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
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
First, we calculate the total volume of water added to the tank while it is filling. The tank fills at a rate of 3 L/min for 4 minutes: \[ \text{Volume filled} = \text{Rate} \times \text{Time} = 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, we calculate the total volume of water drained from the tank. The tank drains at a rate of 500 mL/s for 10 seconds. First, we convert the draining rate to liters: \[ 500 \, \text{mL/s} = 0.5 \, \text{L/s} \] Now we calculate the volume drained: \[ \text{Volume drained} = \text{Rate} \times \text{Time} = 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \] Now we find the net volume in the tank by subtracting the volume drained from the volume filled: \[ \text{Net volume} = \text{Volume filled} - \text{Volume drained} = 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] Thus, the final result is: Answer: 7 L
detector fmi_4_4_v1: correct unit L
First, we calculate the volume of water that fills the tank: Filling rate = 3 L/min Time = 4 minutes Volume filled = Filling rate × Time Volume filled = 3 L/min × 4 min = 12 L Next, we calculate the volume of water that drains from the tank: Draining rate = 500 mL/s Time = 10 seconds Convert the draining rate to liters: 500 mL/s = 0.5 L/s Volume drained = Draining rate × Time Volume drained = 0.5 L/s × 10 s = 5 L Now, we find the net volume in the tank: Net volume = Volume filled - Volume drained Net volume = 12 L - 5 L = 7 L Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · openai/gpt-4o-mini as served · closed weights · released 2024-07 · set s1.1-live · cases v1.3 · conditions v1.0 · ground truth as of 2026-08-09
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.
Nothing supplied. The baseline.
First, we calculate the total volume of water added to the tank while it is filling. The tank fills at a rate of 3 L/min for 4 minutes: \[ \text{Volume filled} = \text{Rate} \times \text{Time} = 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, we calculate the total volume of water drained from the tank. The tank drains at a rate of 500 mL/s for 10 seconds. First, we convert the draining rate to liters: \[ 500 \, \text{mL/s} = 0.5 \, \text{L/s} \] Now we calculate the volume drained: \[ \text{Volume drained} = \text{Rate} \times \text{Time} = 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \] Now we find the net volume in the tank by subtracting the volume drained from the volume filled: \[ \text{Net volume} = \text{Volume filled} - \text{Volume drained} = 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] Thus, the final result is: Answer: 7 L
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–40% · whisker = 95% interval
never pooled · a trap ratio would skew the delta
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.
- 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