EWU – ECON 457 - Natural Resource Economics
The simple mathematics of static and dynamic efficiency: the 2-period model.
Static Efficiency.
Assume that that the demand curve (or marginal benefit curve) for a depletable resource is linear. The
equation of the (inverse) demand curve can be written as:
p  a  bq
(1)
where, p denotes the marginal willingness to pay for quantity q of the resource.
The total benefits (or total willingness to pay) from consuming (or extracting) an amount q0 of the
resource are measured by the area under the demand curve, up to quantity q0. This is the integral of
equation (1):
q0
TB 
 a  bqdq
0
=>
TB  aq 
b 2
q
2
Further assume that the supply curve (or marginal opportunity cost curve) of extracting the resource is as
follows:
(2)
pc
where, p, the marginal cost of extracting the resource at any level q, is constant and equal to c.
The total cost of extracting an amount q0 of the resource is measured by the area under the supply curve,
up to quantity q0. This is the integral of equation (2):
q0

TC  cdq
0
=>
TC  cq
The net benefit, or the difference between willingness to pay and cost, can then be written as:
TNB  TB  TC  aq 
b 2
q  cq
2
Our objective is to choose a level of output in such a way as to maximize the total net benefits from
consuming (extracting) the resource -- this will be the efficient amount of resource extracted:
Max TNB  aq 
q
b 2
q  cq
2
This is a non-constrained optimization problem, which we can readily solve. Recall from calculus that
when seeking the extremum (maximum, or minimum) of a single variable function, a necessary condition
(first order condition) requires that the first derivative of the function, when evaluated at a candidate
extremum, be equal to zero:
F.O.C.:
TNB
 0 => a  bq  c  0 => a  bq  c
q
To insure that we have a maximum and not a minimum, a sufficient (second order condition) requires that
the second derivative of the function, when evaluated at a candidate maximum, be negative:
S.O.C.:
 2 TNB
q 2
 0 => b  0
Dynamic Efficiency.
Consider a 2-period model. Assume that the demand curve and the marginal cost curve in period 1 are as
specified above. Further assume that the demand curve and the marginal cost curve in period 2 are
unchanged. Dynamic efficiency requires that the amount extracted in each period is such that the present
value of the total net benefits over the two periods is maximized.
The total net benefits over period 1 is:
TNB1  TB1  TC 1  aq1 
b 2
q1  cq1
2
The present value of net benefits over period 2 is:
TB  TC 2
TNB 2  2

1 r
aq 2 
b 2
q 2  cq 2
2
1 r
The sum of the amount of resources extracted in period 1 and in period 2 can not exceed the total amount
of resources available over the two periods:
q1  q 2  q
Our optimization problem for this 2-period model is slightly more complex than the static efficiency
optimization problem. We have to determine the optimal level of output in period 1, and the optimal level
of output in period 2, which will maximize the present value of total net benefits over the two periods. We
also have a constraint on the overall level of output. We can however follow a procedure similar to the
one adopted above after forming the appropriate Lagrangian expression for our constrained maximization
problem:
b
Max L  aq1  q 1 2  cq1 
q1 ,q2
2
aq 2 
b 2
q 2  cq 2
2
  q  q 1  q 2 
1 r
where,  is the Lagrangian multiplier.
The first order condition requires taking the first order derivative of the Lagrangian expression, with
respect to each decision variable and the Lagrangian multiplier:
F.O.C.:
L
 0 => a  bq1  c    0
q 1
=> a  bq1  c  
(3)
a  bq2  c
a  bq 2  c
L
   0 =>

 0 =>
1 r
1 r
q 2
(4)
=> a  bq1  c 
a  bq 2  c
1 r
and
L
0

=> q  q1  q 2   0
=> q  q1  q 2
S.O.C.: The second order necessary condition requires that the Hessian matrix of the Lagrangian must be
negative semi-definite at the candidate maximum.
EWU – ECON 477 - Natural Resource Economics
Deriving the solutions to a 2-period model.
Below we are going to derive the static and dynamic solutions to the 2-period model, using the solution
equations derived above.
Static Efficiency.
Assume that the marginal willingness-to-pay curve (or demand curve) is given by the following equation:
P = 8 - 0.4 Q, and the marginal cost is constant at $2 per unit.
Using the notations of the solution equations derived above, we then have the following parameter values:
a  8 , b  0.4 , c  2
Using these we obtain:
a  bq  c => 8  0.4q  2 => 0.4q  6 => q 
6
0.4
=> q*  15
Dynamic Efficiency.
Further assume that the reserve of depletable resource is fixed at 20 units, and the discount rate is 10%.
Using the notations of the solution equations derived above, we then have the following parameter values:
a  8 , b  0.4 , c  2 , q  20 , r  0.10
Using these we obtain:
a  bq1  c 
a  bq 2  c
8  0.4 q 2  2
=> 8  0.4 q1  2 
1  0.1
1 r
=> 8  0.4 q1  2 
8 0.4
2

q2 
1.1 1.1
1.1
(5)
and
q  q1  q 2
=> 20  q1  q 2
=> q1  20  q 2
(6)
Substituting equation (6) into equation (5), we obtain:
8  0.420  q 2   2 
8 0.4
2

q2 
1.1 1.1
1.1
=> 8  8  0.4 q 2  2 
8 0.4
2

q2 
1.1 1.1
1.1
=> 0.4 q 2  2 
6 0.4

q2
1.1 1.1
Collecting the terms with q 2 on the left hand side of our equation and the others on the right hand side of
the equation, and expressing each with the common denominator, we obtain:
0.4 q 2 
0. 4
6
q2 
2
1.1
1. 1
=>
0.44
0.4
6
2.2
q2 
q2 

1.1
1.1
1.1 1.1
=>
0.84
8.2
q2 
1.1
1.1
Finally simplifying (7), we solve for q 2 :
q2 
8.2
.84
(7)
(8)
=> q2  9.762
Substituting (8) back into eq. (6), we solve for q1 : q1  20  9.762 => q1  10.238
Finally using (8) and (9) in equations (3) or (4), we solve for the marginal user cost:   $1.905
(9)
EWU – ECON 477 - Natural Resource Economics
The simple mathematics of static and dynamic efficiency: extension to the n-period model.
Below we are going to derive the dynamic solutions to a n-period model.
The total net benefits from extracting an amount qt in year t are:
TNB t  aqt 
b 2
q t  cqt
2
The present value of the total net benefits from extracting an amount qt in year t is:
b 2
q t  cq t
2
1  r t
aqt 
PVTNB t 
The sum of the amount of resources extracted over n period can not exceed the total amount of resources
available:
n
q
i
q
i 1
The dynamically efficient allocation of the resource over n years is then the one which satisfies the
following maximization problem:
n
Max L 
q1 ,..., qn 

b 2
q i  cq i

2
q 



1  r i 1

aqi 
i 1

n
 q 
i
i 1
Assuming the S.O.C. is satisfied, and taking the first order derivatives of the Lagrangian with respect to
each of the decision variable, the dynamic efficient allocation must satisfy:
a  bqi  c
1  r 
i 1
 0 ,
for i  1,..., n =>
a  bqi  c
1  r 
i

a  bq j  c
1  r  j
, where i  j
and
n
q

i 1
n
q i  0 => q 
q
i
i 1
Note that the following results have been extended to the n-period case:
-
The resource is allocated so that the present value of marginal net benefits is equalized across
periods.
-
The above implies that the marginal user cost is increasing at a rate equal to r percent. This rise
reflects increasing scarcity and the accompanying rise in the opportunity cost of consumption.
-
In response to rising marginal user costs, the quantity extracted falls over time.