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

The maximum difference between two successive real numbers will occur at extremes. This is because numbers are represented up to mantissa bits and as the exponent grows larger, the difference gets multiplied by a larger value. (The minimum difference happens for the least positive exponent value).
Biasing will be done by adding $31$ as given in the question. So, actual value of exponent will be represented value $- 31.$ Also, we can not have exponent field as all $1$’s as given in question (usually taken for representing infinity, NAN etc). So, largest value that can be stored is $111110 = 62.$
Largest number will be $1.111111111 \times 2^{62-31} = \left(2 - 2^{-9}\right) \times 2^{31}$
Second largest number will be $1.111111110 \times 2^{62-31} = \left(2 - 2^{-8}\right) 2^{31}$
So, difference between these two numbers $=  \left(2 - 2^{-9}\right) \times 2^{31} - \left(2 - 2^{-8}\right) \times 2^{31}\\= 2^{31} \left[\left(2 - 2^{-9}\right) - \left(2 - 2^{-8}\right) \right] \\= 2^{31} \left[ 2^{-8} - 2^{-9} \right] \\= 2^{31} \times 2^{-9} = 2^{22}$

Correct Answer: C.

Comments

If the minimum difference was asked it would be 2^(-40), right?

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No it is not necessary to take two maximum numbers, we can take any two consecutive numbers and get the same result as far as  our exponent is maximum.

Say example we first take mantissa(m) as all 0’s m=000 000 000 and m= 000 000 001.  Then num1 = $(1+0*2^{-9})*2^{31}$ and num2 = $(1+1*2^-9)*2^{31}$.Then  num1-num2 = 2^22.

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@aishwary191 Small edit bro : it should be num2-num1 right?

Answer 2

The gap between two succesive numbers is small towards the minimum value and maximu towards the maximum value.

Fist Maximum:

0(sign)

111110 (Biased Exponent)

111111111 mantissa

Second Maximum

0(sign)

111110 (Biased Exponent)

111111110 mantissa

E=62 that is biassed exponent since said not all 1's .

DIfference   = (1+1-2^-9)*2^E-31 - (1+1-2^-8)*2^E-31

                   =(2-2^-9)*2^E-31 - (2-2^-8)*2^E-31

                      = (2-2^-9)*2^62-31 - (2-2^-8)*2^62-31

                      =2³¹ * [(2-2^-9) - (2-2^-8) ]

                    =2²² should be the ans.

Comments

Here form given is : (1+ m*$10^{-9}$) which denotes normalized form.

Exponent = 000000 should not be taken as it denotes denormalized form .

For minimum Exponent should be: 000001 which represents $2^{1-31} $ ( 31 is bias)

So ,exponent = $2^{-30}$

Successive mantissa = 000000001 – 000000000 ( $2^{-9}$ in decimal)

Final minimum difference should be $2^{-30}$ * $2^{-9}$ (Sign is taken as 0)

= $2^{-39}$

Reference : IEEE 754

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@Arpit Patel But if it is not explicitly mentioned in the question, do we need to consider 0 as special or not?
Because nowhere in the question is it mentioned that they are following the IEEE format,
if it was mentioend then 2^−39 is correct.
But in this case, it should be 2^−40?
Correct me if I'm wrong.

 

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best answer

Answer 3

Exponent grows and makes no. larger, so we take adjacent values at extremes. Exponent have to be taken with all 1's except LSB as 0(as all 1's makes no. 0)

	Sign(1 bit)	Exponent(6 bits)	Mantissa(9 bits)
1st largest no	0=>+ve	111110=>(62)10	111111111=>(511)10
2nd large no	0=>+ve	111110=>(62)10	111111110=>(510)10

diff. = [(1+511*2^-9)2^62-31]   -  [(1+510*2^-9)2^62-31]
      = [1+ 511*2^-9 - 1 - 510*2^-9 ] 2^{62-31
        =} [1* 2^-9]  2^62-31 
        = 2^-9+62-31 = 2²²

Comments

Nice explanation

Answer 4

Here's simple answer:

Answer 5

Largest positive number m all 1's exponent=111110=62

Second largest number m 111111110 exponent=111110=62 

Difference=2³¹(2-2⁹-2+2^-8)=2³¹*2^-9=2²²

Answer 6

largest value = 2³¹+511.2²²

2nd largest value = 2³¹+510.2²²

so largest difference between them = 2²²

C . $2^{22}$