Answer the following questions related to IEEE754 single-precision floating-point numbers: a. Encode the following number: -7.4375 s = , e = , f = _ (base 2 for all fields) b. Suppose a floating point number (float x) contains a value of
. What would be the smallest value that can be added to x without a loss of precision?
The Correct Answer and Explanation is:
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IEEE 754 Single-Precision Floating-Point Number Analysis
(a) Encode the number: -7.4375
IEEE 754 single-precision format consists of 32 bits, divided as follows:
- Sign bit (s): 1 bit
- Exponent (e): 8 bits (with a bias of 127)
- Fraction/Mantissa (f): 23 bits
Let’s break down the encoding of -7.4375 step by step:
Step 1: Determine the sign bit (s)
Since the number is negative,
s = 1
Step 2: Convert the absolute value (7.4375) to binary
Split it into integer and fractional parts:
- Integer part:
710=11127_{10} = 111_2710=1112 - Fractional part:
0.4375100.4375_{10}0.437510:
0.4375×2=0.875→00.4375 \times 2 = 0.875 → 00.4375×2=0.875→0
0.875×2=1.75→10.875 \times 2 = 1.75 → 10.875×2=1.75→1
0.75×2=1.5→10.75 \times 2 = 1.5 → 10.75×2=1.5→1
0.5×2=1.0→10.5 \times 2 = 1.0 → 10.5×2=1.0→1
→ So, 0.4375=0.011120.4375 = 0.0111_20.4375=0.01112
Thus,
7.4375 = 111.0111 in binary
Step 3: Normalize the binary number
Convert 111.0111 to scientific form:1.110111×221.110111 \times 2^21.110111×22
- Mantissa (f): Take the bits after the decimal point:
f = 11011100000000000000000 (23 bits) - Exponent (e):
E=2+127=129→e=100000012E = 2 + 127 = 129 → e = 10000001_2E=2+127=129→e=100000012
Final IEEE754 Encoding:
- s = 1
- e = 10000001
- f = 11011100000000000000000
(b) What is the smallest value that can be added to a float x without loss of precision?
This smallest increment is known as machine epsilon (ε) and depends on the current value of x. In general, for IEEE 754 single precision:ϵ(x)=2e−23\epsilon(x) = 2^{e – 23}ϵ(x)=2e−23
Where e is the exponent of the normalized float. This value is the distance between x and the next representable float.
For example, if:
- x = 1.0, its binary is 1.0=1.00000000000000000000000×201.0 = 1.00000000000000000000000 \times 2^01.0=1.00000000000000000000000×20
- Then smallest addable value is: ϵ=2−23≈1.1920929×10−7\epsilon = 2^{-23} ≈ 1.1920929 \times 10^{-7}ϵ=2−23≈1.1920929×10−7
This is because IEEE754 can represent 23 fractional bits, and 2−232^{-23}2−23 is the smallest change we can make at that magnitude without being rounded off.
Summary Answers
a. For -7.4375:
- s = 1
- e = 10000001
- f = 11011100000000000000000
b. Smallest value that can be added to x (e.g., x = 1.0):
- ε = 2⁻²³ ≈ 1.1920929 × 10⁻⁷
This is the unit in the last place (ULP) and indicates floating-point precision.
