The Encyclopedia of Economic Principals

Chapter 82

Exhaustible Resources, Biodiversity, and Intergenerational Stewardship

Treat a resource or a species as an asset, and price the timing.

Four of the chapter's worked examples, made interactive: Hotelling arbitrage for a quarry, the extinction margin for a slow-breeding animal, efficiency rebound, and the quasi-option value of waiting. Change one value at a time and watch the figure, the numbers and the hand calculation respond.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

Hotelling arbitrage: extract now or wait

Does a tonne left underground earn as much as the money from selling it today?

The owner treats the deposit as an asset. Selling now yields today's rent, which can earn r; waiting yields next year's rent. When expected rent grows faster than r, extraction is postponed; when slower, it moves toward the present, until the two match.

Equation, written in LaTeX: \lambda_t=p_t-c_t

Equation, written in LaTeX: \lambda_{t+1}=(1+r)\lambda_t.

Equation, written in LaTeX: \lambda_1=1.08(60)=\$64.80.

Equation, written in LaTeX: 68.04=1.08(63)

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lambda is scarcity rent, price minus marginal extraction cost, in dollars per tonne. Today's rent is $63 after the market adjusts. Next year's price is $99 and c1 is next year's extraction cost. r is the comparable return.

Predict first. If next-year cost rises to $36 and r falls to 4 percent, does waiting become attractive?

Your prediction

Choose an example

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Figure: Hotelling arbitrage: extract now or wait. Bars for today's rent $63, the required next-year rent $68.04 at 8% and the expected next-year rent $68.04 with cost $30.96. Decision: no profitable shift.
Required return: 8%, Next-year extraction cost: $30.96
Constructed example: the chapter's hypothetical quarry (rent $63, price $99, costs $27, $30.96 and $36, returns 8 and 12 percent); a 4 percent return is added for comparison.

Calculated values

Required next-year rent
$68.04
Expected next-year rent
$68.04
Underground return
8.0%
Required return
8%
Decision
No profitable shift

A unit left in the ground must earn 8% to match the financial return: 1.08 x 63 = 68.04. Next year's expected rent is 99 - 30.96 = 68.04: the two returns are equal and no marginal shift is profitable.

Worked steps

  1. Required rent = (1 + 0.08) x 63 = 1.08 x 63 = 68.04
  2. Expected rent = 99 - 30.96 = 68.04
  3. Underground return = 68.04 / 63 - 1 = 8.0%
  4. Compare: 68.04 = 68.04

Use the idea

Compare expected net rent next year, after extraction cost, with today's rent grown at a comparable return; gross price growth alone is not enough.

Where the conclusion applies

Two dates, homogeneous tonnes, known prices and costs, and no capacity limit, tax or externality. Discoveries, market power and unpriced damage change the comparison.

Check your understanding: At r = 0.04 and next-year cost $36, which is larger, the required or the expected next-year rent?
Required 1.04 x 63 = 65.52; expected 99 - 36 = 63; 63 < 65.52, so extraction shifts to the present.

Chapter 82 source: section "Hotelling rule".

Demonstration 2 of 4

The Clark extinction margin

When does a private owner sell a slow-breeding animal rather than keep it?

A slow-growing stock is capital that earns its own growth rate. When the owner's discount rate exceeds that return, selling and investing the money wins, which can drive the stock to zero. A lower rate or a payment tied to survival raises the value of keeping it.

Equation, written in LaTeX: PV_K=\frac{14{,}700}{1.07}=\$13{,}738.32.

Equation, written in LaTeX: \frac{14{,}700}{1.04}=\$14{,}134.62,

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Selling today yields $14,000 net. Keeping the animal yields $14,700 next year (offspring and later sale) plus any payment s per surviving adult. r is the owner's discount rate; PV_K is the present value of keeping it.

Predict first. Is a $900 survival payment enough to save the animal if the owner's rate rises to 10 percent?

Your prediction

Choose an example

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Figure: The Clark extinction margin. Bars of the present value of keeping the animal at 4, 7 and 10 percent with a survival payment of $0, against a line at the $14,000 sale value. At 7% the value is $13,738.32.
Owner's discount rate: 7%, Payment per surviving adult: $0
Constructed example: the chapter's hypothetical antelope (sale $14,000, retained value $14,700, rates 7 and 4 percent, a $900 payment); a 10 percent rate and a $450 payment are added.

Calculated values

Payoff next year
$14,700
PV of retention
$13,738.32
Sale value today
$14,000
Margin
liquidation by $261.68
Decision
Sell now

Retention pays $14,700 next year. Discounted at 7%: 14,700 / 1.07 = 13,738.32, against a sale value of 14,000. Selling now wins by $261.68; if this holds for every animal that can be caught, the private program tends toward extinction.

Worked steps

  1. PV = 14,700 / 1.07 = 13,738.32
  2. 13,738.32 - 14,000 = -261.68

Use the idea

Compare the discounted value of keeping a living asset, including any survival payment, with what it sells for today.

Where the conclusion applies

One period, a certain next-year value, no search cost and no other terminal value. Unpriced benefits to neighbours are left out of the private comparison.

Check your understanding: At r = 0.10 and a $900 payment, which wins and by how much?
15,600 / 1.10 = 14,181.82; minus 14,000 = 181.82, so retention still wins.

Chapter 82 source: section "Clark extinction condition".

Demonstration 3 of 4

Efficiency rebound and backfire

When does a more efficient technology end up using more of the resource?

Efficiency lowers use per task, but cheaper service draws more tasks. Rebound measures how much of the engineering saving the extra tasks take back. When growth exceeds e / (1 - e), the saving is more than erased.

Equation, written in LaTeX: R'=R(1-e)(1+g).

Equation, written in LaTeX: \rho=\frac{R'-R(1-e)}{eR}=\frac{(1-e)g}{e}.

Equation, written in LaTeX: (1-e)(1+g)>1, g>\frac{e}{1-e}.

Equation, written in LaTeX: R'=3.4(6)=20.4\text{ million kilowatt-hours}.

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R is baseline use, 24 million kilowatt-hours a month. e is the proportional cut in intensity and g the induced growth in tasks. rho is rebound as a share of the engineering saving eR.

Predict first. With only a 25 percent intensity cut, does 70 percent task growth cause backfire?

Your prediction

Choose an example

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Figure: Efficiency rebound and backfire. Bars of baseline use 24, engineering use 12.0 and actual use 20.4 million kilowatt-hours for an intensity cut of 50% and task growth of 70%.
Intensity cut e: 50%, Induced task growth g: 70%
Constructed example: the chapter's hypothetical computing platform (24 million kWh, e = 0.5, g = 0.7 and 1.3); cuts of 25 and 75 percent and growth of 30 and 200 percent are added.

Calculated values

Engineering use R(1 - e)
12.0
Actual use R'
20.4
Rebound rho
0.70
Backfire threshold g >
1.00
Backfire
no

Cutting intensity by 50% would bring use to 24 x 0.50 = 12.0 million kWh, but tasks grow 70%, so use is 12.0 x 1.70 = 20.4. Rebound is 0.50 x 0.70 / 0.50 = 0.70, or 70% of the engineering saving. Growth 0.70 is below the threshold 1.00, so use stays 3.6 million below the baseline and part of the saving survives.

Worked steps

  1. R(1 - e) = 24 x 0.50 = 12.0
  2. R' = 24 x 0.50 x 1.70 = 20.4
  3. rho = 0.50 x 0.70 / 0.50 = 0.70
  4. Threshold e / (1 - e) = 0.50 / 0.50 = 1.00

Use the idea

Before counting an efficiency gain as a resource saving, estimate how much service demand the gain itself causes.

Where the conclusion applies

g is the causal growth from the efficiency change, not growth that would have happened anyway. The identity says nothing about where g comes from.

Check your understanding: At e = 0.25 and g = 0.7, what are R' and rebound?
R' = 24 x 0.75 x 1.7 = 30.6 million kWh, above 24, so backfire (0.7 > 0.25 / 0.75 = 0.333); rho = 0.75 x 0.7 / 0.25 = 2.10.

Chapter 82 source: section "Jevons-Like Demand Expansion and the Capex Buildout".

Demonstration 4 of 4

Quasi-option value of waiting to learn

When is it worth delaying an irreversible development to learn what the estuary is worth?

Development is irreversible but waiting is not. Delay lets the decision follow the study: preserve if the estuary proves valuable, develop if not. The value of that flexibility, not preservation itself, can make waiting win.

Equation, written in LaTeX: V_P=8+0.4(230)+0.6(120)=\$172\text{ million}.

Equation, written in LaTeX: 8+0.4(230)+0.6(50)=\$130\text{ million},

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All values are present values in $ million. Developing now yields 140. Waiting preserves 8 of interim services; after the study, development yields 120. The intact estuary is worth the high-state value with probability p and 50 otherwise. V_P is the value of waiting and learning.

Predict first. If the high state has only a 20 percent chance, does learning still beat immediate development?

Your prediction

Choose an example

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Figure: Quasi-option value of waiting to learn. Bars for developing now (140), waiting and learning (172) and waiting then preserving (130), in $ million, with high-state probability 0.4 and high-state value 230.
Probability of the high state: 0.4, High-state intact value ($ million): 230
Constructed example: the chapter's hypothetical estuary (140 now, 8 interim, 120 later, 230 with p = 0.4, 50 otherwise); probabilities 0.2 and 0.6 and high values 150 and 310 are added.

Calculated values

Wait and learn V_P
172
Wait, then preserve
130
Develop now
140
Waiting minus developing now
32
Best choice
wait and learn

Waiting keeps both options: 8 + 0.4 x 230 + 0.6 x 120 = 172 million. A fixed commitment to preserve gives 8 + 0.4 x 230 + 0.6 x 50 = 130. Waiting to learn beats developing now by 32 million; committing to preserve after waiting would not beat developing now. The gap of 42 between the two waiting values is the value of being able to develop in the low state.

Worked steps

  1. V_P = 8 + 0.4(230) + 0.6(120) = 172
  2. Preserve = 8 + 0.4(230) + 0.6(50) = 130
  3. V_P - develop now = 172 - 140 = 32

Use the idea

Before an irreversible commitment, value the option to act after new information, and compare it with the cost of delay.

Where the conclusion applies

An informative study, a fixed later development value, and no partial reversibility. An uninformative study or a large delay cost removes the advantage.

Check your understanding: At p = 0.2 and a high-state value of 230, what are V_P and the fixed-preservation value?
V_P = 8 + 0.2(230) + 0.8(120) = 8 + 46 + 96 = 150 > 140; fixed = 8 + 46 + 0.8(50) = 94.

Chapter 82 source: section "Arrow-Fisher quasi-option value".