GATE CSE 2005 | Question: 85-b

Question

Consider the following floating-point format.

 

Mantissa is a pure fraction in sign-magnitude form.

The normalized representation for the above format is specified as follows. The mantissa has an implicit $1$ preceding the binary (radix) point. Assume that only $0’s$ are padded in while shifting a field.

The normalized representation of the above number $(0.239 \times 2^{13})$ is:

• A. $0A\;20$
• B. $11\;34$
• C. $49\;D0$
• D. $4A\;E8$

Comments

|0|1001010|11101000|

|sign bit|biased exponent|normalized mantissa|

(D) is the correct answer!

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Refer -https://gateoverflow.in/1407/gate2005-85a

Answer 1

For finding normalised representation we need to find unnormalised one first. So, we have:

$0.239 \times 2^{13}$ as the number. So, we find the binary equivalent of $0.239$ till $8$ digits as capacity of mantissa field is $8$ bits. We follow the following procedure:

• $0.239 \times 2 = 0.478$
• $0.478 \times 2 = 0.956$
• $0.956 \times 2 = 1.912$
• $0.912 \times 2 = 1.824$
• $0.824 \times 2 = 1.648$
• $0.648 \times 2 = 1.296$
• $0.296 \times 2 = 0.592$
• $0.592 \times 2 = 1.184$

We stop here as we have performed $8$ iterations and hence, $8$ digits of mantissa of unnormalised number is obtained. Now we have:
Mantissa of given number  $=$  $0011$  $1101$

So, the number can be written as: $0.00111101\times 2^{13}$

Now we need to align the mantissa towards left to get normalised number. And in the question it is mentioned that during alignment process $0's$ will be padded in the right side as a result of mantissa alignment to left.

So, to get normalised number, we align to left $3$ times, to get new mantissa $= 11101 000$

Exponent will also decrease by $3$ hence, new exponent  $=  10$

So, normalised number$ = 1.11101000 \times 2^{10}$

Actual exponent $= 10$

Given excess $64$ is used which means bias value $= 64$

So, exponent field value $=10+64 =  74$

And of course sign $bit = 0$ being a positive number.

Thus the final representation of number 
$\qquad\qquad\qquad=  0\;  1001010\;11101000$
$\qquad\qquad\qquad=  0100\;1010\;1110\;1000$
$\qquad\qquad\qquad=  (4AE8)_{16}$

Hence, (D) should be the correct answer.

Comments

It should be E8

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Can anyone explain why cant we convert the decimal to a number where the point is after one digit.

i.e. 0.239 X 2^13 = (0.239x2x2x2) X 2^10 = 1.912 x 2^10.

Now if we try to solve it exponent part is same as of ur answer , but the mantissa is different

1 = 1

0.912 = 1110 1001(our mantissa)

thus making the mantissa to E 9

so, our final answer came out 4A E9. which is not same as yours. Can anyone help me here if my approach is wrong!?

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If you do not understand this, it is not detailed. read my answer to understand it in easy manner

Answer 2

• Option d is right

 

Comments

this is perfect!

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While converting .239 to binary, the 11th position after the decimal point will be 1, Not 0. So, by your solution, answer will be 4AE9.

Answer 3

The best answer is so confusing

He skipped so many points

 

Solution Approach

1. first multiply with 2 repeatedly to get 0.00111101 $\times 2^{13}$
2. which is equivalent to write as $1.11101 \times \space 2^{10}$
3. now as mantissa is 8 bits, add 0’s in the RHS of mantissa, three 0’s will be appended (given in the question)
4. $1.11101000 \times 2^{10 + 64}$ 64 is added as it is the biased exponent.
5. write it in sign exponent and mantissa form
6. $0 \space 1001010 \space 11101000$
7. group it as hexadecimal
8. $0100 1010 1110 1000$
9. $4AE8$ which the answer