Farm Machinery Management & Cost Analysis Notes

Last reviewed 1 Oct 2026 · 7 min read

Objectives of machinery management

Farm machinery management is the selection, scheduling, operation, maintenance and replacement of machines so that crop operations are done on time at the lowest total cost. Its aims:

Right machine, right size — neither under-powered nor idle.
Timeliness — sowing and harvest at the best date.
Low cost per hectare (or per tonne).
Safe, reliable operation with planned maintenance.
Correct replacement at the economic age.

Field capacity and efficiency (recap)

FormulaCapacity and efficiency

Theoretical field capacity: $T F C = \frac{W \times S}{10}$ ha/h (W in m, S in km/h).

Effective field capacity: $E F C = T F C \times η_{f}$ , where $η_{f}$ = field efficiency (time lost in turning, filling, adjustments).

Material capacity (e.g., harvest): $t/h = E F C \times yield (t/ha)$ .

Typical field efficiency: ploughing 75–85 %, harrowing 80–90 %, seed drilling 65–80 %, spraying 60–75 %, combine harvesting 65–80 %.

Selection of machine size

Required EFC $= \frac{area to be covered}{available working hours in the timely period}$ .
Available hours $=$ (days in the period) × (probability of a workable day) × (hours per day).
Tractor power is matched to the drawbar pull and PTO need of the largest implement, with a reserve of about 10–20 % and traction slip within limits.
Match implement width to the tractor so the tractor works near its rated load; an oversized tractor wastes fuel and an undersized one gives low capacity and wear.

Worked ExampleExample — required machine size

A farmer must sow 60 ha of wheat in a 12-day period. Only 70 % of the days are workable and the machine works 8 h/day. The drill's field efficiency is 75 % at 5 km/h.

Available hours $= 12 \times 0.70 \times 8 = 67.2$ h. Required EFC $= \frac{60}{67.2} = 0.893$ ha/h. $T F C = \frac{0.893}{0.75} = 1.19$ ha/h, so the working width $W = \frac{10 \times T F C}{S} = \frac{10 \times 1.19}{5} = 2.38 m$ — choose the next standard size (for example a 2.5 m drill or 15 rows at 17 cm spacing ≈ 2.55 m).

Cost components

Fixed (ownership) costs — independent of use

Item	Notes
Depreciation	Fall in value with age and use
Interest on investment	Opportunity cost of capital: rate × average investment
Taxes, insurance and registration	Often 1–2 % of the cost per year
Housing (shelter)	Often 0.5–1 % of the cost per year

(Typical combined rule: taxes, insurance and housing ≈ 1–2 % of the purchase price per year.)

Variable (operating) costs — depend on use

Item	Notes
Repair and maintenance	Increases with accumulated hours; often estimated as a percentage of the list price per 1000 hours
Fuel	Diesel use per hour ≈ 0.15–0.25 litre per kW (PTO) per hour at full load (often taken ≈ 0.2 L/kW-h; 1 hp-h ≈ 0.15–0.2 L diesel) — check with the test report
Lubricants	About 10–15 % of the fuel cost
Operator labour	Wage per hour (with allowance for travel and servicing)
Repair parts	Included above

Depreciation

Let $P$ = purchase price, $S$ = salvage (resale) value, $n$ = life in years.

Method	Formula	Remark
Straight line	$D = \frac{P - S}{n}$ per year (constant)	Simple; most used
Declining balance	Each year a fixed fraction $d$ of the book value; $d = 1 - (\frac{S}{P})^{1/ n}$ for the exact case, or $2/ n$ for the double declining balance	Higher early depreciation
Sum of years' digits	Year $k$ : $D_{k} = \frac{( n - k + 1 )}{\sum n} (P - S)$ , with $\sum n = \frac{n ( n + 1 )}{2}$	Decreasing in steps
Sinking fund	Annual deposit that grows to $(P - S)$ with interest	Used in economic studies

Worked ExampleExample — depreciation

A tractor costs ₹ 8,00,000, has a life of 10 years and a salvage value of ₹ 1,00,000.

Straight line: $D = \frac{800000 - 100000}{10} = 70, 000$ rupees per year. Sum of years' digits: $\sum n = \frac{10 \times 11}{2} = 55$ ; year-1 depreciation $= \frac{10}{55} \times 700000 = 127, 273$ rupees (₹ 1,27,273); year-2 $= \frac{9}{55} \times 700000 = 114, 545$ rupees (₹ 1,14,545).

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