GATE CSE 2003 | Question: 43

Question

The following is a scheme for floating point number representation using $16$ bits.

Let $s, e,$ and $ m $ be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:

$\begin{cases}(-1)^s \left(1+m \times 2^{-9}\right) 2^{e-31}, & \text{ if the exponent } \neq 111111 \\  0, & \text{ otherwise} \end{cases}$

What is the maximum difference between two successive real numbers representable in this system?

• A. $2^{-40}$
• B. $2^{-9}$
• C. $2^{22}$
• D. $2^{31}$

Comments

in floating point representation as we move AWAY from zero, numbers are sparsely spaced. This means the largest difference between 2 successive numbers will be at the VERY ends. 
For better explanation →https://gateoverflow.in/934/gate-cse-2003-question-43?show=410241#a410241

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can we use differentiation to calculate the maxima or minima? if we can what is the variable that we will be doing differentiation with respect to?

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hint :

given form in question is the Normalised form of Floating point number(one of the forms) when +ve .

Answer 1

I had confusions in interpretation of above solution. So this might help you.

111111111 can be written as $2^{10}-1$

So Largest No. 1.111111111 x $2^{31}$ equivalent to  ($2^{10}-1$) x $2^{-9}$ x $2^{31}$
Second largest No. 1.111111110 x $2^{31}$ equivalent to  ($2^{10}-2$) x $2^{-9}$ x $2^{31}$
difference of above equation:
= [($2^{10}-1$) x $2^{-9}$ x $2^{31}$ ] - [($2^{10}-2$) x $2^{-9}$ x $2^{31}$]
= $2^{22}$

Answer 2

in floating point representation as we move AWAY from zero, numbers are sparsely spaced. This means the largest difference between 2 successive numbers will be at the VERY ends.
For better explanation → https://youtu.be/4_cbc9Q7b8c 
we can check the right end (+ve side) or left end (-ve side). 
If we go with the +ve side →
the right most numbers will be for exponent 111111 but we cannot choose these because ALL numbers with this exponent have to be considered 0 (according to question)
so max under this exponent → 0 111111 111111111 [is actually zero] 
and 2^nd max under this exponent → 0 111111 111111110 [is actually zero]
DIFF = 0 → Therefore DON’T take 111111

What is the next maximum exponent → 111110
so max number under this exponent → 0 111110 111111111 = $(2^{10}-1) * 2^{-22}$
and 2^nd max number under this exponent → 0 111110 111111110 = $(2^{10}-2)* 2^{-22}$
DIFF = $2^{22}$

NOTE : as long as we are under exponenet 111110 we can take any 2 successive numbers and result will be same

Answer 3

for same exponent any two successive real numbers can differ maximum by magnitude of the least significant matissa. Here the least representable matissa would be 2^-9. the max difference will hence occur for the highest exponent cases. Here 31 (62-31 as explained in other answers)... so the magnitude of max difference will be 2^-9 *2^31 = 2^22