Economically Optimal Use of Discrete Inputs
AAE 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
 noBt  pYnoBt  SeedCostnoBt  K
 noBt  7.00(170)  100  600  490
 Bt  pYBt  SeedCostBt  K
 Bt  7.00(175)  110  600  515
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:
 t  PYt  Ct  K
Suppose have a base treatment (no control or the “base” or “standard” system to compare to):
 0  PY0  K
Can then calculate the change in profit for using the new treatment t
 t   t   0  PYt  Ct  K  ( PY0  K )  PYt  Ct  K  PY0  K
 t  P(Yt  Y0 )  Ct  PY  Ct
If want to know what it takes for the gain in profit to be positive, then
  PY  Ct  0
C
Y  t , which looks a lot like MP > r/P
P
To find the optimal system, choose the treatment t that has the largest profit
 t*  max  t  t  0,..., T
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 system CFP, 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 2 for 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: Y ( N )  Ymax 1  exp(0  1 N ) 
Multiple Inputs: Y ( N , t , m)  Ymax (m) 1  exp(0 (t )  1 N )  , where t denotes the discrete tillage
system used and m denotes the discrete corn hybrid maturity planted
  0FP if fall plow (t  FP)
 FC
90
Ymax
if plant 90 d corn (m  90)
  if fall plow (t  FC )
t
 0   0RT
Ymax   95
Ymax if plant 95 d corn (m  95)
  0 if fall plow (t  RT )
  0NT if fall plow (t  NT )
How we wrote this for estimation using dummy variables:
y   ymax90 D90  ymax95 D951  exp(0 FC DFC  0 FP FFP  0 NT DNT  0 RT DRT  1N ) 
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:
  PYmax 1  exp(0  1 N )   rN  K
FOC: 1PYmax exp( 0  1 N )  r / p . Solve for N: N 
 r
ln 
1  P1Ymax
1
 0

 1
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
1  r   0
solution N  ln 
to find the yield Y and profit  associated with each hybrid

1  P1Ymax  1
maturity and tillage system.
Note that we have two discrete inputs: tillage t and maturity m
Assume price and cost parameters
Parameters
P ($/bu)
r ($/lbs)
C_90 ($/ac)
C_95 ($/ac)
C_FC ($/ac)
C_FP ($/ac)
C_NT ($/ac)
C_RT ($/ac)
K ($/ac)
7.50
0.65
35
40
100
130
35
60
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
90
90
90
90
95
95
95
95
Tillage Ymax beta_0 beta_1 N*
Y*
FC
126 -0.573 -0.026 119 122
FP
126 -0.849 -0.026 108 122
NT
126 -0.427 -0.026 124 122
RT
126 -0.667 -0.026 115 122
FC
165 -0.573 -0.026 129 162
FP
165 -0.849 -0.026 119 162
NT
165 -0.427 -0.026 135 162
RT
165 -0.667 -0.026 126 162
N
Cost
77.17
70.16
80.86
74.77
84.10
77.09
87.79
81.70
Seed
Cost
35.00
35.00
35.00
35.00
40.00
40.00
40.00
40.00
Till
cost
100
130
35
60
100
130
35
60
K
600
600
600
600
600
600
600
600
Pi
104.95
81.96
166.27
147.36
389.09
366.10
450.40
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
Max
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