Planning of farm work system newly, the following procedure for each crop system will be useful.

1. Pre-conditions
1. Region and farm scale
2. Management system
3. Crops
2. Planning table of work system
1. Sequence of farm work
2. Cultivation standard: Period
3. Operation standard

At first, select the basic machinery like as tractor, combine and drying system.

The second, select each machine width or capacity. (See 5-2-3.)

The third, the Rate of work will be calculated by using the empirical data or the official data as followings.

1) Effective Field Capacity(EFC) is calculated by Operation width(W), Operation speed(V) (Table A-215b.) and Field efficiency(EF) (Table A-216.).

2) Coverage(CA) of each work is calculated by a) Working hour per day(Dt), b) Daily net working rate(NWR), c) Days of work period(DWP), d) Rate of available work days(ADR) (Table 33a.).

3. Selection of machinery in planning stage
1. Selection of machine

Economic selection finds that capacity which produces the lowest net cost. The increased ownership cost of high capacity machines are balanced against the increased operation costs and timeliness costs of low capacity machines.

(See ASAE P496)

Size selection of machinery is based on a combination of expected performance and expected costs. Both capacity and capital costs increase with size. At the same time performance improves, particularly with critical operations such as planting and harvesting. Delays in planting can reduce yields. Delays in harvest can reduce both quantity and quality of production.

See [Selection of tractor size] in reference 5)

 

2. Machinery capacity

Simple capacity selection is made by estimating the number of days in the time span within which the operation should be accomplished, and by determining the probability of a working day in this time span. The required capacity of machinery or farm work for an area is

 

EFC = A / (AWD * Dn) Eq. 5-23

or C = A / (D * H * pwd)

Where:

symbol	term	unit	example
EFC	Effective Field Capacity	ha/h	0.139
C	Required machine capacity or farm work	ha/h
A	Area	ha	10
AWD	Available work days	d	12
D	Number of days within the time span within which the operation should be accomplished	d
pwd	Probability of a working day	in decimal
Dn	Net work hours per day	h/d	6
H	Expected time available for field work each day	h/d

The width of machinery will be shown by next equation.

 

Wt = 10 * EFC / (Vt * ef) Eq. 5-24

Where: example is rotary tillage.

symbol	term	unit	example
Wt	Theoretical operation width	m	0.93
EFC	Effective Field Capacity	ha/h	0.14
Vt	Theoretical operation speed	km/h	2.0
ef	Field efficiency	in decimal	0.75

 

3. Low cost machinery by using cost data base

Select minimum cost correspond to the planning farm scale, using the data of annual operation area vs. cost per ha.

See Table A-523. in Appendix

 

4. How many set of tractor and implement is necessary in plural work

Implements are mounted on tractor and the capacities of them are not same, therefore it is not necessary to prepare the same number of these kind of implement. Required number of tractors and implement is obtained by next equation.

 

M >= A / CAS Eq. 5-25

Where:

symbol	term	unit	example
M	Number of machine set	-	3
A	Area	ha	27
CAS	Coverage of one set	ha	9.1

 

Table 523. Required number of implement: Example

(Area: A = 27 ha and Available net working hour: ANWH=157.5 h)

Farm work with tractor	Implement	Work capacity: WC	Number of operation: N	Required number of implement
		h/ha	times
Tillage	Rotary	4.3	1	2
Harrow	Rotary	3.7	2
Leveling	Tooth harrow	0.6	2	1
Seeding	Grain drill	2.9	1	1
Pressing	Roller	1.5	1	1

Work capacity of these five farm work is required as followings:

WCp = ΣWCi * Ni =4.3 + 3.7*2 +0.6*2 +2.9 +1.5 = 17.3 h/ha

Coverage and number of tractor are shown as following from equation 5-12 and the above equation.

CAS = ANWH / WCp = 157.5 /17.3 = 9.1 ha

Required number of tractor: M = 3 >= A/CAS =27/9.1 = 2.97

Required number of each implement or work is obtained by next equation.

Mi >= A / CASi = A * WCi * Ni / ANWH Eq. 5-26

M1 >= 27*4.3/157.5 = 0.74

M2 >= 27*3.7*2/157.5 =1.27

M3 >= 27*0.6*2/157.5 =0.21

M4 >= 27*2.9/157.5 = 0.50

M5 >= 27*1.5/157.5 =0.26

Work-1 and 2 are operated by same implement, therefore number of rotary is M1+M2 >= 2.01. Required numbers of implement are shown in above table.

Exercise 5-10., 5-11., 5-12.

1. Economical optimization of farm work system
1. Optimization by system analysis method

Seek optimal investment by simulation

Select maximum d Profit / d capital

See Steepest gradient method: SGM-001.htm

 

2. Modifying several factor
1. Cropping system or varieties
2. Farm work period
3. Farm scale
4. Machinery

 

3. Improvement by replacing machinery
1. By replacing machine which shows excessive capacity

Machinery cost is decreased by replacing machine, which shows excessive capacity to lower capacity machine.

1. Select maximum CA of farm work operated by machinery:

Example = 21-Baler = 51.5 ha

2. Replace baler to smaller one in sheet step-02 from machinery table step-03:

Example 1.4m ->0.73m

3. Change EFC in sheet 1.field-capacity, and FRh in 2.Variable-cost

4. Calculate new CA, total cost per ha etc. in sheet fwtotal

See Table A-531-i. Improvement by replacing machinery in Appendix.

2. Improvement the system coverage by increasing machine which shows the lowest capacity

See Table A-531-ii. Improvement coverage of system in Appendix.

Exercise 5-8., 5-9.

2. Energetic or environmental evaluation
1. Energetic evaluation

Energy consumption will show the important index of energetic evaluation of farm work systems.

LCA (Life Cycle Assessment) is acceptable method of calculating energy consumption in industry or daily life.

Calculation of energy consumption of farm machinery or facilities by using input-output analysis of inter-industry (SANGYOU RENKANBUNSEKI) is also effective method for energetic evaluation.(ECU in the following table)

Total annual energy consumed for manufacturing will be obtained as following equations. (ECU in next table)

AEG = AFC * Yrate* ECU/1000 Eq. 5-27

where,

symbol	term	unit
AEG	Total annual energy consumed	MJ
AFC	Annual fixed cost ($): Total	$
Yrate	Yen exchange rate	Yen/$
ECU	Energy conversion unit	kJ / Yen

Variable energy per ha(MJ/ha) will be calculated by using conversion factor of next table.

Table 532a. Conversion factor

Output energy of per ha = Yield * RCF =4500(kg/ha) * 14.9(MJ/kg) = 67 GJ/ha

symbol	Conversion factor		unit
ECU	Energy conversion unit for manufacturing the machinery by using input-output table of inter-industry	48.1	kJ / Yen
RCF	Rice grain Conversion factor	14.9	MJ/ kg
GCF	Gasoline Conversion factor	35.2	MJ/ L
KCF	Kerosene Conversion factor	37.3	MJ/ L
DCF	Diesel light oil Conversion factor	38.5	MJ/ L
ECF	Electric power Conversion factor	9.4	MJ / kWh

by handbook of energy save: 1996 and food handbook: 1 kWh = 3.6 MJ

Example See Rice-erg.xls: Step-C1, fwtotal-erg

 

2. Environmental evaluation

Method of environmental evaluation is not yet completed now, but CO2 exhaust amount will show some index of environmental situation.

Table 532b. CO2 gas generation ratio

CO2 gas generation ratio		unit
Gasoline	2.3587	kg / L
Kerosene	2.5284	kg / L
Diesel light oil	2.6444	kg / L
Electric power	0.42	kg / kWh

 

3. Total evaluation See Fmech-10.doc, fm-12.xls

1. Investment, Revenue and Profit

Profit will be calculated simply as next equation.

 

B = NI – I Eq. 5-28

where,

symbol	term	unit
B	Profit	$
NI	Net income	$
I	Investment	$

Example

If we invest to the combine shown in the following table, then the profit of n year will be obtained as follows.(Assume interest rate = 0)

term	Investment: Price	Economic Life	Operating Cost	Custom Charge	Effective Field Capacity	Net work hour	Daily Capacity	Available work days	Coverage
symbol	I	L	OC	CC	EFC	Dn	DC	AWD	CA
unit	$	year	$/ha	$/ha	ha/h	h/d	ha/d	d	ha/year
data	130,000	8	300	1,800	0.2	5.0	1.0	30	30.0

Net income of the year: NIi = (CC-OC)*CA

n	year	1	2	3	4	5	6	7	8
NIi	Net income	45,000	45,000	45,000	45,000	45,000	45,000	45,000	45,000
NI: ($)	Total Net income	45,000	90,000	135,000	180,000	225,000	270,000	315,000	360,000
B=NI-I: ($)	Profit	-85,000	-40,000	5,000	50,000	95,000	140,000	185,000	230,000

See fm-541.xls: Ex-9

This table shows that the profit will be obtained after 3 years and total profit of 8 years later is $230,000, and rate of profit (B / I) is 1.77.

 

2. Recover period

Recovery period will be obtained by next equation.

 

n = I / NI Eq. 5-29

where,

symbol	term	unit	Example
n	Recovery period	year	2.9
NI	Net income of one year	$/year	45,000
I	Investment	$	130,000

This means we will get profit after 3 years use of the combine.

5-4-3.  Machine renewal (or replacement)

1. Replacement

Machine employed in production may need to be replaced for one or more reasons.

1. A machine suffers accidental damage such that the cost of renewal is so great that a new machine is more economical.

2. The capacity of the existing machine is inadequate because of increased scale of production.

3. The machine is obsolete (see ASAE S495)

4. The machine is not expected to operate reliably. (Suffers considerable unanticipated downtime from random part failures.)

5. The cost of making an anticipated repair would increase the average unit accumulated cost above the expected minimum. Only capital costs and actual repair and maintenance costs need be accumulated.

For example, a $3000 machine is used 100 ha annually. It experiences the following end-of-year depreciation, interest (8% simple interest on average investment), and actual repair and maintenance costs in next table. Year 9 has the lowest unit cost and indicates the machine should be replaced with a similar machine at the end of year 9 if not before for other reasons. Inflation effects must be considered in making replacement decisions. Annual depreciation charges may be quite low or even negative in times of rapid inflation producing a premature minimum unit accumulated cost. In such instances replacement is better indicated by comparing the unit accumulated cost of the present machine with the projected costs for a potential successor machine. Optimum replacement time may be delayed beyond that time determined under more stable economic conditions. (See ASAE-P496)

Table 542. Average unit accumulated costs

year	R&M costs	Depr.	Int.	Tot. acc. Costs	Acc. Use, ha	Unit acc. Costs
	$	$	$	$	ha	$/ha
1	10	1000	200	1210	100	12.10
2	50	600	136	1996	200	9.98
3	70	400	96	2562	300	8.54
4	100	300	68	3030	400	7.58
5	200	200	48	3478	500	6.96
6	300	150	34	3962	600	6.60
7	350	125	23	4460	700	6.37
8	450	100	14	5024	800	6.28
9	550	25	9	5608	900	6.23
10	600	25	7	6240	1000	6.24

 

2. Annual payment of worth

Total cost of several years for machinery is calculated as next equation , and machinery should be replaced at the year so that annual payment is minimum, which is called as economical life of devices.

AP(n) = { P + Σ[(Rj + Qj) / (1+i)^j] - [Sn / (1+i)^n]} * {[i * (1+i)^n] / {(1+i)^n - 1} Eq. 5-30

Where:

symbol	term	unit
AP(n)	Adjusted annual payments of worth after n year usage	$ / year
P	Purchase price	$
Rj	Repairing cost in j year	$
Qj	Timely cost etc. in j year	$
Sn	Remaining value after n years	$
i	Annual interest	in decimal

Equation above 5-30 is induced from next equation.

AP(n) = P(n) + R(n) + Q(n) - S(n) Eq. 5-31

E = P * [(1+i)^n] Eq. 5-32

symbol	term	unit
P(n)	Annual present worth after n year	$ / year
R(n)	Annual repairing cost after n year	$ / year
Q(n)	Annual timeliness cost after n year	$ / year
S(n)	Annual remaining value after n year	$ / year
E	Final worth after n years	$
P	Present worth	$
(1+i)^n	Final worth factor or compound amount factor	-
1 / [(1+i)^n]	Present worth factor	-

 

Table 542. Annual payment: Example: Combine(P=5 M Yen, i =0.05)

Year	P(n)	Rn	R(n)	Remaining ratio	Sn	AP(n)*	AP(n)**
n	k Yen	k Yen	k Yen	%	k Yen	k Yen	k Yen
1	5,250	50	50	65	3,250	2,050	2,100
2	2,689	100	74	40	2,000	1,788	1,862
3	1,836	150	98	25	1,250	1,538	1,636
4	1,410	1,850	505	15	800	1,729	2,234
5	1,155	800	558	10	500	1,623	2,181
6	9,85	650	572	4	200	1,527	2,099
7	8,64	2,300	784	2	100	1,636	2,420

*: Assume Q(n) =0 **: Assume Q(n) = R(n)

In case of above table, it is recommendable to replace combine after 6 years or 3 years depending on evaluation of timeliness cost.

3. By decision making method
1. AHP (Analytic Hierarchy Process)

See Table A-542c. AHP: Example in replacement of tractor in Appendix.

Assume, Sales per ha = 12,015 $/ha

Variable cost per ha(VCa)=3,439$/ha

Exercise 5-10. Obtain the required capacity of machinery and the width of machine（Wt）, when data are given in next table.