CS 6553 Performance Evaluation, Spring 2004 Assignment 2 Comments

Each problem was worth 20 points.

Problem 1: 13.1
The answers in the back (after being corrected by the errata sheet) are correct. Remember that in calculating the variance of a linear combination, you must square the coefficients.

Problem 2: 13.2
For part e) there are two one-sided confidence intervals that can be calculated. Note the correction to the answer that makes the intervals go from or to infinity.

Problem 3: 13.3

• These are paired observations.
• Look at the 11 differences.
• You should have gotten a mean of 219.8 and s = 581.8.
• For a 90% confidence interval, use the 95%-ile of the t distribution with 10 degrees of freedom.
• Use t = 1.812
• The semi-width of the confidence interval is (581.8)(1.812)/sqrt(11) = 317.9.
• At what confidence level can you say one is better than the other?
  • st/sqrt(n) = 219.8, so t = 1.25
  • What values of p does this correspond to in the t-dist for 10 dof?
  • p = .8 gives .879
    p = .9 gives 1.372
    so this corresponds to about p = .87.
  • So, with 87% confidence I can say that the difference is > 0.
  • Note that this uses a one-sided confidence interval.
• How many workloads would you need to decide that one is better than the other at a 90% confidence level?
  • Note that we already have 87% with 11 workloads.
  • Not a simple matter of looking in the table since the number of degrees of freedom depends on the number of workloads so we don't know what t-value to use.
  • However, the t-value does not change much for p=.9 above 10 dof and it is decreasing.
  • Solve: 219.8(sqrt(n))/581.8 = 1.372, giving n = 13.1.
  • Can we get by with 13 or do we need 14?
    For n=13, use 12 degrees of freedom, t = 1.350 and the semi-width is (581.8)(1.350)/sqrt(13) = 217.8 so this is good enough (barely).

Problem 4: Redo 13.3 eliminating one of the runs.

• Everyone was given full credit fo this as it is done exactly the same way.
• Differences:
  mean = 102, s = 454, use 9 degrees of freedom.

Program 5: 30.4

• Everyone got the correct answer of 1 request per second.
• This is done using Little's Law.
• The only assumption is that everything that comes in is serviced.
• It does not require exponential service time.