[30 points] You have been asked to evaluate server performance, so you decide to

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Question

[30 points] You have been asked to evaluate server performance, so you decide to run a benchmark program that finds out memory operations currently take 45% of execution time. Your team is evaluating two server upgrade options that can help improve performance:

Option 1: the new server architecture uses a memory cache which speeds up 75% of memory operations by a factor of 5.

Option 2: same as Option 1 along with an additional level (L2) cache which speeds up 1/2 the remaining 25% of memory operations by a factor of 3.
1. Calculate the overall speedup for Option 1?
2. What is the overall speedup for Option 2?
3. What is the upper bound on overall speed-up from any additional optimizations that could be done for the L2 cache described in Option 2?

Solution:

Suppose that the previous server architecture was executed in 100 unit of time.

then this means that 45 unit of time is consumed by the execution time

Now

Option 1:

75% of the time is improved by the factor of 5

this means that

(45*0.75)/5= 6.75, this is after the improvement

25% of previous is still same

so the effective time= 45*0.25 + 6.75= 11.25+6.75 = 18

Server improved time= 18+55= 73

Option 2:

Now along with the option 1 there is one more cache

so for the remaining 25% half of it is getting improved by the factor of 3.

45*0.25= 11.25

half of it is= 11.25/2= 5.625

now improved by the factor of 3

5.625/3= 1.875

Effective time only because of L2 cache

= 1.875 + 5.625= 7.5

Now 11.75 is improved to 7.5, so

total effective time with option 2

= 7.5+6.75= 14.25 unit of time

Effective time of the server = 14.25+55= 59.25 unit of time

Calculate the overall speedup for Option 1?

Speedup for option 1 = 100/73= 1.3698630137

What is the overall speedup for Option 2?

Speedup = 73/59.25= 1.23206751055

What is the upper bound on overall speed-up from any additional optimizations that could be done for the L2 cache described in Option 2?

Speedup = 100/59.25= 1.68776371308

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