Common-pool resources, expanded in the discount factor

Common-pool resources are a powerful and frequent model in economics. The idea is that nobody can be kept away from the pool, and every unit one person takes is a unit nobody else can. Fish stocks are the textbook case. So are groundwater basins, grazing land, road space at rush hour or the atmosphere, considered as a place to put carbon: no one can be excluded from using it, and every tonne emitted uses up part of a finite budget.

What follows are notes on the general model and then I apply them to the example of fishery. I wrote about the same fishery earlier but this time it is all math.

The commons

A common-pool problem needs very little to write down. There is a stock $s_t$ — fish in the sea, water in the aquifer, room left in the carbon budget. Several users draw on it; user $j$ takes $b_j$, and what happens to the pool depends only on the total draw

Drawing pays. Write $u_{j,t}$ for what user $j$ earns in season $t$, and $u(B_t, s_t)$ for what they earn between them — a season's payoff depends on how hard the pool is being worked and on how much is in it. What makes this a problem rather than a list is that the stock remembers: whatever is left, plus whatever grows back on its own, is the starting point for next season,

Nothing in the derivation depends on the shape of $u$ or $\sigma$, which is the whole reason for writing them this way. Users also care about seasons other than this one, and weight the future geometrically:

A geometric weight in time, exactly like a damping factor. $\beta$ near 1 is a patient user, $\beta = 0$ one who does not think about next season at all.

Three conventions, all to save trouble later. A subscript is always a season — $B_t$, $s_t$ — and the user label $j$ drops away almost at once, because from the planner's point of view only the total matters. Derivatives are written $\partial$: since $u$ and $\sigma$ each take two arguments, a derivative has to say which one it acts on, so $\partial_B u$, $\partial_s u$, $\partial_{BB} u$ — never a prime, never a subscript, the subscript slot being spoken for. And $\big(\partial_B u\big)_t$ means that derivative evaluated in season $t$, which keeps the subscript a season even there.

The third is vocabulary. These are the quantities economists call marginal, and the word is less forbidding than it looks: it is simply their way of saying the derivative of. What it does not say on its own is with respect to what — that lives in the phrase that follows. The noun names what gets differentiated and the "of …" names what it is differentiated by, so $\partial_B u$ is the marginal payoff of one more unit drawn and $\partial_s u$ the marginal payoff of one more unit left in the pool. The "of" and the subscript on $\partial$ carry exactly the same information; the notation is just the more honest of the two, since it can never quietly leave the "of" out.

Independent actors

So far $u(B,s)$ is what the pool pays out as a whole. To say what happens when $N$ users draw on it independently, one thing has to be added: how that payout is divided. Take the obvious rule — each user gets the share matching how hard it worked,

That is the only new assumption in this section.

Each user picks $b_j$ to make its own $u_j$ as large as possible, taking everyone else's draw as given. Raising $b_j$ raises $B$ one for one, so

The $-u/B$ in the bracket is the dilution of your own share: raising $b_j$ raises $B$ too, which splits the pool more thinly. It comes from differentiating the share $b_j/B$, not from $u$.

That bracket is worth a name of its own. Write

for how much less one more unit of effort earns than the units already working — the damage a unit does. $D$ is positive whenever $u$ is concave in $B$ and pays nothing for no effort, $u(0,s) = 0$ — and those are the only properties of $u$ this section needs. Diminishing returns and crowding are not separate ingredients; they are two names for $D > 0$. With that, the condition reads

and if the $N$ users are alike they draw alike, so $b_j/B = 1/N$ and it becomes simply

That one line is the tragedy of the commons. Read it as an instruction: draw until what you collect equals the share of the damage you carry. You are paid $u/B$ for the extra unit — the full going rate, because your share grows with your effort. The damage it does is $D$, but $D$ lands on all $N$ users, so your own bill is $D/N$.

It is a dinner split $N$ ways. You eat the whole extra dish and pay a fraction of it, so you order too much — and so does everyone else, for exactly the same good reason.

The consequence follows at once. Raising $N$ shrinks $D/N$, so $u/B$ has to shrink to match, and $u/B$ only falls as $B$ grows. More users, more effort, less left over — with nobody behaving badly. Each is doing the best available thing given what the others do.

The two ends of that range are worth naming.

• $N = 1$ gives $u/B = D$, which is $\partial_B u = 0$: draw until one more unit adds nothing to the pool. You are dining alone and pay the whole bill, so you order exactly as much as you want.
• $N \to \infty$ sends $D/N$ to zero, so $u/B = 0$ and $u = 0$. Your share of the bill has vanished, and the pool ends up yielding exactly what it costs to work it — no more.

The planner

The first of those limits is worth dwelling on, because it is the benchmark the rest of this post measures against. A planner is usually introduced as somebody from outside — a manager, a regulator, an authority with better information. None of that is needed. The planner is simply $N = 1$: the same first-order condition with the externality switched off.

And he will always try to maximise the total discounted payoff:

with $s_0$ given. Working this out in general is analytically not possible but likely we have a good perturbation parameter in $\beta \ll 1$, with both the fleet and the payoff as polynomials:

Zeroth order

At $\beta = 0$ only $t = 0$ survives and the planner maximises this season alone:

The second equation is the myopic rule — the best fleet if next year did not exist — and it defines a policy $B^{(0)}(s)$, one fleet for each stock. That is the entire zeroth order.

It is worth seeing how far that is from what the independent actors do. Their condition was $u/B = D/N$ with $D = u/B - \partial_B u$; eliminating $D$ between the two gives

which is negative for every $N > 1$. They stop at a point where the pool's total payoff is already falling — past the top of the hill, on the way down. The planner, at $\partial_B u = 0$, stops exactly on the summit. That is the precise sense in which it does better, and the gap between the two is the whole of the tragedy.

Keep the condition $\partial_B u = 0$ in view, though. It is about to do a great deal of work: at every order from here on, some term arrives multiplied by it and vanishes.

First order

Two things feed the $\beta^{1}$ coefficient: season 1 at zeroth order, and the first-order correction to season 0's fleet. Expanding $u$ around $B_0^{(0)}$,

and the first term vanishes — its bracket is the zeroth-order condition. The correction to today's fleet drops out of the first-order payoff.

What survives is the same one-season problem, one season later. Maximising over $B_1^{(0)}$ returns the myopic rule at the new stock:

So through first order

the same expression twice, one season apart.

But notice what we have not got. $B_0^{(1)}$, the correction to today's fleet and the thing we actually want, has cancelled. It appears at second order.

Second order

Three seasons now contribute, season $t$ entering at boat-order $2-t$. Two of the three collapse on sight.

Season 0 contributes its second-order Taylor remainder. The $B_0^{(2)}$ piece arrives multiplied by $\big(\partial_B u\big)_0 = 0$ and dies exactly as $B_0^{(1)}$ did one order ago, leaving only the quadratic $\tfrac{1}{2}\big(\partial_{BB} u\big)_0\big(B_0^{(1)}\big)^2$.

Season 1 carries an explicit $\beta$, so it reaches $\beta^2$ through its own first-order parts. Its $B_1^{(1)}$ piece is multiplied by $\big(\partial_B u\big)_1$ — zero for the same reason. What survives is the dependence on the stock, because $s_1$ is not a constant: it inherits a $\beta$-dependence from $B_0$, namely $s_1^{(1)} = \big(\partial_B \sigma\big)_0\,B_0^{(1)}$.

Season 2 contributes a fresh myopic season. Altogether

$B_0^{(1)}$ now appears twice, and in two structurally different ways: quadratically through season 0, linearly through season 1. That pairing is what makes the problem well posed. A downward parabola plus a line has one interior maximum — with only the line the correction would run away, and with only the parabola it would sit at zero and patience would never move anything. The two terms are the cost of leaving the myopic optimum and the reward for the stock that buys.

Setting $\partial U^{(2)} / \partial B_0^{(1)} = 0$:

That is the whole result, and it is three derivatives of two functions. How hard the control pushes the state, what a richer state is worth next season, and how sharply this season's profit falls away from its peak — nothing else about a problem matters at this order.

Note what $\big(\partial_s u\big)_1$ is: the marginal value of one more unit left in the pool — how much extra next season earns from starting slightly fuller. It is an exchange rate between the stock and money, and that is precisely the number you need in order to weigh a unit taken today against a unit left for tomorrow.

Since $\partial_{BB} u < 0$, the sign of the correction is the sign of $\big(\partial_B \sigma\big)_0\big(\partial_s u\big)_1$: the control moves toward whatever raises tomorrow's value. Both factors can carry either sign, and the two examples at the end of this post are chosen so that each carries the opposite one.

Sustainability

Sustainability sounds like a test a policy passes or fails. In this model it is better read as a destination: a stock is steady when the draw exactly matches what grows back,

That is not one condition but a family of them — for every level of effort, a stock at which the pool stops moving. Draw too hard and the stock falls until the two balance; too gently and it rises until they do. Leach works his whole dynamic chapter from this curve, and the useful fact is that every planner ends up somewhere on it (§8.2 of A Course in Public Economics). So the question is never whether a policy is sustainable in the long run, only which resting point it reaches.

That is what $\beta$ decides. At $\beta = 0$ the planner applies the myopic rule wherever it finds itself and drifts to the resting point that rule implies — sustainable, and poor. As $\beta \to 1$ the journey stops counting against an infinite stay at the destination, so the planner simply picks the resting point with the largest payoff.

Where it lands in between is a solved problem, and the answer is worth having even without the derivation. The move, due to Clark and Munro, is to treat the stock as capital: a fish left in the water is an investment like any other, and the planner holds the pool at the size where the resource's own rate of return matches the discount rate — exactly the rule you would apply to a bond or a factory. Everything in the worked example below is a special case of it. The derivation is optimal control and it is done properly in Colin Clark's Mathematical Bioeconomics; the original argument is Clark and Munro, The economics of fishing and modern capital theory, Journal of Environmental Economics and Management 2 (1975).

Example: Fishery

To put numbers on any of this we need an instance, and the fishery is the one you can picture. John Leach sets it up in §8.1 of A Course in Public Economics, and I used it in an earlier post. Two islands send $b_1$ and $b_2$ boats onto the same ground, so $B_t$ is the fleet on the water and $s_t$ the fish stock. Three equations close it:

• Catch. $y_t = q\,s_t\sqrt{B_t}$ — more boats land more fish, but with diminishing returns, and a richer stock is easier to fish. The catchability $q$ says how good the gear is.
• Renewal. $g(s) = r\,s\,(1 - s/K)$ — the usual logistic regrowth, with intrinsic rate $r$ and carrying capacity $K$. Fastest at half of $K$, zero at both ends.
• What is left. $s_{t+1} = s_t + g(s_t) - y_t$ — this season's catch sets next season's starting point.

The catch is shared in proportion to boats and each boat costs $c$ to send, so island $j$ earns $u_j = b_j (v - c)$ in a season, where $v \equiv y/B = q\,s/\sqrt{B}$ is what one boat-trip is worth. It falls as the ground gets crowded, which is the only reason any of this is interesting. In the general notation of the last section, this fishery is the case

Throughout, $q = 0.01$, $c = 0.125$, $r = 0.03$, $K = 1000$ and $s_0 = 100$.

What the tragedy costs

Take the $N$-user condition first. With $u = q s\sqrt{B} - cB$ the two pieces it balances are

so the catch term appears at half strength in the second — that is what diminishing returns amount to here. Substituting into $\big(1-\tfrac1N\big)\tfrac{u}{B} + \tfrac1N \partial_B u = 0$, the $c$ terms combine to exactly $-c$ and the rest to $\tfrac{q s}{\sqrt B}\big(1-\tfrac{1}{2N}\big)$, giving the equilibrium fleet $B^{*}(N)$ — the total number of boats $N$ independent islands end up sending — in closed form:

Four times the fleet, for nothing at all. That gap is the tragedy of the commons, and it is the distance any institution has to close. Two islands already give away most of it — 36 boats is closer to open access than to the owner — which is worth knowing before assuming that a small group will sort itself out.

The derivatives the expansion needed

Everything in the general derivation ran on three derivatives, and for a fishery all three reduce to derivatives of the catch. Profit is the catch minus a linear cost, $u = y - cB$, so $\partial_B u = \partial_B y - c$ and $\partial_s u = \partial_s y$. The stock loses exactly what is landed, $\sigma = s + g(s) - y$, so $\partial_B \sigma = -\partial_B y$ — which at the myopic optimum, where $\partial_B y = c$, is exactly $-c$. One more boat costs the stock precisely $c$ worth of fish, and the general correction becomes $c\,\partial_s y / \partial_{BB} y$.

Here they are.

The zeroth order is the 16 boats and 2.00 promised at the start. One season of that fleet lands $y_0 = 4.00$ and leaves $s_1^{(0)} = 98.7$.

Notice what is missing from that last row. The renewal function never appears. $g$ enters the correction only by setting what $s_1^{(0)}$ is, and nowhere else. It is tempting to assume a patient planner is one who fishes below the growth rate; it is not. The planner is buying stock, not growth — a fuller sea is cheaper to fish next season, and that is the whole mechanism.

The curvature explains the size of the effect: $-0.0039$ is a very flat peak, so a small price on fish slides the fleet a long way. Concretely, it comes off the impatient 16 boats by about eight percent per unit of patience.

Where the fishery ends up

The expansion says which way the fleet moves; it says nothing about where the fishery settles. That was Sustainability's question, and for this fishery the two ends of the patience range are very far apart:

A quarter of the boats landing four times the fish, on a sea seven times fuller. That is what patience buys here, and it is not austerity — a fuller sea is cheaper to work, so more stock and more catch arrive together.

Solving the dynamic problem numerically shows the fishery walking there as $\beta$ rises: starting from $s_0 = 100$, the long-run stock comes out at 69.7, 89.7, 375.9 and 503.2 for $\beta = 0$, $0.9$, $0.99$ and $0.999$. Almost all of that journey happens in the last stretch of the patience range — which is exactly the ground a small-$\beta$ expansion cannot cover.