Question

1. Consider the following 32-bit floating point representation based on the IEEE floating point standard:
   1. There is a sign bit in the most significant bit.
   2. The next eight bits are the exponent, and the exponent bias is 2^8-1-1 = 127.
   3. The last 23 bits are the fraction bits.
   4. The representation encodes number of the form V = (-1)^S x M x 2^E, where S is the sign, M is the significand, and E is the biased exponent.
   5. The rules for the IEEE standard for normalized, denormalized, representation of zero, infinity, and NaN still apply.

1. What is the valid range of values for the biased exponent E? -126 to ______

2. What is the biased exponent E for denormalized values? ______

3. Match the following values in decimal (base 10 form) or descriptions with the corresponding 32-bit binary representation (fill in with a-h):

1. Negative infinity             _____ 1. 01000000111000000000000000000000
2. Positive zero          _____ 2. 11000000011000000000000000000000
3. 3.5                           _____ 3. 01000000011000000000000000000000
4. 7.0                            _____ 4. 11000000111000000000000000000000
5. -3.5                          _____ 5. 00000000000000000000000000000000
6. -7.0                           _____ 6. 01111111100000000000000000000111
7. Not-a-Number (NaN)            _____ 7. 01111111100000000000000000000000
8. Positive infinity              _____ 8. 11111111100000000000000000000000

Answer 1

Solution:

a) What is the valid range of values for the biased exponent E? -126 to ______

The valid range of values for the biased exponent E is from -126 to +127. These values are for single-precision floating-point representation i.e. for 32-bit number.

b) What is the biased exponent E for denormalized values?

A denormal number is represented with a biased exponent of all 0 bits, this represents a value -126. The smallest biased exponent representing a normal number is 1

c)

1. Negative infinity             ___b__ 1. 01000000111000000000000000000000
2. Positive zero          ___e__ 2. 11000000011000000000000000000000
3. 3.5                           ____c_ 3. 01000000011000000000000000000000
4. 7.0                            ___f__ 4. 11000000111000000000000000000000
5. -3.5                          __b___ 5. 00000000000000000000000000000000
6. -7.0                           ___g__ 6. 01111111100000000000000000000111
7. Not-a-Number (NaN)            __h___ 7. 01111111100000000000000000000000
8. Positive infinity              ___a__ 8. 11111111100000000000000000000000

Infinity is represented with all 1's in the exponent and all 0's in the fraction part. For negative infinity, the sign bit is 1 and for positive infinity, the sign bit is 0.

Zero is indicated with all 0's in the exponent and fraction. For positive 0, the sign bit is 0.

Floating-point exception results if the computation is not a number. For representing these kinds of results we use the Not-a-Number format. It is represented either 0/1 in sign bit, all 1's in the exponent field, and anything other than all 0's in the fraction field.

Representation of 3.5:

(representation of floating-point numbers: 1 sign bit, 8 bits of exponent field, 23 bits of mantissa field.)

3 in binary is 011, 0.5 in binary is 1. Concatenating the result: 11.1 is the binary representation of 3.5

as this is +ve number, the sign bit is 0. 11.1 is represented as 1.11 * 2^1. This indicates, the exponent is 1 i.e 127+1 which is 128. 11 is the mantissa field.

Representation of 128 in binary is 10000000

The mantissa is 11, and its representation is 11000000000000000000000.

So the result is 01000000011000000000000000000000

So for 3.5 the result is 01000000011000000000000000000000

Representation of -3.5:

The result is exactly the same as 3.5 except that the sign bit indicates the value as -ve. So the sign bit is 1 in this case.

So the result is 11000000011000000000000000000000

Representation of 7.0:

(representation of floating-point numbers: 1 sign bit, 8 bits of exponent field, 23 bits of mantissa field.)

7 in binary is 111, 0.0 in binary is 0. Concatenating the result: 111.0 is the binary representation of 7.0

as this is +ve number, the sign bit is 0. 111.0 is represented as 1.11 * 2^2. This indicates, the exponent is 1 i.e 127+2 which is 129. 11 is the mantissa field.

Representation of 129 in binary is 10000001

The mantissa is 11, and its representation is 11000000000000000000000.

So the result is 01000000111000000000000000000000

So for 7.0 the result is 01000000111000000000000000000000

Representation of -7.0:

The result is exactly the same as 7.0 except that the sign bit indicates the value as -ve. So the sign bit is 1 in this case.

So the result is 11000000111000000000000000000000