GATE | GATE CS 1996 | Question 25

Consider the following floating point number representation

floating_p

 

 

 

The exponent is in 2’s complement representation and mantissa is in the sign magnitude representation. The range of the magnitude of the normalized numbers in this representation is
a). 0 to 1
b). 0.5 to 1
c). 2-23 to 0.5
d). 0.5 to (1-2-23)
(A) a
(B) b
(C) c
(D) d


Answer: (D)

Explanation: Normalized numbers are of the form: 1.xxxx xxxx… …xxxx {ignoring sign and exponent part}. They always have a Leading 1 before decimal point.
While storing such a number in this floating point representation two cases can be there:

  • Case 1: We can store most significant 1 in mantissa along with other bits as (while MSB of mantissa is a sign bit as ‘S’) :
    Minimum Mantissa=> S100 0000 0000 0000 0000 0000 = 1.00 0000…0000 = 1
    Maximum Mantissa=> S111 1111 1111 1111 1111 1111 = 1.1111…1111 = 2 – 2-22
  • Case 2: We can ignore to do so assuming an implicit MSB as 1.
    Minimum Mantissa=> S000 0000 0000 0000 0000 0000 = 1.000 0000…0000 = 1
    Maximum Mantissa=> S111 1111 1111 1111 1111 1111 = 1.1111…1111 = 2 – 2-23

Thus none of the options matches.

If instead of Normalized numbers they had asked for Normalized mantissa then following would have been the approach:
minimum value of magnitude of Mantissa=>
“S100 0000 0000 0000 0000 0000”
(ignoring (-1)s as a multiple of the number as only magnitude has been asked)

= 0.100 0000 0000 0000 0000 0000
= 1*2-1
= 0.5
and
maximum value of Mantissa=>
“S111 1111 1111 1111 1111 1111”
= 0.111 1111 1111 1111 1111 1111
= [(223)- 1] * 2-23
= 1-2-23 

In this case option (D) would be the answer.

This explanation is contributed by Pradeep Pandey.

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Improved By : PRADEEP PANDEY



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