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Question

The right sequence of suboperations that are performed in arithmetic pipeline is-
A. Align the mantissas
B. Add or subtract the mantissas
C. Normalize the result
D. Compare the exponents
Choose the correct answer from the options given below:

The correct answer is
D, A, B, C

Arithmetic Pipeline Suboperations Sequence Explained

Arithmetic pipelines are essential components in modern processors designed to significantly speed up complex calculations, particularly floating-point operations. They achieve this by breaking down an operation into a series of smaller, independent steps (suboperations) that can be processed concurrently in different stages of the pipeline. Understanding the correct order of these suboperations is key to how these pipelines function efficiently.

Understanding the Arithmetic Pipeline Suboperations

The question focuses on the sequence of suboperations performed during arithmetic computations, specifically related to floating-point numbers. These operations typically involve manipulating the mantissa (the significant digits) and the exponent (the power of the base).

  • D. Compare the exponents: This is the first crucial step when performing operations like addition or subtraction on two floating-point numbers. The exponents of the numbers are compared to determine their relative magnitudes. This comparison dictates how the mantissas will need to be adjusted.
  • A. Align the mantissas: After comparing the exponents, the mantissa of the number with the smaller exponent needs to be shifted. It is shifted to the right by a number of positions equal to the difference between the exponents. This alignment ensures that both numbers are represented with the same exponent, making them ready for the arithmetic operation.
  • B. Add or subtract the mantissas: Once the mantissas are aligned, the core arithmetic operation (addition or subtraction) is performed on these aligned mantissas. The result of this operation forms the new mantissa.
  • C. Normalize the result: Following the addition or subtraction, the resulting mantissa might not be in the standard normalized format required for floating-point representation (e.g., for binary, the mantissa should typically be between 1 and 2, excluding 2). Normalization involves shifting the resulting mantissa left or right and adjusting the exponent accordingly to bring it back into the required normalized range.

Determining the Correct Pipeline Sequence

To perform an arithmetic operation like floating-point addition or subtraction correctly within a pipeline, the suboperations must be executed in a specific logical order. Let's trace the process:

  1. Compare the exponents (D): The pipeline must first determine the relationship between the exponents of the input numbers. This comparison is fundamental for the subsequent alignment step.
  2. Align the mantissas (A): Based on the exponent comparison, the mantissa of the number with the smaller exponent is shifted right so that both numbers share the same exponent value. For example, if we have $1.101 \times 2^3$ and $1.011 \times 2^1$, we compare exponents (3 and 1). Since 3 > 1, we align the second number: $1.011 \times 2^1$ becomes $0.1011 \times 2^2$ (shift right by $3-1=2$ positions). The common exponent is now 3.
  3. Add or subtract the mantissas (B): With aligned mantissas and a common exponent, the actual addition or subtraction is performed. Using the example above, if adding: $(1.101 + 0.1011) \times 2^3$. This results in $10.0101 \times 2^3$.
  4. Normalize the result (C): The result obtained ($10.0101 \times 2^3$) needs to be normalized. In binary, a normalized mantissa is typically between 1 and 2 (e.g., $1.xxxx$). So, we shift the mantissa right by one position and increment the exponent: $1.00101 \times 2^4$.

This step-by-step process demonstrates that the logical and correct sequence for these suboperations in an arithmetic pipeline is indeed D, A, B, C.

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Important Questions from Instruction Pipelining

  1. A 5-stage instruction pipeline has stage delays of 180, 250, 150, 170, and 250, respectively, in nanoseconds. The delay of an inter-stage latch is 10 nanoseconds. Assume that there are no pipeline stalls due to branches and other hazards. The time taken to process 1000 instructions in microseconds is ____________ (rounded off to two decimal places)
  2. Which one of the following is a technique of decomposing a sequential process into sub-operations, with each sub-process being executed in a special dedicated segment that operates concurrently with all other segments?
  3. The EX stage of a pipelined processor performs the memory read operations for LOAD instructions, and the operations for the arithmetic and logic instructions. Let $t_{EX}$ denote the time taken by the EX stage to perform the operation for an instruction. For each instruction type, the values of $t_{EX}$ and $M$ (the number of instructions of that type in a sequence of 100 instructions for a program P), are given in the table below.
    The duration of the pipeline clock cycle is 1 nanosecond. Assume that the latch time for the interstage buffers in the pipeline is negligible.
    Instruction$t_{EX}$ in nanoseconds$M$
    LOAD1.815
    IMUL1.510
    IDIV2.55
    FADD1.710
    FSUB1.75
    FMUL2.815
    FDIV3.25
    All other instructionsLess than 1.035

    When program P is executed, the number of clock cycles for which the pipeline is stalled due to structural hazards in the EX stage is ______. (answer in integer)
  4. A non-pipelined instruction execution unit that operates at 1.6 GHz clock takes an average of 5 clock cycles to complete the execution of an instruction. To improve the performance, the system was pipelined with a goal of achieving an average throughput of one instruction per clock cycle. However, it could operate only at 1.2 GHz due to pipeline overheads. While executing a program in the pipelined design, 30% of instructions encountered a stall of 2 cycles due to pipeline hazards. The speed-up obtained by the pipelined design over the non-pipelined one for this program is ___________. (rounded off to two decimal places)

    Note: 1G=$10^9$
  5. Pipelining improves performance by :
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