Consider an instruction pipeline in which a total of 5 cycles is requi...
k = 5
Sk = 3.75
ττ = 300 picosecond = 0.3 ns
The time required to execute n instructions without pipeline = T1 = nkτnkτ
The time required to execute n instructions with pipeline = Tk = [k + (n – 1)]τ
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Consider an instruction pipeline in which a total of 5 cycles is requi...
The pipeline. This means that it takes a total of 5 cycles for the first instruction to go through all stages of the pipeline and complete its execution.
Assuming that the pipeline is fully utilized and has a steady stream of instructions, we can calculate the number of instructions that can be completed in a given number of cycles.
If it takes 5 cycles to complete the first instruction, then in the next cycle (cycle 6), the first instruction will be completed and the second instruction will be in its final stage of execution. In cycle 7, the second instruction will be completed and the third instruction will be in its final stage, and so on.
Therefore, in cycle 5, we have completed 1 instruction.
In cycle 6, we have completed 2 instructions.
In cycle 7, we have completed 3 instructions.
In cycle 8, we have completed 4 instructions.
In cycle 9, we have completed 5 instructions.
So, in 5 cycles, we can complete 5 instructions.
In general, if it takes n cycles to complete the first instruction, then in n cycles, we can complete n instructions.
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