Farm Management Chapter 22 Machinery Management Chapter Outline Estimating Machinery Costs Examples of Machinery Cost Calculations Factors in Machinery Selection Alternatives for Acquiring Machinery Improving Machinery Efficiency Chapter Objectives
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Interest = Average Value X interest rate
Average Value =(Cost + Salvage Value)/2
Source: ASAE Standards, 2001
Capital recovery =
[amortization factor x (beginning value – salvage value)]
(interest rate x salvage value)
This is an alternative to calculating
depreciation and interest.
Amortizing factors can be found in
Appendix 1 of your textbook. It depends
on interest rate and life.
Example: 5% interest, 7 year life
Amortizing factor is 0.17282
For a machine purchased for $52,000 with
a $10,000 salvage value, and a 7 year life,
Depreciation is $6,000 per year. Fixed
Interest for this machine is $1550. The
sum of depreciation and interest is $7550.
Using the amortization factor, we get:
.17282x42,000 + .05x10,000 = $7758
Capital recovery is the annual payment
that would recover the initial investment
lost through depreciation, plus interest on
the investment. It relates to investment
analysis and net present value. The capital
recovery amount is usually a little higher
than the sum of depreciation and interest
because it accounts for the time value of
Gives average costs
Over entire life. Will
Normally be less for
Higher for older
Source: Hunt, Donnell; see text
Gallons per hour = 0.060xPTO hp (gas)
Gallons per hour = 0.044xPTO hp (diesel)
Where PTO = maximum power takeoff horsepower of the tractor
For powered machinery the costs for
lubricants and filters average about
10 to 15% of fuel costs. For non-powered
machines, it is generally small enough to
Some machines use twine, plastic wraps, or
bags. These costs must also be included.
When a tractor pulls an implement, costs for both must be
calculated. Don’t overlook the implements. Do not add ownership
costs together because the tractor can be used for other purposes.
Field capacity =
speed (mph) x width (feet) x field efficiency (%)
A 12-foot-wide windrower operating at 8 mph, with a field efficiency
of 82% would have an effective field capacity of 9.54 acres/hour.
Field efficiency is included to recognize
that a machine is not always used at
full capacity because of work overlap and
time spent turning, adjusting, lubricating,
and handling. Planting may have field
efficiencies as low as 50%. Some tillage
operations may have field efficiencies
as high as 85 to 90%, particularly in large
Minimum field capacity =
acres to cover
hours per day x days available
Compare this value to the calculated
Windrower: 180 acres, 2 days, and 10 hours per day = 9.00 acres per hour. Compare to 9.54 calculated.
Field days needed =
acres to cover
hours per day x acres completed per hour
Operator must decide if typical weather
will permit this many days of operation
without risk of serious losses.
Some field operations do not have to
be completed within a fixed time period,
but the later they are performed, the
lower the harvested yield is likely to be.
Annual machinery costs are a large
part of a farm’s total costs. Selection
of optimum machinery size should
consider total costs and the effects on
timeliness. Machinery efficiency can
be improved by proper repairs and
maintenance, by owning equipment
jointly, or by exchanging the use of
individually owned machines.