Question 4 of 7: Precommercial Thinning — Break-Even Stumpage and Interest Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Forest Engineering, 04-For-A4 Forest Management, May 2014. Closed book; approved Casio/Sharp calculator only. 3 hours. Seven questions; the instructions call for Questions 1, 2, 3, 6 and 7 plus EITHER Question 4 or Question 5.
Reference texts: Davis, Johnson, Bettinger & Howard, Forest Management: To Sustain Ecological, Economic, and Social Values (age-class regulation, area/volume control, biodiversity planning); Klemperer, Forest Resource Economics and Finance (discounted cash flow, break-even stumpage/rate analysis); Smith et al., The Practice of Silviculture: Applied Forest Ecology (silvicultural systems, natural disturbance regimes); Van Wagner (1978), “Age-class distribution and the forest fire cycle,” Can. J. For. Res. 8 (negative-exponential fire-origin age structure); BC Forest and Range Practices Act and BC Ministry of Forests guidance (Canadian regulatory context).
Check: the paper prints two different mark totals for the same questions — the page-1 scoring table lists [1]=16, [2]=16, [3]=16, [4]=16, [5]=16, [6]=20, [7]=16 (a 116-mark table), while the mark shown directly beside each question is [1]=14, [2]=14, [3]=12, [4]=12, [5]=12, [6]=20, [7]=16. The per-question values sum to a clean 100-mark paper once one of Q4/Q5 is chosen (14+14+12+12+20+16=88, +12=100), matching the stated 5-of-7-plus-either format exactly, so the headings below use the per-question values and treat the page-1 table as a template artifact.
4A — Minimum break-even stumpage at i = 5% (6 marks)
Given. Precommercial thinning cost = $\$750/\text{ha}$, paid today (stand age 15). Harvest at stand age 50, so the cash flows are 35 years apart. Expected yield = $280\ \text{m}^3/\text{ha}$ at harvest. Interest rate $i = 5\%$. Cost and yield are both per-hectare, so the 10 ha stand area cancels out of the break-even calculation.
Find. The minimum stumpage value $S$ (units $\$/\text{m}^3$) received at harvest for the treatment's net present value to be $\ge 0$.
Approach. At break-even, NPV = 0: the present value of the future stumpage revenue must exactly equal the thinning cost paid today.
Set the NPV = 0 condition. Revenue at year 35 is $S \times 280$ (units $\$/\text{ha}$); discounted to today at 5% it must equal the $\$750/\text{ha}$ cost:
$$750 = S \times 280 \times (1.05)^{-35}$$
Evaluate the compounding factor. Over $n = 50-15 = 35$ years,
$$(1.05)^{35} = 5.516$$
Solve for $S$. Substituting and rearranging,
$$S = \frac{750 \times (1.05)^{35}}{280} = \frac{750 \times 5.516}{280} = \boxed{\$14.78/\text{m}^3}$$
Any stumpage rate at or above $\$14.78/\text{m}^3$ makes the thinning a positive-NPV investment at 5% interest; below it, the discounted cost of thinning exceeds the discounted value of the extra/faster-grown wood it buys.
4B — Break-even interest rate at $20/m3 stumpage (6 marks)
Given. Same cost ($\$750/\text{ha}$ today) and yield ($280\ \text{m}^3/\text{ha}$ at year 35), now with a known stumpage rate $S = \$20/\text{m}^3$.
Find. The interest rate $i$ at which NPV = 0 (the break-even, or internal, rate of return on the thinning investment).
Approach. Set up the same single-cost/single-revenue NPV = 0 relation as 4A, but now solve for $i$ instead of $S$.
Compute the revenue at harvest.
$$\text{Revenue} = 20 \times 280 = \$5{,}600/\text{ha}$$
Set NPV = 0 and isolate the compounding factor.
$$750 = 5{,}600 \times (1+i)^{-35} \quad\Rightarrow\quad (1+i)^{35} = \frac{5{,}600}{750} = 7.467$$
Solve for $i$. Taking the 35th root,
$$i = 7.467^{1/35} - 1 = \boxed{5.91\%}$$
At exactly $\$20/\text{m}^3$ stumpage, the thinning breaks even at an interest rate of about 5.91% — a slightly higher hurdle rate than the $5\%$ used in 4A, consistent with $\$20/\text{m}^3$ being above the $\$14.78/\text{m}^3$ minimum found there (more revenue headroom supports a higher break-even discount rate).
