#### Transcript Economic efficiency criteria  Static efficiency –  Maximize net benefits of one optimal rotation Dynamic efficiency – Maximize net benefits from continuous series of optimal rotations, where the.

```Economic efficiency criteria

Static efficiency
–

Maximize net benefits of one optimal rotation
Dynamic efficiency
–
Maximize net benefits from continuous series
of optimal rotations, where the net benefits
from future rotations are discounted back to
present value terms.
Net benefits maximum where,
level of output where TB - TC is greatest
 i.e. where MC = MB

Define MC
Opportunity cost to hold timber for one
more unit of time
 Opportunity is to harvest timber and use
proceeds for some other purpose

–
–
–
Assume it will earn a guaranteed r% in a bank,
the alternative rate of return
Net Revenue = P * Q - Harvesting Costs
(assumed to be zero)
MC = NR * r
Example of MC calculation for
50 to 60 year period
Let, P = \$5 per ft3, Let r = 4%
 From growth function Q 50 = 1,400 ft3
 Therefore,

–
–
NR50 = \$5/ft3 * 1,400 ft3 = \$7,000
Interest on \$7,000 over 10 years is

–
–
(1.04)10 * \$7,000 - \$7,000 = \$3,361.71
But need to discount back to year 50
MC50 =\$3,361.71/(1.04)10 = \$2,271.05
Time line for MC calculation
\$2,271.05
\$7,000
50
\$3,361.71
earn compound interest on \$7,000for 10 years
55
60
MR calculation for 10 year
period
MR =  NR = P * Q60 - P * Q50
 Assume P = \$5per ft3
 NR50 = \$5 * 1,400 ft3 = \$7,000
 NR60 = \$5 * 2,100 ft3 = \$10,500
 MR = \$5 * 700 = \$3,500 received in year 60
 MR50 = \$3,500 / (1.04)10 = \$2,364.47

Time line for marginal benefit calculation
\$10,500 - \$7,000
\$3,500
\$2,364.47
50
55
60
What rate of return has been
earned by holding for 10 years?
We know that Vn = V0 (1+r)n
 Solve for r,

–
r = (Vn/V0)1/n - 1
r = (10,500/7,000)0.1 - 1
 r = 1.5 0.1 - 1 = 4.14%
 since IRR > discount rate, optimal policy is
to hold the resource and let it grow

Interpretation of rate of return
If alternative rate of return is 4% then can
do better financially by letting stand grow
 If alternative rate of return is greater than
4.14% then should cut stand
 key point: is rate of growth of stock greater
than the rate of growth of an alternative?
(should you cut now and put the money in
the bank where it will grow faster?)

Determine economically
optimal length of one rotation

Assume
–
–
\$100 establishment cost in year zero
\$10 per year annual cost
Components of NVP calculation
Present
Present
Present
Present
Value
Value of
Value of Value
of Timber Establishment Annual
1st rotation
Cost
Cost @\$10
PV(V)
PV(EC)
PV(AC)
NPV
\$
\$
100.00 \$ 81.11 \$ (181.11)
\$ 171.15
\$
100.00 \$ 135.90 \$ (64.76)
\$ 308.32
\$
100.00 \$ 172.92 \$ 35.40
\$ 781.08
\$
100.00 \$ 197.93 \$ 483.16
\$ 984.99
\$
100.00 \$ 214.82 \$ 670.17
\$ 998.13
\$
100.00 \$ 226.23 \$ 671.90
\$ 818.80
\$
100.00 \$ 233.95 \$ 484.85
\$ 629.07
\$
100.00 \$ 239.15 \$ 289.92
\$ 468.94
\$
100.00 \$ 242.67 \$ 126.27
\$ 341.55
\$
100.00 \$ 245.05 \$
(3.50)
\$ 244.12
\$
100.00 \$ 246.66 \$ (102.54)
\$ 171.69
\$
100.00 \$ 247.74 \$ (176.05)
\$ 119.04
\$
100.00 \$ 248.47 \$ (229.43)
\$ 81.45
\$
100.00 \$ 248.97 \$ (267.52)
\$ 55.37
\$
100.00 \$ 249.30 \$ (293.93)
\$ 37.41
\$
100.00 \$ 249.53 \$ (312.12)
Present value of one rotation
\$800.00
\$600.00
\$400.00
\$200.00
\$\$(200.00)
\$(400.00)
1
3
5
7
9
11
13
15
Dynamic efficiency
Optimal rotation length for a perpetual
series of uniform rotations
 Use multiplier for capital value (CV) of a
periodic series

–
Let,
a
= value of periodic payment
 t = length of time between payments
 r = interest rate
–
CV = a/((1+r)t -1)
Perpetual series of rotations of length R
ft3
R1
R2
R3
Ri
time
Soil expectation value (SEV)
Identified by Faustman
 Capital value of a perpetual series of forest
rotations
 Dynamic efficiency is achieved when SEV
is maximized
 Represents the value of the soil to produce
the timber crop

Key point: optimal rotation length is
shorter for SEV than for NPV
\$1,000.00
\$800.00
\$600.00
\$400.00
\$200.00
NPV
SEV
\$\$(200.00) 1
\$(400.00)
3
5
7
9
11 13 15
Three opportunity costs
Timber 1. interest on income from timber revenue
if cut sooner rather than later
2. interest on delay of start of next rotation
if lengthen rotation
Land 3. interest on income from sale of land
if sold sooner rather than later
SEV formula
t
SEV =

(Ij - Cj) (1 + r) t-j
t- 1
(1
+
r)
j=0
j = index on time
t = rotation length
I = revenue
r = interest rate
c = cost
Operation of SEV formula
Discount income minus costs back to year 0
It
Year 0
C0
Ct
Compound costs forward to rotation
R
Sensitivity analysis of SEV
Price of timber
 Planting cost
 Interest rate

```