A computer uses an 8-bit floating-point representation consistent with IEEE 754
ID: 3906600 • Letter: A
Question
A computer uses an 8-bit floating-point representation consistent with IEEE 754 normalized format. It is identical to the 32-bit and 64-bit formats in terms of the meaning of fields and special encodings. The format includes 1 sign bit, 4 exponent bits, and 3 mantissa (magnitude or fraction) bits. As in IEEE 754, 0000 and 1111 in exponent field are reserved.The exponent field employs an excess-7 coding.
The following table gives the 8-bit Excess-7 representation.
Decimal
0
1
2
3
4
5
6
7
Unsigned
0000
0001
0010
0011
0100
0101
0110
0111
Excess-7
Reserved
-6
-5
-4
-3
-2
-1
0
Decimal
8
9
10
11
12
13
14
15
Unsigned
1000
1001
1010
1011
1100
1101
1110
1111
Excess-7
1
2
3
4
5
6
7
reserved
[Help: in normal binary representation 240 = 111100002 = 1.111 * 27 and 15.5 = 1111.12 = 1.1111 * 23]
What is 01001000 in this representation?
a.1
b.2
c.3
d.4
If x = - 2, how x will be represented?
a.11000000
b.01000000
c.10000010
d.11111110
What is the largest number in this representation, and how it will be represented?
a.11111111 = 155
b.01101111 = 15.5
c.01110111 = 240
d.00111111 = 240
What is the third largest number in this representation, and how it will be represented??
a.11111101 = 153
b.01111101 = 14.5
c.01110101 = 237
d.01110101 = 208
Decimal
0
1
2
3
4
5
6
7
Unsigned
0000
0001
0010
0011
0100
0101
0110
0111
Excess-7
Reserved
-6
-5
-4
-3
-2
-1
0
Explanation / Answer
Q1
01001000
The number is 2^(9-7)=4
Q2
if x=-2
Then representation is a.11000000
Q3
c.01110111 = 240 which is 2^7*(1+0.5+0.25+0.125)
Q4
c.01110101 = 208
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