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4 votes
4 votes

Consider a hypothetical $8$-bit IEEE floating point representation:
$$\begin{array}{|l|l|l|} \hline \text{s} & \text{exp(3bits)} & \text{frac(4 bits)} \\ \hline \end{array}$$
What is the largest positive denormalized value?

  1. $\frac{1}{4}$
  2. $\frac{31}{2}$
  3. $\frac{15}{64}$
  4. None of the above
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1 Answer

2 votes
2 votes
$s=0; \: k=3; \:$ and $n=4$
$\text{bias} = 2^{k-1} -1 = 4-1=3$
$\text{exp} = 000$
$\text{frac} = 1111$
$M= frac \times 2^{-n} =.1111$
$E=1-\text{Bias}=1-3=-2$
$V=(-1)^s \times M \times 2^E = 1 \times .1111 \times 2^{-2} = (.001111)_2 = \frac{8+4+2+1}{64} = \frac{15}{64}$

3 Comments

why do we need to do biasing?
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@Applied Course

why exp is not taken as 111 so it will gives us largest positive value?

 

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this answer is perfect. We just need to refer ieee single precision formulae  for non normalized mantissa.

We can easily correlate both

for reference here is formulae

$(-1)^{S}(0.M)(2^{-126})$

Please understand it from some standard source, once understood we can easily remember
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Answer:

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