floating point number

Question

Comments

2^31 is the ans

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can u please explain @asu

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considring the options available the sign bit of exponent is of no use.....why

it includes sign bit means it explicitly wants to say that 1 in signbit will be negative

and 0 in signbit exponents will be positive and max value=31 and bias=31 also

so stored exponent=E-bias=0

and if considering all positive numbers then E=111111=63 and bias=31

then SE=63-31=32

Answer 1

Mantissa bits - 10 including sign bit. So, 9 bits for value. Maximum value

$ = 0.111111111$

And in normalized form there is an implicit 1 before "." and so we get

$1.111111111$.

Now for exponent we have 6 bits and given that it includes a sign bit. So, there is no bias being used as in IEEE 754 representation and so we get maximum possible value = $2^5 = 32.$

So, the maximum value that can be represented

$ = 1.111111111 \times 2^{32} \\= (2 - 2^{-9}) \times 2^{32}.$

Approximately $2^{32}.$

Answer 2

The format is as follows:  1 bit sign |  6 bits exponent | 10 bits for mantissa

Now, lets focus on exponent. The range of exponent (without any bias) will be -32 to 31. But, these extreme values are reserved. So, highest value of exponent will be 30.

Now, lets consider normalized manitissa. The max value will be 1.1111111111.

1) The fraction part of 1.1111111111 evaluates as ($2^{-1}+2^{-2}+2^{-3}...+2^{-10} = 2^{-10} *(2^9+2^8...+2^0) = 2^{-10} *(2^{10}-1) =2^{-10} *(2^{10}) =1$)

2) We also need to add the 1 to the left of the fraction 1.1111111111 which is 1.

So, total = 1) + 2) =2

So, max number that can be represented is = $2^{30}$ * 1.1111111111 = $2^{30}$ * 2  = $2^{31}$