Math Problem Statement

what is the unit roundoff error for a binary machine carrying 48-bit mantissas?

Solution

The unit roundoff error, often denoted as u, is the smallest difference that can be represented in a given floating-point system. For a binary machine, this depends on the number of bits used in the mantissa.

In general, the unit roundoff error for a floating-point system is:

u=12×2(p1)=2pu = \frac{1}{2} \times 2^{-(p-1)} = 2^{-p}

Where:

  • p is the number of bits in the mantissa.

In this case, the machine carries a 48-bit mantissa (p = 48), so the unit roundoff error would be:

u=248u = 2^{-48}

Now, let's calculate 2482^{-48}:

2483.5527136788×10152^{-48} \approx 3.5527136788 \times 10^{-15}

Thus, the unit roundoff error for a binary machine with a 48-bit mantissa is approximately 3.55×10153.55 \times 10^{-15}.

Would you like to go into more details on floating-point precision and how roundoff errors affect numerical computations?

Here are five related questions for deeper understanding:

  1. What is the significance of the mantissa in floating-point arithmetic?
  2. How does roundoff error influence numerical stability in algorithms?
  3. Can you explain the difference between absolute and relative error?
  4. How does increasing the number of bits in the mantissa reduce roundoff error?
  5. What role does the exponent play in floating-point numbers?

Tip: Always account for roundoff error in large numerical computations to ensure the stability and accuracy of results.

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Math Problem Analysis

Mathematical Concepts

Floating-Point Arithmetic
Numerical Analysis
Error Analysis

Formulas

u = 2^{-p}

Theorems

Floating-Point Precision Theorem

Suitable Grade Level

Undergraduate (Math/Computer Science)