Guide-linked quiz · Hard Level
FP8 and Low-Precision Formats Quiz
Practice FP8 exponent and mantissa choices, scaling, and hardware requirements, including why a smaller number format does not guarantee identical outputs.
Question 1 of 8
How many bits does an FP8 number use compared to a standard FP32 number?
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Answers and explanations appear as you take the quiz.
- How many bits does an FP8 number use compared to a standard FP32 number?
- 8 bits vs 16 bits (a half)
- 8 bits vs 64 bits (an eighth)
- 8 bits vs 32 bits (a quarter)
- 8 bits vs 8 bits (the same)
- What do the 'E4M3' and 'E5M2' labels describe about an FP8 format?
- The error rate and model size
- The number of exponent bits versus mantissa bits
- The clock speed and memory channel
- The training epochs and model layers
- Why do FP8 pipelines apply scaling factors to tensors?
- To encrypt the weights for security
- To increase the number of model layers
- To convert the model to a different language
- To keep values inside FP8's narrow representable range and avoid overflow/underflow
- Which trade-off is the main downside of using FP8?
- Reduced numerical precision (rounding errors)
- Higher memory usage
- Slower matrix multiplication
- Inability to run on any GPU
- Which NVIDIA GPU generation first added dedicated hardware support for FP8 matrix math?
- Pascal (P100)
- Hopper (H100)
- Kepler (K80)
- Maxwell (M40)
- What is the trend in precision formats moving beyond FP8?
- A return to 64-bit double precision for inference
- Eliminating floating point entirely in favor of text
- Using only FP32 for all model weights
- Even lower-bit micro-scaling formats like MXFP4 and NVFP4
- A tensor’s values need more dynamic range. Which common FP8 encoding allocates more exponent bits?
- E5M2
- E4M3
- Both allocate zero exponent bits
- Neither encoding uses an exponent
- Why can an FP8 result differ from the same operation using FP32 inputs?
- Eight-bit input storage is automatically exact for every real value
- FP8 removes all mathematical operations
- The lower-precision input representation can introduce rounding error
- Higher-precision accumulation restores every lost input detail
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