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# Principles of Computer Architecture - Arithmetic

## Chapter Contents

3.1 Overview
3.2 Fixed Point Addition and Subtraction
3.3 Fixed Point Multiplication and Division
3.4 Floating Point Arithmetic
3.5 High Performance Arithmetic
3.6 Case Study: Calculator Arithmetic Using Binary Coded Decimal

## Computer Arithmetic

•Using number representations from Chapter 2, we will explore four basic arithmetic operations: addition, subtraction, multiplication, division.
•Significant issues include: fixed point vs. floating point arithmetic, overflow and underflow, handling of signed numbers, and performance.
•We look first at fixed point arithmetic, and then at floating point arithmetic.

## Number Circle for 3-Bit Two’s Complement Numbers

•Numbers can be added or subtracted by traversing the number circle clockwise for addition and counterclockwise for subtraction.

•Overflow occurs when a transition is made from +3 to -4 while proceeding around the number circle when adding, or from -4

## Overflow

•Overflow occurs when adding two positive numbers produces a negative result, or when adding two negative numbers produces a positive result. Adding operands of unlike signs never produces an overflow.
•Notice that discarding the carry out of the most significant bit during two’s complement addition is a normal occurrence, and does not by itself indicate overflow.
•As an example of overflow, consider adding (80 + 80 = 160)10, which produces a result of -9610 in an 8-bit two’s complement format:
01010000 = 80
+ 01010000 = 80
----------
10100000 = -96 (not 160 because the sign bit is 1.)

•Two binary numbers A and Bare added from right to left, creating a sum and a carry at the outputs of each full adder for each bit position.

## Full Subtractor

•Truth table and schematic symbol for a ripple-borrow subtractor:

## Ripple-Borrow Subtractor

•A ripple-borrow subtract or can be composed of a cascade of full subtractors.
•Two binary numbers A and Bare subtracted from right to left, creating a difference and a borrow at the outputs of each full subtract or for each bit position.

•A single ripple-carry adder can perform both addition and subtraction, by forming the two’s complement negative for B when subtracting. (Note that +1 is added at c0 for two’s complement.)

•An example of one’s complement integer addition with an end-around carry:

•An example of one’s complement integer addition with an end-around carry:

## Number Circle (Revisited)

•Number circle for a three-bit signed one’s complement representation. Notice the two representations for 0.

## End-Around Carry for Fractions

•The end-around carry complicates one’s complement addition for non-integers, and is generally not used for this situation.

•The issue is that the distance between the two representations of 0 is 1.0, whereas the rightmost fraction position is less than 1.

## Multiplication Example

•Multiplication of two 4-bit unsigned binary integers produces an 8-bit result.

•Multiplication of two 4-bit signed binary integers produces only a 7-bit result (each operand reduces to a sign bit and a 3-bit magnitude for each operand, producing a sign-bit and a 6-bit result).

## Example of Base 2 Division

•(7 / 3 = 2)10 with a remainder R of 1.
Equivalently , (0111/ 11 = 10)2 with a remainder R of 1.

## Multiplication of Signed Integers

•Sign extension to the target word size is needed for the negative operand(s).

•A target word size of 8 bits is used here for two 4-bit signed operands, but only a 7-bit target word size is needed for the result.

•Carries are represented in terms of G i (generate) and Pi (propagate) expressions.

## Floating Point Arithmetic

•Floating point arithmetic differs from integer arithmetic in that exponents must be handled as well as the magnitudes of the operands.

•The exponents of the operands must be made equal for addition and subtraction. The fractions are then added or subtracted as appropriate, and the result is normalized.

•Ex: Perform the floating point operation: (.101 ×23+ .111 ×24)2

•Start by adjusting the smaller exponent to be equal to the larger exponent, and adjust the fraction accordingly. Thus we have .101×23= .010 ×24, losing .001 ×23of precision in the process.

•The resulting sum is (.010 + .111) ×24= 1.001 ×24= .1001 ×25, and rounding to three significant digits, .100 ×25, and we have lost another 0.001 ×24 in the rounding process.

## Floating Point Multiplication/Division

•Floating point multiplication/division are performed in a manner similar to floating point addition/subtraction, except that the sign, exponent, and fraction of the result can be computed separately.

•Like/unlike signs produce positive/negative results, respectively. Exponent of result is obtained by adding exponents for multiplication, or by subtracting exponents for division. Fractions are multiplied or divided according to the operation, and then normalized.

•Ex: Perform the floating point operation: (+.110 ×25) / (+.100 ×24)2

•The source operand signs are the same, which means that the result will have a positive sign. We subtract exponents for division, and so the exponent of the result is 5 –4 = 1.

•We divide fractions, producing the result: 110/100 = 1.10.

•Putting it all together, the result of dividing (+.110 ×25) by (+.100 ×24) produces (+1.10 ×21). After normalization, the final result is (+.110 ×22).

## The Booth Algorithm

•Booth multiplication reduces the number of additions for intermediate results, but can sometimes make it worse as we will see.

•Positive and negative numbers treated alike.

## A Worst Case Booth Example

•A worst case situation in which the simple Booth algorithm requires twice as many additions as serial multiplication.

## Newton’s Iteration for Zero Finding

•The goal is to find where the function f(x) crosses the x axis by starting with a guess xi and then using the error between f(xi) and zero to refine the guess.

•A three-bit lookup table for computing x0:

•The division operation a/bis computed as a×1/b. Newton’s iteration provides a fast method of computing 1/b.

## Residue Arithmetic

•Implements carryless arithmetic (thus fast!), but comparisons are difficult without converting to a weighted position code.

•Representation of the first twenty decimal integers in the residue number system for the given moduli:

## Examples of Addition and Multiplication in the Residue Number System

•A16-bit GCLA is composed of four 4-bit CLAs, with additional logic that generates the carries between the four-bit groups.

•Each CLA has a longest path of 5 gate delays.

•In the GCLL section, GG and GP signals are generated in 3 gate delays; carry signals are generated in 2 more gate delays, resulting in 5 gate delays to generate the carry out of each GCLA group and 10 gates delays on the worst case path (which is s15–not c16).

## HP 9100 Series Desktop Calculator

•Uses binary coded decimal (BCD) arithmetic.

•Addition is performed digit by digit (not bit by bit), in 4-bit groups, from right to left.

•Example (255 + 63 = 318)10:

## Subtraction Example Using BCD

•Subtraction is carried out by adding the ten’s complement negative of the subtrahend to the minuend.

•Ten’s complement negative of subtrahend is obtained by adding 1 to the nine’s complement negative of the subtrahend.

•Consider performing the subtraction operation (255 -63 = 192)10:

## Excess 3 Encoding of BCD Digits

•Using an excess 3 encoding for each BCD digit, the leftmost bit indicates the sign.

•Circuit adds two base 10 digits represented in BCD. Adding 5 and 7 (0101 and 0111) results in 12 (0010 with a carry of 1, and not 1100, which is the binary representation of 1210).

## Ten’s Complement Subtraction

•Compare: the traditional signed magnitude approach for adding decimal numbers vs.the ten’s complement approach, for (21 -34 = -13)10:

## BCD Floating Point Representation

•Consider a base 10 floating point representation with a two digit signed magnitude exponent and an eight digit signed magnitude fraction. On a calculator, a sample entry might look like:

•We use a ten’s complement representation for the exponent, and a base 10 signed magnitude representation for the fraction. A separate sign bit is maintained for the fraction, so that each digit can take on any of the 10 values 0–9 (except for the first digit, which cannot be zero). We should also represent the exponent in excess 50 (placing the representation for 0 in the middle of the exponents, which range from -50 to +49) to make comparisons easier.

•The example above now looks like this (see next slide):

## BCD Floating Point Arithmetic

•The example in the previous slide looks like this:
Sign bit:1
Exponent:0110 1011
Fraction:0110 1010 0100 0011 0011 0011 0011 0011 0011

•Note that the representation is still in excess 3 binary form, with a two digit excess 50 exponent.

•To add two numbers in this representation, as for a base 2 floating point representation, we start by adjusting the exponent and fraction of the smaller operand until the exponents of both operands are the same. After adjusting the smaller fraction, we convert either or both operands from signed magnitude to ten’s complement according to whether we are adding or subtracting, and whether the operands are positive or negative, and then perform the addition or subtraction operation.