Amdahl\'s Law calculates speedup as the following: using GHz Old Execution Time
ID: 3865286 • Letter: A
Question
Amdahl's Law calculates speedup as the following: using GHz Old Execution Time - New Execution Time/New Execution Time Old Execution Time/New Execution Time New Execution Time/Old Execution Time Execution time of a program on my old processor is 100 seconds, the new processor's time is 80 seconds, what does Amdahls law say the speedup is? 100 80 1.25 8 08 Execution Time of my old computer is 6 seconds, the new computer's time is 2 seconds, what does Amdahls law say the speedup is? Microsoft released a new version of Excel that processes my Macro's in 4 seconds instead of the old program which took 5 seconds. According to Amdahls law, what is the speedup? 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
1.
Speedup= Old exection time / new execution time
Option 3 is correct.
2.
From the question:
execution time of old processor time = 100 seconds
execution time of new processor time = 80 seconds
Speedup = 100 / 80 =1.25
Therefore speedup is 1.25 times.
Option 3 is correct.
3.
From the question:
execution time of old computer = 6 seconds
execution time of new computer = 2 seconds
Speedup = 6 / 2 =3
Therefore speedup is 3 times.
4.
From the question:
old version of excel that process macro = 5 seconds
new version of excel that process macro = 4 seconds
Speedup = 5 / 4 = 1.25
Therefore speedup is 1.25 times.
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