Floating Point squareroot (FPSR) operation is one of the costly operations for p
ID: 3818611 • Letter: F
Question
Floating Point squareroot (FPSR) operation is one of the costly operations for processor. given the FPSR occurs 20% of the total execution time, BSCS4-A students are planning to build a processor which do FPSR operation 10 faster than an ordinary processor. While Bscs4-B students thinks as overall Floating Point instruction occur half of the time, so by enhancing execution time of overall FP instructions by 3 times this will give better speedup than that of enhancing just FPSR operations. Being a student of Computer Architecture you are asked to calculate the overall speedup of both the proposals, so better idea could be finalized?Explanation / Answer
We shall solve this in a non-traditional way.
And our system consumes 100 Clocks,
therefore FPSR takes 20Clocks
and FP takes 50Clocks
Group A:
They make the FPSR operation 10 times faster, so now FPSR takes 2 (20/10) clocks (Therefore we save 18 clocks)
Therefore our system shall now take 82 (100 - 18) clocks
Group B:
They make the FP operation 3 times faster, so now FP takes 16.67 (50/3) clocks (Therefore we save 33.33 clocks)
Therefore our system shall now take 66.67 (100 - 33.33) clocks
So it is very clear that the Group B's processor has better speed up compares to Group A's processor.
I hope the explanation was easy and I made myself clear. If you are still facing any problem understanding you can comment below. I shall be glad to help you with the solution.
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