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What is the overall speedup of a system spending 65% of its time on I/O with a disk upgrade that provides for 50% greater throughput.

The formula should be Amdahl's law: overall speedup = 1 / (1 -fraction enhanced) + fraction enhanced/ speedup enhanced

However, I'm not sure if I am setting up the equation right and if the answer would be 60% faster.

  1 / (1 - .65) + .65/.5
  1 / .35 + 1.3
  1 / 1.65
  = .6 
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  • \$\begingroup\$ Well, for one thing, "50% greater throughput" does NOT mean that the new disk system is twice as fast as the old one. It means that in a given amount of time, it processes 150% (3/2) as much information as the original system, or that the same amount of work takes 2/3 the time (not 1/2). \$\endgroup\$ – Dave Tweed Nov 17 '14 at 2:10
  • \$\begingroup\$ So, would the answer be .65 * 2/3 + .35 * 1 = .43 + .35 = .78 = 22% faster? \$\endgroup\$ – Carlo Nov 17 '14 at 2:52
  • \$\begingroup\$ There isn't a simple answer to your question. If the workload has lots of short reads, better throughput may not help. \$\endgroup\$ – Eric Gunnerson Nov 17 '14 at 3:55
  • \$\begingroup\$ There really isn't any other information or description other than this question. I think the workload is supposed to be "constant" if that makes sense. \$\endgroup\$ – Carlo Nov 17 '14 at 4:06
  • \$\begingroup\$ Perhaps I can provide some more details. So the new speed would be .78 -- a ratio of 1/.78 and to get the percent we would subtract 1 which would give me 28%. Can anyone confirm this? \$\endgroup\$ – Carlo Nov 17 '14 at 16:40
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It looks like you were correct in your last comment. I found this answer here: http://www.scribd.com/doc/248756454/Computer-Architecture-HW-5-Answers-CH-7#scribd

...which says...

Amdahl's Law Equation

where F = 0.65, K = 1.5

so S = 1.28 or 28% speedup

I know this is a year late, but hopefully it helps somebody else too.

Cheers!

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