It has been determined that a certain \"Benchmarking Program\" for a MIPS proces

Question

It has been determined that a certain "Benchmarking Program" for a MIPS processor has the following instruction mix: 14% loads (1w), 9% stores (sw), 21% conditional branches (beq, bne), 6% jumps (j, jr, jal), and R-type for the rest. Our current implementation of the multicycle datapath uses five clocks for lw instructions, 4 clocks for sw instructions, 3 clocks for all branch and jump instructions, and 4 clocks for all R-type instructions. The Benchmarking Program to be run executes 525, 500 instructions. What is the average CPI (clocks per instruction) for the Benchmarking Program under these parameters?

Explanation / Answer

average CPI

= (73570*5 + 47295*4 + (110355+31530)*3 + 262750*4)/525500

= (367850+189180+425655+1051000)/525500

= 2033685/525500

= 3.87

Instruction type % of instruction Instruction count from 525500 total instructions Clock cycle count load 14 73570 5 store 9 47295 4 branches 21 110355 3 jumps 6 31530 3 R types 50 262750 4

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