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ELEC 5200-002/6200-002 Computer Architecture and Design Fall 2006 Computer Arithmetic (Chapter 3)

ELEC 5200-002/6200-002 Computer Architecture and Design Fall 2006 Computer Arithmetic (Chapter 3). Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu.

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ELEC 5200-002/6200-002 Computer Architecture and Design Fall 2006 Computer Arithmetic (Chapter 3)

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  1. ELEC 5200-002/6200-002Computer Architecture and DesignFall 2006Computer Arithmetic(Chapter 3) Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu ELEC 5200-002/6200-002 Lecture 6

  2. What Goes on Inside ALU? • Machine instr.: add $t1, $s1, $s2 • What it means to computer: 000000 10001 10010 01000 00000 100000 Arithmetic Logic Unit (ALU) Control unit Flags Registers Registers ELEC 5200-002/6200-002 Lecture 6

  3. Basic Idea • Hardware can only deal with binary digits, 0 and 1. • Must represent all numbers, integers or floating point, positive or negative, by binary digits, called bits. • Devise electronic circuits to perform arithmetic operations, add, subtract, multiply and divide, on binary numbers. ELEC 5200-002/6200-002 Lecture 6

  4. Positive Integers • Decimal system: made of 10 digits, {0,1,2, . . . , 9} 41 = 4×101 + 1×100 255 = 2×102 + 5×101 + 5×100 • Binary system: made of two digits, {0,1} 00101001 = 0×27 + 0×26 + 1×25 + 0×24 +1×23 +0×22 + 0×21 + 1×20 =32 + 8 +1 = 41 11111111 = 255, largest number with 8 binary digits, 28-1 ELEC 5200-002/6200-002 Lecture 6

  5. Base or Radix • For decimal system, 10 is called the base or radix. • Decimal 41 is also written as 4110 or 41ten • Base (radix) for binary system is 2. • Thus, 41ten = 1010012 or 101001two • Also, 111ten = 1101111two and 111two = 7ten ELEC 5200-002/6200-002 Lecture 6

  6. Signed Integers – What Not to Do • Use fixed length binary representation • Use left-most bit (called most significant bit or MSB) for sign: 0 for positive 1 for negative • Example: +18ten = 00010010two - 18ten = 10010010two ELEC 5200-002/6200-002 Lecture 6

  7. Why Not Use Sign Bit • Sign and magnitude bits should be differently treated in arithmetic operations. • Addition and subtraction require different logic circuits. • Overflow is difficult to detect. • “Zero” has two representations: +0ten = 00000000two - 0ten = 10000000two • Signed-integers are not used in modern computers. ELEC 5200-002/6200-002 Lecture 6

  8. Integers With Sign – Other Ways • Use fixed-length representation, but no sign bit • 1’s complement: To form a negative number, complement each bit in the given number. • 2’s complement: To form a negative number, start with the given number, subtract one, and then complement each bit, or first complement each bit, and then add 1. • 2’s complement is the most preferred representation. ELEC 5200-002/6200-002 Lecture 6

  9. 1’s-Complement • To change the sign of a binary integer simply complement (invert) each bit. • Example: 3 = 0011, -3 = 1100 • n-bit representation: Negation is equivalent to subtraction from 2n – 1 Infinite universe -3 0 3 0 3 6 -6 -3 -0 Modulo-16 universe 0 3 6 9 12 15 0000 0011 0110 1001 1100 1111 ELEC 5200-002/6200-002 Lecture 6

  10. 2’s-Complement • Add 1 to 1’s-complement representation. • Some properties: • Only one representation for 0 • Exactly as many positive numbers as negative numbers • Slight asymmetry – there is one negative number with no positive counterpart ELEC 5200-002/6200-002 Lecture 6

  11. Three Representations Sign-magnitude 000 = +0 001 = +1 010 = +2 011 = +3 100 = - 0 101 = - 1 110 = - 2 111 = - 3 1’s complement 000 = +0 001 = +1 010 = +2 011 = +3 100 = - 3 101 = - 2 110 = - 1 111 = - 0 2’s complement 000 = +0 001 = +1 010 = +2 011 = +3 100 = - 4 101 = - 3 110 = - 2 111 = - 1 (Preferred) ELEC 5200-002/6200-002 Lecture 6

  12. 000 001 0 111 1 -1 2 010 110 -2 -3 3 011 -4 101 100 2’s Complement Numbers ELEC 5200-002/6200-002 Lecture 6

  13. 2’s Complement n-bit Numbers • Range: - 2n-1 through 2n-1- 1 • Unique zero: 00000000 . . . . . 0 • Negation rule: see slide 8. • Expansion of bit length: stretch the left-most bit all the way, e.g., 11111101 is still – 3. • Overflow rule: If two numbers with the same sign bit (both positive or both negative) are added, the overflow occurs if and only if the result has the opposite sign. • Subtraction rule: for A – B, add – B to A. ELEC 5200-002/6200-002 Lecture 6

  14. Converting 2’s Compliment to Decimal n-1 an-1an-2 . . . a1a0 = -2n-1an-1 + Σ 2i ai i=0 8-bit conversion box -128 64 32 16 8 4 2 1 -128 64 32 16 8 4 2 1 1 1 1 1 1 1 0 1 Example -128+64+32+16+8+4+1 = -128 + 125 = -3 ELEC 5200-002/6200-002 Lecture 6

  15. MIPS • MIPS architecture uses 32-bit numbers. What is the range of integers (positive and negative) that can be represented? Positive integers: 0 to 2,147,483,647 Negative integers: - 1 to - 2,147,483,648 • What are the binary representations of the extreme positive and negative integers? 0111 1111 1111 1111 1111 1111 1111 1111 = 231 - 1= 2,147,483,647 1000 0000 0000 0000 0000 0000 0000 0000 = - 231 = - 2,147,483,648 • What is the binary representation of zero? 0000 0000 0000 0000 0000 0000 0000 0000 ELEC 5200-002/6200-002 Lecture 6

  16. Addition • Adding bits: • 0 + 0 = 0 • 0 + 1 = 1 • 1 + 0 = 1 • 1 + 1 = (1) 0 • Adding integers: carry 1 1 0 0 0 0 . . . . . . 0 1 1 1 two = 7ten + 0 0 0 . . . . . . 0 1 1 0 two = 6ten = 0 0 0 . . . . . . 1 (1)1 (1)0 (0)1 two = 13ten ELEC 5200-002/6200-002 Lecture 6

  17. Subtraction • Direct subtraction • Two’s complement subtraction 0 0 0 . . . . . . 0 1 1 1 two = 7ten - 0 0 0 . . . . . . 0 1 1 0 two = 6ten = 0 0 0 . . . . . . 0 0 0 1two = 1ten 1 1 1 . . . . . . 1 1 0 0 0 0 . . . . . . 0 1 1 1 two = 7ten + 1 1 1 . . . . . . 1 0 1 0 two = - 6ten = 0 0 0 . . . . . . 0 (1) 0 (1) 0 (0)1 two = 1ten ELEC 5200-002/6200-002 Lecture 6

  18. Overflow: An Error • Examples: Addition of 3-bit integers (range - 4 to +3) • -2-3 = -5110 = -2 + 101 = -3 = 1011 = 3 (error) • 3+2 = 5011 = 3 010 = 2 = 101 = -3 (error) • Overflow rule: If two numbers with the same sign bit (both positive or both negative) are added, the overflow occurs if and only if the result has the opposite sign. 000 111 0 001 1 -1 010 2 110 -2 3 -3 011 -4 101 100 ELEC 5200-002/6200-002 Lecture 6

  19. a half_sum XOR b carry_out AND Design Hardware Bit by Bit • Adding two bits: a b half_sum carry_out 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 • Half-adder circuit ELEC 5200-002/6200-002 Lecture 6

  20. One-bit Full-Adder • One-bit full-adder truth table a b ci half_sum carry_out sum co 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 0 0 1 1 1 0 0 1 0 1 1 1 1 0 1 1 1 ELEC 5200-002/6200-002 Lecture 6

  21. One-bit Full-Adder Circuit ci FAi XOR sumi ai XOR AND bi AND OR Ci+1 ELEC 5200-002/6200-002 Lecture 6

  22. 32-bit Ripple-Carry Adder c0 a0 b0 sum0 FA0 sum1 a1 b1 FA1 sum2 a2 b2 FA2 sum31 FA31 a31 b31 ELEC 5200-002/6200-002 Lecture 6

  23. How Fast is Ripple-Carry Adder? • Longest delay path (critical path) runs from cin to sum31. • Suppose delay of full-adder is 100ps. • Critical path delay = 3,200ps • Clock rate cannot be higher than 1012/3,200 = 312MHz. • Must use more efficient ways to handle carry. ELEC 5200-002/6200-002 Lecture 6

  24. a0-a15 16-bit ripple carry adder b0-b15 cin Speeding Up the Adder sum0-sum15 a16-a31 16-bit ripple carry adder 0 b16-b31 0 sum16-sum31 Multiplexer a16-a31 16-bit ripple carry adder 1 b16-b31 This is a carry-select adder 1 ELEC 5200-002/6200-002 Lecture 6

  25. Fast Adders • In general, any output of a 32-bit adder can be evaluated as a logic expression in terms of all 65 inputs. • Number of levels of logic can be reduced to log2N for N-bit adder. Ripple-carry has N levels. • More gates are needed, about log2N times that of ripple-carry design. • Fastest design is known as carry lookahead adder. ELEC 5200-002/6200-002 Lecture 6

  26. N-bit Adder Design Options Reference: J. L. Hennessy and D. A. Patterson, Computer Architecture: A Quantitative Approach, Second Edition, San Francisco, California, 1990. ELEC 5200-002/6200-002 Lecture 6

  27. MIPS Instructions (see p. 175) • Arithmetic: add, sub, addi, addu, subu, addiu, mfc0 • Data transfer: lw, sw, lhu, sh, lbu, sb, lui • Logical: and, or, nor, andi, ori, sll, srl • Conditional branch: beq, bne, slt, slti, sltu, sltiu • Unconditional jump: j, jr, jal ELEC 5200-002/6200-002 Lecture 6

  28. Exception or Interrupt • If an overflow is detected while executing add, addi or sub, then the address of that instruction is placed in a register called exception program counter (EPC). • Instruction mfc0 can copy $epc to any other register, e.g., mfc0 $s1, $epc • Unsigned operations, addu, addiu and subu do not cause an exception or interrupt. ELEC 5200-002/6200-002 Lecture 6

  29. Multifunction ALU operation c0 a0 b0 result0 ALU0 result1 a1 b1 ALU1 result2 operation a2 b2 ALU2 ci ai bi FAi 3 NOR 2 resulti mux OR 1 result31 ALU31 a31 b31 AND 0 ELEC 5200-002/6200-002 Lecture 6

  30. Binary Multiplication (Unsigned) 1 0 0 0 two = 8ten multiplicand 1 0 0 1 two = 9ten multiplier ____________ 1 0 0 0 0 0 0 0 partial products 0 0 0 0 1 0 0 0 ____________ 1 0 0 1 0 0 0two = 72ten ELEC 5200-002/6200-002 Lecture 6

  31. Multiplication Flowchart Start Initialize product register to 0 Partial product number, i = 1 LSB of multiplier ? Add multiplicand to product and place result in product register 1 0 Left shift multiplicand register 1 bit Right shift multiplier register 1 bit i = 32 i < 32 i = ? Done i = i + 1 ELEC 5200-002/6200-002 Lecture 6

  32. Serial Multiplication shift left shift right Multiplicand (expanded 64-bits) 32-bit multiplier 64 64 Test LSB 32 times 0 64-bit ALU 1 1 64 64-bit product register write 3 operations per bit: shift right shift left add ELEC 5200-002/6200-002 Lecture 6

  33. Serial Multiplication (Improved) Multiplicand 2 operations per bit: shift right add 32 32 1 Test LSB 32 times 32-bit ALU LSB 1 32 write 64-bit product register shift right 00000 . . . 00000 32-bit multiplier Initialized prod. Reg. ELEC 5200-002/6200-002 Lecture 6

  34. Example: 0010two× 0011two 0010two× 0011two = 0110two, i.e., 2ten×3ten = 6ten ELEC 5200-002/6200-002 Lecture 6

  35. Signed Multiplication • Convert numbers to magnitudes. • Multiply the two magnitudes through 32 iterations. • Negate the result if the signs of the multiplicand and multiplier differed. • Time of serial multiplication: N additions, O(N) clock cycles for N-bit integers. • A better method: Booth’s Algorithm. ELEC 5200-002/6200-002 Lecture 6

  36. Test 1 Statistics Test score (Maximum = 24) ELEC 5200-002/6200-002 Lecture 6

  37. Booth Multiplier Algorithm • A. D. Booth, “A Signed Binary Multiplication Technique,” Quarterly Journal of Mechanics and Applied Math., vol. 4, pt. 2, pp. 236-240, 1951. • Direct multiplication of positive and negative integers using two’s complement addition. ELEC 5200-002/6200-002 Lecture 6

  38. A Multiplication Trick • Consider decimal multiplication: 4 5 7 9 9 9 0 1 4 5 7 4 1 1 3 4 1 1 3 4 1 1 3 4 5 6 5 4 7 5 7 Three additions • Operations for each digit of multiplier: • Do nothing if the digit is 0 • Shift left, i.e., multiply by some power of 10 • Multiply by the digit, i.e., by a number between 1 and 9 ELEC 5200-002/6200-002 Lecture 6

  39. What is the Trick? • Examine multiplier: 99901 = 100000 – 100 + 1 • Multiply as follows: 457 × 100000 = 45700000 457 × ( - 100) = - 45700subtraction 45654300 457 × 1 = 457addition 45654757 Reduced from three to two operations. ELEC 5200-002/6200-002 Lecture 6

  40. Booth Algorithm: Basic Idea • Consider a multiplier, 00011110 (30) • We can write, 30 = 32 – 2, or 00100000 (32) = 25 +11111110 (-2) = - 21 00011110(0) 30 • Interpret multiplier (scan right to left), check bit-pairs: • kth bit is 1, (k-1)th bit is 0, multiplier contains -2k term • kth bit is 0, (k-1)th bit is 1, multiplier contains 2kterm • kth bit is 1, (k-1)th bit is 1, -2k is absent in multiplier • kth bit is 0, (k-1)th bit is 0, 2k is absent in multiplier • Product, M×30 = M×25 - M×21 M: multiplicand • Multiplication by 2k means a k-bit left shift ELEC 5200-002/6200-002 Lecture 6

  41. Booth Algorithm: Example 1 • 7 × 3 = 21 0111 multiplicand = 7 ×0011(0) multiplier = 3 11111001 bit-pair 10, add -7 in two’s com. bit-pair 11, do nothing 000111 bit-pair 01, add 7 bit-pair 00, do nothing 00010101 21 ELEC 5200-002/6200-002 Lecture 6

  42. Booth Algorithm: Example 2 • 7 × (-3) = -21 0111 multiplicand = 7 ×1101(0) multiplier = -3 11111001 bit-pair 10, add -7 in two’s com. 0000111 bit-pair 01, add 7 111001 bit-pair 10, add -7 in two’s com. bit-pair 11, do nothing 11101011 - 21 ELEC 5200-002/6200-002 Lecture 6

  43. Booth Algorithm: Example 3 • -7 × 3 = -21 1001 multiplicand = -7 in two’s com. ×0011(0)multiplier = 3 00000111 bit-pair 10, add 7 bit-pair 11, do nothing 111001 bit-pair 01, add -7 bit-pair 00, do nothing 11101011 - 21 ELEC 5200-002/6200-002 Lecture 6

  44. Booth Algorithm: Example 4 • -7 × (-3) = 21 1001 multiplicand = -7 in two’s com. ×1101(0)multiplier = -3 in two’s com. 00000111 bit-pair 10, add 7 1111001 bit-pair 01, add -7 in two’s com. 000111 bit-pair 10, add 7 bit-pair 11, do nothing 00010101 21 ELEC 5200-002/6200-002 Lecture 6

  45. Booth Advantage Serial multiplication Booth algorithm 00010100 20 ×00011110 30 00000000 00010100 00010100 00010100 00010100 00000000 00000000 00000000________ 000001001011000 600 00010100 20 ×00011110 30 111111111101100 00000010100 __________________ 0000001001011000 600 Four partial product additions Two partial product additions ELEC 5200-002/6200-002 Lecture 6

  46. Adding Partial Products y3 y2 y1 y0 Multiplicand x3 x2 x1 x0 Multiplier ________________________ x0y3 x0y2 x0y1 x0y0 carry← x1y3 x1y2 x1y1 x1y0 Partial carry← x2y3 x2y2 x2y1 x2y0 Products carry← x3y3 x3y2 x3y1 x3y0 __________________________________________________ p7 p6 p5 p4 p3 p2 p1 p0 Notes: Requires three 5-bit adders. Slow. ELEC 5200-002/6200-002 Lecture 6

  47. Array Multiplier: Carry Forward y3 y2 y1 y0 Multiplicand x3 x2 x1 x0 Multiplier ________________________ x0y3 x0y2 x0y1 x0y0 x1y3 x1y2 x1y1 x1y0 Partial x2y3 x2y2 x2y1 x2y0 Products x3y3 x3y2 x3y1 x3y0 __________________________________________________ p7 p6 p5 p4 p3 p2 p1 p0 Note: Carry is added to the next partial product. Adding the carry from the final stage needs an extra stage. ELEC 5200-002/6200-002 Lecture 6

  48. Basic Building Blocks • Two-input AND • Full-adder yi xk kth partial product carry bits from (k-1)th product yi x0 Full adder p0i = x0yi 0th partial product carry bits to (k+1)th product (k+1)th partial product ELEC 5200-002/6200-002 Lecture 6

  49. y3 y2 y1 y0 Array Multiplier x0 ppk yj xi 0 x1 ci 0 0 0 0 0 FA x2 co ppk+1 0 x3 Critical path 0 0 FA FA FA FA p3 p2 p1 p0 p7 p6 p5 p4 ELEC 5200-002/6200-002 Lecture 6

  50. Types of Array Multipliers • Baugh-Wooley Algorithm: Signed product by two’s complement addition or subtraction according to the MSB’s. • Booth multiplier algorithm • Tree multipliers • Reference: N. H. E. Weste and D. Harris, CMOS VLSI Design, A Circuits and Systems Perspective, Third Edition, Boston: Addison-Wesley, 2005. ELEC 5200-002/6200-002 Lecture 6

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