How Do RISC and CISC Architectures Affect Program Speed?

  • Thread starter Thread starter loonny01
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CISC architectures have longer instruction execution times compared to RISC, which can lead to significant differences in program speed. In the provided example, calculating the total execution time for both architectures reveals that RISC is faster due to its lower cycle counts for most operations. The speedup ratio can be determined by comparing the total execution times of both architectures. In the second question, despite computer A achieving higher MIPS, its longer execution time suggests that MIPS does not account for all factors affecting performance, such as I/O operations. Thus, MIPS alone may not accurately reflect a computer's efficiency in executing a program.
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1st ques :
In a CISC computer, the instructions addition, multiplication, memory and branch
all take 20 cycles each while in a RISC computer, the execution time is given
below:
addition - 2 cycles
multiplication - 20 cycles
memory - 10 cycles
branch - 4 cycles
Calculate the speedup ratio for a program with 10 million additions, 30 million
multiplications, 35 million memory instructions and 15million branch
instructions. Explain!...

2nd ques:
When running a particular program, computer A achieves 100 MIPS(million
instructions per second) and computer B achives 75 MIPS. However, computer A
takes 60 seconds to execute a program while computer B takes only 45s. How is
this possible?

Really thanz to any1 who willing 2ans this ques!
 
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Hint(s) for #1:
How long would the program take to run on the CISC computer?
How long would the program take to run on the RISC computer?
Which one is faster? Why? By how much? What is the definition of speedup you are using... a formula, perhaps?

Hint(s) for #2:
What does MIPS measure?
Is this the only thing that a program does... or are there other things a program can do that this measure doesn't take into account (I/O, for instance?)
Could things like this make MIPS more or less irrelevant?
 

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