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Consider a 6-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is ____________
Correct answer is '4'. Can you explain this answer?
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Consider a 6-stage instruction pipeline, where all stages are perfectl...
Time without pipeline = 6 stages=6 cycles
Time with pipeline =1+stall freqency*stall cycle
=1+.25*2
=1.5
Speed up = 6/1.5
=4
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Consider a 6-stage instruction pipeline, where all stages are perfectl...
Understanding the Problem

To solve this problem, we need to understand the concept of instruction pipelines and how they affect the execution time of a program.

Instruction Pipelining

- Instruction pipelining is a technique used in modern processors to improve their performance by allowing multiple instructions to overlap in execution.
- This is achieved by dividing the instruction execution into several stages, where each stage performs a specific operation on the instruction.
- The pipeline is divided into stages such as instruction fetch, decode, execute, memory access, writeback, etc.
- Each stage takes a certain amount of time to complete, and instructions move from one stage to the next in a sequential manner.

Perfectly Balanced 6-Stage Pipeline

In this problem, we are given a 6-stage instruction pipeline that is perfectly balanced. This means that each stage takes the same amount of time to complete.
- Let's assume that each stage takes one cycle to complete its operation.

Impact of Pipeline Stalls

- A pipeline stall occurs when an instruction cannot proceed to the next stage due to a dependency or a data hazard.
- In this problem, 25% of the instructions incur 2 pipeline stall cycles. This means that 25% of the instructions experience a delay of 2 cycles before proceeding to the next stage.
- The remaining 75% of the instructions do not incur any pipeline stalls and proceed smoothly through the pipeline.

Calculating Speedup

To calculate the speedup achieved with respect to non-pipelined execution, we need to compare the execution time of the pipelined and non-pipelined scenarios.

Non-Pipelined Execution

- In a non-pipelined execution, each instruction completes its execution before the next instruction starts.
- The time taken to execute a program with N instructions is N cycles.

Pipelined Execution

- In a pipelined execution, multiple instructions are overlapped in execution.
- Assuming each stage takes one cycle, the execution time of a program with N instructions is equal to the number of stages.
- In our case, the execution time would be 6 cycles.

Calculating Speedup

- Speedup is defined as the ratio of the execution time in the non-pipelined scenario to the execution time in the pipelined scenario.
- Speedup = Non-Pipelined Execution Time / Pipelined Execution Time
- Speedup = N / 6

Given Information

- 25% of the instructions incur 2 pipeline stall cycles.
- This means that 25% of the instructions take 2 extra cycles to complete their execution.

Calculating Effective Pipeline Stages

- Since 25% of the instructions incur 2 pipeline stall cycles, it means that these instructions effectively occupy 2 extra pipeline stages.
- The total number of pipeline stages would be 6 stages + 2 stages = 8 stages.

Calculating Speedup with Effective Pipeline Stages

- Speedup = Non-Pipelined Execution Time / Pipelined Execution Time
- Speedup = N / 8

Substituting the Given Values

- Since the problem does not provide the number of
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An instruction pipeline has five stages where each stage take 2 nanoseconds and all instruction use all five stages. Branch instructions are not overlapped. i.e., the instruction after the branch is not fetched till the branch instruction is completed. Under ideal conditions,a. Calculate the average instruction execution time assuming that 20% of all instructions executed are branch instruction. Ignore the fact that some branch instructions may be conditional.b. If a branch instruction is a conditional branch instruction, the branch need not be taken. If the branch is not taken, the following instructions can be overlapped. When 80% of all branch instructions are conditional branch instructions, and 50% of the conditional branch instructions are such that the branch is taken, calculate the average instruction execution time. Correct answer is 'Each stage is 2 ns.'. Can you explain this answer?

An instruction pipeline has five stages, namely, instruction fetch (IF), instruction decode and register fetch (ID/RF), instruction execution (EX), memory access (MEM), and register writeback (WB) with stage latencies 1 ns, 2.2 ns, 2 ns, 1 ns, and 0.75 ns, respectively (ns stands for nanoseconds). To gain in terms of frequency, the designers have decided to split theID/RF stage into three stages (ID, RF1, RF2) each of latency 2.2/3 ns. Also, the EX stage is split into two stages (EX1, EX2) each of latency 1 ns. The new design has a total of eight pipeline stages. A program has 20% branch instructions which execute in the EX stage and produce the next instruction pointer at the end of the EX stage in the old design and at the end of the EX2 stage in the new design. The IF stage stalls after fetching a branch instruction until the next instruction pointer is computed. All instructions other than the branch instruction have an average CPI of one in both the designs. The execution times of this program on the old and the new design are P and Q nanoseconds, respectively. The value of P/Q is __________. Correct answer is '1.54'. Can you explain this answer?

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Consider a 6-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is ____________Correct answer is '4'. Can you explain this answer?
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