Practice problems with solutions: sawmill calculations and decisions

This set contains 30 worked problems for mill staff, trainees and students. Each problem gives a statement with all numbers, a hint, a step-by-step solution, a final answer and one sentence on what the result means for a mill. The formulas and figures come only from the portal's fact-checked articles and courses; the numbers in the statements are new. Figures are labelled by region where the source is regional (Europe or North America), many shrinkage, moisture and drying figures are North American, and every price, cost and mill figure in the economics section is invented for practice and says so. Answers are rounded for presentation; every solution was recomputed and asserted in a Python script. Work each problem before reading the solution.

Log measurement and volume

Problem 1. Volume of one log under the German framework agreement (RVR)

Statement. A spruce log in Germany (Europe) is traded under the RVR framework agreement (voluntary, 5th edition 2023). The order length is 4.50 m. The log is delivered 4.60 m long, because the contract allows a length allowance of 10 cm that is excluded from volume. The mid-diameter without bark is read as 36.8 cm. Compute the billed volume using V = (pi/4) x d^2 x L x 10^-4 (d in cm, L in m), applying the rule that the diameter is rounded down to whole centimetres. Then compute what the volume would be if the mill wrongly used the unrounded diameter and the full delivered length.

Hint. Round the diameter down first, use only the order length, and square the diameter before multiplying. The second calculation uses 36.8 and 4.60.

Solution.

1. Rounded diameter: 36.8 cm rounds down to 36 cm.
2. Billed length is the order length, 4.50 m; the 10 cm allowance is excluded.
3. V = 0.7854 x 36^2 x 4.50 x 10^-4 = 0.7854 x 1296 x 4.50 / 10,000 = 0.4580 m3, reported as 0.46 m3.
4. Wrong method: 0.7854 x 1,354.24 x 4.60 / 10,000 = 0.4893 m3, reported as 0.49 m3.
5. Overstatement: 0.4893 / 0.4580 - 1 = 6.8 percent.

Final answer. Billed volume 0.46 m3. The unrounded, full-length shortcut would give 0.49 m3, about 6.8 percent too much.

What this tells a mill. Rounding and length rules are part of the price, so a mill should check that its scaling software applies exactly the rule version named in the contract.

Problem 2. A caliper that reads one centimetre high

Statement. A supplier delivers a batch whose true volume, measured correctly under the Huber-type formula, is 1,200 m3. All logs have a true mid-diameter of 24 cm. The mill's caliper reads 25 cm on every log, and nobody notices. (The price of 85 per m3 is invented for practice.) (a) By what percentage is the volume overstated? (b) How many m3 and how much money is that over the batch? (c) Compare with a 30 cm log with the same 1 cm reading error.

Hint. Volume is proportional to d squared and length is unchanged, so the error is (d_read^2 - d_true^2) / d_true^2.

Solution.

1. Ratio of squares: 25^2 = 625, 24^2 = 576.
2. Overstatement: (625 - 576) / 576 = 49 / 576 = 8.51 percent.
3. Billed volume: 1,200 x 625 / 576 = 1,302.08 m3, so the excess is 102.08 m3.
4. Money: 102.08 x 85 = 8,677 (invented currency units).
5. A 30 cm log with the same error: (31^2 - 30^2) / 30^2 = 61 / 900 = 6.8 percent, smaller than for the 24 cm log.

Final answer. (a) 8.5 percent. (b) about 102 m3, about 8,677 at the invented price. (c) 6.8 percent for the 30 cm log.

What this tells a mill. A systematic measuring error never cancels in a large batch, and it bites hardest on small logs, so calibration of the caliper or scanner is a money question.

Problem 3. Stacked beech: from stacked to solid cubic metres and tonnes

Statement. A German (Europe) delivery consists of 120 Rm (stacked cubic metres with bark) of beech cut to 3 m lengths. The RVR average stack factors to Fm o.R. (solid without bark) are 0.70 for 1 m lengths, 0.65 for 2 m and 0.60 for 3 m. For weight, 1 t oven-dry with bark equals 1.5 Fm o.R. for beech. (a) How many Fm o.R. is the delivery? (b) How many tonnes oven-dry? (c) What would the same 120 Rm be if it were 1 m lengths?

Hint. Pick the factor by length, multiply the Rm figure, then divide Fm by the tonne conversion.

Solution.

1. 3 m lengths use the factor 0.60: 120 x 0.60 = 72 Fm o.R.
2. Oven-dry mass: 72 / 1.5 = 48 t.
3. For 1 m lengths the factor is 0.70: 120 x 0.70 = 84 Fm o.R., which is 56 t oven-dry.

Final answer. (a) 72 Fm o.R. (b) 48 t oven-dry. (c) 84 Fm o.R. (56 t).

What this tells a mill. The brochure calls these factors averages for internal and statistical use, so a mill should use them for planning and never as a sales measure.

Problem 4. A Swedish pine stack built from the real dimensions

Statement. A pine stack in Sweden (Europe) measures 5.40 x 3.10 x 1.05 m (external dimensions). The Biometria stack regulation multiplies the external volume (m3t) by a wood-volume percentage; the base percentage for pine is 68 percent, adjusted for bark, stacking, crookedness, snow and ice and other factors. Suppose the measurer's adjustments bring the figure to 59 percent (the adjusted value is invented for practice; the portal's articles give the base values and one regulation example at 55 percent, not the adjustment tables). Compute (a) the external volume, (b) the stack volume in m3sub at 59 percent, (c) the volume at the unadjusted base, and (d) the difference.

Hint. External volume is the product of the three dimensions. Then multiply by the percentage as a decimal.

Solution.

1. External volume: 5.40 x 3.10 x 1.05 = 17.577 m3t.
2. Adjusted: 17.577 x 0.59 = 10.37 m3sub.
3. At the base 68 percent: 17.577 x 0.68 = 11.95 m3sub.
4. Difference: 11.95 - 10.37 = 1.58 m3, which is 13.2 percent below the base-percentage figure.

Final answer. (a) 17.577 m3t. (b) 10.37 m3sub. (c) 11.95 m3sub. (d) 1.58 m3, about 13.2 percent.

What this tells a mill. Because the Swedish figure is built from the actual stack, snow, ice and crooked wood visibly change the payment, which a flat German-style factor would hide.

Sawing, kerf and recovery

Problem 5. Volume yield against value yield

Statement. Two sawing patterns are run on 640 m3 of the same logs (all numbers and prices in this problem are invented for practice). Pattern A makes 345.6 m3 of boards sold at 330 per m3 and 17,500 of by-products. Pattern B makes 332.8 m3 of boards sold at 352 per m3 (wider, higher-grade mix) and 18,200 of by-products. Compute volume yield, the conversion factor of A, and the value yield (per m3 of log) of both. Which pattern earns more per m3 of log?

Hint. Value yield = (lumber revenue + by-product revenue) / log volume. Volume yield = sawn volume / log volume. Conversion factor = 1 / yield.

Solution.

1. Volume yield A: 345.6 / 640 = 54.0 percent; B: 332.8 / 640 = 52.0 percent.
2. Conversion factor A: 1 / 0.54 = 1.85 m3 of logs per m3 of boards.
3. Revenue A: 345.6 x 330 = 114,048; plus 17,500 = 131,548; per m3 of log 205.54.
4. Revenue B: 332.8 x 352 = 117,145.6; plus 18,200 = 135,345.6; per m3 of log 211.48.

Final answer. A: 54 percent volume yield, conversion factor 1.85, value yield 205.5 per m3 of log. B: 52 percent and 211.5. Pattern B earns about 5.9 more per m3 of log even though its volume yield is lower.

What this tells a mill. Volume and value yield can pull in opposite directions, so a mill should report both together, with the price basis and date.

Problem 6. From standing stem to sawn timber

Statement. A felled stem holds 42 m3. In the Romanian silver fir study (Europe, eight stems) the debarked sawlogs averaged 76 percent of stem volume, with a range of 67 to 85 percent. In the Finnish Norway spruce simulation (Europe, normal thinning regime) 584 m3/ha of log volume gave 314 m3/ha of sawn timber. Combining the two sources as an illustration, compute the sawn timber from this stem at the low end (67 percent), the average (76 percent) and the high end (85 percent) of the sawlog range, and the sawn volume as a share of stem volume at both ends.

Hint. First the sawn share from the Finnish table (314 / 584). Then multiply stem volume by sawlog share, then by the sawn share.

Solution.

1. Finnish sawn share: 314 / 584 = 53.8 percent.
2. Debarked sawlogs: low 42 x 0.67 = 28.14 m3; average 42 x 0.76 = 31.92 m3; high 42 x 0.85 = 35.70 m3.
3. Sawn timber: low 28.14 x 0.5377 = 15.13 m3; average 17.16 m3; high 19.19 m3.
4. As a share of the stem: low 0.67 x 0.5377 = 36.0 percent; high 0.85 x 0.5377 = 45.7 percent.

Final answer. About 15.1 m3 (low), 17.2 m3 (average) and 19.2 m3 (high); that is 36 to 46 percent of stem volume.

What this tells a mill. Losses before the saw (crosscutting, grading, debarking) can be as large as machine effects, so a mill should agree where the yield baseline starts before benchmarking.

Problem 7. What taper costs in log volume

Statement. In the Austrian glued-laminated lamella case (Europe, 43 x 146 x 4,000 mm lamella, three-stem pattern) every log yields 0.075 m3 of main product regardless of taper. The log volume is 0.166 m3 at 10 mm taper and 0.212 m3 at 25 mm taper. A mill needs 30 m3 of this lamella. How many logs and how much log volume does it need from each taper class, what is the main-product yield of each, and by what percentage does the high-taper supply need more wood?

Hint. The sawn volume per log is fixed, so the number of logs is the same; only the denominator changes.

Solution.

1. Logs needed: 30 / 0.075 = 400 logs of each class.
2. Log volume at 10 mm taper: 400 x 0.166 = 66.4 m3; at 25 mm: 400 x 0.212 = 84.8 m3.
3. Yields: 0.075 / 0.166 = 45.2 percent; 0.075 / 0.212 = 35.4 percent.
4. Extra wood: 84.8 - 66.4 = 18.4 m3, which is 27.7 percent more.

Final answer. 400 logs either way; 66.4 m3 against 84.8 m3 of logs; yields about 45 and 35 percent; the high-taper supply needs 18.4 m3 (28 percent) more wood.

What this tells a mill. Taper is a purchasing variable: a log supplier with straighter, less tapered logs is worth a price premium that can be calculated from the log volume saved.

Problem 8. Kerf is discontinuous: when does a thinner saw add a board?

Statement. The Polish Scots pine study (Europe) described a band saw with a 1.0 mm plate and 0.75 mm set per side, and a frame saw with a 2.2 mm plate and 0.9 mm set per side. Counting set on both sides, compute the cut widths. A cant is resawn into green boards of 22 mm thickness. Ignoring edging, wane and sawing variation (a simplified geometry), how many boards fit in a cant 245 mm thick with each saw, and in a cant 232 mm thick? In the Polish test, the yield difference between the saws was not statistically significant on six logs; does your result conflict with that?

Hint. n boards need n x thickness + (n - 1) x cut width of space. Test n upwards until it no longer fits.

Solution.

1. Cut widths: band 1.0 + 2 x 0.75 = 2.5 mm; frame 2.2 + 2 x 0.9 = 4.0 mm.
2. Cant 245 mm, band saw: 10 boards need 10 x 22 + 9 x 2.5 = 242.5 mm, which fits; 11 would need 267.0 mm, which does not.
3. Cant 245 mm, frame saw: 10 boards need 10 x 22 + 9 x 4.0 = 256.0 mm, which does not fit; 9 boards need 230.0 mm, which fits.
4. Cant 232 mm: band 9 boards need 218.0 mm, 10 need 242.5 mm (too much); frame 9 boards need 230.0 mm (fits). Both saws give 9 boards.
5. A gain of one board in nine is 11.1 percent for the 245 mm cant and zero for the 232 mm cant. Over many logs of mixed size the average gain is small and depends on the log size mix, which is consistent with a non-significant result on six logs.

Final answer. Cut widths 2.5 mm (band) and 4.0 mm (frame). Cant 245 mm: 10 boards (band) against 9 (frame). Cant 232 mm: 9 against 9. No conflict with the Polish result: the gain appears only when the saved kerf tips the log into an extra board.

What this tells a mill. Kerf savings arrive in jumps, so a mill should test a thinner saw against its own log size distribution before expecting a yield gain, and fix guiding first.

Problem 9. How much premium must quarter-sawn beech earn?

Statement. In the Slovak beech experiment (Europe, six logs, two per pattern) the mean quantitative yield was 72.28 percent for through-and-through (live) sawing and 61.26 percent for quarter sawing; the mean radial share was 19.69 and 55.28 percent. (a) What price premium on all quarter-sawn output equals the revenue per m3 of log of live sawing, ignoring costs? (b) Suppose the non-radial part of the quarter-sawn output is sold at the ordinary live-sawn price, and only the radial part earns a premium. What premium must the radial boards earn? (An ordinary price of 300 per m3 is invented for practice.)

Hint. Equal revenue per m3 of log means yield_live x P = yield_quarter x P_quarter. For (b), split the quarter-sawn output into radial and non-radial shares.

Solution.

1. (a) Required price ratio: 72.28 / 61.26 = 1.1799, so a premium of 18.0 percent. At P = 300 the required price is 354.0 per m3.
2. (b) Let the radial boards sell at k x P. Revenue per m3 of quarter-sawn output = (1 - 0.5528) x P + 0.5528 x k x P = 0.4472 P + 0.5528 k P.
3. Set this equal to the required average price ratio 1.1799 P: 0.5528 k = 1.1799 - 0.4472 = 0.7327.
4. k = 1.3254, a radial premium of about 32.5 percent.

Final answer. (a) about 18 percent on all quarter-sawn output. (b) about 33 percent on the radial boards alone, under the stated assumption.

What this tells a mill. A volume loss translates into a price premium that the sales side must be able to achieve, and with only two logs per pattern the yields themselves are an illustration, not a benchmark.

Moisture and shrinkage

Problem 10. Moisture content from two weighings, and a cold meter reading

Statement. A full-section slice cut from a green board weighs 78.4 g as cut and 52.0 g after oven-drying at 101 to 105 degrees C to constant mass. (a) Compute moisture content on the dry basis and on the wet basis. (b) Is a pin meter suitable for this board? (c) On another, 40 mm thick board that has dried below the fibre saturation point, a resistance meter reads 11.5 percent while the boards are 27 degrees F colder than the meter's calibration temperature (the correction is about 0.5 points per 10 degrees F, added when the wood is colder). What is the corrected value, and how deep should the pins go for an average reading (one fifth to one quarter of the thickness)?

Hint. Dry basis divides the water by the oven-dry mass; wet basis divides by the mass as cut. A pin meter is unreliable above about 30 percent.

Solution.

1. Water: 78.4 - 52.0 = 26.4 g.
2. Dry basis: 26.4 / 52.0 x 100 = 50.8 percent. Wet basis: 26.4 / 78.4 x 100 = 33.7 percent. Check: 50.8 / 150.8 x 100 = 33.7 percent.
3. 50.8 percent is above the fibre saturation point of about 30 percent, so a pin meter is not reliable; the oven-dry result is the right method.
4. Temperature correction: 0.5 x 27 / 10 = 1.35 points, corrected value 11.5 + 1.35 = 12.85 percent.
5. Pin depth: 40 / 5 = 8 mm to 40 / 4 = 10 mm.

Final answer. (a) 50.8 percent dry basis, 33.7 percent wet basis. (b) No, use the oven-dry method. (c) 12.85 percent, pins 8 to 10 mm deep.

What this tells a mill. Every specification and kiln log should state the basis, and meters should be trusted only below about 30 percent and with corrections applied.

Problem 11. Green density, water per cubic metre and a truck load

Statement. The Swedish study of Hajian and co-authors (Europe) measured starting moisture contents (dry basis) of 95 percent for Norway spruce sapwood and 35 percent for Norway spruce heartwood. Assume a basic specific gravity Gb of 0.4 (a round value assumed for practice). Compute the green density and the mass of water per m3 of green wood for both zones, and, for an invented payload of 25,000 kg of wood and water (ignoring bark and the truck's own weight), the green volume per load.

Hint. Green density = Gb x 1000 x (1 + MC/100). Water per m3 = Gb x 1000 x MC/100.

Solution.

1. Sapwood: 0.4 x 1000 x (1 + 0.95) = 780 kg/m3; water 0.4 x 1000 x 0.95 = 380 kg/m3.
2. Heartwood: 0.4 x 1000 x (1 + 0.35) = 540 kg/m3; water 0.4 x 1000 x 0.35 = 140 kg/m3.
3. Load: 25,000 / 780 = 32.1 m3 of sapwood-type wood; 25,000 / 540 = 46.3 m3 of heartwood-type wood.

Final answer. Sapwood 780 kg/m3 with 380 kg of water; heartwood 540 kg/m3 with 140 kg of water. A load carries about 32.1 m3 or 46.3 m3.

What this tells a mill. Sapwood share changes both the freight per cubic metre and the water the yard or kiln must remove, so sorting by sapwood share pays at the log end and at the kiln.

Problem 12. How much narrower does a spruce board get?

Statement. A green flat-sawn board of Sitka spruce is 175 mm wide and is dried to 12 percent moisture content. Wood Handbook values (North America, green to oven-dry, percent of the green dimension): radial 4.3, tangential 7.5. Use shrinkage (%) = table value x (30 - final MC) / 30. (a) Compute the width loss for the flat-sawn board. (b) Compute the loss for a quarter-sawn board of the same width. (c) The linear method underestimates the true change by about 5 percent of the change, and pieces scatter with a coefficient of variation near 15 percent. Give the adjusted flat-sawn figure and one standard deviation of scatter.

Hint. Use the tangential value for flat-sawn width and the radial value for quarter-sawn. The factor (30 - MC) / 30 is the share of the full shrinkage that has happened.

Solution.

1. Flat-sawn: 7.5 x (30 - 12) / 30 = 7.5 x 18 / 30 = 4.50 percent, so 175 x 0.045 = 7.875 mm.
2. Quarter-sawn: 4.3 x 18 / 30 = 2.58 percent, so 175 x 0.0258 = 4.515 mm.
3. Adjusted for the 5 percent underestimate: 7.875 x 1.05 = 8.27 mm.
4. One standard deviation of scatter: 0.15 x 7.875 = 1.18 mm.

Final answer. (a) about 7.9 mm. (b) about 4.5 mm. (c) about 8.3 mm, with scatter of about +/- 1.2 mm (one standard deviation).

What this tells a mill. Planing allowances and customer tolerances should be set against the likely scatter of shrinkage, not against the average alone, and the numbers here are North American, not European species data.

Problem 13. A joinery shop at 60 percent relative humidity

Statement. A joinery shop is held at about 21 degrees C and 60 percent RH. The Wood Handbook EMC table (North America, 21.1 degrees C) gives 11.0 percent at 60 percent RH. Flat-sawn white fir boards, 190 mm wide, arrive at 8 percent moisture content. For the service range of 6 to 14 percent, Bergman's coefficients for white fir are 0.00245 per point (tangential) and 0.00112 per point (radial). (a) How many points will the boards gain, and how much wider do flat-sawn boards become? (b) How much wider would quarter-sawn boards become? (c) The average ratio of adsorption to desorption EMC is about 0.8. If the boards reach equilibrium by gaining water, what changes?

Hint. Change in dimension = initial dimension x coefficient x change in MC. For (c) apply 0.8 to the table EMC.

Solution.

1. Points gained: 11.0 - 8 = 3.0.
2. Flat-sawn: 190 x 0.00245 x 3.0 = 1.40 mm.
3. Quarter-sawn: 190 x 0.00112 x 3.0 = 0.64 mm.
4. Hysteresis: gaining water, the boards would settle at roughly 0.8 x 11.0 = 8.8 percent, a gain of 0.8 points; flat-sawn change 190 x 0.00245 x 0.8 = 0.37 mm.

Final answer. (a) 3.0 points, about 1.4 mm wider. (b) about 0.6 mm. (c) about 0.4 mm, because wet-side equilibrium is lower; the ratio is an average and the table is a compromise.

What this tells a mill. Stock delivered away from the service EMC will move after delivery, so a mill should know the destination climate and the direction of approach before agreeing a target moisture.

Drying planning

Problem 14. Does air drying pay for thick stock?

Statement. A mill air-dries 38 mm oak lumber in 95 days when stacked in July (invented trial value). Drying time rises roughly with thickness to the power 1.5. (a) Estimate the time for 63 mm stock stacked in July. (b) Madison, Wisconsin data (North America, 4/4 northern red oak, days to 20 percent) show 72 days for July stacking and 115 days for January stacking; apply that ratio to the 38 mm lot as an analogue. (c) The stock is worth 320 per m3 (invented) and financing costs 6 percent a year. What is the capital cost of holding the 63 mm stock for the time in (a), and how does it compare with the portal's assumed kiln-heat saving of 3 to 12 per m3? Capital cost = value x rate x days / 365.

Hint. Scale time by (new thickness / old thickness)^1.5. Season is a ratio of days. Interest is value x rate x days / 365.

Solution.

1. Thickness factor: (63 / 38)^1.5 = 1.6579^1.5 = 2.135.
2. (a) 95 x 2.135 = 203 days.
3. (b) Season factor 115 / 72 = 1.597; 95 x 1.597 = 152 days for the 38 mm lot stacked in January.
4. (c) 320 x 0.06 x 203 / 365 = 10.67 per m3.
5. Against a heat saving of 3 to 12 per m3 the margin is negative at the low end and about 1.3 at the best case; the decision turns on heat price, stock value and land cost.

Final answer. (a) about 203 days. (b) about 152 days. (c) about 10.7 per m3, which exceeds most of the assumed 3 to 12 per m3 heat saving.

What this tells a mill. Yard drying earns its place only for valuable, thick or slow stock; for cheap softwood the interest on stock usually cancels the kiln saving, and no European air-drying series was found to replace the US analogue.

Problem 15. Equalising, conditioning and wet-bulb setpoints

Statement. A hardwood charge must finish at an average of 6 percent. A second, softwood appearance charge must finish at 11 percent. Using the FPL rules (North America): equalising starts when the driest sample is 2 points below the desired average, and conditioning sets the EMC 4 points above the target for hardwoods and 3 points for softwoods. (a) At what sample moisture does equalising start in the hardwood charge, and what are the conditioning EMCs of both charges? (b) In schedule C3 for 4/4 sugar maple the depression at the 30 percent step is 19 degrees F and at the 25 percent step 35 degrees F. If the dry bulb were 125 degrees F at both steps (invented for illustration only), what are the wet-bulb setpoints?

Hint. Equalising start = target - 2. Conditioning EMC = target + 4 (hardwood) or + 3 (softwood). Wet bulb = dry bulb - depression.

Solution.

1. Equalising starts at 6 - 2 = 4 percent (driest sample), with the equalising EMC set at 4 percent.
2. Conditioning EMC: hardwood 6 + 4 = 10 percent; softwood 11 + 3 = 14 percent.
3. Wet bulb at the 30 percent step: 125 - 19 = 106 degrees F; at the 25 percent step: 125 - 35 = 90 degrees F.
4. Note that conditioning is generally applied for final contents of about 5 to 11 percent and is not really necessary for softwood dimension lumber dried to 15 percent average, so the softwood case is only an exercise in the rule.

Final answer. (a) Equalise from 4 percent; conditioning EMC 10 percent (hardwood) and 14 percent (softwood). (b) Wet bulb 106 and 90 degrees F.

What this tells a mill. Schedules are written as air states, so the operator sets dry bulb and depression and the wood follows toward the matching EMC; the T/C system is North American and European mills use their own calibrated schedules.

Problem 16. The average is right, the lot is not

Statement. Ten boards from one kiln charge finish at the following moisture contents (percent): 11.2, 9.4, 12.8, 19.5, 13.1, 8.7, 15.4, 12.0, 18.2, 10.0. Classify them with the thresholds of the UBC hemlock study (Canada, a study convention and not a standard): below 10 percent over-dried, 10 to under 19 percent acceptable, 19 percent or more under-dried. Compute the mean and the share in each class. Then apply the EN 14081-1 dry-grading limits as reported (Europe, publisher preview): mean at most 20 percent and no individual reading above 24 percent.

Hint. Count each class separately, remember that 10.0 is acceptable and 19.0 is not, and add all values before dividing by ten.

Solution.

1. Over-dried (below 10): 9.4, 8.7, which is 2 boards, 20 percent.
2. Under-dried (19 or more): 19.5, which is 1 board, 10 percent.
3. Acceptable: the remaining 7 boards, 70 percent.
4. Mean: 130.3 / 10 = 13.03 percent.
5. Dry-grading limits: mean 13.03 is at most 20 and the maximum 19.5 is at most 24, so the criteria are met even though one board is under-dried by the study threshold.

Final answer. Mean 13.03 percent; 20 percent over-dried, 70 percent acceptable, 10 percent under-dried. The dry-grading limits (mean at most 20, no value above 24) are met.

What this tells a mill. A lot should be judged on its average and its spread, and sorting by sapwood share and thickness before drying is cheaper than re-drying afterwards.

Grading and strength

Problem 17. What a characteristic value promises, and what a class step costs

Statement. The indicative strength class table used by the portal (a secondary reproduction of EN 338:2016 values, with conflicts between sources on several other cells; use the current standard for real values) lists C18 with bending strength 18 MPa, mean MOE 9.0 GPa and density 320 kg/m3, and C24 with 24 MPa, 11.0 GPa and 350 kg/m3. (a) A mill ships 3500 boards correctly graded as C18. About how many may legitimately break below 18 MPa? (b) And for 1200 boards of C24, below 24 MPa? (c) By what factor or percentage do strength, mean MOE and density rise from C18 to C24?

Hint. Bending strength and density are 5th percentile values, so 5 percent of a correctly graded population is expected below the characteristic value.

Solution.

1. (a) 3500 x 0.05 = 175 boards.
2. (b) 1200 x 0.05 = 60 boards.
3. (c) Strength: 24 / 18 = 1.333, +33.3 percent. MOE: 11.0 / 9.0 = 1.222, +22.2 percent. Density: 350 / 320 = 1.094, +9.4 percent.
4. Strength is therefore the most demanding part of the step, which is why machines that measure stiffness and density still need a correction for strength-reducing features.

Final answer. (a) about 175 boards. (b) about 60 boards. (c) strength +33 percent, MOE +22 percent, density +9 percent.

What this tells a mill. A single failed test piece does not prove a batch is mis-graded, because the class is a population statement; a mill should keep the population statistics right and tell customers so.

Problem 18. A visual grading exercise and the cost of sloping grain

Statement. The portal reproduces softwood squared-timber limits from an Austrian textbook excerpt citing DIN 4074-1 (undated; an illustration of how a rule is built). Knots as share of cross-section / slope of grain / ring width / wane: S7 up to 3/5, 16 percent, 6 mm, 1/3; S10 up to 2/5, 12 percent, 6 mm, 1/3; S13 up to 1/5, 7 percent, 4 mm, 1/4. Ring shake is not permitted in any grade. Grade three pieces without ring shake or rot: A has knots 1/3, slope 8 percent, ring width 5.0 mm, wane 1/4. B has knots 1/6, slope 9 percent, ring width 3.8 mm, wane 1/5. C has knots 1/6, slope 6 percent, ring width 3.8 mm, wane 1/5. Then, for a piece with a straight-grain bending strength of 36 MPa, use the indicative fractions from a Canadian teaching lecture (property unnamed; North America): 93 percent at slope 1 in 20, 81 percent at 1 in 10, 55 percent at 1 in 5.

Hint. A piece takes the highest grade whose limits it meets on every feature; a single failing feature downgrades the whole piece.

Solution.

1. Piece A: S13 fails on knots (1/3 = 0.33 > 0.20), slope (8 > 7) and ring width (5.0 > 4). S10: knots 0.33 <= 0.40, slope 8 <= 12, ring 5.0 <= 6, wane 0.25 <= 0.33, so A is S10.
2. Piece B: S13 fails only on slope (9 > 7); knots 0.167 <= 0.20, ring 3.8 <= 4, wane 0.20 <= 0.25 would pass. S10 passes everything, so B is S10.
3. Piece C: knots 0.167 <= 0.20, slope 6 <= 7, ring 3.8 <= 4, wane 0.20 <= 0.25: all S13 limits are met, so C is S13.
4. Slope of grain: 1 in 20 gives 36 x 0.93 = 33.48 MPa; 1 in 10 gives 36 x 0.81 = 29.16 MPa; 1 in 5 gives 36 x 0.55 = 19.8 MPa.

Final answer. A = S10, B = S10 (one failing feature, slope), C = S13. Indicative remaining strength 33.5, 29.2 and 19.8 MPa at slopes 1 in 20, 1 in 10 and 1 in 5.

What this tells a mill. One failing feature costs a whole grade, so the grader's attention belongs to the features that actually decide the class, and slope of grain is hard to judge by eye on a rough face.

Problem 19. Dynamic modulus of elasticity and what the machine does not know

Statement. A grading machine excites a board by a short impact and records the fundamental frequency. The board is 3.6 m long, the fundamental frequency is 620 Hz and the density from weight and dimensions is 470 kg/m3 (illustrative values). Use v = 2 x f x L and E = rho x v^2. (a) Compute the dynamic MOE in GPa. (b) The indicative C18 mean MOE is 9.0 GPa; does 9.37 GPa mean this board is C18? (c) In a Swedish, Norwegian and Finnish dataset (Europe) dynamic MOE alone reached R2 = 0.53 with bending strength, while laser scanning plus dynamic excitation reached 0.69. How much variation remains unexplained in each case? (d) With a correlation of 0.75, what share is explained? (e) In that study the C40 share of a C40/C24 double grade was 14.5 percent with dynamic MOE and 30.3 percent with the new method. For a batch of 2400 boards, how many more reach C40?

Hint. Velocity first, then E = rho x v^2 in pascals, then convert to GPa. Explained variation is the square of the correlation.

Solution.

1. (a) v = 2 x 620 x 3.6 = 4,464 m/s. E = 470 x 4,464^2 = 9,365,829,120 Pa = 9.37 GPa.
2. (b) No. A class mean is a population statement, the machine compares the indicating property against thresholds in an approved setting for a machine, species, size range and classes, and the portal gives no thresholds. One board's MOE cannot settle its class.
3. (c) Unexplained: 1 - 0.53 = 0.47, i.e. 47 percent; 1 - 0.69 = 0.31, i.e. 31 percent.
4. (d) 0.75 x 0.75 = 0.5625, about 56 percent explained and 44 percent not.
5. (e) 2400 x 0.145 = 348 boards; 2400 x 0.303 = 727 boards; the new method adds 379 boards, a factor of 2.09.

Final answer. (a) 9.37 GPa. (b) No. (c) 47 and 31 percent unexplained. (d) about 56 percent. (e) about 379 more boards in C40.

What this tells a mill. Extra sensors pay mainly if the mill wants to sell high classes, because the gain in the study appeared in C40 while C24 yield was already near 100 percent.

Economics and break-even

Problem 20. Log cost per cubic metre sawn and the first-order margin

Statement. All numbers in this problem are invented for practice and are not market data. A mill pays 118 per m3 for logs, recovers 47 percent, sells sawn timber at 285 per m3, earns 26 of residue revenue per m3 of sawn output and has 33 of other variable cost per m3 sawn. Use margin = sawn price + residue revenue - log price / recovery - other variable costs. (a) Compute the log cost per m3 sawn and the margin. (b) Logs rise by 6 per m3: new margin, and the equivalent fall in the sawn price. (c) Instead recovery improves to 48.5 percent: saving in log cost per m3 sawn. (d) At what recovery would the margin be zero at the original log price?

Hint. The key term is log price / recovery. A rise in log price is multiplied by 1 / recovery.

Solution.

1. (a) Log cost per m3 sawn: 118 / 0.47 = 251.06. Margin: 285 + 26 - 251.06 - 33 = 26.94.
2. (b) Extra cost per m3 sawn: 6 / 0.47 = 12.77. New margin 26.94 - 12.77 = 14.17. A sawn price fall of 12.77 has exactly the same effect.
3. (c) 118 / 0.485 = 243.30; saving 251.06 - 243.30 = 7.76 per m3 sawn.
4. (d) Zero margin needs log cost = 285 + 26 - 33 = 278; recovery = 118 / 278 = 42.4 percent.

Final answer. (a) 251.1 and 26.9 per m3 sawn. (b) margin 14.2; equivalent to a 12.8 fall in sawn price. (c) saving 7.8 per m3 sawn. (d) 42.4 percent.

What this tells a mill. At 47 percent recovery every currency unit added to the log price costs more than two units per m3 sawn, so recovery is the first lever to look at when logs get dear.

Problem 21. Break-even volume and operating leverage

Statement. Invented numbers for practice. A mill has fixed costs of 2,900,000 a year, an installed capacity of 70,000 m3 of sawn timber a year and a contribution of 52 per m3 sawn. (a) Compute the break-even volume and break-even utilisation. (b) Logs rise by 5 percent from 118 per m3 at 47 percent recovery. What is the new contribution, the new break-even volume and the result at full capacity? (c) Planned output is 60,000 m3; compute fixed cost per m3 and what it becomes if volume falls 10 percent.

Hint. Break-even volume = fixed cost / contribution. A log price rise lowers contribution by the rise divided by recovery.

Solution.

1. (a) 2,900,000 / 52 = 55,769 m3. Utilisation: 55,769 / 70,000 = 79.7 percent.
2. (b) Log rise: 0.05 x 118 = 5.90; cost per m3 sawn rises by 5.90 / 0.47 = 12.55. New contribution 52 - 12.55 = 39.45.
3. New break-even: 2,900,000 / 39.45 = 73,517 m3, which is above the capacity of 70,000 m3 (about 105 percent of capacity), so the mill cannot break even. At full capacity the result is 39.45 x 70,000 - 2,900,000 = about -138,723, a loss; before the rise it was +740,000.
4. (c) Fixed cost per m3: 2,900,000 / 60,000 = 48.33; at 90 percent of the volume: 2,900,000 / 54,000 = 53.70, an increase of 5.37 per m3.

Final answer. (a) 55,769 m3, 80 percent utilisation. (b) contribution 39.4; break-even 73,517 m3, above capacity; the mill makes a loss of about 139,000 even at full capacity. (c) 48.3 per m3, rising to 53.7.

What this tells a mill. A thin contribution means a modest log price rise can push break-even beyond capacity, so a mill should rerun break-even at plus 5 percent logs, minus 5 percent sawn price and minus 10 percent volume.

Problem 22. What finished stock costs to carry

Statement. Invented numbers for practice. A mill sells 36,000 m3 of sawn timber a year at a cost value of 240 per m3 and carries 55 days of finished stock, financed at 7 percent. (a) What is the value of the stock and the annual interest? (b) What do 10 days of stock cost in capital and in interest? (c) By how much does interest fall if stock is cut from 55 to 40 days?

Hint. Stock value = annual cost of output x days / 365. Interest = rate x stock value.

Solution.

1. (a) Annual cost value: 36,000 x 240 = 8,640,000. Stock: 8,640,000 x 55 / 365 = 1,301,918. Interest at 7 percent: 91,134.
2. (b) 10 days: 8,640,000 x 10 / 365 = 236,712 of capital; interest 16,570 a year.
3. (c) 15 days fewer: 355,068 of capital released, interest saved 24,855 a year.

Final answer. (a) stock about 1,301,918, interest about 91,134 a year. (b) about 236,712 of capital and 16,570 of interest per 10 days. (c) about 24,855 a year.

What this tells a mill. Stock also carries price risk, so a mill should track days of stock as a monthly KPI beside contribution, not only at year end.

Problem 23. Simple payback of a scanner upgrade

Statement. Invented numbers for practice, for the mill of the previous problems (sawn price 285, other variable cost 33, residue revenue 26 per m3 sawn). A scanner and optimiser upgrade costs 650,000. The mill processes 60,000 m3 of logs a year and expects the upgrade to raise recovery by 0.8 percentage points. Extra sawn timber comes from the same logs, so the net gain per extra m3 is the sawn price minus the other variable cost minus the residue revenue displaced. (a) Annual gain and simple payback. (b) Payback if the gain is only 0.5 points (300 m3). (c) How many points of recovery gain give a five-year payback?

Hint. Extra sawn volume = log volume x recovery gain. Payback = investment / annual gain.

Solution.

1. (a) Extra sawn timber: 60,000 x 0.008 = 480 m3. Net gain per m3: 285 - 33 - 26 = 226. Annual gain 480 x 226 = 108,480. Payback 650,000 / 108,480 = 6.0 years.
2. (b) 300 x 226 = 67,800; payback 650,000 / 67,800 = 9.6 years.
3. (c) Five years means 130,000 a year; 130,000 / 226 = 575 m3, which is 575 / 60,000 = 0.96 percentage points.

Final answer. (a) gain 108,480 a year, payback about 6.0 years. (b) about 9.6 years. (c) about 0.96 points.

What this tells a mill. Payback is a screening tool that ignores later cash flows, so a mill should follow it with NPV or IRR over a full price cycle and ask the supplier what recovery gain is guaranteed.

Energy and carbon

Problem 24. Kiln heat from the water removed

Statement. A planed softwood product has a density of 470 kg/m3 at 15 percent moisture content (dry basis, illustrative values). It enters the kiln at 85 percent. Use the reference of about 1.6 kWh per kg of water removed (a theoretical value from a trade source, not verified) and 1 kWh = 3.6 MJ. (a) Oven-dry mass and water removed per m3. (b) Heat per m3 and for a charge of 45 m3, in kWh and MJ. (c) The heat meter on that charge reads 24,500 kWh. What is the specific consumption and how far is it above the reference? (d) What would the heat per m3 be if the stock were pre-dried to 60 percent, and how much is saved?

Hint. Oven-dry mass = density at 15 percent / 1.15. Water = oven-dry mass x (MC in - MC out).

Solution.

1. (a) Oven-dry mass: 470 / 1.15 = 408.7 kg. Water: 408.7 x (0.85 - 0.15) = 286.1 kg per m3.
2. (b) Heat: 286.1 x 1.6 = 458 kWh per m3. Charge: 45 x 458 = about 20,598 kWh, or 20,598 x 3.6 = about 74,154 MJ. (Water in the charge: 12,874 kg.)
3. (c) 24,500 / 12,874 = 1.90 kWh per kg; relative to 1.6 that is 18.9 percent higher, a prompt to check sealing, venting and fan control.
4. (d) At 60 percent entry: water 408.7 x 0.45 = 183.9 kg, heat 294 kWh per m3, a saving of 163 kWh per m3.

Final answer. (a) 409 kg oven-dry and 286 kg water per m3. (b) about 458 kWh per m3; about 20,598 kWh (74,154 MJ) per charge. (c) 1.90 kWh/kg, 19 percent above the reference. (d) about 294 kWh per m3, saving 163.

What this tells a mill. Heat demand is nearly proportional to the water removed, so starting moisture and yard or pre-drying matter, and every batch should be logged with moisture in and out.

Problem 25. Specific electricity and a variable speed drive

Statement. A mill used 1,540,000 kWh of electricity last year and sawed 26,000 m3. (a) Compute kWh per m3 and compare it with the 53 to 67 kWh/m3 range of the Norwegian data (Europe, boundary and product basis not defined). (b) Using the 35.3 g CO2 eq per kWh of the Swedish production-mix electricity in one declaration, what is the electricity footprint per m3? (c) A kiln fan draws 55 kW at full flow. According to the fan affinity laws as quoted in an Oregon guide (North America), at 80 percent flow it uses 51 percent of full power and at 50 percent flow about 13 percent. With a variable speed drive holding 80 percent flow for 3500 hours, how much energy is saved, what share of annual mill electricity is that, and what is it worth at an invented 0.11 per kWh?

Hint. Specific electricity is annual kWh divided by annual m3. Fan power at reduced flow is the full power times the quoted fraction.

Solution.

1. (a) 1,540,000 / 26,000 = 59.2 kWh/m3, inside the Norwegian range.
2. (b) 59.2 x 35.3 g = 2,091 g, i.e. 2.09 kg CO2 eq per m3. A different grid or year needs its own factor.
3. (c) Power at 80 percent flow: 55 x 0.51 = 28.05 kW; saving 26.95 kW; over 3500 hours 94,325 kWh.
4. Share of annual electricity: 94,325 / 1,540,000 = 6.1 percent. Value: 94,325 x 0.11 = 10,376.
5. For comparison, at 50 percent flow the fan would draw 7.15 kW, but a real fan rarely holds that flow all year.

Final answer. (a) 59.2 kWh/m3. (b) about 2.1 kg CO2 eq per m3. (c) about 94,325 kWh a year, 6.1 percent of the mill's electricity, worth about 10,376.

What this tells a mill. The figures are comparable only on the same reference unit and boundary, so a mill should fix one unit, meter the big loads and review fans first because airflow savings follow the cube-type affinity law.

Problem 26. Stored carbon, GWP-total and the share of manufacturing

Statement. The product of Problem 24 has a density of 470 kg/m3 at 15 percent moisture. (a) Taking carbon as 50 percent of oven-dry mass and CO2 = C x 44/12, how much CO2 is stored per m3? (b) If the fossil GWP of production were 38.5 kg CO2 eq per m3 (an invented value for practice), what is GWP-total as the sum of fossil and biogenic terms? (c) The British Columbia average dried lumber EPD (North America, 2023) lists A1 7.59, A2 3.99 and A3 34.28 kg CO2 eq per m3. Check that they add to the reported 45.86 and compute the share of A3. (d) Using default half-lives of 35 years for sawnwood and 25 years for panels (Swiss inventory, IPCC 2014 Tier 2), how much of 640 kg CO2 stored in sawnwood remains after 70 years, and in panels after 75 years?

Hint. Oven-dry mass first. Half-life: remaining share = 0.5 to the power (years / half-life).

Solution.

1. (a) Oven-dry mass 408.7 kg; carbon 204.3 kg; CO2 204.3 x 44/12 = 749 kg per m3.
2. (b) GWP-total = 38.5 + (-749) = -711 kg CO2 eq per m3. A strongly negative figure is stored carbon, not proof of a climate benefit.
3. (c) 7.59 + 3.99 + 34.28 = 45.86, matching the reported value. A3 share: 34.28 / 45.86 = 74.7 percent.
4. (d) Sawnwood: 70 / 35 = 2 half-lives, 640 x 0.25 = 160 kg. Panels: 75 / 25 = 3 half-lives, 640 x 0.125 = 80 kg.

Final answer. (a) about 749 kg CO2 per m3. (b) about -711 kg CO2 eq per m3. (c) A3 is 75 percent. (d) 160 kg (sawnwood after 70 years) and 80 kg (panels after 75 years).

What this tells a mill. Manufacturing is the part of the footprint a mill controls, and an internal estimate must not be called an EPD or presented as a climate benefit without the whole product life in view.

Reading research critically

Problem 27. Two machine-vision papers: recompute before you believe

Statement. (1) A 2023 paper on birch boards (China, one factory, one batch of 33.56 m3) reports an optimised value of USD 12,844.52 against USD 11,236.27 for manual marking, a difference of USD 1,608.25, presented as a 14.13 percent improvement. The articles flag that the value terms listed in the paper appear to give about USD 15,400 rather than 12,844.52. The paper also reports CNY 88,241.89 against CNY 77,193.19, a typical board of 4,000 x 58 x 43 mm, scanning 0.311 s plus analysis 0.368 s per board, line speeds of 150 or 214 m/min, and an additional annual capacity of 6,451,200 m equal to 16,089.29 m3. (a) Recompute the percentage in USD and in CNY. (b) What would the improvement be if the value terms really summed to USD 15,400? (c) Does the stated additional capacity agree with the board cross-section? (d) Can analysis keep up with a 4 m board at 214 m/min? (2) A 2026 paper on beech boards (China, public dataset, 100 test images) reports mAP@50 of 91.8 against 90.8 for the baseline, mAP@50-95 of 69.3 against 67.8, parameters of 34.08 M against 25.86 M and FLOPs of 75.4 G against 78.9 G. (e) Compute the gains in points and in relative percent, and the change in parameters and FLOPs.

Hint. Divide the difference by the manual value, not by the optimised value. For (c), volume = length x width x thickness. For (d), time per board = length / speed.

Solution.

1. (a) 1,608.25 / 11,236.27 = 14.31 percent. In CNY: 88,241.89 / 77,193.19 - 1 = 14.31 percent. Both give 14.31, so 14.13 looks like transposed digits.
2. (b) 15,400 / 11,236.27 - 1 = 37.1 percent; the discrepancy of about 2,555 in the value terms is 20 percent of the reported total, so the headline gain is sensitive to it.
3. (c) A metre of board has a volume of 0.058 x 0.043 = 0.002494 m3, so 6,451,200 m x 0.002494 = 16,089.29 m3, equal to the paper's 16,089.29 m3. That figure is internally consistent, though it rests on the assumed speed gain and operating hours.
4. (d) A 4 m board passes at 214 m/min in 4 / (214/60) = 1.12 s. Scan plus analysis take 0.311 + 0.368 = 0.679 s, so a single station keeps up with that board length if the two steps run in series. The paper's 'decision speed of at least 60 m/min' is a separate claim.
5. (e) mAP@50: 91.8 - 90.8 = 1.0 point, 1.1 percent relative. mAP@50-95: 69.3 - 67.8 = 1.5 points, 2.2 percent relative. Parameters: 34.08 - 25.86 = 8.22 M, +31.8 percent. FLOPs: -4.4 percent, which the authors say may not reflect real speed.

Final answer. (a) 14.31 percent in both currencies. (b) about 37 percent. (c) Yes, 16,089.29 m3 is consistent. (d) Yes, 0.68 s against 1.12 s. (e) +1.0 and +1.5 points (about 1.1 and 2.2 percent relative), with 32 percent more parameters. Both papers support a pilot on your own material, not an investment decision.

What this tells a mill. A promising vision result from one batch, one factory, a public dataset and 100 test images calls for a pilot with the mill's own species, line speed and price list before any money is committed.

Problem 28. Cutting force in a circular saw paper: the theory is not the test

Statement. A 2023 paper (Slovakia, Poland, Czech Republic) cut Norway spruce across the grain with a 600 mm high-speed steel (HSS) blade (3.5 mm tooth width, 56 teeth) and a sintered carbide (SC) blade (5.2 mm tooth width, 54 teeth). The portal's own summation of the paper's tables (not stated in the paper) gives measured totals of 40.5 N (HSS) and 43.7 N (SC) and theoretical totals of 51.2 N and 82.7 N. The seventh tooth in the cut gave measured 7.79 N (HSS) and 8.39 N (SC), theoretical 9.83 N and 15.87 N. The abstract gives a theoretical SC total of 76 to 94 N against a measured 40 to 44 N; the article notes this does not exactly match the tables. The first-tooth measured values were 3.43 N (HSS) and 3.67 N (SC). (a) Compute theoretical / measured for each blade. (b) Is the SC ratio inside the abstract's implied range? (c) How much larger is the measured force of tooth 7 than tooth 1? (d) Compare the tooth width ratio with the measured force ratio. What can and cannot be concluded about carbide against HSS?

Hint. Ratios first, then compare. Remember that the two blades differ in material, tooth width and tooth count at once.

Solution.

1. (a) HSS: 51.2 / 40.5 = 1.26. SC: 82.7 / 43.7 = 1.89. At tooth 7: 9.83 / 7.79 = 1.26 and 15.87 / 8.39 = 1.89, which agrees.
2. (b) The abstract implies ratios from 76 / 44 = 1.73 to 94 / 40 = 2.35. The SC ratio 1.89 lies inside that range, so the numbers are consistent in magnitude although the exact range is not reproduced by the tables.
3. (c) HSS: 7.79 / 3.43 = 2.27 times; SC: 8.39 / 3.67 = 2.29 times. The force roughly doubles from the first to the seventh tooth in the cut.
4. (d) Tooth width ratio 5.2 / 3.5 = 1.49, measured force ratio 43.7 / 40.5 = 1.08. Per millimetre of tooth width the portal-sum forces are 8.4 N (SC) and 11.6 N (HSS), but the blades also differ in tooth count and material, so this is no ranking of carbide against HSS.
5. What can be concluded: a sharp-tooth specific-cutting-force model overestimates the load on the wide-toothed carbide blade by roughly a factor of 1.9 and on the HSS blade by about 1.3. What cannot: that carbide needs less or more force, that speed has no effect, or that real blunt blades behave the same.

Final answer. (a) 1.26 (HSS) and 1.89 (SC). (b) Yes, inside 1.73 to 2.35. (c) about 2.3 times for both blades (2.27 and 2.29). (d) widths differ by 1.49, forces by 1.08; the design is confounded, so no material ranking follows.

What this tells a mill. Use the measured force range as an order of magnitude for motor and blade checks, treat idealised calculations with suspicion for worn teeth, and base purchasing on tool life and cut quality, which this paper does not measure.

Problem 29. Machine learning for kiln moisture: what can the paper support?

Statement. A 2023 University of British Columbia study (Canada, western hemlock, 0.73 m3 laboratory kiln) started with 96 timbers, cut four kiln specimens from each, discarded 6 and formed nine batches of 42. Targeted final moisture was 12 percent. Regression with six inputs gave R2 of 73.86 percent on training and 44.81 percent on test data; with the three wood attributes only, 56.80 and 32.10. Classification (below 10 percent over-dried, 10 to under 19 acceptable, 19 or more under-dried) gave 76.19 percent training and 69.05 percent test accuracy. The articles flag that the per-class statements are internally inconsistent (71.64 percent is called both the highest and the lowest). Mean final moisture was 15.59 percent (spread 5.16 as reported) for Schedule I and 10.87 percent (2.19) for the control schedule. (a) Check the specimen counts. (b) Compute the training-test gaps and the test improvement from adding process variables. (c) What share was misclassified, and what does one board mean for a test set of 42? (d) What can the paper support?

Hint. Gaps are differences in percentage points. A percentage that is a multiple of 1/42 hints at the size of the partition, but the article does not state it.

Solution.

1. (a) 96 x 4 = 384; 384 - 6 = 378; 378 / 9 = 42 per batch. Cookies: 96 x 5 = 480. Counts are consistent.
2. (b) Six-input regression: 73.86 - 44.81 = 29.05 points. Three-input: 56.8 - 32.1 = 24.70 points. Process variables raise test R2 by 44.81 - 32.1 = 12.71 points. Classification gap: 76.19 - 69.05 = 7.14 points, much smaller than the regression gap.
3. (c) Misclassified: 100 - 69.05 = 30.95 percent. Reader's check (an inference, because the article does not state partition sizes): 69.05 is exactly 29/42 and 30.95 is 13/42, and 76.19 is 32/42, so the test set may be one batch of 42 boards; each board would then move accuracy by 100 / 42 = 2.4 points.
4. (d) Spread: Schedule I 5.16 / 15.59 = 33 percent of its mean; control 2.19 / 10.87 = 20 percent (the article labels the spread 'as reported', so this ratio is only indicative). The supported claims are: initial weight is the best cheap predictor, schedule and conditioning matter, classes are workable. Not supported: a precise prediction for an individual board (test R2 under 45 percent), a mill-scale benefit, any advice on schedules.

Final answer. (a) consistent. (b) regression gaps 29.05 and 24.70 points, test gain 12.71 points; classification gap 7.14 points. (c) 31.0 percent misclassified; if the test set is 42 boards, one board is worth 2.4 points of accuracy. (d) classification is workable as a first step, regression is not.

What this tells a mill. Collect schedule settings and initial weights with every charge, think in classes, and validate any model on your own species and kiln before using it to sort.

Problem 30. A market review: headline, gap indicator and a flow diagram

Statement. A 2026 review from Poland (Forests, 16-country ISC set, Europe) gives production and reported consumption in million m3: 2021 97.709 and 82.916; 2024 88.472 and 66.87; preliminary 2025 88.325 and 68.627. The paper reports production down 9.5 percent in 2024 and consumption down 19.4 percent, an accounting difference P minus C that rose from 14.793 to 21.602, and for Germany a fall from 25.3 to 21.3 million m3 (2021 to the 2026 outlook) stated as minus 16.1 percent. The Latvian flow diagram shows about 2.6 million m3 of sawnwood from sawlogs, a primary-market flow of about 2.4 and about 0.6 for domestic end use or further processing (the paper does not explain why 2.4 plus 0.6 differs from 2.6). (a) Recompute the 2024 changes and the preliminary 2025 changes against 2021. (b) Recompute the P minus C difference for 2021 and 2024, and the increase. (c) Check the Latvian figures. (d) Recompute Germany and decide whether rounding of the one-decimal inputs can explain 16.1 percent. (e) What can the P minus C figure support?

Hint. Relative change = (value in year t / value in 2021 - 1) x 100. For rounding, test the extreme values the inputs could have had.

Solution.

1. (a) 2024 production: 88.472 / 97.709 - 1 = -9.5 percent; consumption 66.87 / 82.916 - 1 = -19.4 percent. Preliminary 2025: production -9.6 percent, consumption -17.2 percent. These match the reported values; the title's 'about 10 percent' and '17 percent' refer to 2025 and are rounded.
2. (b) 2021: 97.709 - 82.916 = 14.793; 2024: 88.472 - 66.87 = 21.602. Increase 46 percent. The authors stress that this is not a trade balance, export surplus or stock change.
3. (c) 2.4 + 0.6 = 3.0, not 2.6: a gap of 0.4 million m3, about 15 percent of 2.6. The diagram cannot be used to close a balance without the authors' explanation.
4. (d) 21.3 / 25.3 - 1 = -15.8 percent, not minus 16.1. If both numbers are rounded to 0.1, the true change could lie between -16.2 percent (21.25 and 25.35) and -15.4 percent (21.35 and 25.25). Minus 16.1 lies inside, so rounding can explain it.
5. (e) P minus C signals that supply adjusted less than demand. It cannot say how the balance was achieved: exports, imports, stocks and reporting changes are not separated. The same rounding test shows that the review's ETTF sum rising from 18.1 to 18.3 (plus 1.1 percent from the rounded numbers) is consistent with the stated plus 1.4 percent.

Final answer. (a) 2024: -9.5 and -19.4 percent (matching the paper); 2025: -9.6 and -17.2 percent. (b) 14.793 and 21.602, up 46 percent. (c) a 0.4 million m3 gap. (d) -15.8 percent, within the rounding range -16.2 to -15.4 percent. (e) only a signal of slower supply adjustment.

What this tells a mill. Use the review for direction and for the shifting of export destinations, not for exact balances, and always check which scope, product definition and rounding sit behind a headline number.

Limits

• This article is a study aid synthesised from the portal's fact-checked articles and its courses. It is not a standard, a design manual, financial advice or a substitute for your own measurements.
• Regional data: shrinkage coefficients, EMC values, air-drying times, kiln schedule rules, the sloping-grain fractions and the energy references are mostly North American (USDA Forest Products Laboratory, Wood Handbook, US extension material). The log measurement rules are German and Swedish, the sawing examples come from Sweden, Finland, Austria, Romania, Poland and Slovakia, and some are single studies. Do not transfer any figure between regions, species or mills without checking.
• The strength class values (C18, C24) are an indicative secondary reproduction of EN 338:2016 with conflicts between sources on other cells; the DIN 4074-1 limits come from an undated textbook excerpt; the sloping-grain fractions come from a Canadian lecture that does not name the property. Use the current purchased standards for any declaration.
• All prices, costs, capacities, payloads, adjustments and the fossil GWP value in Problem 26 are invented for practice. They show how the arithmetic works and are not market data or benchmarks. Assumed values such as Gb 0.40 and the fan hours are also for practice.
• Some combinations mix sources as an illustration (for example the Romanian fir losses with the Finnish simulation). Linear scaling of the Alaskan oversizing rule, the hysteresis ratio of 0.8 and the 1.6 kWh/kg reference are approximations; the last comes from a trade source.
• In the research-reading problems the recalculations are the portal's own arithmetic on figures the articles quote. The article summaries, not the original papers, were the basis, so the underlying papers and their supplementary files were not re-checked. The observation in Problem 29 that the reported percentages are multiples of 1/42 is an inference and the article does not state the partition sizes.
• Every solution was recomputed and asserted in Python. That checks the arithmetic against the stated inputs, not the truth of the inputs themselves, which are as reliable as the sources they come from.