2.2 Errors

1. 2.2 Errors

2. Why Study Errors First? • Nearly all our modeling is done on digital computers (aside: what would a non-digital analog computer look like?)

3. Analog Computer for Fitting a Line to a Set of Points

4. Why Study Errors First? Nearly all our modeling is done on digital computers (aside: what would a non-digital analog computer look like?) A better set of terms is discrete vs. continuous (recall the discrete/continuous distinction from 1.2). Discrete devices have finite precision - we cannot represent all the different values we’d like. Finite precision leads to errors

5. Finite Precision with Bits • Bit = value of 0 or 1 (aside: why base 2 ?) • With n bits, we can distinguish 2n different values • E.g., with three bits we get eight possible values. • These values can correspond to anything we want to represent… 000 0 A Greg A 001 1 B Wade B 010 2 C Sam C 011 3 D Jordan I 100 4 E Mark E 101 5 F Jason F 110 6 G Gaurav G 111 7 H Alex H

6. Floating-Point Numbers • History: origins of computers in finance (banking) and military (physics). • Financial calculations are (usually) fixed-point: the decimal place is fixed in one position, two digits from right: $129.95, $0.59, etc. • In physical calculations, we need the decimal place to “float”: 3.14159, 0.333, 6.0221415 × 1023 , etc.

7. Exponential Notation • Format: Mantissa X 10Exponent, or a X 10n • Abbreviated a e n ora E n; for example, 6.0221415e23 • Mantissa (a.k.a. significand, a.k.a. fractional part) is a floating-point number; exponent is an integer. • So someone has to decide the precision: • IEEE 754 standard: with 64-bit numbers, use 52 bits for mantissa, 11 for exponent, 1 for sign (+ or −) • Rough conversion: 3 bits  1 decimal digit (why?) E.g., 53 bits  16 decimal digits of precision

8. Normalized Notation • Normalization is a common operation: e.g., percentage normalizes everything to 100: 2/4 = 75/50 = 50% • Scientists usually prefer to normalize to 1: 2/4 = 75/150 = 0.5 • So a normalized number in exponential notation has the decimal place immediately before the first nonzero digit: .314159e1, .60221415e24, etc. • Significant digits of a normalized floating-point number are all digits except leading zeros.

9. Precision and Magnitude . 3 1 4 1 5 9 e 1 6 digits of precision Magnitude = 1 • Precision is the number of significant digits. • Magnitude is the power of 10.

10. Quick Review Question 1 • For the number 0.0004500, give • the significand in normalized exponenent notation • the magnitude in normalized exponenent notation • the precision

11. Absolute and Relative Error absolute error = | correct - result | relative error = (absolute error) | correct | = (absolute error) X 100% | correct | If correct is the answer and result is the result obtained, then

12. Truncation To truncate a normalized number to k significant digits, eliminate all digits of the significand beyond the kth digit. E.g., .314159 truncated to 5 significant digits = .?????

13. Truncation X 100% | correct | • To truncate a normalized number to k significant digits, eliminate all digits of the significand beyond the kth digit. • E.g., .314159 truncated to 5 significant digits = .31415 • Relative error = | correct - result |

14. Truncation X 100% | .314159 | • To truncate a normalized number to k significant digits, eliminate all digits of the significand beyond the kth digit. • E.g., .314159 truncated to 5 significant digits = .31415 • Relative error = | .314159 - .31415 |

15. Truncation X 100% .314159 • To truncate a normalized number to k significant digits, eliminate all digits of the significand beyond the kth digit. • E.g., .314159 truncated to 5 significant digits = .31415 • Relative error = .00009

16. Truncation To truncate a normalized number to k significant digits, eliminate all digits of the significand beyond the kth digit. E.g., .314159 truncated to 5 significant digits = .31415 Relative error =0.0286479 %

17. Rounding To round a normalized number to precision k, consider the (k+1)th significant digit d. If it is less than 5, round down the normalized number by truncating the significand to k significant digits. If d is greater than or equal to 5, round up the normalized number by truncating the significand to k significant digits and then adding 1 to the kth significant digit of the significand, carrying as necessary to digits on the left.

18. Quick Review Question 3 • Round each of the following so that the significand has a precision of 2: • 0.93742e-5 • 0.93472e-5 • 0.93572e-5

19. Assignment • An assignment statement causes the computer to store the value of an expression in a memory location associated with a variable. In most programming languages the assignment statement has a format similar to the following, with the expression always appearing on the right amd the varoable getting the value always being on the left of an assignment operator, here an equal sign: variable = expression

20. Assignment R-value L-value • An assignment statement causes the computer to store the value of an expression in a memory location associated with a variable. In most programming languages the assignment statement has a format similar to the following, with the expression always appearing on the right amd the varoable getting the value always being on the left of an assignment operator, here an equal sign: variable = expression

21. Assignment • An assignment statement causes the computer to store the value of an expression in a memory location associated with a variable. In most programming languages the assignment statement has a format similar to the following, with the expression always appearing on the right amd the varoable getting the value always being on the left of an assignment operator, here an equal sign: X = X + 1

22. Round-off Error • Round-off error is the problem of not having enough bits (or digits) to store an entire floating-point number and approximating the result to the nearest number that can be represented. • E.g., 1/3 = 0.333 ≠ 0.3 ≠ 0.33 ≠ 0.333 ≠ etc.

23. Avoidnig Round-off Error • Some values (1/3) will always result in roundoff; however, we can help reduce error: • When adding numbers whose magnitudes are drastically different, accumulate smaller number before combining them with larger ones. • When multiplying and dividing, perform all multiplcations in the numerator before dividing by the denominator.

24. Avoiding Round-off Error • E.g. consider computing (x/y)z on a machine with three significant digits of precision, where x = 2.41 y = 9.75 z = 1.54 (x / y)z = (2.41 / 9.75)(1.54) = (0.247)(1.54) = 0.380 But (x / y)z = (xz) / y: (xz) / y = ((2.41) )(1.54)) / 9.75 = 3.71 / 9.75 = 0.381

25. Overflow and Underflow • Overflow is an error condition that occurs when there are not enough bits to express a value in a computer. • Underflow is an error condition that occurs when the result of a computation is too small for the computer to represent.

26. Overflow and Underflow in Excel

27. Looping and Error Propagation • A loop is a segment of code [computer instructions] that is executed repeatedly. • Accumulating values in a loop (repeatedly) can produce error • E.g., Patriot Missile failure in 1991 Gulf War

28. Patriot Missile Failure • Patriot’s internal clock measured time in 1/10 of a second. • Each tick of the clock added 1/10 s to current time. • But 1/10 cannot be represented in a finite number of bits (just as 1/3 cannot be represented in a finite number of decimal digits)

29. Representing Fractions in Binary • Binary works the same way; e.g., what is 101.012? • 1 0 1 . 0 1 22 21 20 .2-1 2-2 • = 4 + 1 + 1/4 = 5.025 • Consider ordinary Base-10 notation: what does e.g. 1205.91410 mean? 1 2 0 5 . 9 1 4 103 102 101 100 .10-1 10-2 10-3

30. Patriot Missile Failure = 0.34 sec • 10 hr 60 min 60 sec .000000095 sec • 1 hr 1 min 1 sec • Scud travels 1,676 meters / sec. So error = • 1676 meters 0.34 sec • 1 sec = 570 meters • Result Patriot missed Scud , 28 Americans died • Each one-tenth increment produced an error of about .000000095 seconds. • After ten hours of running Patriot’s computer:

31. Error in the Text! During that third of a second, a Scud flew about 1,676 meters. Shiflet & Shiflet p. 24 A Scud travels at about 1,676 meters per second http://www.ima.umn.edu/~arnold/disasters/patriot.html (FYI: speed of sound at sea level = 340 m / s)

32. Looping and Error Propagation in Excel* + See apparent uniform sequence: Highlight; click corner; drag downward: Double click for actual values: Enter values: *Courtesy of Bob Panoff

33. Avoiding Error Propagation in Excel* Enter first value; highlight rest: Do Fill/Series: *Courtesy of Bob Panoff