10. Computer Arithmetic

충남대학교 정보통신공학과박 종 원

Chapter 10 Computer Arithmetic

Four types of data for add, subtract, multiply, divide

1. Fixed-point binary data in signed-magnitude representation2. Fixed-point binary data in signed-2’s complement representation3. Floating-point binary data4. Binary-Coded Decimal(BCD) data

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Add

Subtract

(+3) + (+2)(+6) + (–3)(–3) + (+2)(–3) + (–6)(+3) – (+2)(+3) – (–6)(–3) – (+2)(–3) – (–6)

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A - B A + B’ + 1 = A - B + 2n

( B + B’ + 1 = 2n)

If A B 2n(E = 1)If A < B E = 0

Flowchart for add and subtract operations(signed magnitude representation)

A < B 인 경우에EA A – B하였기때문에A 를 2 의 보수로해주고부호를 반대로 함

(+3) + (–2)(+3) + (+6)(+3) – (–2)(–3) – (–6)

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10-3 Multiplication Algorithms

23 10111 Multiplicand 19 x 10011 Multiplier------ -----------

10111 10111 00000 00000 10111 -------------------437 110110101

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1 0 1 1 1

1 0 0 1 10 0 0 0 0

10111 x 10011 ----------- 10111 10111 00000 00000 10111 ------------------- 110110101

AQ B x Q for multiplication

1 0 1 1 10 1 0 1 1 1 1 0 0 1

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Booth Multiplication Algorithm

Multiplication withbinary integers in signed-2’s complement representation

Ex) (-9) x (-13) = + 117

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컴퓨터구조 1(-9) x (-13) = + 117BR = 10111(-9)BR’ + 1 = 01001QR = 10011(-13)

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(-9) x (-13) = + 117BR = 10111(-9)BR’ + 1 = 01001QR = 10011(-13)QnQn+1 1001 10 10 - : -1 11 01 + : +4 00 10 - : -16

-13

Proof of the Booth Algorithm

111111111 = -1 10 = -1

111011111 = - 25 –1 10 = - 20 01 = + 25 10 = - 26 - 25 –1

111001111 = - 25 – 24- 1 = - 49 10 = - 20 01 = + 24 = 16 10 = - 26 = - 64 - 49

Proof of the Booth Algorithm

If the number is –1, it adds –1, which is the same as the Booth algorithm.If there is a 0 in the position 2k then it adds –2k .Also, the Booth algorithm adds 2k – 2k+1, which is –2k. k 543210111…..1111111101111…..111111 = - 000…..0000000010000…..000001

k+m-1If there is m 0’s from the position 2k then it adds – 2l

l=k k+m-1 k 543210111…111100….0011111…111111 = - 000…000011….1100000…000001

Also, the Booth algorithm adds 2k – 2k+m, which is k+m-1 k 543210 k+m-1 k+m-1 000…000011….1100000…000000- 2l because –( 2l + 2k ) = -2k+m + 000…000000….0100000…000000 l=k . l=k 000…000100….0000000…000000

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b3 b2 b1 b0 a2 a1 a0--------------------------------------------- a0b3 a0b2 a0b1 a0b0 a1b3 a1b2 a1b1 a1b0a2b3 a2b2 a2b1 a2b0

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multiplicationdivision

Q AQ / B for divisionAQ B x Q for multiplication

SubtractionAddition

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AQ 에서 AB 이면 Overflow 임 , EA A + B: 원 상태 A Q

B

A<B 에서 shl EAQ 한 후 E=1 이면 EA > B 임 1 n-1

E A

(EA –2n-1) + (2n-1 – B) = EA – B

Flowchart for divide operation(signed magnitude representation)Q AQ / B

1 0 1 0 0 1 1 0

0 0 1 1

Addition and Subtraction of floating-point operands

1. Check for zeros2. Align the mantissas3. Add or subtract the mantissas4. Normalize the result

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0 0 0 1 0 1 1

1 0 0 0 1 1 0 0 1

1 0 1 0 1 1 0 1 1

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Figure 10-15 Addition and subtraction of floating-point numbers

Align mantissas : 큰 쪽의 지수에 맞춤

Mantissa addition and subtraction : Fig. 10-2 의 signed magnitude 방식임

Check for zeros: A 에 결과가 있음

Normalization: A1 = 1 이 되도록 shift

A x 2a A x 2a B x 2b

AC BR

A < B 인데A – B했으므로 A B

1 n-1

E AC

Multiplication of

floating-point numbers

1. Check for zeros2. Add the exponents3. multiply the mantissas4. Normalize the product

컴퓨터구조 1Figure 10-16

Flowchart for multiplication of

floating-point numbers

지수 : - max1 ~ + max2

0 ~ + max2 + bias

Ex) 5 bit : - 16 ~ + 15 0 ~ + 31

a = a’ + bias(16 in this example)

b = b’ + bias

a + b = a’ + b’ + 2 x bias

0.1 x 0.1 = 0.01

A x 2a B x 2b x Q x 2q

Division of

floating-point numbers

1. Check for zeros2. Initialize registers and evaluate the sign3. Align the dividend4. Subtract the exponents5. Divide the mantissas

Figure 10-17Flowchart for division of floating-point numbers

Overflow checkIf E = 1 then A B

shr A a a+1 하여 Fig. 10-13 의 divide magnitude of mantissa 에서 overflow 를 방지한 후 적용하도록 함 .

a - b = a’ + bias – (b’ + bias) a a + bias

Q x 2q A x 2a / B x 2b

1 0 1 0 0 1 1 0

1 0 0 1

A Q

B

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Table 10-4Derivation of BCD Adder

최소값 : 0최대값 : 9+9+1=19

0 1 0 9+ 0 + 9-------- ----------- 0 19

C = k + Z8Z4 + Z8Z2

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C = k + Z8Z4 + Z8Z2

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BCD 9’s complementer

M B8 B4 B2 B1 x8 x4 x2 x1

0 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 10 0 0 1 0 0 0 1 0. . 0 1 0 0 0 1 0 0 00 1 0 0 1 1 0 0 11 0 0 0 0 1 0 0 11 0 0 0 1 1 0 0 01 0 0 1 0 0 1 1 1..1 1 0 0 0 0 0 0 11 1 0 0 1 0 0 0 0

X1 = B1 M’ + B1 ’ M X2 = B2

X4 = B4 M’ + (B4 ’ B2 + B4 B2 ’ ) M

X8 = B8 M’ + B8 ’ B4 ’ B2 ’ M

그대로

보수로

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X1 = B2 M’ + B1 ’ M X2 = B2

X4 = B4 M’ + (B4 ’ B2 + B4 B2 ’ ) M

X8 = B8 M’ + B8 ’ B4 ’ B2 ’ M

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Figure 10-21Registers for decimal arithmetic multiplication and division

Ae : to accommodate an overflow while adding the multiplicand to the partial product during multiplication

Be : to form the 9’s complement of the devisor when subtracted from the partial remainder during the division operation

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AQ B x Q by adding A A + B and decrementing QL

Q AQ / B by subtracting A A - B and incrementing QL

4 0 5 7

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AQ B x Q by adding A A + B and decrementing QL

4 0 5 7 6

3 2 1 8

0 0 0 0

3 2 1 8

+

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Q AQ / B by subtracting A A - B and incrementing QL

4 0 5 0 1

3 2 1 8

2 8 3 4

0 3 8 4

-

0 3 8 4 4 0 5 1 0