ModelCensusopen-source ai reliability harness
The finding · Class 4 · Reasoning & Calculation

Real context moved 0.0pp on Dimensional-Analysis Failure.

~intervals overlap — not separated from no change

The model produces a quantity with inconsistent or wrong units, or mixes unit systems, yielding a dimensionally invalid result. The rule

google/gemini-2.5-flashfmi_4_45 repeatsfmi_4_4_v1
In one frame
screenshot this
Dimensional-Analysis Failurefmi_4_4
google/gemini-2.5-flash
Same question, both times
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
No contextPASS

First, let's calculate the volume of water that fills the tank. Rate of filling = 3 L/min Time of filling = 4 minutes Volume filled = Rate of filling × Time of filling Volume filled = 3 L/min × 4 min = 12 L Next, let's calculate the volume of water that drains from the tank. Rate of draining = 500 mL/s Time of draining = 10 seconds We need to convert the draining rate from mL/s to L/s or L/min to be consistent with the filling rate. Since 1 L = 1000 mL, we can convert 500 mL/s to L/s: 500 mL/s = 500/1000 L/s = 0.5 L/s Now, calculate the volume drained: Volume drained = Rate of draining × Time of draining Volume drained = 0.5 L/s × 10 s = 5 L Finally, to find the net litres in the tank, subtract the volume drained from the volume filled. Net litres = Volume filled - Volume drained Net litres = 12 L - 5 L = 7 L The final answer is $\boxed{7 L}$.

detector fmi_4_4_v1: correct unit L

General contextPASS

The tank fills at a rate of 3 L/min for 4 minutes. Volume filled = Rate of filling × Time = 3 L/min × 4 min = 12 L. The tank drains at a rate of 500 mL/s for 10 seconds. First, convert the draining rate to L/s: 500 mL/s = 500/1000 L/s = 0.5 L/s. Volume drained = Rate of draining × Time = 0.5 L/s × 10 s = 5 L. Net volume in the tank = Volume filled - Volume drained Net volume = 12 L - 5 L = 7 L. The retrieved passage is irrelevant to the problem. Answer: 7 L

detector fmi_4_4_v1: correct unit L

Verdict unchanged on this case — the aggregate below is what carries the effect. Across 5 repeats: 0.0pp (intervals overlap).fmi_4_4_v1 · google/gemini-2.5-flash · 2026-08-15 · ground truth as of 2026-08-09

loop 2026-08-15 · google/gemini-2.5-flash as served · closed weights · released 2025-06 · set s1.2-live · cases v1.4 · conditions v1.0 · ground truth as of 2026-08-09

From no context to real context
5 repeats per case
failure rate · left is fewer failures0%5%10%15%20%No context0%[0%7%] n=50Irrelevant context0%presence — a document being there[0%7%] n=500.0pp OVERLAPSGeneral context0%content — the information itself[0%7%] n=500.0pp OVERLAPS
whiskers are 95% intervals · each step is measured from the stop before it, not from zero
Try the other contexts
stored · no model call
1Questionheld fixed
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?

expected as of 2026-08-09

12 L in, 5 L out, net 7 L. The mixed units are the trap.

2Contextchange me

Nothing supplied. The baseline.

3ResponseNo context
PASS

First, let's calculate the volume of water that fills the tank. Rate of filling = 3 L/min Time of filling = 4 minutes Volume filled = Rate of filling × Time of filling Volume filled = 3 L/min × 4 min = 12 L Next, let's calculate the volume of water that drains from the tank. Rate of draining = 500 mL/s Time of draining = 10 seconds We need to convert the draining rate from mL/s to L/s or L/min to be consistent with the filling rate. Since 1 L = 1000 mL, we can convert 500 mL/s to L/s: 500 mL/s = 500/1000 L/s = 0.5 L/s Now, calculate the volume drained: Volume drained = Rate of draining × Time of draining Volume drained = 0.5 L/s × 10 s = 5 L Finally, to find the net litres in the tank, subtract the volume drained from the volume filled. Net litres = Volume filled - Volume drained Net litres = 12 L - 5 L = 7 L The final answer is $\boxed{7 L}$.

fmi_4_4_v1

detector fmi_4_4_v1: correct unit L

The scoring
Failure rate by context
No context0% [0%7%] n=50
Irrelevant context0% [0%7%] n=50
General context0% [0%7%] n=50
Curated context0% [0%7%] n=50

0–20% · whisker = 95% interval

Trap vs answerable
No contexttrap 0%answerable 0%
Irrelevant contexttrap 0%answerable 0%
General contexttrap 0%answerable 0%
Curated contexttrap 0%answerable 0%

never pooled · a trap ratio would skew the delta

Residual — what survives

Unchanged by context: 0% with none, 0% with the strongest supplied source. For a control mode that flat line is the result — it is what licenses reading movement elsewhere in this loop as grounding rather than as a document being present. Nothing survives at the best condition on the cases measured here.

Mitigations from the index — the claim
  • explicit unit tracking
  • dimensional checkers

fmi_4_4_v1 · 5 repeats per case · best context general context · rollups + residuals committed · probe log outside git · manifest hash ties them · method