Architecture: The Representation For Floating Point

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Evaluate and represent floating point numbers in IEEE 754 single-precision format, analyze floating point addition outcomes, discuss strategies for accurate summation of a wide range of values, and explore performance implications of pipeline modifications and hardware enhancements in a MIPS architecture context.

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In this paper, we explore the representation of floating point numbers in IEEE 754 single-precision format, focusing on specific decimal values and the implications of floating point arithmetic limitations. We analyze how the IEEE 754 standard encodes decimal numbers, with attention to precision and biases, and discuss the impact of floating point addition involving vastly different magnitudes. Furthermore, we delve into strategies to improve the accuracy of summing large datasets spanning multiple orders of magnitude, emphasizing numerical stability techniques. Lastly, the paper investigates pipeline performance factors, calculating clock cycle times for pipelined and non-pipelined processors, evaluating instruction latency, considering pipeline stage splitting, and examining the effects of hardware improvements such as increased register counts and their influence on processor efficiency and instruction count.

Representation of Decimal Values in IEEE 754 Single Precision

To represent a decimal number in IEEE 754 single-precision format, the number is expressed as a binary fraction (mantissa) multiplied by a power of two (exponent), with specific bits assigned to sign, exponent, and significand (mantissa). The single-precision format uses 32 bits: 1 for sign, 8 for exponent, and 23 for mantissa. The bias for the exponent is 127.

1. Represent the decimal value 63.25 in IEEE 754 single-precision format

Step-by-step:

- Convert 63.25 to binary:

- 63 in binary: 111111

- 0.25 in binary: 0.01 (since 0.25 = 1/4)

- Combined: 111111.01

- Normalize: 1.1111101 × 2^5

- Sign bit: 0 (positive)

- Exponent: 5 + 127 = 132 → binary: 10000100

- Mantissa: take bits after the leading 1 (which is implicit): 1111101, padded with zeros to 23 bits: 11111010000000000000000

- Final IEEE 754 representation in binary:

0 | 10000100 | 11111010000000000000000

- Hexadecimal: 0x427FA000

2. Represent the decimal value -1.125 in IEEE 754 single-precision format

Step-by-step:

- Binary form:

- 1.125 in binary: 1.001

- Normalize: 1.001 × 2^0

- Sign bit: 1 (negative)

- Exponent: 0 + 127 = 127 → binary: 01111111

- Mantissa: 001 (after the implicit 1), padded to 23 bits: 00100000000000000000000

- IEEE 754 binary:

1 | 01111111 | 00100000000000000000000

- Hexadecimal: 0xBF800000

Analysis of Floating Point Addition: 1E10 + 1E-32

Performing addition of 1×10^10 and 1×10^-32 in IEEE 754 single precision highlights the limitations of floating point precision and the concept of significant digits. In single precision, the precision is approximately 7 decimal digits, which means that when adding a very large number (1E10) and an extremely tiny number (1E-32), the tiny one may be lost due to rounding, resulting in no change in the large number.

Implication:

- The sum will likely be approximately 1E10, because 1E-32 is too small to affect the significant digits of 1E10 in single-precision floating point. This illustrates the phenomenon of floating point underflow and limited precision, where adding numbers of vastly different magnitudes causes the smaller to have no effect.

Strategies for Accurate Summation of a Large List of Values

When summing a large set of floating point values varying across many orders of magnitude, naive addition can cause significant numerical errors due to loss of significance. An effective approach is to perform pairwise or incremental summation in a manner that minimizes rounding errors, such as the Kahan summation algorithm or summing from smallest to largest values. These techniques help preserve numerical precision by compensating for small errors accumulated during the summation process.

Specifically, one simple method is to sort the list of values by magnitude and sum from the smallest to the largest, reducing the risk of the small numbers being "washed out" by larger numbers. Using compensated summation methods like Kahan summation further improves accuracy by tracking small errors during the process, which is invaluable when dealing with data spanning many orders of magnitude.

Pipeline Performance Analysis in a MIPS Architecture

Given a 5-stage pipeline with known latencies:

- IF: 250 ps

- ID: 350 ps

- EX: 150 ps

- MEM: 300 ps

- WB: 200 ps

and a program with instruction mix: 45% arithmetic, 20% branch, 35% load/store.

a. Clock cycle time for pipelined and non-pipelined processors

For a non-pipelined processor, the clock cycle must be at least as long as the longest stage: 350 ps (ID stage). For the pipelined processor, the cycle time is dictated by the slowest stage: 350 ps. Therefore:

- Non-pipelined cycle time: 350 ps

- Pipelined cycle time: 350 ps

b. Total latency of a load instruction

In a non-pipelined processor, total latency sums all stages: 250 + 350 + 150 + 300 + 200 = 1,350 ps.

In a pipelined processor, the latency is the sum of stages for the load instruction, but because subsequent instructions can overlap, the throughput depends on cycle time, not total latency; however, the individual instruction latency remains 1,350 ps.

c. Splitting one pipeline stage into two

To improve performance, splitting the EX stage (150 ps) into two stages, each approximately 75 ps, is optimal. The new cycle time becomes the maximum of the existing stages: the largest original latency remains 350 ps (ID stage), so the cycle time remains at 350 ps, but stages now are more balanced, potentially reducing stall hazards and increasing throughput.

d. Hardware improvements: doubling register count

Doubling the number of registers reduces load/store instructions by 10% and increases register latency by 50 ps.

- New register latency: 200 ps + 50 ps = 250 ps.

- Reduced load/store instructions decrease memory traffic, potentially improving performance.

i. Speedup calculation:

Assuming a simplified model, the primary benefit is reduced stalls and memory delays. The speedup (S) can be approximated by the reduction in instructions (10%) and decreased memory bottlenecks, combined with the slight increase in register latency.

S ≈ 1 / [(1 - 0.10) + (additional delays due to increased register latency)] ≈ 1 / (0.90 + small increase).

Exact quantification requires detailed cycle time and instruction mix modeling, but a rough estimate suggests around a 10-15% performance improvement.

ii. Effect on instruction count:

Increasing register count typically reduces the need for instruction serialization and memory accesses, potentially decreasing total instruction count due to more efficient register usage and fewer memory operations, fostering better compiler optimizations and code efficiency.

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