13.3 Floating-point numbers, representation and manipulation (3)
Resources |
Revision Questions |
Computer Science
Login to see all questions
Click on a question to view the answer
1.
Consider a scenario where you are performing complex mathematical calculations using a computer with a limited number of bits to represent real numbers. Explain how the limitations of binary representation could lead to unexpected results or errors. Provide a specific example illustrating this issue.
When performing complex mathematical calculations with limited-precision binary representation, the accumulation of rounding errors can lead to significant deviations from the expected results. These errors can propagate through the calculations, eventually resulting in inaccurate or even incorrect outcomes. This is particularly problematic in iterative algorithms or calculations involving many steps.
Example: Subtracting Nearly Equal Numbers
Consider the calculation: (1.0 + 1.0e-17) - 1.0. Ideally, the result should be 1.0e-17. However, due to the limited precision of binary representation, the subtraction might result in a value very close to 1.0e-17, but not exactly equal. The subtraction essentially loses the small 1.0e-17, and the result is rounded to the nearest representable value. This is because the difference between 1.0 + 1.0e-17 and 1.0 is smaller than the smallest representable difference.
Further Explanation:
- Each operation (addition, subtraction, multiplication, division) introduces a small rounding error.
- These errors accumulate over multiple operations.
- The magnitude of the error is proportional to the magnitude of the numbers being processed and the precision of the representation.
- In some cases, the accumulated error can be significant enough to completely alter the outcome of the calculation.
This issue is a fundamental limitation of floating-point arithmetic and is a common source of errors in scientific computing, financial modeling, and other applications where high precision is required. Techniques like interval arithmetic and arbitrary-precision arithmetic are used to mitigate these errors, but they come with increased computational cost.
2.
Explain the concept of 'normalization' in the context of binary floating-point numbers. Why is normalization necessary, and what are the consequences of not normalizing a number?
Normalization is the process of adjusting the binary representation of a floating-point number so that it is in the form 1.xxxx... (where xxxx... represents the fractional part). This involves shifting the binary point until only one non-zero digit precedes the binary point.
Normalization is necessary for several reasons:
- Increased Precision: Normalization allows for a more precise representation of numbers with a wider range of magnitudes. It ensures that the most significant digit is always '1', maximizing the use of the available bits for the fractional part.
- Avoiding Redundant Representation: Without normalization, the same number could be represented in multiple ways, leading to inconsistencies and potential errors.
- Efficient Storage: Normalization allows for efficient storage of floating-point numbers by ensuring a consistent format.
If a number is not normalized, it can lead to:
- Loss of Precision: Numbers with leading zeros after the binary point will occupy extra bits, reducing the precision available for the fractional part.
- Multiple Representations: The same decimal number might have different binary representations, which can cause problems in calculations and comparisons.
- Increased Storage Requirements: Unnormalized numbers require more bits to represent accurately.
3.
Consider a decimal number 65536. Convert this number to its binary floating-point representation using the single-precision (32-bit) IEEE 754 format. Show each step of the conversion, including the calculation of the exponent and mantissa.
1. Convert 65536 to binary: 65536 in decimal is 10000000000 in binary.
2. Normalize the binary number: The normalized form is 1.10000000000000000000000 (we've added an implicit leading '1').
3. Determine the sign bit: Since 65536 is positive, the sign bit is 0.
4. Calculate the exponent: The exponent in the normalized form is 10. The biased exponent is (10 - 127) = -117. The exponent in binary is 01110101.
5. Determine the mantissa: The mantissa is 10000000000000000000000. Since we are using 23 bits for the mantissa, we only take the first 23 bits: 10000000000000000000000.
6. Combine the bits: The complete 32-bit representation is: 0 01110101 10000000000000000000000
7. Represent in hexadecimal: The binary representation is 0x40000000.