Optimization in farm work system

　

1. Modeling
1. Analytical model (Mathematical model)

Coding by software (LP, etc.)

1. Rice production farming system by Farm Mechanization Planning

See TE2002.xls or Rice-sy1.xls

　

2. Dynamic model

Coding by function of time

1. Rice production farming system by Dynamic model



Fig. S-1. Flow chart of dynamic model

See reference 6.

　

2. Objective function
1. Cost of the farm work system
2. Profit of the farm work system
3. Total benefit of the farm work system

Economical, ecological and healthy (metabolic energy index) benefit

　

3. Optimization
1. Improvement by empirical knowledge
1. By checking coverage etc. of each farm work

Example 1.

2. By using database

Example 2.

2. Simulation: Experiment in computer

We may test the model like machine experiment by Experiment Design.

Mesh type simulation: Example 3.

1. Select variables
2. Set function level of each variable
3. Calculate objective function under above conditions
4. Obtain optimal result

　

3. Improvement step by step

We may apply the analytical method like Steepest Gradient Method etc. Example 4 and 5.

1. Steepest Gradient Method
1. Analysis by differential equation

Increment step: q = S * p

where, p= grad(C), that is, = [δC/δx , δC/δy], C = cost, S = constant, x and y = factor

2. Simulation of steepest gradient direction

　

2. Hill climbing simplex method

　

4. Optimization by multi evaluation criteria including feeling evaluation

We may use optimization method for sustainable farming system based on multi evaluation criteria including feeling evaluation method like AHP, CP etc. Example 6.

　

　

　

4. Exercise
1. Example 1 by empirical knowledge
1. Modeling

Rice production farming system in planning stage

FS01-J by EXCEL file (TE2002p1.xls)

2. Objective function

Make total cost of the system minimum at 10 ha

3. Optimization

Improvement by empirical knowledge by checking coverage etc. of each farm work

Process under Conditions: Farm scale = 10 ha

	Stage	Coverage (ha)	Total cost at 10 ha ($/ha)	Total cost at 10 ha ($)	Cost ($) at CA
1	First planning	6.5	-		6,116
2	By new combine	13.3	5,149	51,490	4,685
3	By new sprayer 61	13.3	5,190	51,900	4,706
4	By new sprayer 62	13.3	5,137	51,370	4,680

　

1. Check coverage: 6.5 ha by 7-Harvest at sheet [Summary-fw-system]
2. Replace machine: Combine from 2 row to 3 row at sheet [step 3]

Input new combine with machine code 71

Replace machine code from 7 to 71 at sheet [step-2], [Machinery-cost]

3. Check price, EFC, FRh of new machine at sheet [step-3]

EFC = EFCold * Wnew / Wold = 0.06 * (3^2) / (2^2) = 0.09 ha/h [Field capacity]

FRh = FRh * HPnew / HPold = 1.33 * 20 /15 = 1.77 L/h [Variable-cost]

4. Confirm new coverage of 7-Harvest = 14.5 ha from old 6.5 ha at sheet [Coverage]

Total cost at 10 ha ($) = 51,490

-------------------------------------------

5. Check coverage again: 13.3 ha by 5-Caring crop is larger than 10 ha at sheet [Summary-fw-system]
6. Search maximum excessive coverage: 35.1 ha by 6-Chemical application
7. Replace machine in order to decrease the coverage: Sprayer from HP-2 toHp-17 at sheet [step 3]

Input new sprayer with machine code 61

Replace machine code from 6 to 61 at sheet [step-2] , [Machinery-cost]

8. Check price, EFC, FRh of new machine at sheet [step-3]

EFC = EFCold * Wnew / Wold = 0.53 * (20) / (40) = 0.27 ha/h [Field capacity]

20,40 L/min

FRh = FRh * HPnew / HPold = 0.26 * 20 /40 = 0.13 L/h [Variable-cost]

9. Confirm new coverage = 17.9 ha from 35.1 ha of 6-Chemical application

Total cost at 10 ha ($) = 51,190 : This is more than before, that is, smaller machine is not always less cost than larger one.

-------------------------------------------

10. Replace machine to increase the coverage little bit: Sprayer from HP-17 toHp-303 at sheet [step 3]

Input new sprayer with machine code 62

Replace machine code from 61 to 62at sheet [step-2] , [Machinery-cost]

11. Check price, EFC, FRh of new machine at sheet [step-3]

EFC = EFCold * Wnew / Wold = 0.53 * (30) / (40) = 0.40 ha/h [Field capacity]

30,40 L/min

FRh = FRh * HPnew / HPold = 0.26 * 30 /40 = 0.20 L/h [Variable-cost]

12. Confirm new coverage = 26.5 ha from 35.1 ha of original 6-Chemical application

Total cost at 10 ha ($) = 51,370 : This is less than before.

-------------------------------------------

2. Example 2 by using database

Select combine by using farm work database.

Farm scale	Cost per ha (ACa: $/ha)
	Head feeding riding type combine				Normal combine
ID	83	84	85	86	87	88	89
A: ha	3 row	4 row	5 row	6 row	1.5m	2.4m	3.0m
1	7,606	11,393	12,959	18,537	8,722	23,795	25,837
2	3,931	5,794	6,559	9,336	4,469	11,973	12,972
3	2,705	3,927	4,425	6,269	3,051	8,033	8,684
4	2,093	2,994	3,358	4,735	2,342	6,062	6,540
5	1,725	2,434	2,718	3,815	1,917	4,880	5,254
10	990	1,314	1,438	1,975	1,066	2,516	2,681
15	745	940	1,011	1,361	783	1,728	1,823
20	622	754	798	1,055	641	1,334	1,394
25	549	642	670	871	556	1,097	1,137
30	500	567	584	748	499	940	965

Coverage of combine

	3 row	4 row	5 row	6 row	1.5m	2.4m	3.0m
CA: ha	15.4	20.5	25.6	30.7	20.5	25.6	30.7

Combine selected for each farm scale

Farm scale: A (ha)	Combine selected
1	Head feeding type 3 row
10	Head feeding type 3 row
20	Head feeding type 4 row
30	Head feeding type 6 row

See DB-FW.XLS

　

3. Example 3 by mesh type simulation
1. Modeling

Rice production farming system in improving stage

FS01-J by EXCEL file (TE2002p3.xls sheet-simulation)

2. Objective function

Make Total profit(P) maximum

P = Total sale - Total cost + Capital remained

3. Optimization

Apply Simulation: Experiment in computer by Experiment Design

1. Select variables and constraints

Assume we can invest $10,000 in the farming system, and starting coverage 6.5 ha, and profit is 39,364$ at coverage.

X = Land rent ($), Y = Purchase machine (combine) ($) and X + Y =< 10,000

2. Set function level of each variable

Level of X and Y are 2,000, 4,000, 6,000, 8,000, 10,000.

　

3. List of conditions and constraint

symbol	term	default	simulation	unit
dX	Mesh of X	1,000	2,000	$
dY	Mesh of Y	1,000	2,000	$
LR	Land rent	585	585	$/ha
dL	Land area by 1 $		0.00171	ha/$
dE	EFC increase by 1000 $		0.00796	(ha/h) / 1000$
CAP	Capital initial	10,000	10,000	$
Ainitial	Land initial	6.5	6.5	ha
PRinitial	Profit initial	39,375	39,375	$
CAinitial	Coverage initial	6.5	6.5	ha
EFC	Effective field capacity	0.06	0.06	ha/h
P	Total Profit	49,375	49,375	$
X	Invest to land			$
Y	Invest to machine			$

　

4. Calculate objective function under above conditions
5. Step 1: Mesh = 2000
1. Calculation land area: A A = Ainitial + dL * X

X: $	0	2,000	4,000	6,000	8,000	10,000
A: ha	6.50	9.92	13.34	16.76	20.18	23.59

2. Calculation EFC and CA

EFC = EFCinitial +dE * Y / 1000

CA = CAinitial * EFC /EFCinitial

Y: $	0	2,000	4,000	6,000	8,000	10,000
EFC: ha/h	0.060	0.076	0.092	0.108	0.124	0.140
CA: ha	6.50	8.22	9.95	11.67	13.40	15.12

3. Calculation PR

PR by EXCEL

input EFCnew and Anew at sheet [field capacity] in this color

IF CA>A, obtain PRnew of at A at sheet[Summary-fw-system]

IF CA<A, obtain PRnew of at CA at sheet[Summary-fw-system]

			Y: $
X: $	0	2,000	4,000	6,000	8,000	10,000
0	39,375	40,131	40,425	40,632	40,785	40,904
2,000	39,375	54,640	69,905	70,444	70,678	70,859
4,000	39,375	54,640	69,905	85,169	99,872	100,114
6,000	39,375	54,640	69,905	85,169	99,872	100,114
8,000	39,375	54,640	69,905	85,169	99,872	100,114
10,000	39,375	54,640	69,905	85,169	99,872	100,114

4. Calculation P

P = Total sale - Total cost + Capital remained

P = PR + CAP - X - Y

			Y: $
X: $	0	2,000	4,000	6,000	8,000	10,000
0	49,375	48,131	46,425	44,632	42,785	40,904
2,000	47,375	60,640	73,905	72,444	70,678	68,859
4,000	45,375	58,640	71,905	85,169	97,872	96,114
6,000	43,375	56,640	69,905	83,169	95,872	94,114
8,000	41,375	54,640	67,905	81,169	93,872	92,114
10,000	39,375	52,640	65,905	79,169	91,872	90,114

5. Obtain optimal result

Pmax (Mesh = 2000): Maximum profit = 85,169 $

　

6. Step 2: Mesh = 500

　

			Y: $
X: $	5,000	5,500	6,000	6,500	7,000	7,500
2,500	80,200	79,777	79,353	78,924	78,489	78,051
3,000	79,733	82,853	86,169	85,926	85,497	85,063
3,500	79,233	82,353	85,669	88,985	92,302	91,988
4,000	78,733	81,853	85,169	88,485	91,802	95,118
4,500	78,233	81,353	84,669	87,985	91,302	94,618
5,000	77,733	80,853	84,169	87,485	90,802	94,118

Pmax (Mesh = 500): Maximum profit = 88,985 $

　

7. Step 3: Mesh = 100

Pmax (Mesh = 100): Maximum profit = 89,949 $

　

4. Example 4 by Steepest gradient method
1. Modeling

Rice production farming system in improving stage

FS01-J by EXCEL file (TE2002p3.xls sheet SGM)

2. Objective function

Make Total profit(P) maximum

P = Total sale - Total cost + Capital remained

3. Optimization
1. Initial conditions

Assume we can invest $10,000 in the farming system, and starting coverage 6.5 ha, and profit is 39,375$ at starting point.

　

2. Select variables and constraint

X = Land rent ($), Y = Purchase combine ($) and G = X + Y - 10,000 =< 0

　

3. Obtain steepest gradient

Seek steepest gradient by setting dX and dY as followings:

　

4. Calculation step by step

Conditions and factors:

symbol	term	default	simulation	unit
dX	Incremnt X	1,000	1,000	$
dY	Incremnt Y	1,000	1,000	$
S	Constant S	0.001	0.001
LR	Land rent	585	585	$/ha
dL	Land area by 1 $		0.00171	ha/$
dE	EFC increase by 1000 $		0.00796	(ha/h) / 1000$
CAP	Capital initial	10,000	10,000	$
Ainitial	Land initial	6.5	6.5	ha
PRinitial	Profit initial	39,375	39,375	$
CAinitial	Coverage initial	6.5	6.5	ha
EFC	Effective field capacity	0.06	0.06	ha/h
P	Total Profit	49,375	49,375	$
X	Invest to land			$
Y	Invest to machine			$
dP	Increment P			$
Pmax	Maximum profit		88,436	$

EFC = EFCinitial +dE * Y / 1000

CA = CAinitial * EFC /EFCinitial

　

5. Result
1. Step 1 (dX, dY = 1000) and direction of (dX, dX) = (10, 0), (6, 4), (4, 6), (0, 10)

Pmax: Maximum profit = 88,436 $

2. Step 2 (dX, dY = 100) and direction of (dX, dX) = (10, 0), (9, 1), (8, 2), ----, (2, 8), (1, 9), (0, 10)

Pmax: Maximum profit = 89,905 $

　

	Land invest	Machine invest	Land rent	Machine	Land area	Coverage	Machine	Profit	Total profit	Increment P	Select	Constraint G
	dX	dY	X	Y	A	CA	EFC	PR	P	dP		G
	$	$	$	$	ha	ha	ha/h	$	$	$
Initial	0	0	0	0	6.50	6.50	0.06	39,375	49,375	0		-10,000
1a	1000	0	1,000	0	8.21	6.50	0.060	39,375	48,375	-1,000		-9,000
1b	600	400	600	400	7.53	6.84	0.063	42,428	51,428	2,053		-9,000
1c	400	600	400	600	7.18	7.02	0.065	43,955	52,955	3,580	*	-9,000
1d	0	1000	0	1,000	6.50	7.36	0.068	39,375	48,375	-1,000		-9,000
1	400	600	400	600	7.18	7.02	0.065	43,955	52,955			-9,000
2a	1000	0	1,400	600	8.89	7.02	0.065	43,955	51,955	-1,000		-8,000
2b	600	400	1,000	1,000	8.21	7.36	0.068	47,008	55,008	2,053		-8,000
2c	400	600	800	1,200	7.87	7.53	0.070	48,534	56,534	3,579	*	-8,000
2d	0	1000	400	1,600	7.18	7.88	0.073	45,835	53,835	880		-8,000
2	400	600	800	1,200	7.87	7.53	0.070	48,534	56,534			-8,000
3
4
5
6
7
8	400	600	2,800	5,200	11.29	10.98	0.1014	79,063	81,063			-2,000
9a	1000	0	3,800	5,200	13.00	10.98	0.101	79,063	80,063	-1,000		-1,000
9b	600	400	3,400	5,600	12.31	11.33	0.105	82,116	83,116	2,053		-1,000
9c	400	600	3,200	5,800	11.97	11.50	0.106	83,643	84,643	3,580	*	-1,000
9d	0	1000	2,800	6,200	11.29	11.85	0.109	82,416	83,416	2,353		-1,000
9	400	600	3,200	5,800	11.97	11.50	0.1062	83,643	84,643			-1,000
10a	1000	0	4,200	5,800	13.68	11.50	0.106	83,643	83,643	-1,000		0
10b	600	400	3,800	6,200	13.00	11.85	0.109	86,696	86,696	2,053		0
10c	400	600	3,600	6,400	12.65	12.02	0.111	88,282	88,282	3,639		0
10d	0	1000	3,200	6,800	11.97	12.36	0.114	88,436	88,436	3,793	*	0
10	0	1000	3,200	6,800	11.97	12.36	0.1141	88,436	88,436			0

	dX	dY	X	Y	A	CA	EFC	PR	P	dP		G
8	400	600	2,800	5,200	11.29	10.98	0.1014	79,063	81,063			-2,000
9a	1000	0	3,800	5,200	13.00	10.98	0.101	79,063	80,063	-1,000		-1,000
9n	800	200	3,600	5,400	12.65	11.16	0.103	80,590	81,590	527		-1,000
9c	600	400	3,400	5,600	12.31	11.33	0.105	82,116	83,116	2,053		-1,000
9d	400	600	3,200	5,800	11.97	11.50	0.106	83,643	84,643	3,580		-1,000
9e	200	800	3,000	6,000	11.63	11.67	0.108	85,169	86,169	5,106	*	-1,000
9f	0	1000	2,800	6,200	11.29	11.85	0.109	82,416	83,416	2,353		-1,000
9	400	600	3,000	6,000	11.63	11.67	0.1078	85,169	86,169			-1,000
10a	1000	0	4,000	6,000	13.34	11.67	0.108	83,643	83,643	-2,526		0
10b	800	200	3,800	6,200	13.00	11.85	0.109	85,381	85,381	-788		0
10c	600	400	3,600	6,400	12.65	12.02	0.111	86,696	86,696	527		0
10d	400	600	3,400	6,600	12.31	12.19	0.113	88,282	88,282	2,113		0
10e	300	700	3,300	6,700	12.14	12.28	0.113	89,905	89,905	3,736	*	0
10f	200	800	3,200	6,800	11.97	12.36	0.114	88,436	88,436	2,267		0
10g	100	900	3,100	6,900	11.80	12.45	0.115	86,967	86,967	798		0
10h	0	1000	3,000	7,000	11.63	12.54	0.116	85,497	85,497	-672		0
10	0	1000	3,300	6,700	12.14	12.28	0.1133	89,905	89,905			0

　

　

5. Example 5 by factors more than two
1. Modeling
2. Objective function

Make maximum the total profit of farm work system using simulation or Steepest Gradient System

Assume capital available 15,000 $ in FS-01-J

3. Optimization
1. Initial conditions

Assume we can invest $10,000 in the farming system, and starting coverage 6.5 ha, and profit is 39,375$ at starting point.

　

2. Select variables and constraint

X = Land rent ($),

Y1 = Purchase machine 1 ($)

Y2 = Purchase machine 2 ($)

Y3 = Purchase machine 3 ($)

and X + Y1 + Y2 + Y3 =< 15,000

Term	Land	Machine 1: Combine	Machine 2: Puddler	Machine 3:
	dX	dY1	dY2	dY3
Increment
Total profit ($)

This is 4 dimensional problem, therefore, combination of these factors will increase tremendously.

　

　

6. Example 6 for sustainable farming system
1. Modeling
2. Objective function

B = k1 * Y1 + k2 * Y2 + k3 * Y3 + k4 * Y4 + k5 * Y5 + k6 * Y6 + k7 * Y7

where,

Symbol	Term	unit	Example: Weight
B	Total benefit of system	-
Y1	Economical profit of system	$		41
Y2	Ecological pollution of system	kg		11
Y3	Human health (Ergonomics) of system	-		13
Y4	Energetic evaluation	-		9
Y5	Cultural evaluation	-		7
Y6	Social evaluation	-		10
Y7	Political evaluation	-		9
Y1i	Profit of system: initial or basic	$
Y2i	Ecological pollution of system: initial or basic	kg
Y3i	Human health (Ergonomics) of system: ditto	-
Y4i etc.	Initial or basic data, respectively	-
k1	Coefficient of Y1	1 / $	41 / Y1i
k2	Coefficient of Y2	1 / kg	11 / Y2i
k3	Coefficient of Y3	-	13 / Y3i
k4	Coefficient of Y4	-	9 / Y4i
k5	Coefficient of Y5	-	7 / Y5i
k6	Coefficient of Y6	-	10 / Y6i
k7	Coefficient of Y7	-	9 / Y7i

3. Optimization
1. Evaluation of each term by respective unit
1. Economical profit: by $

Example TE2002.xls

2. Ecological pollution: by CO2 pollution etc.

We may evaluate CO2 input/output by CO2 gas generation ratio or photosynsesis.

3. Human health (Ergonomics): Metabolic energy required to farm works

Try to evaluate each work by feeling heavy or light

4. Energy input and output

Example Rice-erg.xls

5. Cultural, Social and Political evaluation will be done by AHP etc.

　

2. Normalization of each term

Normalization of B: by arranging weight of k1,k2,k3.

B will be changed by modifying Y1, Y2, , Y7.

　

3. Optimization

return

2004/6/6