An arithmetic pipeline separates a given arithmetic problem into subproblems that can be executed in different pipeline segments. It’s used for multiplication, floating-point operations, and a variety of other calculations.
In this article, we will dive deeper into the Arithmetic Pipeline in Computer Architecture according to the GATE Syllabus for (Computer Science Engineering) CSE. Keep reading ahead to learn more.
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Arithmetic Pipeline in Computer Architecture
Arithmetic Pipelines are commonly used in various high-performance computers. They are used in order to implement floating-point operations, fixed-point multiplication, and other similar kinds of calculations that come up in scientific situations.
Let’s look at an example to better understand the ideas of an arithmetic pipeline. We perform addition and subtraction of floating points on a unit of the pipeline here.
The inputs in the floating-point adder pipeline refer to two different normalized floating-point binary numbers. These are defined as follows:
A = X * 2x = 0.9504 * 103
B = Y * 2y = 0.8200 * 102
Where x and y refer to the exponents and X and Y refer to two fractions representing the mantissa.
The floating-point addition and subtraction process is broken into four pieces. The matching sub-operation to be executed in the specified pipeline is contained in each segment. The four segments depict the following sub-operations:
1. Comparing the exponents using subtraction
2. Aligning the mantissa
3. Adding or subtracting the mantissa
4. Normalizing the result
Later in this section, we’ll go through each sub-operation in greater depth. The sub-operations conducted in each pipeline section are depicted in the block diagram below:
Note: The registers are placed after every sub-operation in order to store the intermediate results.
1. Comparing Exponents by Subtraction
The difference between the exponents is calculated by subtracting them. The result’s exponent is chosen to be the larger exponent.
The exponent difference, 3 – 2 = 1, defines the total number of times the mantissa associated with the lesser exponent should be shifted to the right.
2. Aligning the Mantissa
As per the difference of exponents calculated in segment one, the mantissa corresponding with the smaller exponent would be moved.
A = 0.9504 * 103
B = 0.08200 * 103
3. Adding the Mantissa
Both the mantissa would be added in the third segment.
C = A + B = 1.0324 * 103
4. Normalizing the Result
After the process of normalization, the result would be written as follows:
C = 0.1324 * 104
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