Economically Optimal Use of Discrete InputsAAE 575 Paul D. Mitchell

Many inputs are discrete:

Tillage system used: Conventional tillage, conservation tillage or no till

Hybrid type planted: 100 day corn or 105 day corn

Pest control: Bt or non-Bt, RR or non-RR crop, seed treatment or no seed treatment

How do you choose the economically optimal use of discrete inputs? Calculus no longer works!

Method: Calculate net returns (profit) for each level or type of discrete input and choose the input that gives you the highest return

(Super) Simplified example: Bt and non-Bt corn

Production Function: yield with Bt trait = 175 bu/ac, yield without Bt trait = 170 bu/ac

Cost: Bt seed costs $110/ac and non-Bt seed costs $100/ac

Price of corn = $7.00/bu and other costs of production = $600

Make more money with Bt under these price and yield assumptions, so choose Bt trait.

With only two levels (Bt or non-Bt), very much like a Partial Budget Analysis (Google it, it’s very simple)

More general

Suppose have treatment t and net returns with this treatment system are:

Suppose have a base treatment (no control or the “base” or “standard” system to compare to):

Can then calculate the change in profit for using the new treatment t

If want to know what it takes for the gain in profit to be positive, then

, which looks a lot like MP > r/P

To find the optimal system, choose the treatment t that has the largest profit

Tillage (Multiple levels or types of the discrete input)

Fall Plow (FP): moldboard plow and disk in fall, then disk and field cultivate and plant in spring

Fall Chisel (FC): chisel plow in fall, then disk and field cultivate and plant in spring

Reduced Tillage (RT): disk and field cultivate and plant in spring

No-Till (NT): no till plant in spring

Yield: assume varies by tillage system: YFP, YFC,YRT,and YNT

Cost: also varies by tillage systemCFP, CFC,CRT,and CNT

Net Returns: t = PYt – Ct – K, where t = {FP, FC, RT, NT} is the tillage system

FP = PYFP – CFP – K, FC = PYFC – CFC – K, RT = PYRT – CRT – K, NT = PYNT – CNT – K

Plug in the yield and cost numbers and see which one has the highest net returns

Multiple Inputs: Mix of Discrete and Continuous

Method: For each level or type of the discrete inputs, have to optimize the amount of the continuous input.

General model:

Yield depends on X and Z inputs where X is a continuous input and Z is a discrete input with levels Za, Zb, Zc, ..., or Y = f(X, Z)

Profit: p(X, Z) = Pf(X, Z) –rX – c(Z) – K, where c(Z) = Ca if Z = Za, Cb if Z = Zb, …

Solution Process

1)Set Z = Za and use calculus to find optimal level of X, given Z = Za. Define this as X*(Za).

2)Substitute this X*(Za) back into the profit function and calculate profit, given that X = X*(Za) and Z = Za. Define this as *(X*(Za), Za) = a.

3)Repeat steps 1 and 2for each level of Z to find a, b, c, …

4)Optimal Z is the Z that that gives the highest 

Example to Illustrate

Problem set #2: Negative exponential production function for corn yield and nitrogen, with the Ymax and 0 parameters depend on the hybrid maturity and/or the tillage system

Base model:

Multiple Inputs: , where t denotes the discrete tillage system used and m denotes the discrete corn hybrid maturity planted

How we wrote this for estimation using dummy variables:

Given the tillage system and hybrid maturity, we know Ymax and 0, and then the optimal N rate is found by setting MP = r/p and solving for N:

FOC: . Solve for N:

SOC: satisfied as long as 0 and 1 are negative

Each combination of tillage and hybrid maturity determine Ymax and 0, then use the optimal solution to find the yield Y and profit  associated with each hybrid maturity and tillage system.

Note that we have two discrete inputs: tillage t and maturity m

Assume price and cost parameters

Parameters
P ($/bu) / 7.50
r ($/lbs) / 0.65
C_90 ($/ac) / 35
C_95 ($/ac) / 40
C_FC ($/ac) / 100
C_FP ($/ac) / 130
C_NT ($/ac) / 35
C_RT ($/ac) / 60
K ($/ac) / 600

Based on these parameters, for each hybrid maturity and tillage system, can calculate Ymax and 0, then use these to calculate N*, Y*, costs, and *

Maturity / Tillage / Ymax / beta_0 / beta_1 / N* / Y* / N Cost / Seed Cost / Till cost / K / Pi / Max
90 / FC / 126 / -0.573 / -0.026 / 119 / 122 / 77.17 / 35.00 / 100 / 600 / 104.95 / -
90 / FP / 126 / -0.849 / -0.026 / 108 / 122 / 70.16 / 35.00 / 130 / 600 / 81.96 / -
90 / NT / 126 / -0.427 / -0.026 / 124 / 122 / 80.86 / 35.00 / 35 / 600 / 166.27 / -
90 / RT / 126 / -0.667 / -0.026 / 115 / 122 / 74.77 / 35.00 / 60 / 600 / 147.36 / -
95 / FC / 165 / -0.573 / -0.026 / 129 / 162 / 84.10 / 40.00 / 100 / 600 / 389.09 / -
95 / FP / 165 / -0.849 / -0.026 / 119 / 162 / 77.09 / 40.00 / 130 / 600 / 366.10 / -
95 / NT / 165 / -0.427 / -0.026 / 135 / 162 / 87.79 / 40.00 / 35 / 600 / 450.40 / ****
95 / RT / 165 / -0.667 / -0.026 / 126 / 162 / 81.70 / 40.00 / 60 / 600 / 431.49 / -

With these prices and costs, optimal hybrid maturity is 95 days, optimal tillage is no-till and optimal N rate is 135, generating a yield of 162 bu/ac and profit of $450.40/ac