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Computer ArchitectureFall 2007 © September 5, 2007 Karem Sakallah CS 447 – Computer Architecture.

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Presentation on theme: "Computer ArchitectureFall 2007 © September 5, 2007 Karem Sakallah CS 447 – Computer Architecture."— Presentation transcript:

1 Computer ArchitectureFall 2007 © September 5, 2007 Karem Sakallah ksakalla@qatar.cmu.edu www.qatar.cmu.edu/~msakr/15447-f07/ CS 447 – Computer Architecture Lecture 3 Computer Arithmetic (2)

2 Computer ArchitectureFall 2007 © Last Time °Representation of finite-width unsigned and signed integers °Addition and subtraction of integers °Multiplication of unsigned integers

3 Computer ArchitectureFall 2007 © Today °Review of multiplication of unsigned integers °Multiplication of signed integers °Division of unsigned integers °Representation of real numbers °Addition, subtraction, multiplication, and division of real numbers

4 Computer ArchitectureFall 2007 © Multiplication °Complex °Work out partial product for each digit °Take care with place value (column) °Add partial products

5 Computer ArchitectureFall 2007 © Multiplication Example ° 1011 Multiplicand (11 dec) ° x 1101 Multiplier (13 dec) ° 1011 Partial products ° 0000 Note: if multiplier bit is 1 copy ° 1011 multiplicand (place value) ° 1011 otherwise zero ° 10001111 Product (143 dec) ° Note: need double length result

6 Computer ArchitectureFall 2007 © Unsigned Binary Multiplication

7 Computer ArchitectureFall 2007 © Execution of Example

8 Computer ArchitectureFall 2007 © Flowchart for Unsigned Binary Multiplication

9 Computer ArchitectureFall 2007 © Multiplying Negative Numbers °This does not work! °Solution 1 Convert to positive if required Multiply as above If signs were different, negate answer °Solution 2 Booth’s algorithm

10 Computer ArchitectureFall 2007 © Observation °Which of these two multiplications is more difficult? 98,765 x 10,001 98,765 x 9,999 °Note: 98,765 x 10,001 = 98,765 x (10,000 + 1) 98,765 x 9,999 = 98,765 x (10,000 – 1)

11 Computer ArchitectureFall 2007 © In Binary Let Q = d d d 0 1 1... 1 1 0 d d QL = d d d 1 0 0... 0 0 0 d d QR = 0 0 0 0 0 0... 0 1 0 0 0

12 Computer ArchitectureFall 2007 © In Binary Let Q = d d d 0 1 1... 1 1 0 d d QL = d d d 1 0 0... 0 0 0 d d QR = 0 0 0 0 0 0... 0 1 0 0 0 Then Q = QL – QR And M x Q = M x QL – M x QR

13 Computer ArchitectureFall 2007 © A bit more explanation P = M x Q where M and Q are n-bit two’s complement integers e.g., Q = -2 3 q 3 + 2 2 q 2 + 2 1 q 1 + 2 0 q 0 which can be re-written as follows: Q = 2 0 (q -1 –q 0 )+2 1 (q 0 -q 1 )+2 2 (q 1 -q 2 )+2 3 (q 2 -q 3 ) leading to P =M x 2 0 x (q -1 - q 0 ) + M x 2 1 x (q 0 - q 1 ) + M x 2 2 x (q 1 - q 2 ) + M x 2 3 x (q 2 - q 3 )

14 Computer ArchitectureFall 2007 © Finally! P =M x 2 0 x (q -1 - q 0 ) + M x 2 1 x (q 0 - q 1 ) + M x 2 2 x (q 1 - q 2 ) + M x 2 3 x (q 2 - q 3 ) qiqi q i-1 q i-1 – q i Action 000Noop 011Add M 10Subtract M 110Noop

15 Computer ArchitectureFall 2007 © Example of Booth’s Algorithm

16 Computer ArchitectureFall 2007 © Booth’s Algorithm

17 Computer ArchitectureFall 2007 © Division °More complex than multiplication °Negative numbers are really bad! °Based on long division

18 Computer ArchitectureFall 2007 © Division of Unsigned Binary Integers 001111 1011 00001101 10010011 1011 001110 1011 100 Quotient Dividend Remainder Partial Remainders Divisor

19 Computer ArchitectureFall 2007 © Flowchart for Unsigned Binary Division

20 Computer ArchitectureFall 2007 © Real Numbers °Numbers with fractions °Could be done in pure binary 1001.1010 = 2 4 + 2 0 +2 -1 + 2 -3 =9.625 °Where is the binary point? °Fixed? Very limited °Moving? How do you show where it is?

21 Computer ArchitectureFall 2007 © Floating Point °+/-.significand x 2 exponent °Misnomer °Point is actually fixed between sign bit and body of mantissa °Exponent indicates place value (point position) Sign bit Biased Exponent Significand or Mantissa

22 Computer ArchitectureFall 2007 © Floating Point Examples

23 Computer ArchitectureFall 2007 © Signs for Floating Point °Mantissa is stored in two’s complement °Exponent is in excess or biased notation e.g. Excess (bias) 128 means 8 bit exponent field Pure value range 0-255 Subtract 128 to get correct value Range -128 to +127

24 Computer ArchitectureFall 2007 © Normalization °FP numbers are usually normalized °i.e. exponent is adjusted so that leading bit (MSB) of mantissa is 1 °Since it is always 1 there is no need to store it °(c.f. Scientific notation where numbers are normalized to give a single digit before the decimal point °e.g. 3.123 x 10 3 )

25 Computer ArchitectureFall 2007 © FP Ranges °For a 32 bit number 8 bit exponent +/- 2 256  1.5 x 10 77 °Accuracy The effect of changing lsb of mantissa 23 bit mantissa 2 -23  1.2 x 10 -7 About 6 decimal places

26 Computer ArchitectureFall 2007 © Expressible Numbers

27 Computer ArchitectureFall 2007 © Density of Floating Point Numbers

28 Computer ArchitectureFall 2007 © IEEE 754 Formats

29 Computer ArchitectureFall 2007 © FP Arithmetic +/- °Check for zeros °Align significands (adjusting exponents) °Add or subtract significands °Normalize result

30 Computer ArchitectureFall 2007 © FP Addition & Subtraction Flowchart

31 Computer ArchitectureFall 2007 © FP Arithmetic x/  °Check for zero °Add/subtract exponents °Multiply/divide significands (watch sign) °Normalize °Round °All intermediate results should be in double length storage

32 Computer ArchitectureFall 2007 © Floating Point Multiplication

33 Computer ArchitectureFall 2007 © Floating Point Division


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