UGC NET CSE | December 2018 | Part 2 | Question: 16

Question

The decimal floating point number $-40.1$ represented using $\textsf{IEEE-754} \: 32$-bit representation and written in hexadecimal form is

• A. $\textsf{0x}C2206666$
• B. $\textsf{0x}C2206000$
• C. $\textsf{0x}C2006666$
• D. $\textsf{0x}C2006000$

Comments

3.

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I also got 3 )

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should be A.

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One can take help from this example also. 

Example is ok for -176.375 but for -40.1 still I am unable to find it.

Answer 1

IEEE 754, 32 bit representation,  1 bit for sign, 8 bits for Exponent and 23 bits for mantissa

Note that Exponent represented in Excess-127 format.

We have to represent -40.1, So

sign bit = 1

40 = 32+8 = (101000)$_2$

0.1 = (00011 0011 0011 0011 0011 00)$_2$ .......... represented with 23 bits

40.1 = 101000 . 00011 0011 0011 0011 0011 00    ----------------- in binary form.

       = 1 . 01000 00011 0011 0011 0011 0011 00 * 2$^5$ ----- NOW it is in 1.M format, Keep only 23 bits in Mantissa

       = 1 . 01000 00011 0011 0011 0011 0 * 2$^5$ ----- NOW it is in 1.M format, and Mantissa have 23 bits only.

True exponent is 5 ( which is in power of 2) ==> Excess Exponent value = 5+127 = 132 = 128 + 4 = 10000100

total representation is like:

sign	Exponent	Mantissa
1	10000100	01000000110011001100110

Finally it is like (11000010001000000110011001100110)$_2$

1100 0010 0010 0000 0110 0110 0110 0110 ==> 0xC2206666

Comments

This is wrong answer. It should be 0x 2206666, option A is correct

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did you mean, the solution is wrong ? or last line missing 6 ?

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just the last 6 is missing, else everything is fine

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then it's typing mistake but not "wrong answer".

Answer 2

Option A is right.

Answer 3

Option 3. 0xC2206666 right

Answer 4

Given $-(40.1)$, convert it into binary we get

$(00101000.0001101100110011001100..)$

IEEE $754$ notation which will be used is:

$V= (-1)^{S} * (1.M) * 2^{E-127}$ where $S$ is sign bit, $M$ is mantissa and $E$ is exponent

Converting our given number into the above format we get

$1.010000001100110011.... *2^{5}$, we know sign take 1 bit, exponent 8 and mantissa 23 bit 

$E$-127=$5$, $E$= $132$($10000100$)

SEM: $11000010001000000110011001100110$

Convert into hexadecimal:

$1100 0010 0010 0000 0110 0110 0110 0110$

$C2206666$

Option d) is correct

Comments

Absolutely correct sir ! Thank you.

I was doing it wrong for recursive binary conversion of .1 , actually I was stopping it till non recursive part and then put zeros.

I forgot that it could repeat it's value till we have exhausted our mantissa limit.

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Yes u r also correct

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Option D is correct