Lecture 1 - Numerical Methods
- 1. Numerical Methods Lecture 1 Representationof Numerical Values in the Computer
- 2. Overview – Number Systems –Number Conversion – Representation of Numbers – Computer Arithmetic – Errors in Arithmetic Tashreef Muhammad 2
- 3. Number Systems Let usdiscuss mainly about “Positional Number Systems” Tashreef Muhammad 3
- 4. Number Systems – Non-positionalNumber Systems – Positional Number Systems – Decimal – Binary – Hexadecimal Tashreef Muhammad 4
- 5. Number Conversion Converting numbersrepresented in one format, to other Tashreef Muhammad 5
- 6. Number Conversion – Decimalto Non-decimal Number – Non-decimal to Decimal Number – Binary to Hexadecimal Number – Hexadecimal to Binary Number Tashreef Muhammad 6
- 7. Representation of Numbers How moderncomputers store and represent numbers Tashreef Muhammad 7
- 8. Representation of Numbers –Dependent on words – Words are a number of bits – Word size varies from machine to machine – Computers process using binary – Humans easily understand decimal – Hexadecimal expresses more in less digits Tashreef Muhammad 8
- 9. Representation of Numbers –Integer Representation – For word size n, maximum possible number to store is 2n – 1 – Negative numbers are stored as 2’s complement – The use of sign bit makes the highest possible number 2n-1 – 1 – First toggle all the bits of the number – Add 1 to the whole number – Use the sign bit as it is Tashreef Muhammad 9
- 10. Representation of Numbers –Floating Point Representation – Exponential form, 𝑥 = 𝑓 × 10𝐸 – 𝑓 is mantissa and 𝐸 is exponent – The entire memory location is divided to three parts – Sign, mantissa and exponent – Typically 24 bits for mantissa and 7 bits for exponent are used on a 32 bit word representation system Tashreef Muhammad 10
- 11. Representation of Numbers A samplefloating point representation from Numerical Methods textbook (Figure 3.1) Tashreef Muhammad 11 SIG N EXPONENT MANTISSA 1 bit 7 bits 24 bits 0 24 23 31 30
- 12. Representation of Numbers –Floating Point Representation – Shifting of decimal pints Normalisation – Numbers in normalized form Normalised Floating Point Numbers – Two possible notations: 0.596 x 10-2 or .596E-2 – Conditions to be satisfied by mantissa, 𝑓 – For positive numbers: 0.1 ≤ 𝑓 < 1.0 – For negative numbers: −1 < 𝑓 ≤ −0.1 – In general: 0.1 ≤ |𝑓| < 1.0 Tashreef Muhammad 12
- 13. Computer Arithmetic How computersprocess arithmetic operations Tashreef Muhammad 13
- 14. Computer Arithmetic – IntegerArithmetic – Integer arithmetic always results in integer – Exception: – Cannot represent infinite numbers, bounded above and below – Integer division provides two results: quotient and remainder – Sample integer arithmetic: – 25 + 12 = 37, 25 - 12 = 13, 12 – 25 = -13 – 25 x 12 = 300, 25 ÷ 12 = 2, 12 ÷ 25 = 0 – Show the following is incorrect for integer arithmetic – 𝑎+𝑏 𝑐 = 𝑎 𝑐 + 𝑏 𝑐 Tashreef Muhammad 14
- 15. Computer Arithmetic –Floating PointArithmetic: Addition – Let 𝑥 and 𝑦 be added to result in 𝑧; where 𝑥 ≥ 𝑦 – The fractional parts: f𝑥, f𝑦, f𝑧 – The exponent parts: E𝑥, E𝑦, E𝑧; where E𝑥 ≥ 𝐸𝑦 – Algorithm: – Set E𝑧 = max(E𝑥, E𝑦) E𝑧 = E𝑥 – Shift f𝑦 to the right by E𝑥 − E𝑦 places – Set f𝑧 = f𝑥 + f𝑦 – Normalize if necessary Tashreef Muhammad 15
- 16. Computer Arithmetic –Floating PointArithmetic: Subtraction – Addition with different sign bits – May result in mantissa overflow – Left shift f𝑧 and decrease E𝑧 accordingly Tashreef Muhammad 16
- 17. Computer Arithmetic –Floating PointArithmetic: Multiplication – Multiply mantissa, f𝑧 = 𝑓𝑥 × 𝑓𝑦 – Add the exponents, E𝑧 = 𝐸𝑥 + 𝐸𝑦 – Normalise if necessary Tashreef Muhammad 17
- 18. Computer Arithmetic –Floating PointArithmetic: Division – Divide mantissa, f𝑧 = 𝑓𝑥 ÷ 𝑓𝑦 – Subtract the exponents, E𝑧 = 𝐸𝑥 − 𝐸𝑦 – Normalise if necessary Tashreef Muhammad 18
- 19. Computer Arithmetic –Floating PointArithmetic: Practice – Add 0.964572 E2 and 0.586351 E5 – Add 0.735816 E4 and 0.635742 E4 – Subtract 0.994576 E-3 from 0.999658 E-3 – Multiply 0.20 E4 and 0.40 E-2 – Divide 0.876543 by 0.200000 E-3 Tashreef Muhammad 19
- 20. Computer Arithmetic –Floating PointArithmetic: Practice Answers – Add 0.964572 E2 and 0.586351 E5 0.587315 E5 – Add 0.735816 E4 and 0.635742 E4 0.1371558 E5 – Subtract 0.994576 E-3 from 0.999658 E-3 0.508200 E-5 – Multiply 0.20 E4 and 0.40 E-2 0.80 E1 – Divide 0.876543 by 0.200000 E-3 0.438271 E-1 Tashreef Muhammad 20
- 21. Errors in Arithmetic Errorsthat are seen in arithmetic operations Tashreef Muhammad 21
- 22. Errors in Arithmetic –FloatingPoint Arithmetic – Inexact binary representation of decimal number: 0.1 to 0.0001100110011… – Errors due to rounding-off a number: Add the numbers 0.500000 E1 and 0.100000 E-7 – Subtractive cancellation: Subtract 0.499998 from 0.500000 – Overflow or underflow of numbers Tashreef Muhammad 22
- 23. Errors in Arithmetic –Laws of Arithmetic – Errors in arithmetic induce error in associative and distributive law errors – 𝑥 + 𝑦 + 𝑧 ≠ 𝑥 + 𝑦 + 𝑧 and 𝑥 × 𝑦 × 𝑧 ≠ 𝑥 × 𝑦 × 𝑧 – 𝑥 × 𝑦 + 𝑧 ≠ 𝑥 × 𝑦 + 𝑥 × 𝑧 – Provide an example where associative law of addition fails due to an arithmetic error of floating point number – 𝑥 = 0.456732 × 10−2 , 𝑦 = 0.243451, 𝑧 = −0.248000 – 𝑥 + 𝑦 + 𝑧 ≠ 𝑥 + 𝑦 + 𝑧 Tashreef Muhammad 23
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