Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
Arithmetic pipeline
Arithmetic pipeline divides an arithmetic operation into
sub-operations for execution in the pipeline segments, it's used
to implement the floating point operation , multiplication of
fixed point numbers and similar computations encountered in
scientific problems.
Arithmetic pipe with fixed point
Suppose that we want to perform the combined multiply
and add operations with a stream numbers:
Ai * Bi +Ci
for i=1,2,3,…, 7
The first rule: there is no fixed number of segments ,How?
The No. of segment depends on the arithmetic expression, i.e.
each sub-operation is implemented in a segment.
The second rule : we need registers for holding inputs and the
results of segments. In addition, we need a combinational
circuit to perform required operation. return to previous
example , we have three segments:
seg1: R1← Ai , r2← Bi
(loading of inputs Ai and Bi)
seg2: R3 ← R1 *R2, R4 ← Ci
(multiplying and load C)
seg3: R5 ← Ci
(adding Ci to the product)
The No. of tasks = 7 because these segments repeated 7 times
(i=1,2,...,7) .
The combinational circuit that perform this instruction is
consist of multiplier and adder as shown below :
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
Now, if tp= 25 ns:
n * k* tp
speed up= ‫ـــــــــــــــــــــــــــــــــــــــــ‬
(n+k-1)*tp
= (7*3*25)/ (9*25) = 2.333
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
The content of the registers will be:
Clock
SEGMENT1
SEGMENT2
cycle
R1
R2
R3
R4
1
A1
B1
2
A2
B2
A1 *B1
C1
3
A3
B3
A2*B2
C2
A1*B1+C1
4
A4
B4
A3*B3
C3
A2*B2+C2
5
A5
B5
A4*B4
C4
A3*B3+C3
6
A6
B6
A5*B5
C5
A4*B4+C4
7
A7
B7
A6*B6
C6
A5*B5+C5
A7*B7
C7
A6*B6+C6
8
SEGMENT3
9
R5
A7*B7+C7
Question for discussion :
If the previous example was :
x= Ai * Bi +Ci
for i=1,2,3,…, 7
1- How many segments we need?
2- Design a hardware to perform this problem.
3- show the content of the registers during the execution.
Pipeline for floating point operation
Representation of floating point number is not unique.
For
example,
the
number 55.66 can
be
represented
as 5.566×10^1, 0.5566×10^2,0.05566×10^3, and so on.
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
It is important to note that floating-point numbers suffer
from loss of precision when represented with a fixed number of
bits
(e.g.,
32-bit
or
64-bit).
This
is
because
there
are infinite number of real numbers (even within a small range
of says 0.0 to 0.1). On the other hand, a n-bit binary pattern can
represent a finite 2^n distinct numbers. Hence, not all the real
numbers can be represented. The nearest approximation will be
used instead, resulted in loss of accuracy.
It is also important to note that floating number
arithmetic is very much less efficient than integer arithmetic. It
could be speed up with a so-called dedicated floating-point coprocessor. Hence, use integers if your application does not
require floating-point numbers.
floating point operations are easily decomposed (‫ )ﺗﺠ ﺰأ‬into
sub-operations. We will now show an example of a pipeline unit
for floating point addition and subtraction :
the inputs to the floating point adder pipeline are two
normalized floating point binary numbers :
X= A × 2a
Y= B ×2b
A and B are fractions (‫ )ﻛﺴ ﻮر‬that represent the mantissa (‫)اﻟﻤﻘ ﺪار‬
and a and b are the exponents. The floating point addition and
subtraction can be performed in four segment , the registers
labeled
R
are
placed
between
the
segments
to
store
intermediate results. The Four segments are :
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
1- compare the exponents and find the difference (d) by
subtracting them:
- choose the larger exponent to be the exponent of the result.
-the difference (d) is important for shifting operation in the next
step.
2- Align the mantissa: use the difference (d) to shift the
mantissa associated with the smallest exponent d times to the
right , each right shift increases the exponent by one.
3- add or subtract the mantissa.
4- normalize the result so that it has a fraction with a nonzero
first digit this is done by shifting the mantissa to right and
increasing the exponent by one for each shifting to obtain the
normalized result .
The following example may clarify the sub-operations
performed in each segment. For simplicity, we use decimal
numbers.
Consider the two normalized floating point numbers :
x= 0.9504 ×103
y= 0.8200 ×102
1- comparison between the exponents:
d= 3-2=1
this mean that the alignment done by shifting y(number with
smallest exponent) once to the right.
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
2- Alignement:
x= 0.9504 ×103
y= 0.082 ×103
3- addition :
z=x+y
z= 0.9504 ×103 + 0.082 ×103
z= 1.0324 ×103
4- result normalization by shifting the mantissa to the right
and increasing the exponent :
z= 0.10324 ×104
To build a hardware for this operation we need :
1- comparator (subtructer).
2- shifter .
3- adder (subtracter).
4- normalizer.
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Segment 1 :
Lecturer: Noor Kadhum
Exponents
a
b
Mantissas
A
B
R
R
Compare
exponents
by subtraction
Difference
R
Segment 2 :
Choose exponent
Align mantissas
R
Add or subtract
mantissas
Segment 3 :
Segment 4 :
R
R
Adjust
exponent
Normalize
result
R
R
u Vector Instruction Format : Fig. 9-11
Babylon university - Sciences College For Women - Architecture – Third class
Lecture 15: pipeline (1)
Lecturer: Noor Kadhum
Babylon university - Sciences College For Women - Architecture – Third class