A CPU has a clock rate of 125Ghz Let us assume that the CPU

A CPU has a clock 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.1. 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) (i.e., no pipeline) execution? What will be the instruction throughput (instructions/sec) 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 integer) What will be the maximum speed-up achieved with this pipeline? Given that the branch penalty is 10 cycles, the probability of a branch instruction is 15% and the probability that a branch is taken is 50%, what will be the average CPI for this CPU? What will the execution efficiency? (up to two decimal places)

Solution

1. clock period = 1/ frequency = 1/1250000000 = 0.8ns

2. Execution Time = clock period * total number of cycles = 0.8 * 4 = 3.2 ns

3. Instruction throughput = instruction/ clock period
For no pipeline execution, instruction = 1 ,
=> Instruction throughput = 1/(0.8 ns) = 125000000 instruction/sec

4. Instruction throughput = instruction/ clock period
For pipeline execution, instruction = 4 ,
=> Instruction throughput = 4/(0.8 ns) = 5000000000 instruction/sec