How Far Ahead Is Far Enough

Demand for something you make grows, year on year, and will not stop. You meet it by building plant. Plant comes in a few fixed sizes, and bigger is cheaper per unit, so there is a real argument for overbuilding now and coasting. Against that: money spent later is worth less, so there is an equally real argument for building the smallest thing that gets you through the year.

You have to build something, this year. How much of the future do you have to work out before you can know what it is?

The lower panel is the whole subject. It plots the size you would build now against how far ahead you let yourself look. For short horizons it jumps about — of course it does, there is no sense buying plant you will never use. Then it locks, and looking further changes nothing.

Past that point the future stops mattering to today’s decision. A finite calculation has settled a question about an infinite one.

The dilemma, stated by the man who solved it

One of the planner’s greatest dilemmas is how long a horizon time to select so that end of horizon effects do not distort the first capacity deployment decision. It is of course this first decision which is the only one implemented and therefore the critical one.

— Robert L. Smith, Planning Horizons for the Deterministic Capacity Problem, 1981

Pick a horizon too short and you are answering a different question: one where the world ends in year ten and there is no point owning anything that lasts longer. Pick one too long and you are burning effort forecasting a decade whose details cannot change what you do this afternoon.

Notice on the panel that the decision does not have to settle tidily. Bend the demand curve away from a straight line and the first build can rise, drop back and rise again before it locks. That is why “look a good long way ahead” is not a substitute for knowing when you are allowed to stop looking.

Where it comes from

The model goes back to Alan Manne: capacity for steadily growing demand over an unlimited horizon, weighing economies of scale in big increments against the value of deferring the spend. It is a standard shape in process industries, communications and electric power — anywhere plant is lumpy, durable and expensive.

Two papers by Robert Smith and James Bean at Michigan carry it further:

• Smith (1981), Planning Horizons for the Deterministic Capacity Problem, solves the problem for all horizon times by solving a single finite one, and bounds the first expansion as an explicit function of horizon length. Facility sizes there are continuous.
• Bean and Smith (1985), Optimal Capacity Expansion Over an Infinite Horizon, Management Science 31(12), 1523–1532, takes the sizes from a finite menu of durable facilities. The continuous and discrete faces of one question.

There is also a stochastic companion with Julia Higle, and it has the loveliest result of the three: when demand is genuinely uncertain, the problem collapses back to a deterministic one. Replace the random demand by its trend, and discount at a lower rate, roughly in proportion to how uncertain things are. Uncertainty in, patience out. Not knowing what is coming makes you behave as though you cared more about the future.

My part in it

The 1985 result is Bean and Smith’s. The program that ran it was mine — Fortran, on an IBM 360, around 1980. That is the whole of my contribution and I want it stated exactly: I implemented their algorithm, I am not an author of it.

They were both later my thesis advisors.

The machine is worth a sentence. An algorithm whose entire point is knowing when it may stop computing, first run on hardware where computing was expensive enough that you felt every extra year of horizon. The code is long gone. What you are running above is the same act, forty-odd years on, in a browser tab, fast enough to redraw while you drag a slider.

Why an optimum exists at all

Before you can ask which infinite policy is best you have to know that a best one exists. As I remember the argument:

• Discounting makes the cost of every policy a convergent sum, so each has a finite value.
• Put a metric on the space of policies that discounts disagreements further out. Under it the space is bounded — and compact.
• Cost is continuous in that metric.
• A continuous function on a compact space attains its minimum. So an optimal policy exists.

(Recalled rather than checked against the paper — read it as the shape of the argument, not a transcription.)

The discounting is doing two jobs at once, and they are the same job. It makes the sum converge, and it makes the distant future negligible enough to measure with. That second one is why a finite horizon can settle the first decision: if the far future barely registers in the total, it cannot swing what you do today.

What this page actually computes

Honesty about the model. What runs above is the simplest thing that shows the phenomenon, not Bean and Smith’s algorithm:

• Demand is $g\,t^{\,p}$. Capacity is bought from the menu 1, 2, 3, 5, 8.
• Building size $s$ at time $t$ costs $s^{\alpha}e^{-rt}$ in present value, with $\alpha < 1$ for economies of scale.
• Plant is bought exactly when existing capacity binds. Costs are discounted, so buying earlier than you must is never worth it — and buying later is not allowed.
• The finite-horizon problem is a dynamic program over standing capacity, which is all the state you need: how much capacity you have tells you both what demand you face and what the clock says.

At $p = 1$ the problem looks identical from every capacity level, so the optimal infinite policy just repeats one size forever and there is nothing for a horizon to settle. Bending the demand breaks that self-similarity, which is why the curvature slider is there.

And that is the gap between this page and the paper. Here you can watch the first decision settle for one tidy family of models. Bean and Smith prove you can always find such a horizon, and that its first decision is optimal for the infinite problem, without leaning on any of the structure above.