Question
Pipelining improves performance by :
The correct answer is
exploiting instruction level parallelism
Pipelining Performance Improvement Explained
Pipelining enhances processor performance by enabling the overlap of instruction execution stages. Instead of completing one instruction entirely before starting the next, pipelining divides instruction processing into distinct stages (like fetch, decode, execute). These stages operate concurrently on different instructions.
Key Mechanism: Exploiting Instruction-Level Parallelism
- Pipelining breaks down the execution of a single instruction into multiple sequential steps or stages.
- Multiple instructions are processed simultaneously, with each instruction residing in a different stage of the pipeline. For example, while instruction n is executing, instruction n+1 might be decoding, and instruction n+2 might be fetching.
- This overlap effectively allows different parts of multiple instructions to be processed in parallel within the CPU.
- This parallelism, derived from the instruction stream itself, is called Instruction-Level Parallelism (ILP). Pipelining is a primary technique for exploiting ILP.
- By keeping the pipeline stages busy, the overall rate at which instructions are completed (throughput) increases significantly, leading to better performance, even though the time taken for any single instruction (latency) might slightly increase due to pipeline overhead.
Why Other Options Are Less Accurate:
- Decreasing instruction latency: Pipelining aims to increase instruction *throughput*, not necessarily decrease the latency of a single instruction.
- Eliminating data hazards: Pipelining actually introduces hazards (like data dependencies) that require special handling (stalling, forwarding). It does not eliminate them inherently; managing them is crucial for pipeline efficiency.
- Decreasing the cache miss rate: Cache performance is related but distinct. Pipelining doesn't directly alter the cache's fundamental design or its miss rate calculation.
Therefore, the main way pipelining improves performance is by effectively exploiting instruction-level parallelism.
Was this answer helpful?
Important Questions from Instruction Pipelining
- 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: - 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)
- 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?
- 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 $M$ LOAD 1.8 15 IMUL 1.5 10 IDIV 2.5 5 FADD 1.7 10 FSUB 1.7 5 FMUL 2.8 15 FDIV 3.2 5 All other instructions Less than 1.0 35
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) - 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=$10^9$
Need Expert Advice?
