Answer Key: 0.8|0.8ns|0.8 ns, 3.2|3.2ns|3.2 ns, 312.5, 1250, 3.99, 4, 4, 1.75, 0

Question

Answer Key: 0.8|0.8ns|0.8 ns, 3.2|3.2ns|3.2 ns, 312.5, 1250, 3.99, 4, 4, 1.75, 0.57

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A CPU has a dock rate of 1.25Ghz. Let us assume that the CPU has 4 pipeline stages and processing delay of each stage is exactly one clock cycle period. What is the clock period in ns? What is the execution time (in ns) of each instruction passing through all stages? What will be the instruction throughput (instructions/sec) in MIPS with sequential (serial e.. no pipeline) execution? What will be the instruction throughput (instructionssec in MIPS with pipeline operation? What is the speed-up achieved when 500,000 instructions are processed using the pipeline operation up to two decimal places) What is the speed-up achieved when 1,000,000 instructions are processed using the pipeline operation? (round it to nearest intege What will be the maximum speed-up achieved with this pipeline? and the probability that a branch is taken is 50%, what will be the average CPI for this CPU? Given that the branch penalty is 10 cycles, the probability of a branch instruction is 15% What will the execution efficiency? (up to two decimal places)

Explanation / Answer

1) clock period = 1/clockrate = 1/1.25 = 0.8

2) execution time : 0.8ns

3) 0.8 * 4(pipeline) = 3,2

4) 0.8 + 0.8 + 0.8 + 0.8 = 3.2ns

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