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Consider the following $32\text{-bit}$ 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 $24 \;\text{bit}$ fraction field (with the binary point is at the left end of the fraction bits), and a $7\;\text{bit}$ exponent field (in $\text{excess-}64$ signed integer representation, with $16$ is the base of exponentiation). The sign bit is the most significant bit.

  1. 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.
  2. What is the largest value that can be represented using this format? Express your answer as the nearest power of $10$.
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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

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

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Hemant Parihar

Is it possible to represent part a  in normalized form ?

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@rishi71662data4 in the last part, equation 1 becomes $-0.1111 * 16^{65 - 64}$ right?

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4 Answers

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Best answer

Here, mantissa is represented in normalized representation and exponent in $\text{excess-}64$ (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 $\text{IEEE}\; 754$ as this is not mentioned in question) which in $\text{excess-}64$ will be $64 + 1 = 65.$ Number being negative sign bit is $1.$ So, we get

$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (\text{BC}000041)_{16}$

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 

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

$($Again we did not add implicit $1$ as in $\text{IEEE}\; 754)$

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

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

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@Arjun sir 

How

1-$2^{24}$ = 1 – $2^{-24}$

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edited by

@Overflow04 Yes, the answer needs some edit.

The largest number is $0.1111…1 $(24 1’s to the right of $.$)$*16^{63}$

Shifting the point (.) towards the right 24 times:-

$= 11111...1.0 * 2^{-24} * 16^{63}$ 

$ = (2^{24}-1)*2^{-24} * 16^{63}$

$= (1 – 2^{-24}) * 16^{63}$ 

and then the remaining.

@Lakshman Patel RJIT Sir, it needs an edit:-

It should be $  (2^{24}-1)*2^{-24} *16^{63} = (1-2^{-24})*16^{63} $

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@Abhrajyoti00 "0." followed by n 1s $= 1-2^{-n} . $ I have fixed the typo now. 

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15 votes
15 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 $7.237006 \times 10^{75}$.
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Maximum value possible is (1−2−24)×16(127−64).

WHen converted into base 10 we get 7.237006×1075.
plzz xplain dis part...
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As Base is 16 & We need to normalize we need to at least normalize to 16^1

Why? Why can't we keep it as 16^0 itself? 0 is a valid exponent right?

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@Sambhrant Maurya  yes 0 is a valid exponent but mantissa is represented as M in 0.M , that's why we do normalization.

Normalized form is $(-1)^S (0.M)\times 16^{E-64}$

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9 votes
9 votes

First express (-7.5) into binary

How to write (-7.5) in binary

first write 7.5 into binary = (111.1)

Do not worry about negative sign here sign bit will contain 1 because above number is negative.

And base 16 is given hence we will convert into power of 16(we know 2^4=16)

So we can write 111.1*2^0 = 0.01111*2^4

0.01111*2^4 = 0.01111*16^1( here 16 is a base and 1 is exponent)

Excess 64 is given hence 1+64=65

65 in binary= 1000001(we will represent 65 in exponent field)

Now we have to represent into floating point

Sign bit = 1

Mantissa = 011110...0(19 zero consecutive)

Exponent = 1000001 

( 1  011110....0   1000001 ) in floating point

convert it into hexadecimal and make a pair of 4 digit from right side. ( for example 0001=1 in hexadecimal )

then you will get (BC000041) in hexadecimal

 

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nice !
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1 vote
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

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edited by
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 :)
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