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

2. Homework and Lab

3. 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

4. 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

5. 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

6. 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

7. Depreciation Method

8. 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

9. 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

10. 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[ (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

11. 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

12. 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

13. 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

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

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

16. 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 = X1.4173
DT = accumulated hours of downtime
X = accumulated hours of use

17. 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

18. 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

19. 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

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

21. 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

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

23. Machinery Costs
Variable/Operating Costs
Dependent only on machine use

24. 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

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

26. 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

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

28. 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= *P (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: L/h (ASAE EP496.3)

29. 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

30. Example X=75/90=0.83 N=0.8 PTM=1- (N - 1)(0.45*X - 0.877)
SFC=( /X)PTM
SFC=( /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= *P = *
Oil consumption= L/h
Csi=1.50*0.0748/6= $/ha

31. 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