Issue 010 - Public administration - Operational throughput
How many officer-days does one immigration arrest take?
Axios reported on an ACLU report describing a broadened U.S. deportation apparatus, including a claimed 50,000-person enforcement network. AP reported that ICE arrests continued at high levels, including 10,000 arrests over five days in late June. Convert those headline totals into labor throughput.
The problem
Estimate how many officer-hours are required per immigration arrest under a large-scale enforcement campaign.
If a deportation force produces 10,000 arrests in five days, is that operationally closer to one arrest per officer-day, one arrest per ten officer-days, or one arrest per hundred officer-days?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
There are not a lot of assumptions to be made for the basic version because the critical numbers are given in the scenario.
If 50,000 officers are responsible for 10,000 arrests in 5 days, then those arrests used 250,000 officer-days.
officer-days ~= 5.0 x 10^4 officers x 5 days
~= 2.5 x 10^5 officer-days
officer-days per arrest ~= (2.5 x 10^5 officer-days) / (1.0 x 10^4 arrests)
~= 25 officer-days/arrestI think this is a high-tempo period for those officers' activities, so the arrest rate is likely high compared with other times.
Second, if the ICE structure works anything like what I am familiar with in the military, the administrative-role-to-field-worker ratio might be something like 5 or 10 to 1. That would mean 5,000 to 10,000 people are the front-line personnel directly involved in arrests.
Using the larger field-personnel number:
field officer-days ~= 1.0 x 10^4 field personnel x 5 days
~= 5.0 x 10^4 field officer-days
field officer-days per arrest ~= (5.0 x 10^4 officer-days) / (1.0 x 10^4 arrests)
~= 5 officer-days/arrestThat is closer to my intuition for what a high-tempo field-agent experience should look like, and it also emphasizes how administratively demanding the process of immigration enforcement can be.
Calibration Score
Matt's Calibration Score: 90 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 20/30. The denominator definitions were uncertain, but the supplied anchors were used well.
Model: 30/30. Officer-days per arrest is exactly the right throughput model.
Math: 10/10. The arithmetic was clean.
Result: 30/30. The corrected bracket closely matched the first pass.
Grounding facts
Ten thousand arrests in five days equals about 2,000 arrests per day. With a 50,000-person denominator, that is only 0.04 arrests per person per day, or one arrest for every 25 person-days. With a 10,000-person denominator, it is 0.2 arrests per person per day, or one arrest for every 5 person-days.
That is why the denominator matters so much. The same arrest total can look like rapid field throughput or a large administrative system, depending on what labor is included.
After checking sources
Check and recalibrate
The arithmetic is straightforward. The key uncertainty is definitional: does the denominator mean every person in the reported enforcement network, only people assigned to the surge, or only front-line field teams?
full-network scenario:
officer-days per arrest ~= (5.0 x 10^4 personnel x 5 days) / (1.0 x 10^4 arrests)
~= 25 officer-days/arrestIf one officer-day is 8 to 10 hours, the full-network scenario is:
officer-hours per arrest ~= 25 officer-days x 8 to 10 hours/day
~= 2.0 x 10^2 to 2.5 x 10^2 officer-hours/arrestThat is closest to the one arrest per ten officer-days option, though it is drifting toward the hundred-officer-day scale.
If only 10,000 to 20,000 people are directly involved in arrest operations and immediate support, the result falls sharply:
partial-force scenario:
(1.0 x 10^4 to 2.0 x 10^4 personnel) x 5 days / (1.0 x 10^4 arrests)
~= 5 to 10 officer-days/arrest
~= 40 to 100 officer-hours/arrest at 8 to 10 hours/dayIf only a few thousand field officers are the arrest teams and everyone else is support infrastructure, the narrow field-team rate can approach one or a few officer-days per arrest:
field-team scenario:
(2.0 x 10^3 to 5.0 x 10^3 field officers) x 5 days / (1.0 x 10^4 arrests)
~= 1 to 2.5 field officer-days/arrestThe best Fermi answer is therefore not one number but a bracket: about 25 officer-days per arrest for the full reported network, about 5 to 10 officer-days per arrest for a partial-force denominator, and about 1 to 3 field officer-days per arrest if only arrest teams are counted.
That makes the headline operationally ambiguous. It can describe both a very high-volume enforcement surge and a process that still consumes many person-hours per arrest once support, transport, detention, legal processing, and administration are included.
Post-check reflection
Matt's reflection
I did not have any major errors working through this one. The corrected number was closely aligned with my answer.
Regarding the news item, I do not think this deeper dive tells me much that I was not already aware of. It does suggest that there is a lot of effort entailed in every arrest.
Recommended memory peg
Remember that throughput = outcomes / resource-time, so resource-days per outcome = resource count x days / outcomes. For labor estimates, an officer-day is often usefully converted with 8 to 10 hours per day.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.