What causes the round-off error? What kind of calculations are most susceptible to round-off error?

Short Answer

Expert verified
Question: What causes round-off errors, and provide an example of a calculation that is highly susceptible to these errors. Answer: Round-off errors are caused by representing numbers with a finite number of bits and performing arithmetic operations involving numbers with limited precision. An example of a calculation highly susceptible to round-off errors is the subtraction of two nearly equal numbers, also known as catastrophic cancellation. For instance, subtracting 2.0000001 and 2.000000001 with limited precision could result in an erroneous result of 0.000000 instead of the correct difference of 0.000000099.

Step by step solution

01

Definition of Round-off Error

Round-off error is the difference between the exact value of a number and its approximate representation due to limitations in a computer's storage capacity. Computers use a finite number of bits to represent real numbers, so there will always be some loss of precision when dealing with irrational numbers (such as pi) or very large or very small numbers.
02

Causes of Round-off Error

Round-off error is mainly caused by two factors: 1. Representing numbers with a finite number of bits: Computers often represent real numbers using a fixed number of bits, and this limits the precision of the numbers it can represent. For example, the standard floating-point format (IEEE 754) uses 32 or 64 bits to represent a number. 2. Arithmetic operations: Round-off errors can accumulate and propagate through calculations as a result of addition, subtraction, multiplication, or division involving numbers with limited precision.
03

Example of Calculations Susceptible to Round-off Error

Subtraction of two nearly equal numbers is an example of a calculation that is highly susceptible to round-off error. This is known as catastrophic cancellation. Suppose we have two numbers \(x = 2.0000001\) and \(y = 2.000000001\). Using a limited-precision format such as single-precision floating-point (which has 7 decimal digits of precision), both numbers will be stored in the computer's memory, approximately as: \(x \approx 2.000000\), \(y \approx 2.000000\). Now, when we subtract them, we get the erroneous result: \(x - y \approx 2.000000 - 2.000000 = 0.000000\). This error occurred because we lost the precision due to the limited number of bits used to represent the numbers. The correct difference between the numbers is: \(x - y = 2.0000001 - 2.000000001 = 0.000000099\). So, the round-off error in this case is \(0.000000099\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Computer Numerical Precision
Computers are incredibly powerful tools for processing and calculating numerical data. However, they have a fundamental limitation: they cannot represent most real numbers exactly. This limitation comes from the need to store values using a finite number of bits, leading to a constraint known as computer numerical precision.

In the context of numerical computing, precision refers to the level of detail in representing a number. For example, if we can only use two decimal places to represent the value of pie, it would be roughly 3.14 instead of its true infinite decimal representation. In a computer, we use binary digits (bits) instead of decimal places, and this representation can only approximate real numbers up to a certain point. When performing arithmetic operations like addition and subtraction, this precision issue becomes evident, especially when dealing with very small differences in large numbers, or numbers that would require more bits than are available to represent their exact decimal or binary value.

Improvement in numerical precision can be made by using more bits to represent numbers (going from 32-bit to 64-bit representations, for example), but there is always a balance to be struck between the computational resources used and the precision required. For most applications, standard levels of precision are adequate, but for calculations requiring high precision, such as scientific modeling or financial computations, even the slightest inaccuracy could lead to significantly different outcomes.
IEEE 754 Standard
When discussing the precision of floating-point numbers, the IEEE 754 standard often comes into the conversation. This standard is a widely-adopted system for computer arithmetic on floating-point numbers. The IEEE, or the Institute of Electrical and Electronics Engineers, set forth this standard to unify the approach to binary floating-point computation to minimize discrepancies across different computing systems.

The standard defines both the representation and behavior of floating-point numbers, including formats for different precision levels and rules for rounding, exceptions, and operation methods. The most commonly used formats are the 32-bit (single precision) and 64-bit (double precision) representations. Single precision offers about 7 decimal digits of precision, while double precision allows for about 16 decimal digits.

This unified approach allows for the reliable exchange of floating-point data between different applications and hardware without unexpected changes in the results. However, even with the IEEE 754 standard, the issue of round-off error cannot be entirely eliminated, and understanding its causes and effects remains crucial for developers and numerical analysts.
Catastrophic Cancellation
One specific type of round-off error that can greatly impact calculations is known as catastrophic cancellation. This situation arises most frequently during the subtraction of two nearly equal floating-point numbers. When such numbers are subtracted, the significant digits of the smaller number are largely canceled out by the corresponding digits of the larger number. This event leads to a result where the digits that remain are primarily the insignificant, unreliable digits that were introduced by round-off error.

For example, let's consider the exercise provided, where we have two very close numbers, and their subtraction in single-precision floating-point format leads to a complete loss of significance. The end result is effectively zero, rather than the very small difference that should theoretically exist between them. This scenario is a classic demonstration of catastrophic cancellation.

To minimize the risk of catastrophic cancellation, it is crucial to use algorithms that maintain numerical stability and to be mindful of the order of operations in calculations. Certain mathematical techniques, such as Kahan summation algorithm or compensated arithmetic, can help account for and correct some of the round-off errors that occur in computations sensitive to catastrophic cancellation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How do numerical solution methods differ from analytical ones? What are the advantages and disadvantages of numerical and analytical methods?

In the energy balance formulation of the finite difference method, it is recommended that all heat transfer at the boundaries of the volume element be assumed to be into the volume element even for steady heat conduction. Is this a valid recommendation even though it seems to violate the conservation of energy principle?

The wall of a heat exchanger separates hot water at \(T_{A}=90^{\circ} \mathrm{C}\) from cold water at \(T_{B}=10^{\circ} \mathrm{C}\). To extend the heat transfer area, two-dimensional ridges are machined on the cold side of the wall, as shown in Fig. P5-76. This geometry causes non-uniform thermal stresses, which may become critical for crack initiation along the lines between two ridges. To predict thermal stresses, the temperature field inside the wall must be determined. Convection coefficients are high enough so that the surface temperature is equal to that of the water on each side of the wall. (a) Identify the smallest section of the wall that can be analyzed in order to find the temperature field in the whole wall. (b) For the domain found in part \((a)\), construct a twodimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\) and write the matrix equation \(A T=C\) (elements of matrices \(A\) and \(C\) must be numbers). Do not solve for \(T\). (c) A thermocouple mounted at point \(M\) reads \(46.9^{\circ} \mathrm{C}\). Determine the other unknown temperatures in the grid defined in part (b).

A DC motor delivers mechanical power to a rotating stainless steel shaft ( \(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with a length of \(25 \mathrm{~cm}\) and a diameter of \(25 \mathrm{~mm}\). The DC motor is in a surrounding with ambient air temperature of \(20^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the base temperature of the motor shaft is \(90^{\circ} \mathrm{C}\). Using a uniform nodal spacing of \(5 \mathrm{~cm}\) along the motor shaft, determine the finite difference equations and the nodal temperatures by solving those equations.

Consider steady one-dimensional heat conduction in a pin fin of constant diameter \(D\) with constant thermal conductivity. The fin is losing heat by convection to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\). The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of this problem to determine \(T_{1}\) and \(T_{2}\) for the case of specified temperature at the fin base and negligible heat transfer at the fin tip. All temperatures are in \({ }^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free