard plow: 1 h/400 ha (1000 acres) Row planters: 1 h/250 ha (600 acres) SP combines: 0 h for first 365 ha (900 acres), 1 h/30 ha (70 acres) thereafter Tractors: DT = 0.0003234 X1.4173 DT = accumulated hours of downtime X = accumulated hours of use

Fixed Costs: Interest on Investment Depreciation Interest on Investment Taxes, Insurance, Shelter Fixed/Ownership Costs Money spent on purchasing machine unavailable for other productive enterprises Cost of ownership includes interest on money invested in machine If loan used to purchase machine, interest rate is known If machine purchased for cash, interest rate based on prevailing rate that could have been obtained if money had been invested instead With straight line depreciation, calculate average investment=(P-S)/2 Annual interest charge=i x average investment

Fixed Costs: Taxes, Insurance and Housing Depreciation Interest on Investment Taxes, Insurance, Shelter Fixed/Ownership Costs Estimated as percentage of purchase price Sales tax and property tax spread over life of machine (no tax for some states) Reasonable to assume annual tax charge at 1% of purchase price of machine Machines may be insured against loss by fire or other causes Reasonable estimate= 0.25% of purchase price of machine Storing machines in a shelter can be expected to increase machinery life and resale value Estimate=0.75% of purchase price Total cost of taxes, insurance and shelter=2% of purchase price unless more accurate data available

Capital Recovery Factor (CRF) Capital Recovery Factor (CRF) – method for determining capital costs of ownership, including time value of money Combines total depreciation and interest charges into series of equal annual payments at compound interest Total annual ownership costs Coa ($/yr) Coa/Pu=(1-Sv)*CRF + Ktis/100 Pu=purchase price,$ Sv=salvage value as fraction of Pu CRF=i*(1+i)L/[(1+i)L-1] i=real interest rate, % L=economic life of machine, years Ktis=annual cost of taxes, insurance, shelter as % of Pu

Real Interest Rate If no inflation, i=ip If ig >=ip, i=0, Coa=cost of taxes, insurance, and shelter

Class Problem Case AFX8010 Calculate total annual ownership costs Purchase price is $250,000 Economic life of 10 years Salvage value is 10% of new cost Interest rate at time of purchase=6.5% General inflation rate=3.5% Calculate total annual ownership costs

Class Problem Assume taxes, etc to be 2% of purchase price, Ktis=2.0

Machinery Costs Variable/Operating Costs Dependent only on machine use

Machinery Operating Costs Labor Hired labor: constant hourly cost Owner: cost determined from alternative uses of owner’s time UI Extension 2010: $16/h charge and labor time 10% more than machine operating time Divide hourly labor cost by effective field capacity for labor cost per hectare

Operating Costs: Fuel and Oil Labor Fuel & Oil Repairs and Maintenance Variable/Operating Costs For any given operation: Qi determination most difficult

Estimation of Fuel Consumption, Qi Two methods (ASAE EP496.3 see 6.3.2) Average annual diesel consumption (L/h) Qi=(0.73)*0.305*Ppto=0.223* Ppto where Ppto = maximum pto power (kW) Consumption for a specific operation Qi=Qs*PT where Qs=specific fuel consumption for particular operation (L/kWh) determined from ASAE D497.6 Clause 3.3.3) PT =Total PTO equivalent power for particular operation , kW

Estimation of FC Qs=specific fuel consumption (SFC) for particular operation (ASABE D497.7)

Estimation of Oil Consumption ASAE D497.6 Defined as volume per hour of engine crankcase oil replaced at manufacturer’s recommended change interval For diesel engines: Qi=0.00059*P+0.02169 (L/h) where P=rated engine power (kW) Oil added between changes and other lubricants not included in equation Total cost of all lubricants is about 10-15% of fuel costs Typical range: 0.0378-0.0946 L/h (ASAE EP496.3)

Example Tractor with rated pto power 90kW performs tillage operation requiring 75kW equivalent pto power at an engine speed averaging 80% of rated engine speed. Effective field capacity is 6 ha/h. Fuel costs are $0.75/L and oil cost is $1.50/L. Calculate: Specific fuel consumption in L/kWh Fuel consumption in L/h Per-hectare fuel costs Oil consumption in L/h Per-hectare oil costs

Example X=75/90=0.83 N=0.8 PTM=1- (N - 1)(0.45*X - 0.877) SFC=(0.22+0.096/X)PTM SFC=(0.22+0.096/0.83)(0.8993)=0.302 L/kWh fuel consumption, Qi=Qs*PT=0.302*75=22.6 L/h Csf=0.75*22.6/6= 2.83 $/ha Qio=0.00059*P+0.02169=0.00059*90+0.02169 Oil consumption=0.0748 L/h Csi=1.50*0.0748/6= 0.019 $/ha

Summary Machinery costs can be divided into fixed, operating and timeliness costs ASAE standards EP496.3 and D497.7 can be used as a basis for machinery costing An optimum machine size can be determined based on timeliness costs and machinery costs