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ICAR AICE-JRF/SRF Growth & Yield Modelling: practice questions

ICAR AICE-JRF/SRF (PhD) — Subject 59 Forestry · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
41written to this exam in this subject
10syllabus units

What this subject covers

Predicting how stands grow and what they will yield: yield tables, stand density, diameter-distribution and individual-tree models, and their use in planning.

In the exams: ICAR AICE-JRF 59 Unit 4 (growth and increment, yield) and Unit 9 (growing stock, site quality); IFoS Forestry Paper II Section A; SAF Certified Forester

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF (PhD) — Forestry, Units 4 and 9

MCQs on growth and increment, yield tables (normal, empirical, variable-density), growing stock and site quality; definitions of stand density measures and the self-thinning rule.

Where the 41 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Growth & Yield Modelling. Difficulty: 6 easy, 17 medium, 18 hard.

Growth curves and growth functions6
Stand density measures: stand density index, relative spacing, crown competition factor5
Self-thinning rule and density management diagrams5
Individual-tree distance-independent and distance-dependent models5
Diameter-distribution models4
Process-based and hybrid models4
Model fitting, validation and projection4
Whole-stand models3
Using growth models in harvest scheduling3
Normal, empirical and variable-density yield tables2

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Growth curves and growth functions · medium

A Chapman-Richards height model is H = 30(1 − e^(−0.05t))^1.5. What height does it predict at age 20?

  1. 19.0 m
  2. 15.1 m
  3. 6.7 m
  4. 30.0 m
Answer and working

B. 15.1 m

H = 30 × (1 − e^(−1))^1.5 = 30 × 0.6321^1.5 = 30 × 0.5026 = 15.1 m. Forgetting the exponent 1.5 gives 19.0 m; 30.0 m is the asymptote, approached only at great age.

2 · Growth curves and growth functions · medium

The Gompertz function Y = A exp(−b e^(−kt)) has its inflection point at:

  1. A/2, exactly half of the asymptote
  2. A/e, or roughly 37% of the asymptote
  3. A, the asymptote itself, reached at infinite age
  4. A/4, a quarter of the asymptote at all parameter values
Answer and working

B. A/e, or roughly 37% of the asymptote

Setting the second derivative to zero gives Y = A/e for the Gompertz. The logistic inflects at A/2.

3 · Growth curves and growth functions · hard

Why are polynomial equations poor choices for projecting stand yield beyond the ages in the data?

  1. They can turn downward or rise without limit outside the data range
  2. They cannot be fitted by least squares to forest growth data of any kind
  3. They always pass exactly through every data point, leaving no residuals
  4. They require site index as a variable, which is rarely measured
Answer and working

A. They can turn downward or rise without limit outside the data range

Polynomials fit flexibly within the data but have no asymptote and no biological constraint, so extrapolations behave unrealistically. Growth functions with asymptotes are preferred.

4 · Growth curves and growth functions · hard

For a Chapman-Richards curve with c = 3, the inflection lies at A(1 − 1/c)^c. What fraction of the asymptote is this?

  1. 0.50
  2. 0.37
  3. 0.67
  4. 0.30
Answer and working

D. 0.30

(1 − 1/3)³ = (2/3)³ = 0.296 ≈ 0.30. 0.50 is the logistic inflection; 0.37 the Gompertz; 0.67 forgets to cube.

5 · Self-thinning rule and density management diagrams · medium

Along Reineke's self-thinning line, if quadratic mean diameter doubles, stems per hectare fall by a factor of about:

  1. 2.0
  2. 3.0
  3. 1.6
  4. 4.0
Answer and working

B. 3.0

N falls by 2^1.605 = 3.04, about 3. 2.0 assumes a slope of −1; 4.0 a slope of −2; 1.6 confuses the slope with the factor.

6 · Diameter-distribution models · medium

A three-parameter Weibull with a = 5 cm, b = 15 cm and c = 2 describes a stand's diameters. What proportion of trees is smaller than 20 cm?

  1. 36.8%
  2. 83.1%
  3. 50.0%
  4. 63.2%
Answer and working

D. 63.2%

F(20) = 1 − exp(−((20 − 5)/15)²) = 1 − e^(−1) = 63.2%. e^(−1) = 36.8% is the proportion larger than 20 cm; ignoring the location a gives 83.1%.

7 · Diameter-distribution models · medium

A Weibull shape parameter c less than 1 produces a diameter distribution that is:

  1. Nearly symmetric, as in even-aged stands
  2. Nearly symmetric, as in even-aged stands of one species
  3. Reverse-J shaped, as in uneven-aged stands
  4. Bimodal with two distinct peaks
Answer and working

C. Reverse-J shaped, as in uneven-aged stands

c < 1 gives a steeply declining (reverse-J) density; c = 1 the exponential; c near 3.6 a near-symmetric shape.

8 · Diameter-distribution models · easy

In the three-parameter Weibull, the location parameter a represents:

  1. The most common (modal) diameter class in the stand
  2. The quadratic mean diameter of the stand
  3. The smallest possible diameter in the distribution
  4. The largest diameter in the stand
Answer and working

C. The smallest possible diameter in the distribution

a shifts the distribution's lower bound; b scales it and c sets its shape.

9 · Individual-tree distance-independent and distance-dependent models · hard

A common finding when distance-dependent and distance-independent models are compared for stand totals is that:

  1. Spatial models are only slightly more accurate, despite costly stem maps
  2. Spatial models are always several times more accurate for stand totals
  3. Distance-independent models cannot predict stand totals at all
  4. Spatial models cannot be used in any planted forest
Answer and working

A. Spatial models are only slightly more accurate, despite costly stem maps

Many comparisons show modest gains for spatial models at stand level, while stem mapping is expensive. They remain valuable for research and irregular stands.

10 · Individual-tree distance-independent and distance-dependent models · hard

Errors in which component of individual-tree simulators most often accumulate over long projections?

  1. Conversion of diameters from inches to centimetres
  2. Calculation of basal area from diameter
  3. The species codes in the input tree list
  4. Mortality and ingrowth predictions
Answer and working

D. Mortality and ingrowth predictions

Mortality is a rare, variable event and ingrowth is hard to predict; their errors compound over decades. Unit conversions and basal area are exact.

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Read the subject

Growth & Yield ModellingThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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Growth & Yield Modelling for other exams

Indian Forest Service Examination · IndiaSAF Certified Forester · USA

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