Returns to Scale vs Experience

Shrinkage.md

Returns to Scale vs Experience

How many wristwatches are cheaper than Big Ben?

Abstract

Big machines are sometimes more efficient. But they cost more, so fewer can be produced with a finite budget. Small machines are cheaper and may benefit from improvement over time, driven by experience in building more units. When does this experience lead to greater overall efficiency? We derive an approximation which, given a learning rate, tells how much smaller a machine must be to overcome an initial efficiency disadvantage.

Background

Learning curves were characterized in the context of industrial production in the 1930s by Wright.¹ The production cost of a machine follows a power law in the number of units made so far

$$ \large{} Cn = C_1 n^{-a} \tag{1} $$

where C₁ is the cost of the first unit, C_n is the cost of the nth unit, and a is the learning rate.

Across diverse industries, Wright and others have documented cost reductions of around 20% per doubling in the number of units. This corresponds to a learning rate of nearly

$$ \large{} a = -\log_{2} 0.8 \approx 0.32 \tag{2} $$

and in the following we assume 0 < a < 1.

Notably, Moore's law may be a special case of Wright learning when unit production increases exponentially.²

Machines

A machine is a device that emits some commodity at a constant rate. We call this rate the machine's size. For any commodity of interest, there may be machines of different sizes

$$ \large{} U_b - \text{rate of output of a big machine} $$

$$ \large{} U_s - \text{rate of output of a small machine} $$

$$ \large{} U_b > U_s $$

Each machine must itself be manufactured at a one-time cost in dollars

$$ \large{} C_b - \text{cost for first unit of a big machine} $$

$$ \large{} C_s - \text{cost for first unit of a small machine} $$

$$ \large{} C_b > C_s $$

and its size is fixed at time of manufacture. Its efficiency can be defined by a cost per rate of commodity output

$$ \large{} E_b = \frac{C_b}{U_b} $$

$$ \large{} E_s = \frac{C_s}{U_s} $$

and we may assume larger machines have an initial advantage

$$ \large{} E_b < E_s $$

Finite Budgets

For any finite budget B and learning rate a, we can afford to manufacture N_b units of a big machine or N_s units of a small one, N_b < N_s

$$ \large{} B = \sum_{n=1}^{N_b} C_b n^{-a} = \sum_{m=1}^{N_s} C_s m^{-a} \tag{3} $$

Simplifying gives

$$ \large{} C_b \sum_{n=1}^{N_b} n^{-a} = C_s \sum_{m=1}^{N_s} m^{-a} \tag{4} $$

$$ \large{} \frac{C_b}{C_s} = \frac{H(N_s, a)}{H(N_b, a)} \tag{5} $$

where H(x, y) is the generalized harmonic number of order y of x.

In the limit as N_b goes to infinity, the right side of equation (5) admits the following approximation³

$$ \large{} \frac{H(N_s, a)}{H(N_b, a)} \approx \left(\frac{N_s}{N_b}\right)^{1-a} \tag{6} $$

This approximation seems quite good even at small orders, and is fine for our purposes since the learning rate a is statistical in nature anyway

Ns	Nb	Sum/Sum	(Ns/Nb)0.68
50	49	1.01	1.01
10	9	1.08	1.07
50	39	1.19	1.18
3	2	1.39	1.32
5	3	1.49	1.41
50	25	1.63	1.60

Substituting it into equation (5) gives

$$ \large{} \frac{C_b}{C_s} \approx \left(\frac{N_s}{N_b}\right)^{1-a} \tag{7} $$

From our definitions, we can further substitute the product of size and efficiency for initial cost

$$ \large{} \frac{U_b E_b}{U_s Es} \approx \left(\frac{N_s}{N_b}\right)^{1-a} \tag{8} $$

Wright Sizing

The total output of a fleet of machines is the product of the number built and their size, N_sU_s for small machines and N_bU_b for big ones. Setting these equal, we get

$$ \large{} \frac{N_s}{N_b} = \frac{U_b}{U_s} \tag{9} $$

Using this to substitute in (8) gives

$$ \large{} \frac{U_b E_b}{U_s Es} \approx \left(\frac{U_b}{U_s}\right)^{1-a} \tag{10} $$

$$ \large{} \left(\frac{U_b}{U_s}\right)^{a} \approx \frac{E_s}{E_b} \tag{11} $$

And finally

$$ \large{} a \log \frac{U_b}{U_s} \approx \log \frac{E_s}{E_b} \tag{12} $$

which relates the ratio of sizes to the ratio of initial efficiencies in the case where fleets of each machine type produce the same total output for the same total cost.

Nuclear Fission

This framework can be applied to the question of ideal reactor size in the nuclear energy industry. The United States operates a modest fleet of large reactors, which produce about U_b = 1 GW of electricity per unit. In recent years, it has been claimed that mass production of much smaller reactors may be cheaper in the long run.

Though the question of capital cost for reactors is murky, it appears large reactors can be built in some markets for nearly E_b = $4/W.⁴

On the other hand, a company called Oklo plans to build U_s = 1.5 MW reactors at an initial cost C_s = $10M, corresponding to an initial efficiency E_s = $6.7/W.⁵

Which type of reactor will produce more electricity when built into fleets of equal cost? We'll assume the learning rate from (2) for both types and solve (12) for U_s. If the solution is larger than Oklo's 1.5 MW, we may conclude the Oklo fleet will produce more electricity

$$ \large{} 0.32 \log \frac{1000 MW}{U_s} \approx \log \frac{6.7}{4} $$

$$ \large{} \log \frac{1000 MW}{U_s} \approx \frac{0.52}{0.32} $$

$$ \large{} U_s \approx \frac{1000 MW}{exp(0.52/0.32)} $$

$$ \large{} U_s \approx 200 MW $$

Open Questions

An initial efficiency advantage of a big machine must be due to efficiencies of scale, so it is not independent of size. Can this dependency be added here, to allow a closed-form solution for optimal machine size?

Acknowledgements

Thanks to Ed Pheil (@EdPheil) for many engaging discussions on Twitter.

References

Footnotes

1. Wright (1936) Factors Affecting the Cost of Airplanes ↩

2. Nagy et al (2013) Statistical Basis for Predicting Technological Progress ↩

3. Alexandre et al (2020) Limit of a ratio of harmonic numbers? ↩

4. Energy Technologies Institute (2018) The ETI Nuclear Cost Drivers Project: Summary Report ↩

5. Adams (2020) Oklo has filed first COLA with the NRC since 2009 ↩