I found that I can replace one of my instructions that takes 300 ns, with a diff

Question

I found that I can replace one of my instructions that takes 300 ns, with a different one that takes 200 ns. My program runs this instruction 50% of the time. What is the speedup on the entire program by using this instruction I'm buying a new computer in order to run a floating point intensive program called RUNIT. What program would be the best benchmark to run in order to determine which computer to buy? My speedup factor is two. My old execution time was 100ns. What is my new execution time in ns? (enter number only) My speedup factor is two. My new execution time after speedup is 100s. What was my old or original execution time in seconds? (answer is number only)

Explanation / Answer

5.
speedup = oldtime/newtime * pecent = 300/200 * 0.5 = 0.75

6.
Running RUNIT would be the best option

7.
speedup = oldtime/newtime ; so new time = oldtime/speedup = 50ns

8.
speedup = oldtime/newtime ; so oldtime = newtime*speedup = 200ns

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