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02 A CISC (Complex Instructions set Computer) machine has CPls of 4 for loadstore, 3 for ALU/branch, 10 for call/return and C
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Answer #1

formula used :- Effective CPU time = (Number of Instructions * CPI ) / Clock rate

note :- Here , according to the question , the benchmark is divided into several instruction count i.e Loads , stores , calls , returns , ALU operations and branches.

  • For CISC we have to calculate CPU effective time for these instructions because Each set of instructions have different CPI (Cycle per instructions) that's why

E.CPU time (CISC) = (0.38 * 4 ) / 1.75 for loads + (0.10 * 4 ) / 1.75 for store + (0.35 * 3 ) /1.75 for ALU + (0.06 * 10) / 1.75 for calls and returns + (0.11 * 3)/1.75 for branches .

Total CPU time (CISC) = (1.52 + 0.4 + 1.05 + 0.6 + 0.33 ) /1.75 = 3.9 / 1.75 = 2.2285 seconds.

  • For RISC we don;t have to calculate CPU effective time separately because RISC follows a Pipe-lined Structure and it has only one CPI (Number of cycles per Instruction ) for all Loads , Stores , Calls /returns , ALU /Branches... I.E. 1.2 And CPU Clock Rate is 1 GHz..

Now , Effective CPU TIME (RISC) = (1 * 1.2 ) / 1 = 1.2 seconds.. You probably thinking why No. of Instructions = 1 because (loads + store + calls + returns +branches + ALU ) % = 100% a Complete benchmark.

So, by comparing both CPU Effective times of CISC and RISC , we can say that CPU Time of RISC is less than CPU time of CISC ..

Therefore , RISC can run the benchmark in less time..

I hope you will be able to understand and do please leave a feedback about whether you are satisfied with the answer or not.

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