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Chapter 3 Review Questions
1. Assume 185 and 122 are signed 8-bit decimal integers stored in sign-magnitude format. Calculate 185 + 122. Is there overflow, underflow, or neither?
2. Assume 185 and 122 are signed 8-bit decimal integers stored in sign-magnitude format. Calculate 185 – 122. Is there overflow, underflow, or neither?
3. List four floating-point operations that cause NaN to be created?
5. The floating-point format to be used in this problem is an 8-bit IEEE 754 normalized format with 1 sign bit, 4 exponent bits, and 3 mantissa bits. It is identical to the 32-bit and 64-bit formats in terms of the meaning of fields and special encodings. The exponent field employs an excess- 7coding. The bit fields in a number are (sign, exponent, mantissa). Assume that we use unbiased rounding to the nearest even specified in the IEEE floating point standard.
a) Encode the following numbers to the 8 bit IEEE format 1) 0.0011011)binary
2) 16.0)decimal
4. Assuming single precision IEEE 754 format, what decimal number is represent by this word:
1 01111101 00100000000000000000000
b) Perform the computation 1.011binary + 0.0011011binary showing the correct state of the guard, round bits and sticky bits. There are three mantissa bits.
c) Decode the following 8-bit IEEE number into their decimal value: 1 1010 101
d) Decide which number in the following pairs are greater in value (the numbers are in 8-bit IEEE 754 format):
1) 0 1000 100 and 0 1000 111
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e) In the 32-bit IEEE format, what is the encoding for negative zero?
f) In the 32-bit IEEE format, what is the encoding for positive infinity?
6. Using 32-bit IEEE 754 single precision floating point with one(1) sign bit, eight (8) exponent bits and twenty three (23) mantissa bits, show the representation of -11/16 (-0.6875).
7. What is the smallest positive (not including +0) representable number in 32-bit IEEE 754 single precision floating point? Show the bit encoding and the value in base 10 (fraction or decimal OK)
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1. Assume 185 and 122 are signed 8-bit decimal integers stored in sign-magnitude format. Calculate 185 +
122. Is there overflow, underflow, or neither?
Neither (65)
2. Assume 185 and 122 are signed 8-bit decimal integers stored in sign-magnitude format. Calculate 185 – 122. Is there overflow, underflow, or neither?
Overflow (result =-179, which does not fit into 8-bit format)
3. List four floating-point operations that cause NaN to be created?
Four operations that cause Nan to be created are as follows:
(1) Divide 0 by 0
(2) Multiply 0 by infinity
(3) Any floating point operation involving Nan
= (-1)* (2^(125-127))*(1.001)2 = (-1)*(0.25)*(1.125)
5. The floating-point format to be used in this problem is an 8-bit IEEE 754 normalized format with 1 sign bit, 4 exponent bits, and 3 mantissa bits. It is identical to the 32-bit and 64-bit formats in terms of the meaning of fields and special encodings. The exponent field employs an excess- 7coding. The bit fields in a number are (sign, exponent, mantissa). Assume that we use unbiased rounding to the nearest even specified in the IEEE floating point standard.
g) Encode the following numbers to the 8 bit IEEE format 1) 0.0011011)binary
This number = 1.10112 * 2-3
= 0 0100 110 in the 8-bit IEEE 754
(after applying unbiased rounding to the nearest even)
2) 16.0)decimal
This number = (1000.0)2 = (1.000)2 * 2(11-7)
(4) Adding infinity to negative infinity
4. Assuming single precision IEEE 754 format, what decimal number is represent by this word:
1 01111101 00100000000000000000000
The decimal number
= -0.28125
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= 0 1011 000 in the 8-bit IEEE 754
h) Perform the computation 1.011binary + 0.0011011binary showing the correct state of the guard, round bits and sticky bits. There are three mantissa bits.
1.0110
+ 0.0011 (The least two significant bits are the guard and round bits)
1.1001
After rounding with sticky bit, the answer then is (1.101)2
i) Decode the following 8-bit IEEE number into their decimal value: 1 1010 101
The decimal value is (-1)* 1.625 *(2)(10-7) = – 13
j) Decide which number in the following pairs are greater in value (the numbers are in 8-bit IEEE 754 format):
1) 0 1000 100 and 0 1000 111
The first number = 2(8-7)*(1.5) = 1.5
The second number = 2(8-7)*(1.875) = 1.875
The representation of negative zero in 32-bit IEEE format is
The representation for positive infinity in 32-bit IEEE format is
6. Using 32-bit IEEE 754 single precision floating point with one(1) sign bit, eight (8) exponent bits and twenty three (23) mantissa bits, show the representation of -11/16 (-0.6875).
The representation of -0.6875 is:
1 01111110 01100000000000000000000000
7. What is the smallest positive (not including +0) representable number in 32-bit IEEE 754 single precision floating point? Show the bit encoding and the value in base 10 (fraction or decimal OK)
The smallest positive representable number 32-bit IEEE 754 single precision floating point value is:
0 00000001 00000000000000000000000
The second number is greater in value
k) In the 32-bit IEEE format, what is the encoding for negative zero?
1 00000000 00000000000000000000000
l) In the 32-bit IEEE format, what is the encoding for positive infinity?
0 11111111 00000000000000000000000
• Its value is = ±1.0 × 2–126 ≈ ±1.2 × 10–38