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Economic efficiency criteria n Static efficiency – Maximize net benefits of one optimal rotation n 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.

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Net benefits maximum where, n level of output where TB - TC is greatest n i.e. where MC = MB

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n Opportunity cost to hold timber for one more unit of time n 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 Define MC

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n Let, P = $5 per ft 3, Let r = 4% n From growth function Q 50 = 1,400 ft 3 n Therefore, – NR 50 = $5/ft 3 * 1,400 ft 3 = $7,000 – Interest on $7,000 over 10 years is n (1.04) 10 * $7,000 - $7,000 = $3,361.71 – But need to discount back to year 50 – MC 50 =$3,361.71/(1.04) 10 = $2,271.05 Example of MC calculation for 50 to 60 year period

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50 60 55 $7,000 earn compound interest on $7,000for 10 years $3,361.71 $2,271.05 Time line for MC calculation

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n MR = NR = P * Q 60 - P * Q 50 n Assume P = $5per ft 3 n NR 50 = $5 * 1,400 ft 3 = $7,000 n NR 60 = $5 * 2,100 ft 3 = $10,500 n MR = $5 * 700 = $3,500 received in year 60 n MR 50 = $3,500 / (1.04) 10 = $2,364.47 MR calculation for 10 year period

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505560 $3,500 $2,364.47 Time line for marginal benefit calculation $10,500 - $7,000

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n We know that V n = V 0 (1+r) n n Solve for r, – r = (V n /V 0 ) 1/n - 1 n r = (10,500/7,000) 0.1 - 1 n r = 1.5 0.1 - 1 = 4.14% n since IRR > discount rate, optimal policy is to hold the resource and let it grow What rate of return has been earned by holding for 10 years?

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n If alternative rate of return is 4% then can do better financially by letting stand grow n If alternative rate of return is greater than 4.14% then should cut stand n 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?) Interpretation of rate of return

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n Assume – $100 establishment cost in year zero – $10 per year annual cost Determine economically optimal length of one rotation

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Components of NVP calculation

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Present value of one rotation

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n Optimal rotation length for a perpetual series of uniform rotations n Use multiplier for capital value (CV) of a periodic series – Let, n a = value of periodic payment n t = length of time between payments n r = interest rate – CV = a/((1+r) t -1) Dynamic efficiency

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R1R2R3 Ri Perpetual series of rotations of length R ft 3 time

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Soil expectation value (SEV) n Identified by Faustman n Capital value of a perpetual series of forest rotations n Dynamic efficiency is achieved when SEV is maximized n Represents the value of the soil to produce the timber crop

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Key point: optimal rotation length is shorter for SEV than for NPV

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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 ofland if sold sooner rather than later Three opportunity costs

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SEV formula SEV = j=0 t (I j - C j ) (1 + r) t-j (1 + r) t - 1 j = index on time t = rotation length I = revenue r = interest rate c = cost

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R ItIt CtCt C0C0 Year 0 Operation of SEV formula Compound costs forward to rotation Discount income minus costs back to year 0

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n Price of timber n Planting cost n Interest rate Sensitivity analysis of SEV

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