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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T17:04:34+00:00 2026-05-25T17:04:34+00:00

I have been working on these three lab questions for about 5 hours. I’m

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I have been working on these three lab questions for about 5 hours. I’m stuck on the last question.

Consider the following floating point number representation which
stores a floating point number is 16 bits. You have a sign-bit, a six
bit (excess-32) exponent, and a nine bit mantissa.

  1. Explain how the 9-bit mantissa might get you into trouble.

Here is the preceding question. Not sure if it will help in analysis.

  1. What is the range of exponents it supports?

000000 to 111111 or 0 to 63 where exponent values less
than 32 are negative, and exponent values greater than 32 are
positive.

I have a pretty good foundation for floating points and converting between decimals and floating points. Any guidance would be greatly appreciated.

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  1. Editorial Team
    Editorial Team
    2026-05-25T17:04:35+00:00Added an answer on May 25, 2026 at 5:04 pm

    To me, the ratio mantissa to exponent is a bit off. Even if we assume there is a hidden bit, effectively making this into a 10 bit mantissa (with top bit always set), you can represent + or – 2^31, but in 2^31/2^10 = 2^21 steps (i.e. steps of 2097152).

    I’d rather use 11 bits mantissa and 5 bit exponent, making this 2^15/2^11 = 2^4, i.e. steps of 16.

    So for me the trouble would be that 9+1 bits is simply too imprecise, compared to the relatively large exponent.

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