Off-Road Equipment Management TSM 262: Spring 2016 LECTURE 26: Machinery Costing II Off-Road Equipment Engineering Dept of Agricultural and Biological Engineering achansen@illinois.edu

Class Objectives Students should be able to: Determine the costs of owning and operating machines Evaluate the effect of machine size on timeliness and overall machinery costs

Timeliness Costs Cost related to a timeliness penalty A reduction in return, not an out-of-pocket expense Examples of penalties Unfavorable weather, effective field capacity of machine too low for timely completion of operation Machine breakdowns, shortage of machine operators, low work-hours per day on operation

Timeliness Costs (cont.) Timeliness cost can be calculated from (ASAE EP496.3): How can this cost be reduced?

Timeliness Costs (cont.) KT = fraction of crop yield lost for each day of delay of an operation (depends on operation) Planting crop earlier or later than optimum planting date diminishes crop yield Actual planting period can be balanced around optimum date, λo=4 For tillage, no timeliness coefficient, provided no delays to planting For most harvesting operations, λo=2, because have to wait until crop is mature Denominator = rapidity with which operation can be completed

Timeliness Costs (cont.) pwd dependent on: geographical location time of year Refer to ASAE D497.7 Table 5 Probabilities given at both 50% and 90% confidence levels Probabilities are averages for biweekly periods Pwd = 0.4 means 0.4*14=5.6 working days in 2-week period If probability at 50% level, 5.6 days exceeded in 5 out of 10 years At 90% level, 5.6 day level exceeded in 9 out of 10 years

Example A 12-row conventional planter is to be used to plant 180 ha of soybeans with 75-cm row spacing in early June in Central Illinois. The soybeans have an anticipated yield of 2.7 Mg/ha and an anticipated selling price of $250/Mg. Using typical travel speed and field efficiency for the planting operation from Table 3 (D497.6), calculate field capacity Calculate timeliness cost assuming farmer works 10-hour days and wants 90% confidence of having the required number of good working days

Solution (a) Effective field capacity (b) Timeliness cost penalty

Machinery Selection

Machinery Selection (cont.) Optimum machine size for given field area and crop yield is one that minimizes sum of timeliness costs and machinery costs including labor

Class Problem A self-propelled combine rated at a maximum power of 300kW (400Hp) with a sixteen-row corn head, is capable of harvesting corn at 7.2 km/h (4.5 mph). The corn is yielding 10 Mg/ha (160 bushels/ac) with 0.762 m (30 inch) row spacing. The field efficiency of the combine is 65%. Calculate the following: a) Theoretical field capacity (ha/h) b) Actual field capacity of the combine (ha/h). The combine purchase price is $450,000 and has an expected life of 8 years. The interest rate is 8% and inflation is at 1%. The salvage value of the machine after 8 years is 10% of the purchase price. Cost of taxes, insurance and housing is 2% of the purchase price per annum. The labor costs are $16/h and diesel fuel cost is $0.75/L($3.00/gallon). You may assume that oil and lubrication costs may be ignored. The combine operates on average 250 hours a year, with 10-hour harvest days. Assume that the timeliness coefficient is 0.002 and that harvest operations are evenly balanced about the optimum harvest date. The price of corn is $180/Mg ($4.50/bu). The probability of a working day is 0.65 during harvest. Calculate the following: c) Fixed costs of ownership per annum (per ha) d) Fuel costs per annum assuming average annual diesel consumption based on maximum engine power(Qi =0.223*Pengine)(L/h) e) Labor costs per annum f) Estimated repair and maintenance costs for each year assuming RF1 = 0.04 and RF2 = 2.1 g) Total operating costs per annum, excluding timeliness costs h) Timeliness costs ($/ha) i) Average combine harvest costs ($/ha)

Solution (a) (b) = (c) Coa = Pu[(1-Sv)CRF + Ktis/100) Pu = Sv= i=(ip-ig)/(1+ig) =

Solution (d) Qi=(0.223) Pengine= (e) Labor cost= Cost of fuel per hour= Cost per ha= Cost per yr= (e) Labor cost= Cost per year= (f) Accumulated R&M costs

Solution (g) Total operating costs = + + = $ /yr (h) (i) Cave = ($ + $ )/( x ) + $ Cave=