Finding the Optimal Economic Threshold
Some inputs are discrete, but manager collects information (a signal) to decide if/when to use
them. IPM is the most common example. The issue: what do you use for the threshold? How
many insects is enough to treat?
Basic Idea: Define profit as a function of the observed signal (i.e., pest population density N)
both with and without the treatment.
What do we expect (properties of returns)
1) Intercept (Level at N = 0): because expect same loss with no pests (N = 0), but the
treatment is costly:  0 ( N  0)   t ( N  0) .
2) Slope: Expect negative slope (assuming monotonic functions): as pest density increases,
loss increases and profit decreases and more negative slope for returns without treatment
d 0
d
(N )  t (N )  0 .
because greater losses for each increase in the pest density:
dN
dN
These two mean there must exist a pest density N* such that  0 ( N *)   t ( N *) and for all N <
N*,  0 ( N *)   t ( N *) and for all N > N*,  0 ( N *)   t ( N *) .
This N* is the threshold. If N < N*, do not treat and N > or = N*, treat.
Linear Case
Analyzing Risk for IPM: works out the case for linear (N)
no = V(1 – IDN) – G
t = V(1 – IDN(1 – K)) – C – G
C
EIL nr  N * 
VIDK
Non-Linear Cases
Cousens’ loss model
Profit in N: can solve analytically, but messy
aN


 0  PY 1 

 1  aN / B 
asN


 t  PY 1 
C
 1  asN / B 
P is price, Y is the pest free yield, a and B are parameters of the loss function, N is the pest
population density, s is the survival rate of the pest after treatment, and C is the cost of treatment.
Equate and solve for N
aN
asN




PY 1 
  PY 1 
C
 1  aN / B 
 1  asN / B 
aN
asN
C
aN
asN
C
1
 1



1  aN / B
1  asN / B PY
1  aN / B 1  asN / B PY
BaN
BasN
C
BaN
BasN
C




B  aN B  asN PY
B  aN B  asN PY
BaN ( B  asN )  BasN ( B  aN ) C
B 2 aN  Ba 2 sN 2  B 2 asN  Ba 2 sN 2
C


( B  aN )( B  asN )
PY
( B  aN )( B  asN )
PY
2
2
2
B aN  B asN
C
B aN (1  s)
C


( B  aN )( B  asN ) PY
( B  aN )( B  asN ) PY
C
C
B 2 aN (1  s ) 
( B  aN )( B  asN ) 
( B 2  BasN  BaN  a 2 sN 2 )
PY
PY
C
( B 2  BasN  BaN  a 2 sN 2 )  B 2 a (1  s ) N  0
PY
C 2 C
C 2 2
B 
Ba (1  s ) N  B 2 a (1  s ) N 
a sN  0
PY
PY
PY
Ca 2 s 2  CBa(1  s)
C 2

N 
 B 2 a(1  s)  N 
B 0
PY
PY
PY


Solve using quadratic formula: ax 2  bx  c  0 , solution x 
b  b2  4ac
2a
Empirically, find the lower root is the one we want
Graphics, PY = 100, a = 0.01, B = 0.25, s = 0.10, C = 10, N* = 23.691
Exponential Loss
Can’t solve analytically, must use numerical solution algorithm
 0  PY exp(  N )
 t  PY exp( sN )  C
Equate and solve for N
PY exp(  N )  PY exp(  sN )  C
exp(  N )  exp(  sN ) 
C
0
PY
exp(  N )  exp(  sN ) 
F ( N |  , s , C , P, Y )  0
Gives Implicit function F(N) = 0. Use solver in Excel
Find the value of N that sets the implicit function F(N) equal to 0
Parameters:  = -0.2, PY = 100, s = 0.1, C = 30
Solver gives N* = 2.0838
C
PY