Issue 038 - Military logistics - Mass flow
How much logistics does a major military exercise consume?
The annual U.S.-South Korea Ulchi Freedom Shield exercise began this week with about 18,000 South Korean troops participating. It was originally scheduled to run for 11 days, but the countries agreed to shorten it to five days and reduce field training.
The problem
Estimate the total mass of food, drinking water, and fuel required to support 18,000 troops during the originally planned 11-day exercise.
Then estimate how much logistical effort is avoided by cutting six days from the exercise, expressed as supply-truck loads, total material mass, and person-days of military activity.
When governments describe a large military exercise as expensive, how much of the physical burden comes simply from sustaining thousands of people and vehicles in the field for days at a time?
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
I assumed it was safe to use a 2,500 calorie daily consumption by each soldier, which comes to about 2 kg of food per soldier per day. With 18,000 soldiers, that is about 3.6 x 10^4 kg of food per day of the exercise.
I assumed the soldiers would each drink about 3 L of water daily, or 3 kg. With 18,000 soldiers, that is about 5.4 x 10^4 kg of water per day of the exercise.
food/day ~= 18,000 troops x 2 kg/troop-day
~= 3.6 x 10^4 kg/day
water/day ~= 18,000 troops x 3 kg/troop-day
~= 5.4 x 10^4 kg/dayFor vehicles, I assumed the average military vehicle would be holding 4 soldiers. Troop carriers hold many more; some Humvees might have only two. So 18,000 soldiers at 4 per vehicle gives about 4,500 vehicles.
I assumed vehicles might idle or sit with the engine off while others drive great distances. I used an average of 12 miles per gallon fuel efficiency, about 5 km/L, and assumed an average of 20 km traveled per day. That gives 4 L used while in motion and an assumed additional 1 L idling, for a total of 5 L per day for each vehicle.
vehicles ~= 18,000 / 4
~= 4.5 x 10^3 vehicles
fuel/day ~= 4.5 x 10^3 vehicles x 5 L/vehicle-day
~= 2.3 x 10^4 L/dayI wrote 2.7 x 10^4 L/day at the time, but the multiplication is closer to 2.3 x 10^4 L/day.
Using those numbers, an 11-day exercise would require:
food ~= 3.6 x 10^4 kg/day x 11
~= 4 x 10^5 kg
water ~= 5.4 x 10^4 kg/day x 11
~= 5.9 x 10^5 kg
fuel ~= roughly 3 x 10^5 kgIf the exercise was cut by 6 days, that would save about 2.2 x 10^5 kg of food, 3.2 x 10^5 kg of water, and 1.4 x 10^5 kg of fuel.
That avoids moving a total of about 7 x 10^5 kg of materials, or 700 tons. Assuming a large cargo truck can easily transport about 1 metric ton of materials, that would save 700 truckloads of materials.
Calibration Score
Matt's Calibration Score: 70 / 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: 10/30. Food was reasonable, water was low, and truck-payload assumptions were debatable.
Model: 30/30. Troop-days times daily sustainment load was the right model.
Math: 0/10. There were minor fuel calculation slips.
Result: 30/30. The total avoided-mass conclusion was close enough for the problem.
Grounding facts
The six-day cut removes about 108,000 troop-days from the schedule. Even before ammunition, spare parts, medical support, communications gear, lodging, aircraft, ships, and administrative work, the basic sustainment mass is already in the thousand-ton range.
This is also why the phrase "military exercise" can hide very different physical realities. A command-post simulation, a short live-fire range event, and a dispersed field maneuver all have different fuel, water, transport, and maintenance profiles.
After checking sources
Check and recalibrate
Matt's person-days structure is the right spine. The hardest assumption is not the troop count; it is what fraction of those troops are actually being sustained like a field force on each day.
Start with the calendar math:
original troop-days ~= 18,000 x 11
~= 2.0 x 10^5 troop-days
avoided troop-days ~= 18,000 x 6
~= 1.1 x 10^5 troop-daysFor food, DLA says a case of 12 MREs weighs about 22 lb, or about 10 kg. Three MRE-equivalent meals are about one quarter of a case:
food package mass/day ~= 10 kg/case x 3 meals / 12 meals
~= 2.5 kg/troop-dayFor water, Matt's 3 L/day is a decent low drinking-only estimate, but field water planning can easily climb once you include heat, food preparation, hygiene, and safety margin. Use a central planning figure around 10 kg per troop-day for drinking plus simple field support water, with a range from 5 to 15 kg.
food + water ~= 2.5 + 10
~= 1.25 x 10^1 kg/troop-day
original food/water mass ~= 2.0 x 10^5 x 1.25 x 10^1
~= 2.5 x 10^6 kg
~= 2,500 tons
avoided food/water mass ~= 1.1 x 10^5 x 1.25 x 10^1
~= 1.4 x 10^6 kg
~= 1,400 tonsFuel is more scenario-dependent. UFS is partly a command-post exercise, and field training was being reduced, so do not assign every soldier a continuously operating vehicle. A rough mixed-support assumption might be 1,000 to 2,000 actively supported vehicles using 20 to 50 L/day each, plus generators and support equipment. That gives maybe:
fuel/day ~= 1,500 vehicles x 40 L/day
~= 6 x 10^4 L/day
fuel mass/day ~= 6 x 10^4 L/day x 0.8 kg/L
~= 5 x 10^4 kg/day
~= 50 tons/dayAcross the original 11 days:
original fuel mass ~= 50 tons/day x 11
~= 550 tons
avoided fuel mass ~= 50 tons/day x 6
~= 300 tonsPutting it together:
original total ~= 2,500 tons food/water + 550 tons fuel
~= 3,000 tons
avoided total ~= 1,400 tons food/water + 300 tons fuel
~= 1,700 tonsA reasonable answer is therefore around 3,000 tons of food, water, and fuel for the originally planned 11-day schedule, and around 1,500 to 2,000 tons avoided by cutting six days.
Truckloads depend on what you mean by a truck. Oshkosh lists modern FMTV payload variants around 2.5 to 8 tons, while older FMTV descriptions commonly use 2.5-ton and 5-ton classes. If all material could be packed ideally into 5-ton loads:
ideal avoided loads ~= 1,700 tons / 5 tons/load
~= 340 truckloadsBut food, water, and fuel move differently, in different containers, to different places, often below ideal payload. If the effective mixed-delivery load is closer to 1 to 3 tons, Matt's hundreds to about 1,000 truck movements is very plausible.
Post-check reflection
Matt's reflection
Looks like I made some minor calculation errors for fuel, but otherwise some decent assumptions. Actual vehicular need might have been lower, since many soldiers are in admin roles and not actively needing transport during the exercise. Actual water need was probably much higher: active soldiers might require more like 5 to 12 L per day instead of 3, and doctrine suggests multiples of the drinking volume need to be available for food prep and hygiene purposes. The number I finished on was probably correct.
There was some pushback on the number of vehicles necessary to transport that mass, because military cargo trucks may carry 2 to 5 tons. But those three materials are going to be transported in very different vehicle types, and we will not see ideal mass transport, so I bet my answer is closer to correct than the ideal calculation of a few hundred cargo trucks.
I do not know how much the added cost of extending the exercise by almost a week factors into the decision to cut it shorter. I think this is much more a political news item and those political motivators are the main thrust of the decision. That said, this sure does help frame the real-world scope of these exercises.
Recommended memory peg
For field-sustainment estimates, remember troop-days = troops x days, food is roughly 2 to 3 kg per troop-day with packaging, field water can be 5 to 15 kg per troop-day, and medium tactical truck payloads are often a few tons, not one perfect universal truckload.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.