Macrodata Refinement Production
Severance: S1, E2 “Half Loop”
Production and Costs: Production Functions
Helly R is the newest hire on Lumon’s severed floor. As part of her training, she begins to learn what her new department is up to. Macrodata Refinement has four computers and Helly makes four refiners. Mark S describes what happens at each workstation.
“Okay, this is the Siena file. Now, all the data you see falls into one of four essential categories. And we group each line of code and then sort it evenly between five digital buckets.”
The sorting process occurs as each refiner scans through numbers in a data file and organizes them according to the emotions the numbers elicit. The end goal? That’s unclear, but progress is measured by the completion of files like the Siena file and the Tumwater file. How can this production process be modeled? And what can the model tell us about why Lumon decides to employ four refiners and install four computers in MDR?
First, the notation. The variable Q (quantity) is the total number of files MDR completes each day, where decimal points measure percent progress towards the next completed file. The variable L (labor) is the number of refiners employed in MDR. The variable K (capital) is the number of computer stations. A production function describes the mathematical relationship between the inputs (L and K) and output (Q). We can build a production function for MDR step by step using a few observations.
Observation 1: a computer with no refiner seems to provide no additional production value and likewise for a refiner with no computer. This can be modeled as follows:
Q = min{K,L}
where in this function, the number of completed files equals whichever number is smaller between K and L.1 Note that with this function, if MDR has four computers and four refiners, adding another refiner keeps output constant at four. Same for another computer.2
Observation 2: the function as is greatly exaggerates the productivity of MDR. It says that with four refiners and four computers, four files are completed each day.
“My current file’s called ‘Tumwater’ … which I started 11 weeks back.” – Dylan G
Dylan G also mentions that he has the file 96% sorted, so 12 weeks is a realistic timeline for a refiner with a computer to finish a file. Assuming a refiner works on one file at once, the math works out to a combination of four refiners and four computers completing roughly 8% of a file each day. The function can be adapted for this production rate as follows:
Q = 0.02 × min{K,L}
where daily output is now scaled to 2 percent daily progress per refiner-computer pair.
Observation 3: the function as is suggests no collaboration between refiners. Each time a new refiner-computer pair is added, output increases by 0.02 files as if each pair is independent and equally productive no matter the size of MDR.
But in the previous episode, Dylan G makes it clear that the pairs are reliant on each other. He laments the possibility of Petey missing work again, since he needs Petey to process his almost-completed file. And outies falling ill, leading the refiners to be absent, indeed happens. Fellow refiners are helpful for completing files and time is of the essence.
“But why don’t we always finish the files?” – Helly R
“Cause they only keep so long … You know, we finish, on average, one in five files before they expire.” – Dylan G
The production function needs to reflect that as MDR grows, the boost to file completion isn’t the same with each additional refiner-computer pair. Each additional boost should be bigger than the last. A bigger department reduces the probability that files expire before completion.
In technical terms, the MDR production function should have increasing returns to scale: doubling each of the inputs more than doubles total output. This can be accomplished with the following update:
Q = 0.005 × [min{K,L}]1.5
where the exponent being greater than one creates increasing returns to scale and the multiplier in front is reduced to maintain a realistic output with four refiners and four computers.
Imagine that Ms. Cobel, Mr. Milchick, and The Board are considering how large of a department MDR should be. They are thinking in terms of file progress. How much would they like to be completed each day? 2%? 4%? 10%? We can visualize what these outputs would require in inputs in the following figure.
Each line is an isoquant: a line connecting all input combinations that produce the same output. The L-shaped isoquants here are distinctive of inputs that are perfect complements, as refiners and computers are.
Moving up and to the right, the isoquants get closer and closer. It takes smaller and smaller increases in refiner-computer pairs to get equal size boosts in output. That’s the result of a production function with increasing returns to scale.3
And what do the decision-makers conclude? That’s a matter of cost.4 More daily progress means more refiners and more computers. And that means more cost each day. But one thing is clear. They will design MDR with an equal number of refiners and computers. Additional refiners added to the department with no additional computers adds cost without speeding progress. Same for adding computers with no additional refiners.
Given that MDR has four refiners and four computers, the task is clear: 4% daily progress.
“If we hit our numbers by quarter’s end, one of us gets named refiner of the quarter, and that shit gets you a waffle party.”
1 This is a Leontief production function and is used for inputs that are perfect complements.
2 In other words, the marginal product of a 5th refiner (with the number of computers held at 4) is 0. The marginal product of labor measures the additional output from adding an additional worker, holding constant the level of capital. With differentiable production functions, this can be found by taking the partial derivative of Q with respect to L.
3 One could also visualize this by showing isoquants for whole number pairs of refiners and computers. One of each completes 0.5% of a file each day. Two of each completes 1.4% each day. Three of each completes 2.6%. Four of each completes 4%. The isoquants are now similarly spaced, but their associated quantities increase at an increasing rate as you move up and to the right. There are infinitely many isoquants, since any combination of labor and capital is on one.
4 We could further this model by adding isocost lines which connect all input combinations with the same cost. The problem is then solved as a cost-minimization problem for a given level of output. With isoquants that are smoothed curves, the cost-minimizing combination occurs where an isocost line is tangent to the chosen isoquant line. Or more formally, where the marginal rate of technical substitution equals to the ratio of input prices.
More from Severance:
¹ This is a Leontief production function and is used for inputs that are perfect complements.
2 In other words, the marginal product of a 5th refiner (with the number of computers held at 4) is 0. The marginal product of labor measures the additional output from adding an additional worker, holding constant the level of capital. With differentiable production functions, this can be found by taking the partial derivative of Q with respect to L.
3 One could also visualize this by showing isoquants for whole number pairs of refiners and computers. One of each completes 0.5% of a file each day. Two of each completes 1.4% each day. Three of each completes 2.6%. Four of each completes 4%. The isoquants are now similarly spaced, but their associated quantities increase at an increasing rate as you move up and to the right. There are infinitely many isoquants, since any combination of labor and capital is on one.
4 We could further this model by adding isocost lines which connect all input combinations with the same cost. The problem is then solved as a cost-minimization problem for a given level of output. With isoquants that are smoothed curves, the cost-minimizing combination occurs where an isocost line is tangent to the chosen isoquant line. Or more formally, where the marginal rate of technical substitution equals to the ratio of input prices.