Is it necessary for the MSB of the Mantissa to be non zero in Normalized Form ?

The Gateway to Computer Science Excellence

First time here? Checkout the FAQ!

x

+19 votes

Consider the following $32-bit$ floating-point representation scheme as shown in the format below. A value is specified by $3$ fields, a one bit sign field (with $0$ for positive and $1$ for negative values), a $24 bit$ fraction field (with the binary point is at the left end of the fraction bits), and a $7 bit$ exponent field (in $excess-64$ signed integer representation, with $16$ is the base of exponentiation). The sign bit is the most significant bit.

- It is required to represent the decimal value $- 7.5$ as a normalized floating point number in the given format. Derive the values of the various fields. Express your final answer in the hexadecimal.
- What is the largest value that can be represented using this format? Express your answer as the nearest power of $10$.

0

yes, in binary it must always be '1'. But IEEE 754 format hides this one and not explicitly represent this for normalized numbers.

+1

Sir,

But then when you converted **111.1 * 16^0** to **0.0111*16^1 **

**0.0111*16^1 **, is not in Normalized Form ( and the question says we want it in Normalized Form)

the defination of Normalised Number -

**A number in Scientific Notation with no leading 0s is called a Normalised Number: 1.0 × 10**

where am I going wrong ?

0

The given ans in part 1 is not in normalized form.

(0.01111000...) * (16^1).

(0.0000011110...) * (16^2).

both these have the same decimal value which is 15 / 4.

(0.01111000...) * (16^1).

(0.0000011110...) * (16^2).

both these have the same decimal value which is 15 / 4.

0

Not being able to understand part b) answer. The calculation part which results into 10^{76} and why 2^{24} is subtracted from 1. Please help.

+6

Re-writing the same answer as that of Arjun Sir (correct me if I am wrong anywhere)

The Floating point representation given in the question is not that of IEEE-754. The equation of the given representation is:

$0.M X 2^{Exponent-Base}$

**The 0 in 0.M is not implicit, but explicit**

Now, given number -7.5

$(-7.5)_{10} = (-111.1)_{2} = -0.1111 * 2^{3} \rightarrow\left[1\right]$

Now, base is given as 16 with excess 64, that is

$16^{x-64} = 2^{4(x-64)} = 2^{4x-256} \rightarrow\left[2\right]$

Here, x is the exponent. Now from equation 1 and 2, we get

$4x-256=3$

$x\approx 65$

Now, equation 1 becomes:

$-0.1111 * 2^{65-64}$

This gives S=1, M = 011110000000000000000000 and E = 1000001

$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (BC000041)_{16}$

+20 votes

Best answer

Here, mantissa is represented in normalized representation and exponent in excess-64 (subtract 64 to get actual value).

a. We have to represent $-(7.5)_{10} = -(111.1)_2$.

Now we are using base 16 for exponent. So, mantissa will be .01111 and this makes exponent as 1 (4 bit positions and no hiding first 1 as in IEEE 754 as this is not mentioned in question) which in excess-64 will be 64 + 1 = 65. Number being negative sign bit is 1. So, we get

$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (BC000041)_{16}$

b. Largest value will be with largest possible mantissa, largest possible exponent and positive sign bit. So, this will be all 1's except sign bit which will be

$0.\underbrace{111\dots 1}_{\text{24 ones}} \times 16^{127-64} = \left( 1 - 2^{24}\right) \times 16^{63} = \left(1- 2^{-24}\right) \times 16^{63}$

(Again we did not add implicit 1 as in IEEE 754)

$2^x = 10^{y} \implies y = \log 2^x = x \log 2$

So, $(1-2^{-24}) \times 16^{63} \\= \left(1 - 10^{-24 \log 2}\right) \times 10^{63 \log 16}\\ \approx \left(1 - 10^{-7}\right) \times 10^{76}\\ = 10^{76}$

Not directly relevant here, but a useful read: https://jeapostrophe.github.io/courses/2015/fall/305/notes/dist/reading/help-floating-point.pdf

0

But sir my question was how (111.1)_{2} become (0.01111)_{16}

why decimal point is shifted 4 points left ?

0

@ G.K.T

Because in base 16, 4 bits are used to represent each Byte.

Each byte is shown as a pair of nibbles (*4 bits*) .

+2

This is what I got from your answer please verify

111.1 x 2^{0}

11.11 x 2^{1}

1.111 x 2^{2}

0.1111 x 2^{3}

0.01111 x 2^{4} that is same as 0.01111 x 16

and as you have already mentioned above that in IEEE 754 msb 1 is implicit so thats why we aren't mentioning in the representation

0

@Arjun sir and bikram sir

In ques https://gateoverflow.in/2729/gate1996_1-25

we have used first bit after decimal as 1 because it was normalized form but not IEEE.And same is the case here.So why in current question we are able to store fraction as .0111 as normalized form with first bit after decimal as 0

+1

Sir if i try to make first bit after decimal as 1,then i am not able to adjust getting exponent as base 16

111.1 * 16^0

=.1111 *16^0*2^3, but if i do one more right shift as in the answer ,then we can get 16^1 ,but is'nt is required in normalized form to have first bit of mantissa as 1?

111.1 * 16^0

=.1111 *16^0*2^3, but if i do one more right shift as in the answer ,then we can get 16^1 ,but is'nt is required in normalized form to have first bit of mantissa as 1?

0

@rahul

" The leading bit of mantissa is 1 ( this 1 is implicit in IEEE representation) "

but it is only normalized representation, that is why it differs

but without MSB bit 1 how a number can be normalized? that is question for me too

" The leading bit of mantissa is 1 ( this 1 is implicit in IEEE representation) "

but it is only normalized representation, that is why it differs

but without MSB bit 1 how a number can be normalized? that is question for me too

+9 votes

It is clearly given here that this is excess 64 & Base is 16. So we need to use that.

a)

-7.5 Due to minus sign we will get sing bit as 1.

7.5 = 111.1 * 16^0

As Base is 16 & We need to normalize we need to at least normalize to 16^1. (This answer maybe different if we choose not to normalize, Though I think we can not use either implicit or explicit normalization correctly here.

It will be 0.0111*16^1.

So now our biased exponent is 1+64 = 65.

Mantissa is (0.) 0111000...

When we save it we get hex as BC000041

2. Maximum value possible is $(1- 2^{-24}) \times 16^{(127-64)}$.

WHen converted into base 10 we get $7.237006 \times 10^{75}$.

a)

-7.5 Due to minus sign we will get sing bit as 1.

7.5 = 111.1 * 16^0

As Base is 16 & We need to normalize we need to at least normalize to 16^1. (This answer maybe different if we choose not to normalize, Though I think we can not use either implicit or explicit normalization correctly here.

It will be 0.0111*16^1.

So now our biased exponent is 1+64 = 65.

Mantissa is (0.) 0111000...

When we save it we get hex as BC000041

2. Maximum value possible is $(1- 2^{-24}) \times 16^{(127-64)}$.

WHen converted into base 10 we get $7.237006 \times 10^{75}$.

+1 vote

a) 7 bit exponent means biasing of = 2^(7-1) -1 = 63

-7.5= - 111.1 =- 1.111 * 2^2

sign bit =1

biased exponent = 65 as (65 - 63 = 2) = 0010001

fraction = 111000000000000000000000 (three 1s followed by twenty one 0s)

normalized representation = 1 111000000000000000000000 0010001

hexadecimal representation = F0000011

b) largest value :

sign bit = 0 (positive)

exponent = 1111110

fraction = 111111111111111111111111 ( twenty four 1s )

normalized number = 0 111111111111111111111111 1111110

= 2147483646

= 2.147483646 * 10^9

CORRECT ME IF I AM WRONG

-7.5= - 111.1 =- 1.111 * 2^2

sign bit =1

biased exponent = 65 as (65 - 63 = 2) = 0010001

fraction = 111000000000000000000000 (three 1s followed by twenty one 0s)

normalized representation = 1 111000000000000000000000 0010001

hexadecimal representation = F0000011

b) largest value :

sign bit = 0 (positive)

exponent = 1111110

fraction = 111111111111111111111111 ( twenty four 1s )

normalized number = 0 111111111111111111111111 1111110

= 2147483646

= 2.147483646 * 10^9

CORRECT ME IF I AM WRONG

+1

for 7 bit the biasing is 64 don't know why i saw this in many answers!!!

BTW..65 binary is 1000001 which brings ur answer to F0000041

if we use 64 biasing i am getting F0000042

can some confirm which biasing to use??

for part b)

the mantissa will look like

1.1...24 1's...1 which is 2.16777215

exponent will be (126-64 or 63) (say 126-64 = 62)

so..=2.16777215 X 10^62

correct me also if wrong :)

BTW..65 binary is 1000001 which brings ur answer to F0000041

if we use 64 biasing i am getting F0000042

can some confirm which biasing to use??

for part b)

the mantissa will look like

1.1...24 1's...1 which is 2.16777215

exponent will be (126-64 or 63) (say 126-64 = 62)

so..=2.16777215 X 10^62

correct me also if wrong :)

- All categories
- General Aptitude 1.6k
- Engineering Mathematics 7.5k
- Digital Logic 3k
- Programming & DS 4.9k
- Algorithms 4.3k
- Theory of Computation 6k
- Compiler Design 2.1k
- Databases 4.2k
- CO & Architecture 3.5k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.5k
- Admissions 584
- Exam Queries 566
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,108 questions

53,221 answers

184,627 comments

70,462 users