Issue 024 - Supply chains - Backlog recovery
How long does a three-day factory shutdown echo?
Reuters reported that a magnitude 7.1 earthquake in the Kumamoto region disrupted auto and semiconductor production. Toyota suspended three Fukuoka plants, while Sony and other semiconductor facilities halted operations for safety checks, damage assessment, or phased restarts.
The problem
Estimate the total value of automotive and semiconductor production delayed during the first three days after the earthquake.
Then estimate how long manufacturers would need to clear that backlog while also meeting new demand.
Use the result to answer: how can a factory shutdown lasting only a few days disrupt downstream supply chains for much longer?
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
This one does not feel like a traditional Fermi problem, because we are not dealing with very large or very small figures. But if I think about this in the abstract, the key is capacity.
I believe most manufacturing firms want to operate near full capacity, but not quite at 100% capacity. Some flexibility is needed or the system becomes too brittle. I assumed most semiconductor and auto manufacturers, especially in Japan, are targeting and usually operating somewhere near 97% capacity.
Use a manufacturing plant with capacity to make 100 units per day as an example. If it ordinarily operates at 97% capacity, then in 3 days it would normally manufacture:
normal output ~= 100 units/day x 0.97 x 3 days
~= 291 unitsA 3-day stoppage means a backlog of about 291 units.
Now suppose that, to catch up, the plant increases from 97% to 99% of capacity. That creates only 2 extra units per day beyond normal demand:
catch-up capacity ~= 99 units/day - 97 units/day
~= 2 units/day
catch-up time ~= 291 units / 2 units/day
~= 146 daysThat is almost 5 months to catch up on a 3-day stoppage, assuming no facility improvements that increase overall capacity and no other interruptions.
Calibration Score
Matt's Calibration Score: 75 / 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 capacity-utilization peg may have been high, but it was plausible.
Model: 30/30. Backlog divided by spare recovery capacity is the right supply-chain model.
Math: 10/10. The arithmetic was clean.
Result: 15/30. The result is scenario-dependent, so the count lands between close and order-of-magnitude rather than exact.
Grounding facts
The counterintuitive part is that the recovery time is controlled by the spare margin, not by the shutdown length alone. A plant running at 95% has only 5% of capacity left for catch-up unless it can add overtime, add shifts, outsource, pull from inventory, or skip some normal demand.
Semiconductor plants add a second complication: "production delayed" is not always the same as "finished units delayed." Inspections, wafer holds, contamination checks, recalibration, and scrapped work-in-progress can make the calendar impact larger than the number of dark factory days.
After checking sources
Check and recalibrate
Matt's abstract capacity model is the most important part of the problem. The production-value estimate gives scale, but the backlog equation explains the long tail.
Start with Toyota Motor Kyushu. If annual capacity is about 430,000 vehicles and we use 300 production days per year, then:
vehicles per day ~= 4.3 x 10^5 / 3 x 10^2
~= 1.4 x 10^3 vehicles/day
3-day delayed vehicles ~= 4.3 x 10^3 vehiclesIf those are Lexus-scale vehicles and the wholesale/factory value is roughly $40,000 to $50,000 each, then vehicle assembly alone is:
vehicle value delayed ~= (4.3 x 10^3) x ($4 x 10^4 to $5 x 10^4)
~= $1.7 x 10^8 to $2.2 x 10^8Kanda and Kokura add engine and hybrid-unit output. If their combined daily capacity is roughly 3,000 units/day and the factory value per unit is a few thousand dollars, then a 3-day delay adds perhaps another $2 x 10^7 to $5 x 10^7. A Toyota-only central estimate is therefore roughly $2 x 10^8 to $3 x 10^8 of delayed production value for three days, before counting Nissan, Honda, suppliers, logistics, and unfinished work.
Semiconductors are harder because reports distinguish inspections, suspended production, phased restarts, and unknown damage. A wafer-fab shortcut gives the right order of magnitude. A large 12-inch fab doing about 50,000 to 100,000 wafers per month is producing:
wafers per day ~= (5 x 10^4 to 1 x 10^5 wafers/month) / 30
~= 1.7 x 10^3 to 3.3 x 10^3 wafers/dayIf processed wafer value is roughly $5,000 to $10,000 each, then one big fab's delayed output is:
daily wafer value ~= (2 x 10^3 wafers/day) x ($5 x 10^3 to $1 x 10^4)
~= $1 x 10^7 to $2 x 10^7/day
3-day wafer value ~= $3 x 10^7 to $6 x 10^7 per major fabFor Sony image sensors, a unit shortcut can be higher: millions of sensors per day times a few dollars to tens of dollars per sensor can easily land in the tens of millions of dollars per day. A reasonable Fermi range for affected semiconductor output across one major halted site plus inspections or partial slowdowns elsewhere is about $1 x 10^8 to $5 x 10^8 over three days.
Putting autos and semiconductors together:
auto delayed value ~= $2 x 10^8 to $3 x 10^8
semiconductor delayed value ~= $1 x 10^8 to $5 x 10^8
combined 3-day delayed value ~= $3 x 10^8 to $8 x 10^8A good central estimate is about $5 x 10^8, or a few hundred million dollars of delayed production value. The plausible range is wide enough that $1 billion is not absurd if more plants are included or high-value work-in-progress is scrapped.
Now comes the important backlog math. If normal demand uses fraction u of capacity and the plant can recover at 100% of capacity, the spare catch-up fraction is 1 - u. So:
catch-up time ~= shutdown days x u / (1 - u)For three lost days:
if u = 0.90: catch-up ~= 3 x 0.90 / 0.10 ~= 27 days
if u = 0.95: catch-up ~= 3 x 0.95 / 0.05 ~= 57 days
if u = 0.97: catch-up ~= 3 x 0.97 / 0.03 ~= 97 daysMatt's 146-day result comes from a slightly different recovery assumption: instead of going from 97% to 100%, he assumed the plant only rises from 97% to 99%, leaving just 2 percentage points of spare recovery capacity. That is a perfectly coherent scenario:
catch-up ~= 3 days x 0.97 / (0.99 - 0.97)
~= 146 daysThe best approximation is therefore not one number. If plants normally run around 90% to 95% and can add overtime, the backlog may clear in a few weeks to two months. If they run closer to 97%, have fragile supplier constraints, or need inspections/recalibration before ramping, a few days of lost output can echo for three to five months.
Post-check reflection
Matt's reflection
The way I handled the problem was fine. My assumptions about percent capacity were probably a bit high. My guess about time to resolve the backlog is fine for the capacity numbers I used, but could be lowered to a few weeks if the standard operating percent of capacity is lower.
This all makes sense to me and is consistent with my prior worldview. I already had an appreciation for how a stoppage of a few days could have consequences that last months.
Recommended memory peg
Remember the backlog shortcut: catch-up time ~= lost production / spare recovery capacity. If a factory loses 3 days while normally running at 95% of capacity, full-speed recovery gives roughly 3 x 0.95 / 0.05 ~= 60 days of catch-up time.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.