Stand-Level Management Planning ISA Presentation Stand-Level Management Planning Lecture 4 –Financial Analysis and Uneven-aged Management Wednesday, November 23, 2016

Lecture Outline Discussion of uneven-aged management What is an uneven-aged stand? Advantages and disadvantages of uneven-aged management Key decisions in uneven-aged management Selecting a target diameter-class distribution Selecting the cutting cycle The Forest Value for uneven-aged stands

What is an Uneven­aged Stand? Uneven-aged stands have, by definition, at least three age groups (or cohorts) Theoretically, all age groups (up to some maximum) are represented in an uneven-aged stand However, in practice there usually are identifiable cohorts of trees that are about the same age the stand is considered uneven-aged if there are at least three Still, it is useful to think of the idealized uneven-aged stand as a model (simplification of the real world)

Uneven-aged Management and Even-Aged Management …can be viewed as the same thing, practiced at different scales. A landscape made up of even-aged stands can be considered uneven-aged at the landscape scale. As the patches become smaller, you go through patch clearcutting (even-aged) to group selection (uneven-aged) to individual tree selection (uneven-aged).

Diameter Class Distribution Transition

Advantages of Uneven-­aged Management Mimics small-patch (gap phase) disturbance patterns typical of natural late-successional forests Diverse vertical and horizontal structure Small, medium and large trees provide a multi-layered canopy (vertical structure) Small patches of trees of different ages provide horizontal diversity Ideal when the desired species mix includes shade-tolerant tree species Many people are offended by the sight of a clearcut With uneven-aged management the stand is never clearcut (socially or aesthetically beneficial)

Advantages of Uneven-aged Management (continued) Provide continuous cover on a site, reducing problems with erosion and excessive run-off after heavy rains More compatible with many owners’ financial constraints more frequent cash flow relatively low investment requirements In theory, could provide more growth than even-aged management In practice, however, the densities required to maintain adequate regeneration are relatively low, so growth rates are typically lower with uneven-aged management than with even-aged management

Disadvantages of Uneven-aged Management Generally does not work well when shade-intolerant species are desired Some wildlife species prefer or require habitat provided by even-aged stands e.g., large openings with early successional vegetation Uneven-aged management is complex and difficult to implement More detailed information about the stand is required Growth and yield models are less likely to be available Requires the ability to obtain more frequent “pulses” of regeneration Difficult where deer browsing pressure is high

Disadvantages of Uneven-aged Management (Continued) The frequent, lighter harvests required for uneven-aged management create numerous problems: Since less volume is removed per unit of area, average woods hauling distances and logging costs will be greater More area must be disturbed to obtain a given amount of wood More roads are needed It is difficult to avoid some damage to the residual stand More frequent entries into the stand increase the potential for site degradation during harvesting such as compaction and rutting May provide a pseudo-scientific rationale for high-grading

Key Decision Parameters in Uneven-aged Management The three basic decision parameters in uneven-aged management that we will consider are: the target diameter-class distribution, the cutting cycle, and individual tree harvesting decisions. Could also consider patch size to be a decision parameter.

An “Ideal” Diameter Class Distribution Figure 9.1 – An “ideal” diameter-class distribution for an uneven-aged stand

The Q-factor A widely-used simplifying assumption in uneven-aged management is that the ratio of the number of trees in any diameter class to the number of trees in the next larger diameter class should be constant This ratio is called the “Q factor,” where Q stands for “quotient” If n(d) is the number of trees in diameter class d, then the Q factor can be expressed as follows:

The Negative Exponential Diameter Class Distribution The negative exponential function is useful for modeling uneven-aged diameter class distributions because it has the property that the Q-factor is constant for all diameter classes. A negative exponential diameter class distribution function can be written as follows: where k and a are parameters, and e is the base of the natural logarithm (in Excel, exp()).

Properties of the Negative Exponential Function Proof that the a negative exponential function has a constant Q-factor: First, plug the exponential function into the definition of Q: Now, use the fact that e a+b = e a e b : Since the Q-factor determines the slope of the negative exponential function and the a parameter is directly related to Q, a is like the slope of the function

Properties of the Negative Exponential Function (continued) We just saw that a is like the slope of the negative exponential function The k parameter is the intercept of the negative exponential function Proof:

Selecting a Target Diameter-Class Distribution Specifying a target diameter class distribution with the negative exponential function requires that you specify the values of three parameters: a, k, and dmax (the maximum diameter). How does one decide which values of these three parameters are best for a given stand?

Selecting k, Q and dmax One approach is: First identify the maximum diameter, Then select a Q factor – which implies a specific value for the a parameter; Finally, the k value should be selected to ensure that the stand density is consistent with obtaining an appropriate level of regeneration.

Selecting dmax Selecting the maximum diameter is essentially the same as choosing a rotation age.

Identifying the Maximum Diameter Financial Maturity (William Duerr) The idea that the rate at which the value of a tree increases should be at least equal to the alternate rate of return (ARR) is called the financial maturity principle Determine the guiding rate of return (i.e., the ARR) Calculate the current stumpage value of the tree Estimate the stumpage value of the tree at the next point in time when the tree could be cut Compare the rate of value increase over the interval between this cutting and the next possible cutting with the guiding rate of return The tree is financially mature when the projected rate of value increase is less than the guiding rate of return

Financial Maturity Example Consider a tree whose stumpage value now (SV0 ) is $80. We estimate that the stumpage value of the tree after 8 years (SV8 ) will be $110. If the ARR is 4%, is the tree financially mature? Answer: Conclusion: since the annual rate of value increase (4.06%) is greater than the alternate rate of return (4%), the tree is not yet financially mature, so we should keep it.

Relationship Between Choosing dmax and R* The stock holding cost is like the inventory cost in the even-aged management rotation decision. The tree value growth is analogous to the annual increment value. The Financial Maturity approach misses the land holding cost which is analogous to the rent on the land in the optimal even-aged rotation decision. The Financial Maturity approach can be modified to include the land rent, but we won’t go into that here. The land holding cost is typically small compared to the stock holding cost. The two decisions are essentially the same, only the spatial and time scales are changed.

Selecting a Target Q-Factor The Q-factor determines the slope of the diameter-class distribution. As the Q-factor is decreased the number of small trees decreases and the number of large trees increases. As the Q-factor is increased the number of small trees increases and the number of the number of large trees decreases.

Selecting a Target Q-Factor So, what are the implications of choosing different values of Q (when the other parameters are held constant)? As the Q-factor is decreased the number of small trees decreases and the number of large trees increases. As the Q-factor is increased the number of small trees increases and the number of the number of large trees decreases. Thus, economically the best value of Q is the smallest one (as close to 1 as possible). Silvicultural constraints will limit how low the Q factor can go.

Selecting k The k parameter is like the intercept of the diameter class distribution. Increasing k while holding everything else constant increases the number of trees in all diameter classes. Therefore, for given values of Q and dmax , increasing k increases the basal area of the stand.

Selecting k The amount or rate of regeneration must be sufficient to sustain the target diameter-class distribution. Assume that the amount/rate of regeneration is a function of the basal area of the stand. Is this relationship positive or negative? How does the basal area of the stand vary with different values of k (holding dmax and Q constant)? Increasing k increases the basal area. There will be only a small range of possible values of k for each value of Q for which regeneration will be sustainable. k is selected to give a particular basal area that will result in the desired amount/rate of regeneration.

Selecting k, Q and dmax Selecting the maximum diameter is essentially the same as choosing a rotation age. The Q factor is largely determined by silvicultural constraints. There will be only a small range of possible values of k for each value of Q for which regeneration will be sustainable. k is selected to give a particular basal area that will result in the desired amount/rate of regeneration.

Individual Tree Selection Keep the trees with the most potential to increase in value: desirable species, trees with healthy crowns and no defects. Choose the following for cutting: undesirable species, trees with diseases or insect damage, trees that will never produce a high-quality sawlog, trees with defects, such as low forks, sweep, crook, fire scars, frost cracks, too many large branches, etc. spacing considerations: when two healthy, vigorous trees clearly are competing with each other, one should go, creating “regeneration patches.”

Selecting the Cutting Cycle and the Residual BA 160 1st entry 2nd entry 3rd entry 4th entry 5th entry Volume per acre (tons) Reserve growing stock level 45 90 Time (years)

Selecting the Cutting Cycle and the Residual BA To calculate the optimal residual basal area and cutting cycle, calculate the uneven-aged management version of the Forest Value for a range of basal areas and cutting cycles. where: NetRevI = the initial harvest net revenue, NetRevcc = the net harvest revenue at the end of each cutting cycle, A = the net annual revenue (negative for costs), t = the cutting cycle, and r = the real interest rate.

Uneven-Aged Forest Value Example You have a 200-acre uneven-aged forest stand, and you need to determine the best cutting cycle and residual basal area for the stand. You are considering three residual basal area levels – 50, 60, and 70 square feet, and three cutting cycles – 5, 10, and 15 years The volume from future harvests, however, will depend on the cutting cycle and the residual basal area following each harvest, as shown on the next screen... The stumpage price for all harvests is $220/mbf. There are fixed costs for sale preparation, permitting, and moving harvesting equipment to the site totaling $2,000 (for the whole stand) each time a harvest is made.

Uneven-Aged Forest Value Example (continued) Harvest volumes for uneven-aged management cutting cycle and residual basal area example. Residual Basal Area Initial Harvest Volume (mbf per acre) Future Harvest Volumes (mbf/acre) 5-year Cycle 10-year Cycle 15-year Cycle 50 1.98 0.88 2.42 3.00 60 1.76 1.10 2.53 3.10 70 1.21 1.30 2.64 3.63

Uneven-Aged Forest Value Example (continued) Consider the case where you leave a residual basal area of 50 and you harvest on a 5-yr cutting cycle: NetRevI = $220/mbf×1.98 mbf/ac - $10/ac = $425.60/ac NetRevcc = $220/mbf×0.88 mbf/ac - $10/ac = $183.60/ac Now, plug these values into the uneven-aged management Forest Value equation:

Uneven-Aged Forest Value Example (continued) Forest Values for uneven-aged management cutting cycle and residual basal area example (200 acre tract) What would happen to these values if the tract was only 50 acres? Residual Basal Area Forest Value (per acre) 5-year Cycle 10-year Cycle 15-year Cycle 50 $1,148.00 $1,388.40 $1,112.10 60 $1,323.00 $1,390.40 $1,091.20 70 $1,405.10 $1,319.80 $1,115.80

Uneven-Aged Forest Value Example (continued) Forest Values for uneven-aged management cutting cycle and residual basal area example (50 acre tract) Residual Basal Area Forest Value (per acre) 5-year Cycle 10-year Cycle 15-year Cycle 50 $1,063.80 $1,342.10 $1,078.40 60 $1,238.80 $1,344.10 $1,057.50 70 $1,320.90 $1,273.50 $1,082.10