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AGM 2060 Final Project

Jared Marshall

12/10/14

Submitted To:

H. F. Massey

AGM 2060

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Introduction

Moo-Moo Milk Markers, a dairy farming operation owned by Mortimer McMaster, is expanding its operations to include row crops as a supplement to the farm’s revenue. After researching the market and discovering the potential profit, owner Mortimer McMaster has decided that he wants to double crop soybeans in the spring and wheat in the fall.

As the new Machinery Manager at Moo-Moo Milk Markers, the task at hand is to determine the cost to run and the gross and net income of a 7800 acre farm. A conventional double crop planting system for wheat and soybeans is to be used. For wheat, a burn down application of herbicide (glyphosate) is to be put down at least 14 days prior to planting, followed by two diskings to bury previous crop residue, then a subsoiling operation. This is followed by a smoothing operation and finally the planting of the wheat. During planting, fertilizers (P2O5 and K2O) are to be applied. Halfway through the growing season, a post-emergent herbicide (Harmony Extra) is to be sprayed on the wheat. Once the wheat has been harvested, a pre-plant herbicide (2,4-D and Glyphosate) application is to be applied for a burn down at least 14 days before planting. While or after the herbicide is being applied, a single disking operation is to be completed to bury the small grain and weed residue. During the disking operation, lime is to be applied, followed by a subsoiling operation, and then the planting of the soybeans. One-fourth way through the growing season, a post-emergent spraying operation (Reflex) is to be applied to the beans and then halfway through the growing season another post-emergent spraying operation (Cobra) is to take place.

All equipment has to be sized to complete the projects on time and still make a profit. Two systems are to be made, one that operates multiple smaller implements and tractors and one that operates with fewer but larger implements and tractors. A detailed cost analysis of both systems has to be made to show cost versus profit of each system for ten years so the best system can be chosen for this size of an operation.

Methods

The first step Mortimer assigned was to size and select the equipment needed to complete all the farming operations. All implements and machinery were selected for primary tillage, planting, cultivation, spraying, and harvest. Because of this, a tractor was then sized and matched to each implement. When sizing the equipment, eleven hour workdays and 50% probabilities of suitable fieldwork days were used. All seed and chemical costs were obtained from “Chemical and Seed Spreadsheet” and the prices of soybeans and heat were obtained from www.cmegroup.com. The prices were current future prices for 2015 in the correct harvest month.

CalendarThe calendar was the first step to solving this problem because it incorporated the suitable fieldwork days and made it possible to see how long you have for each operation. Suitable fieldwork days were the only way to farm on paper since they are statistically based on historical data, and they vary as a function of temporal field conditions such as: region, weather, time of year, and soil type.

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Therefore, the Suitable Fieldwork Days is calculated bySFWD = (Sum of days)*0.86*TC

Where, Sum of Days = use SC Suitable Fieldwork Days chart and find sum of probabilities for

the days of your operation

TC =Tractability Coefficients

Once the Suitable days have been calculated, you then need to take that number and put it into a modified EFC equation.

EFCEFC stands for Effective Field Capacity, also known as Actual Field Capacity, and it is the next step in solving the problem at hand. EFC determines the Acres per Hour of a machine with its efficiency, but in this project it was modified to use with the Suitable Fieldwork Days. Therefore, EFC is calculated by,

EFC = [ac/SFWD]*[day/hrs.]Where,

Ac= Total acres of farmSFWD = Suitable Fieldwork Days calculated in last problemDays/hrs. = How many hours you are working in 1 day (ex. 11-hour days would be

represented as [1day/11Hrs])

Once the EFC has been calculated, it can now be placed into a modified TFC equation.

TFCTFC stands for Theoretical Field Capacity and it is also expressed in acres per hour like EFC, but it doesn’t have the implements efficiency built into the normal equation like EFC does. To calculate TFC, the following equation is used:

TFC = [EFC/e]Where,

EFC = Effective Field Capacity that was calculated in the previous probleme = efficiency of the implement, as a decimal

Now that the TFC has been calculated, the width of the implement that is needed to finish the operation on time is calculated.

Width SizingThe size of the implements has a huge role in determining if the tasks will be completed or not. As a result, width sizing is so important. You normally know your width before you find the TFC and EFC, but since TFC and EFC are known, we need to solve for width in feet or inches depending on which implement we are sizing. Our equation for width is,

W = [TFC/S]*[8.25 mi ft./1Ac]

Where,

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TFC = Theoretical field capacity from previous equationS = Speed in MPH

The width found from the equation only tells the width if that was one machine or implement. Therefore, you have the total width needed to complete that single operation, and from there you have to look at the available widths of that implement and decide how many different implements and sizes it will take to complete the task.

Horsepower SizingNow that all the widths have been calculated, you have find out how much horsepower it will take to pull or operate that implement. There are a few different types of power that have to be found to be able to accurately make a decision on what size tractor you need for each implement. For the pull behind implements, you only have to calculate drawbar horsepower; however, to calculate drawbar horsepower you first have to calculate the draft. Draft is the force acting on the drawbar in pounds.

D=Fi∗(A+B∗( S )+C∗(S2 ))∗wp∗TWhere,

F = Soil texture adjustment parameter, (see table 3.3), unit lessI =1 for fine, 2 for medium, 3 for coarse textured soils

A, B, C = machine specific parameters (see table 3.3), unit lessS= field speed, mph

We = width of implement, ft.T= tillage depth, in (1 for minor tillage and seeding implements)

Now that the draft has been found, we can calculate the drawbar power in horsepower. This is done by,

Pdb=

D∗S∗5280 ftmi

∗1hr

60min∗1hp

33,000 ft lbsmin

Where,Pdb= drawbar power, hp D= draft, lbs. S= Speed, mph

For implements that are strictly pull behind, this is all that needs to be calculated at this time. Even though the sprayer and combine do not pull anything, they still have a drawbar horsepower that needs to be calculated. To do this, you have to find the rolling resistance which is done by calculating the force of friction. Force of friction is calculated by,

F f=wt∗μ f

Where,Ff = force of friction (or rolling resistance), lb.Wt = weight (of trailer, equipment, etc.), lb.

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μf = Coefficient of friction, unitless (see table 3.1)

Since Force of Friction is the only major resistive force for the sprayer and the combine, you can use Ff in the place of draft in the drawbar horsepower equation. Drawbar power is not the only thing that has to be found for the combine. PTO power is what runs the combines head so that is the next thing to be calculated.

Ppto=a+b+w p+c∗FRWhere,

Ppto = PTO power, hp a = machine specific parameter, hp (see table 3.4) b = machine specific parameter, hp/ft. c = machine specific parameter, hp*hr./ton wp = width of implement, ft. FR = material feed rate, ton/hr.

PTO and force of friction is all that has to be calculated for the combine at the moment. However, the sprayer doesn’t have a PTO but it has a pump so you have to find the horsepower of the pump, which is done by calculating the fluid power.

P=

p∗Q∗2.3 ftps i

∗8.3 lb

gal∗1hp

33,000 ft∗lbmin

Where,P = fluid power, hpp = gage pressure, psiQ = flow rate, gpm

This equation will find the fluid power for one nozzle on the sprayer, therefore once this has been calculated you have to find the number of nozzles on the boom and multiply that by the hp to get the correct horsepower for the sprayer.

Each separate power has been found so now they all have to be combined to find the total implement power for each separate piece of equipment. This is found by,

Ptot=Pdb

e t+Ppto+Phyd

Where,Ptot = total implement power requirement, hpPdb = drawbar power requirement, hpEt = traction efficiency, unit lessPpto = PTO power, hpP hyd = hydraulic power (total fluid power), hp

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Now that the total power for each implement is known, the engine power required can be estimated. The engine power is more than the total implement power because of power losses in the tractors transmission, the traction of the tractor, and other power losses. The engine power requirement is found by,

Peng=P tot

0.86Where,

Peng = engine power requirement, hpPtot = total implement power requirement, hp

The engine power requirement represents the minimum horsepower required to satisfy implement power needs. The actual engine power used will always need to be higher than this number so the tractor can overcome changes in topography and soils, power to accelerate, and friction forces exerted by the tractor its self.

CostsFor every operation there is always a cost to run that operation which then goes into an even bigger system that has all the costs of everything that is needed to be bought for the farm to properly run. The most costly thing in any farming system is going to be the costs of the implements and the machinery. Each implement and machine that is bought has a fixed cost and a variable cost.

Fixed CostsFixed costs depend on how long a machine is owned rather than how much it is used. The fixed cost of a machine is 56% of the total cost. Depreciation is the loss in value of a machine due to age, wear, and other obsolescence’s. It is greatest during the early stages of a machine’s life. Other fixed costs include taxes, shelter, insurance, and interest. The first step in finding the fixed cost is to find the retail value of the machine. This is found by looking at farm machinery costs (Hunt 1999) this shows the value of the machine in 1999; however, since the prices rise every year, we have to use,

RV=1999 price∗1.03n

Where, RV= retail value for any given year n = the number of years since 1999

The retail value of the machine is now known so the depreciation can be found for each year by using the declining balance equation, which is,

D=P∗(1− xL )

n

−P∗(1− xL )

n+1

Where,D = annual depreciation charge for year n+1, $N = age of machine at beginning of year in question, yr.X = rate of depreciation

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L = time between buying and selling, yr.

The depreciation is now known for the year in question. To find the resale value of that year, take the retail value of the year before and subtract it from the depreciation that was calculated. The retail value is what is used to find all other fixed costs for the machinery (interest, insurance, and shelter) which is found by,

Intrest=RV∗IR

Insurace=RV∗coverage∗premium

Shelter=RV∗0.03Where,

RV = retail value of year in questionIR= interest rate Coverage = insurance coverage % as a decimalPremium = insurance premium % as a decimal

Variable CostsVariable costs vary in proportion to the amount of time the machine is used, as acreage or tonnage rise so does the cost. Variable costs include fuel, lubrication, maintenance, repair, and labor. Labor is the first thing to find because it will give us the hours of operation of that piece of equipment. The hours of operation is found by looking back at the suitable fieldwork days and amount of days the equipment is being used. The equation looks like:

Hours per year=Dop∗hd

Where,Dop = days of operationhd = hours of work day

Once hours per year is found, simply multiply it by labor cost per hour to get total labor costs for the implement.

Fuel cost is the most expensive of the variable costs and it a four step equation. First the % power the tractor is using needs to be found. This is done by,

P x%=hpneededhp actual

∗100

Where,Px% = % power the tractor has to exert to pull the implement Hp needed= engine power requirementHp actual = the Hp the tractor actually has

Now that % power has been determined, the rate of fuel consumption is found by,

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Qf=Px %

fe x%

Where, Qf = gal/hr.

Fex% = fuel efficiency, hp∗hrgal (see table)

Once the fuel consumption is calculated, the total volume of fuel consumed per year is found by,

V f=Qf∗tWhere,

Vf = gal/yeart = hours per year the equipment is used

The gallons per year is now known for the single piece of equipment. The final step is to find the cost of fuel for it, which is done by,

Fuel=V f∗P f

Where,Fuel = total fuel cost, $Pf = price of 1 gallon of fuel, $

Lubrication is equal to 10% of the fuel costs for any given piece of machinery so the equation is,Lubrication=Fuel∗0.10

Repair and maintenance is the last of the variable costs. Routine wear, accidental breakage and damage, operator neglect, and routine overhauls are all different types of things that will call for repairs. Repair and maintenance costs can be found by,

Repair=RV∗RF 1∗( thr1000 )

RF2

Where,Repair = cost of repair for a given yearRV= retail or resale value for a given yearT= total hours on machineRF1= see table 7.3RF2= see able 7.3

Now that all fixed and variable costs have been found for one implement, the total cost can be found per year and for the life of that machine.

Alternatives to PurchasingRenting of equipment can be advantageous in some cases, and the price per year is found by taking the retail value of the implement and multiplying it by .15. This gives you the cost to rent. Then go back and add rental cost to variable costs minus the repair (instead of adding fixed and variable costs). Keep in mind that rental cost goes up by 3% a year after the first year.

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Chemical and SeedFor the chemical and seed costs, the prices were found from the chemical and seed spreadsheet. The spreadsheet gave prices per gallon, per ounce, or per pound. Then using common conversion factors, the amount per acre was found. Once amount per acre was found, the following equation was used to find price,

Cost=amountacre

∗7800acres∗price of chemical

ResultsTwo systems were found and evaluated to discover what the best and most profitable system would be.

Chemical and SeedEven though there are two systems (large, and small scale), the price of chemicals and seed will be the same for both systems. The chemicals to be applied before wheat operations is glyphosate at 0.9 lb.-active ingredient per acre, then P2O5 at 30 lb./ac and K2O at 100 lb./ac during planting, and Harmony extra at 0.6 oz./ac half way through the growing season. For soybeans 2,4-D at 1.6 pt./ac and glyphosate at 2.5 lb.- active ingredient per acre before planting, during disking lime is to be applied at 0.6 tons/ac, then ¼ way through the growing season reflex at 1.35 pt./ac is to be applied, and ½ way through cobra at 1.55 oz./ac. The seeding rate for wheat is 38 seeds/ft2 and 3.8 seeds/ft. for soybeans.

2,4-D Glyphosate

Harmony

Reflex Cobra Lime P2O5 K2O Soybeans

Wheat

29,359.2

75,562.5

59,342.4

123,108.9

12,177.84

187,200 81,900 246,870 4,412.1 279.41

This is the amount each chemical and seed will cost per year once added. The total amount for seeds and chemicals is $820,212.35.

Miscellaneous CostsLand rent and my total salary are the only miscellaneous costs. Land rent is $50/ac which comes out to be $390,000 per year. My salary is $150,000 per year.

Gross ProfitFor the futures price of soybeans on cme.com, the price is $10.31. Wheat is $6.43 per bushel. The average soybean yield is 33bu/ac, and wheat is 75bu/ac; therefore, the gross profit for 7800 acres is $2,653,794 for soybeans and $3,761,550 for wheat per year.

Suitable Fieldwork DaysThe first thing to de determined was the suitable fieldwork days and the calendar. In the spring, we have wheat harvest and soybean planting, we are not working on Sundays, and our

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tractability coefficient is 0.96 for both because we have a sandy loam soil. In the fall, we have soybean harvest and wheat planting, we are still not working on Sundays, and our tractability coefficient is 1.06 for soybean harvest and 1.03 for soybean planting. The suitable days were calculated.

Large Scale OperationFor the large scale operation, I found the biggest implement available then found the TFC, EFC, and days for 1 implement to do 7800 acres.

Width (ft. or rows)

Speed (mph)

TFC (ac/hr.)

EFC (ac/hr.)

Workday (ac/day)

Days for 7800 acres

Combine 36 3 13.09 9.16 100.80 77.38Sprayer 120 6.5 94.55 61.45 676 11.53

Disk Harrow

40.5 6 29.45 23.56 259.20 30.09

Field Cultivator

60 7 50.91 43.27 476 16.39

Chisel plow

40 5 24.24 20.61 226.67 34.41

Row crop planter

24 (rows) 5.5 31.77 20.65 227.18 34.33

Grain drill 40 5 24.24 16.97 186.67 41.78

For the spring operations, I knew my wheat had to be out of the field before I could start spraying, disking, and subsoiling, and planting my soybeans. Through some guess and check I figured out how many implements I would need to properly run my system:

# of implements Days for 7800(SWFD)

Actual days of operation

Combine 6 12.89 5/26-6/16Sprayer 1 11.53 6/1-6/17

Disk Harrow 2 15.05 6/2-6/23Chisel Plow 2 17.21 6/3-6/27

Row crop planter 2 17.17 6/16-7/111st post emergent

spray2 5.77 8/5-8/14

2nd post emergent 2 5.77 9/8-9/17

For the fall operations, the beans had to be picked before I could spray. After I started spraying, there had to be two disking operations, subsoiling, smoothing, and then planting. To complete all the fall operations I figured out how many implements would be needed to run the system properly was,

Wheat Harvest Soybean planting Soybean Harvest Wheat plantingTypical dates May 26-July 1 May 20- July 11 October 20-

December 30October 15- December 31

SFWD 21.34 days 35.28 days 15.77 days 18.01 days

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# of implements Days for 7800 ac(SFWD)

Actual days of operation

Combine 6 12.89 10/20-11/17Spray 1 11.53 10/23-11/18

1st Disking 3 10.03 10/26-11/192nd Disking 3 10.03 10/27-11/22Chisel plow 3 11.47 11/1-12/3

Field cultivator 2 8.2 11/5-12/4Grain drill 5 8.36 11/7-12/31

Post emergent spray 2 5.77 3/12-4/21

Now that the number of implements was found, the next thing to find was the required engine power for each implement. Below is the calculated drawbar hp for each implement or machine.

Draft(Lb.)

Traction coefficient Total implement power at drawbar (hp)

Combine 2500 0.78 26Sprayer 760 0.78 16Disking 17,235 0.78 354

2nd fall Disking 11,490 0.75 245Chisel plow 20,808 0.75 379

Field cultivator 12,892 0.75 321Grain drill 4,000 0.78 68

Row crop planter 4,800 0.78 91

For all of these calculations, draft was found by using 6 inches deep for primary tillage, 4 inches deep for secondary tillage, and 8 inches deep for the chisel plow. The soil type was sandy loam and medium texture. The traction coefficient 0.78 was used for firm soil and 0.75 was used for soft or tilled soil.

Along with the drawbar horsepower, the combine has PTO power that needs to be calculated. Below is the calculated PTO power.

PTO power (hp)Combine 184

For PTO power, the material feed rate was found by finding the EFC and multiplying it by the maximum wheat yield (130 bu/ac) then multiplying it by the weight of a bushel of wheat (60 lbs.) and finally diving by 2,000 to get the tons/hour of the head which is 35.7 tons/hr.

Total engine power required was then calculated and the hp requirements are,

Combine

Sprayer Disking 2nd fall Disking

Chisel plow

Field Cultivator

Grain drill

Row crop

PlanterPeng (hp) 209 340 411 285 430 373 79 105

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Since the minimum horsepower was been found, I then went on to size tractors for the implements, I only needed four different size tractors for my large operation,

6 tractors with 450 horsepower 2 tractors with 400 horsepower 3 tractors with 300 horsepower 5 tractors with 120 horsepower

Then to complete all other operations I needed, 2 120ft. sprayers 6 combines with 36 foot head

For the spring and fall seasons, six combines will be run and 1 sprayer for pre emergent operations. For all post emergent spraying operations, two sprayers will be used. In the spring, two 450 hp tractors will pull the disk, two 450 hp tractors will pull the chisel plows, and two 120 hp tractors will pull the row crop planters. In the fall, there will need to be three 450 hp tractors pulling disks for primary tillage, three 300 hp tractors pulling disks for the secondary tillage operation, three 450 hp tractors to pull the chisel plow, two 400 hp tractors pulling the field cultivators, and finally five 120 hp tractors pulling the grain drills. Part load fuel and hours of operation were then found for each implement, as seen below.

Hp∗hrgal

Gal/hr. Hours/year Gal/year

Combine 14.81 14.1 284 4004Sprayer 15.68 21.7 319 6922

Row crop planter

15.1 6.9 189 1311

Grain drill 13.84 5.8 92 534Disk (both seasons)

15.39 26.7 277 7382

Disk (fall only)

15.39 17.9 111 1709

Chisel Plow (both seasons)

15.68 27.4 190 5187

Chisel plow (fall only)

15.39 24.7 127 3480

Field cultivator 15.39 24.24 91 2187

Hours of operation for tractorsTractor

(hp)120

(2 tractors)120

(3 tractors)300 400 450

(4 tractors)450

(2 tractors)Hours 281 92 111 91 301 126

Gallon per year is found for each implement not each tractor. Once the hours and gal/year were found for each implement, we can now go into cost for this system.

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The total costs of the system are the equipment cost per year plus chemical and seed cost, plus land rent, and my salary. The total cost per year can be seen below,

With the total costs for all bought and all rented machinery, we can now analyze the profit or losses.

 Machinery bought

Costs Profit Profit per Year

Profit per ac

Year 0 6710047.881 -6,710,047 -860.26

Equipment Costs total Rented

Year 0 0Year 1 2570795.871Year 2 2647919.747Year 3 2727357.34Year 4 2809178.06Year 5 2893453.402Year 6 2980257.004Year 7 3069664.714Year 8 3161754.655Year 9 3256607.295Year 10 3354305.514

Total Costs (rented)Year 0 0Year 1 3931008.221Year 2 4008132.097Year 3 4087569.69Year 4 4169390.41Year 5 4253665.752Year 6 4340469.354Year 7 4429877.064Year 8 4521967.005Year 9 4616819.645Year 10 4714517.864Total 43073417.1

Total Costs (Bought)Year 0 6710047.881Year 1 3727579.856Year 2 3758080.69Year 3 3470014.158Year 4 3242887.977Year 5 3066226.628Year 6 2930235.069Year 7 2826963.63Year 8 2749998.116Year 9 2694175.456Year 10 2655344.391Total 37831553.85

Equipment Costs total Bought

Year 0 6,710,047.881Year 1 2,367,367.506Year 2 2,397,868.34Year 3 2,109,801.808Year 4 1,882,675.627Year 5 1,706,014.278Year 6 1,570,022.719Year 7 1,466,751.28Year 8 1,389,785.766Year 9 1,333,963.106Year 10 1,295,132.041

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Year 1 3727579.856 6415344 2687764.14 344.58Year 2 3758080.69 6415344 2657263.31 340.67Year 3 3470014.158 6415344 2945329.84 377.60Year 4 3242887.977 6415344 3172456.02 406.72Year 5 3066226.628 6415344 3349117.37 429.37Year 6 2930235.069 6415344 3485108.93 446.80Year 7 2826963.63 6415344 3588380.37 460.04Year 8 2749998.116 6415344 3665345.88 469.91Year 9 2694175.456 6415344 3721168.54 477.07Year 10 2655344.391 6415344 3759999.60 482.05

Total 37,831,553.85 64,153,440 26,321,886.15 3374.60

 Machinery Rented

Costs Profit Profit per Year

Profit per ac

Year 0 0 0 0Year 1 3931008.221 6415344 2484335.779 318.50Year 2 4008132.097 6415344 2407211.903 308.61Year 3 4087569.69 6415344 2327774.31 298.43Year 4 4169390.41 6415344 2245953.59 287.94Year 5 4253665.752 6415344 2161678.248 277.13Year 6 4340469.354 6415344 2074874.646 266Year 7 4429877.064 6415344 1985466.936 254.54Year 8 4521967.005 6415344 1893376.995 242.74Year 9 4616819.645 6415344 1798524.355 230.58Year 10 4714517.864 6415344 1700826.136 218.05Total 43,073,417.1 64,153,440 21,080,022.9 2702.56

Small scaleFor small scale operations, all the parameters are the same; therefore, I found a smaller implement available then found the TFC, EFC, and days for 1 implement to do 7800 acres.

Width (ft. or rows)

Speed (mph)

TFC (ac/hr.)

EFC (ac/hr.)

Workday (ac/day)

Days for 7800 acres

Combine 36 3 13.09 9.16 100.80 77.38Sprayer 120 6.5 94.55 61.45 676 11.53

Disk Harrow

21 6 15.27 12.22 134.4 58.03

Field Cultivator

30 7 25.45 21.64 238 32.7

Chisel plow

20 5 12.12 10.3 113.3 68.8

Row crop planter

12 (rows) 5.5 15.77 10.25 112.78 69.16

Grain drill 25 5 15.15 10.61 116.67 66.85

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Spring operations and # of implements needed,

# Of implements Days for 7800(SWFD)

Actual days of operation

Combine 6 12.89 5/26-6/16Sprayer 1 11.53 6/1-6/17

Disk Harrow 5 11.6 5/30-6/15Chisel Plow 5 13.76 5/31-6/20

Row crop planter 4 17.29 6/17-7/111st post emergent

spray2 5.77 8/5-8/14

2nd post emergent 2 5.77 9/8-9/17

Fall operations and # if implements needed,# Of implements Days for 7800 ac

(SFWD)Actual days of

operationCombine 6 12.89 10/20-11/17

Spray 1 11.53 10/23-11/181st Disking 5 11.61 10/24-11/202nd Disking 5 11.61 10/25-11/23Chisel plow 5 13.8 10/26-12/16

Field cultivator 3 10.9 10/31-12/17Grain drill 7 9.55 11/4-12/31

Post emergent spray 2 5.77 3/12-4/21

Calculated drawbar hp for each implement or machineDraft(Lb.)

Traction coefficient Total implement power at drawbar (hp)

Combine 2500 0.78 26Sprayer 760 0.78 16Disking 8937 0.78 183

2nd fall Disking 4154 0.75 103Chisel plow 13005 0.75 231

Field cultivator 6446 0.75 120Grain drill 2500 0.78 45

Row crop planter 2400 0.78 46

Total engine power required Combin

eSprayer Disking 2nd fall

DiskingChisel plow

Field Cultivator

Grain drill

Row crop

PlanterPeng 209 340 213 103 268 186 51 54

Since the minimum horsepower was been found I then went on to size tractors for the implements, I only needed five different size tractors for my small operation,

5 tractors with 300 horsepower

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5 tractors with 250 horsepower 3 tractors with 200 horsepower 5 tractors with 120 horsepower 7 tractors with 70 horsepower

Then to complete all other operations I needed 2 120ft. sprayers 6 combines with 36 foot head

For the spring and fall season, six combines will be run and one sprayer for pre emergent operations. For all post emergent spraying operations, two sprayers will be used. In the spring, five 250 hp tractors will pull the disk, five 300 hp tractors will pull the chisel plows, and four 70 hp tractors will pull the row crop planters. In the fall, there will need to be five 250 hp tractors pulling disks for primary tillage, five 120 hp tractors pulling disks for the secondary tillage operation, five 300 hp tractors to pull the chisel plow, three 200 hp tractors pulling the field cultivators, and finally seven 70 hp tractors pulling the grain drills.

Part load fuel and hours of operation small scale:

Hp∗hrgal

Gal/hr. Hours/year Gal/year

Combine 14.81 14.1 284 4004Sprayer 15.68 21.7 319 6922

Row crop planter

14.52 3.8 191 726

Grain drill 14.18 3.6 106 382Disk (both seasons)

15.1 14.4 256 3687

Disk (fall only)

14.81 7 128 896

Chisel Plow (both seasons)

15.1 17.8 304 5411

Field cultivator 15.39 12.1 120 1452

Tractor hours per year.Tractor hp

70(4 tractors)

70(3 Tractors)

120 200 250 300

Hours 297 106 128 120 256 304

Costs bought and rented.

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The total costs of the system are the equipment cost per year plus chemical and seed cost, land rent, and my salary. The total cost per year can be seen below,

Equipment Costs total rented

Year 0 0Year 1 2563866.14Year 2 2640782.12Year 3 2720005.59Year 4 2801605.76Year 5 2885653.93Year 6 2972223.55Year 7 3061390.25Year 8 3153231.96Year 9 3247828.92Year 10 3345263.79

Equipment Costs total Bought

Year 0 6849456.75Year 1 2182612.86Year 2 2164915.08Year 3 1860638.77Year 4 1619110.61Year 5 1429248.62Year 6 1280931.92Year 7 1165997.26Year 8 1077877.68Year 9 1011295.92Year 10 962013.27

Total Costs boughtYear 0 6849456.75Year 1 3542825.21Year 2 3525127.4Year 3 3220851.12Year 4 2979322.96Year 5 2789460.97Year 6 2641144.27Year 7 2526209.61Year 8 2438090.03Year 9 2371508.27Year 10 2322225.62Total 35,206,222.29

Total Costs rentedYear 0 0Year 1 3924078.494Year 2 4000994.479Year 3 4080217.942Year 4 4161818.11Year 5 4245866.283Year 6 4332435.901Year 7 4421602.608Year 8 4513444.315Year 9 4608041.274Year 10 4705476.142Total 42,993,975.55

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With the total costs for all bought and all rented, we can now analyze the profit or losses. Bought Costs Profit Profit per

YearProfit per ac

Year 0 6849456.752 -6849456.75 -878.13Year 1 3542825.215 6415344 2872518.785 368.27Year 2 3525127.43 6415344 2890216.57 370.54Year 3 3220851.127 6415344 3194492.873 409.55Year 4 2979322.966 6415344 3436021.034 440.51Year 5 2789460.976 6415344 3625883.024 464.85Year 6 2641144.276 6415344 3774199.724 483.87Year 7 2526209.617 6415344 3889134.383 498.60Year 8 2438090.037 6415344 3977253.963 509.90Year 9 2371508.276 6415344 4043835.724 518.44Year 10 2322225.621 6415344 4093118.379 524.75

Total 35,206,222.29

64,153,440

28,947,217.71

3711.18

 Rented Costs Profit Profit per Year

Profit per ac

Year 0 0 0 0Year 1 3924078.49

46415344 2491265.50

6319.39

Year 2 4000994.479

6415344 2414349.521

309.53

Year 3 4080217.942

6415344 2335126.058

299.37

Year 4 4161818.11 6415344 2253525.89 288.91Year 5 4245866.28 6415344 2169477.71 278.13

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3 7Year 6 4332435.90

16415344 2082908.09

9267.03

Year 7 4421602.608

6415344 1993741.392

255.60

Year 8 4513444.315

6415344 1901899.685

243.83

Year 9 4608041.274

6415344 1807302.726

231.70

Year 10 4705476.142

6415344 1709867.858

219.21

Total 42,993,975.55

64,153,440

21,159,464.45

2712.75

Custom opsCustom ops goes off the acreage and normally has a fee per acre for each different implement. Below is a chart that has total costs for custom ops with my salary and chemical added together:

Custom ops profit

Total CostsYear 0 0Year 1 2307419.897Year 2 2347536.123Year 3 2388855.836Year 4 2005159.951Year 5 2475251.225Year 6 2520402.391Year 7 2566908.092Year 8 2614808.964Year 9 2664146.863Year 10 2714964.898Total 24605454.24

Costs Profit Profit per Year

Profit per ac

Year 0 0 0 0Year 1 2307419.897 6415344 4107924.103 526.65Year 2 2347536.123 6415344 4067807.877 521.51Year 3 2388855.836 6415344 4026488.164 516.21Year 4 2005159.951 6415344 4410184.049 565.40Year 5 2475251.225 6415344 3940092.775 505.14Year 6 2520402.391 6415344 3894941.609 499.35Year 7 2566908.092 6415344 3848435.908 493.38Year 8 2614808.964 6415344 3800535.036 487.24Year 9 2664146.863 6415344 3751197.137 480.92Year 10 2714964.898 6415344 3700379.102 474.40

Total 24,605,454.24 64,153,440 39,547,985.76 5070.25

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ConclusionMoo-Moo Milk Markers row crop operations with 7800 acres if very profitable in every system that was analyzed. With 50% suitable fieldwork days on 7800 acres on a farm in South Carolina Mortimer McMaster best option for his farm is to hire out custom ops for every operation, he only has to buy chemical and seed every year and pay the custom ops fee. For custom ops he will bring home $39,547,985 in a 10-year period he will be more profitable at first but each year the profit declines. If he doesn’t want to go with custom ops his next best thing is to buy all the equipment in small scale operation if he does this he will bring home $28,947,217 in a 10 year period. Next most profitable is buying all equipment up front in large scale and he will make $26,321,886 in 10 years. However the problem with this system is some of the machinery has to be replaced around year 7 and he will need almost 7 million for an initial investment. If he doesn’t want to buy all the equipment and instead rent it small scale renting will make $21,159,464 in 10 years. Lastly the least profitable system is renting large scale in 10 years he will still make $21,080,022.

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Appendix A 

Table 3.4

Table 3.1 Typical values for f in agricultureGround conditions f

Smooth, firm ground 0.05Soft, tilled ground 0.075Loose sand or gravel 0.10Soft, muddy ground 0.15

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