Homework SourseConsider a hypothetical limited accuracy rounding computer t
Consider a hypothetical limited accuracy rounding computer that stores floats with 6-bit mantissas.
Find the smallest real number that may be legally stored as this same value. Write the answer in 2 ways:
i) in base 2 floating point form (?.?? ......2 x 2?)
ii) as a base 10 decimal value
0.0001100Solution
0.099609375 will be represented as
(0.099609375)10 = (0)10+(0.09961)10
= (1.0)2 + (0.0111)2 * 2 -4
0
1`
1
1
0
1
1
0
0
ó = 0
m = 1101
e = 100
The first bit is 0, so the number is positive.
The second bit is 1, so the exponent is negative.
The next four bits, 1101, are the magnitude of the mantissa
M = ( 0.0111)2 = ( 0 * 20 + 0 * 2-1 + 1 * 2-2+ 1 * 2-3 + 1 * 2-4 )
The last three bits, 100, are the magnitude of the exponent
e= ( 1 * 22 + 0 * 21 + 1 * 20 )
To understand the values better let us take an example:
Old time cash register that would ring any purchase between 0 and 999.99 units of money. There are five (not six) working spaces in the cash register
0
0
0
.
0
0
9
9
9
.
9
9
sign *mantissa * 10exponent
or for a number y ,
y = ×m×10e
Where,
Sign of the number, + 1 or -1
m = mantissa, 1 m < 10
e = integer exponent (also called fecund)
Represent (54.75 )10 in floating point binary format .
the number is written to a hypothetical word that is 9 bits long where the first bit is used for the sign of the number, the second bit for the sign of the exponent, the next four bits for the mantissa, and the next three bits for the exponent,
(54.75)10 = (110110.11)2 = (1.1011011)2 * 2(5)10
The exponent 5 is equivalent in binary format as
(5 )10 = ( 101)2
Hence
(54.75 )10 = (1.1011011) 2 * 2(101)2
The sign of the number is positive, so the bit for the sign of the number will have zero in it.
ó = 0
The mantissa
m = 1011
(There are only 4 places for the mantissa, and the leading 1 is not stored as it is always
Expected to be there)
e = 101.
we have the representation as
0
0
1
0
1
1
1
0
1
Hope this Helps!!
| 0 | 1` | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
