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SIMULATION OUTPUT ANALYSIS

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Title: SIMULATION OUTPUT ANALYSIS


1
SIMULATION OUTPUT ANALYSIS
  • Week 8
  • Kelton text (Ch. 6) and prepared notes
  • GE-703 Kumpaty

2
Statistical Analysis of Simulation Output
  • Based on inferential/ descriptive statistics
  • The population is the entire set of items or
    outcomes. Applied to simulation, it is the entire
    set of observations or outcomes that a simulation
    model can produce. A sample is at least one
    unbiased observation from the population. The
    sample size is the number of samples that have
    been collected.

3
Statistical Analysis of Simulation Output
  • Statistical concepts
  • Samples are generated by running the experiment.
  • The population size is often very large or
    infinite.
  • A random variable is used to express each
    possible outcome of an experiment as a continuous
    or discrete number.
  • Experimental samples (replications) are
    independent.

4
Simulation Replications
  • One run constitutes one replication of the
    experiment. To obtain a sample size n, we need
    to run n independent replications of the
    experiment.
  • Random Number Generation (Ch. 3) Initial Seed
    Values
  • With adequate collection of independent
    observations, statistical methods could be
    applied to estimate output response.

5
Point Estimates
  • Population Mean
  • Mean of a Distribution
  • Continuous case
  • m -?? ? x p(x) dx where p(x) is the
    probability density function
  • Discrete case
  • m i1?n pi xi where pi is the probability of
    i th outcome

6
Point Estimates
  • Population Variance
  • Variance calculated from a distribution
  • Continuous Case
  • s2 -?? ? (x m)2 p(x) dx where p(x) is the
    probability density function
  • Discrete Case
  • s2 i 1?n (xi m)2 pi where pi is the
    probability of i th outcome

7
Estimators for the population
  • Sample
  • Collection of n observations (outcome of a
    process) X1, X2, X3 --------- Xn
  • Observations has randomness that can be modeled
    by normal distribution
  • Note Size of n may not be large enough to prove
    that n observations form a normal distribution

8
Estimators for the population
  • Sample Mean
  • It is an estimator for the population mean (i.e.
    the mean value of the probability distribution
    that is modeling the outcome)
  • m i1 ? n Xi/n where Xi is the ith
    observation
  • Sample Variance
  • It is an estimator for the population variance
    from n observations X1, X2, X3, Xn
  • s2 i 1?n (Xi m)2 /(n-1)

9
Theorem
  • If a population has a finite mean m and finite
    variance s2 then the distribution of sample mean
    m approaches normal distribution with variance
    s2/n and mean m as the sample size n increases.
  • The central limit theorem relates that if a
    variable is defined as the sum of several
    independent and identically distributed (IID)
    random values (observations), that the variable
    representing the sum of the observations tends to
    be normally distributed.

10
Properties of m and s2
  • If X1, X2,X3, ..,Xn are samples (observations or
    outcome) from a population of normal distribution
    of mean m and variance s2. Then
  • The sample mean m i1 ? n Xi/n is normally
    distributed with mean m and variance s2/n
  • The quantity s2 i 1?n (Xi m)2 /(n-1)
    follows a Chi Square distribution with n-1
    degrees of freedom (?)
  • The quantity t? (m-m)/(s/?n) follows Students
    t distribution with n-1 degrees of freedom (?)

11
Range of m
  • The expression of t? can be written as
  • m m - t? (s/?n)
  • An item of interest is the calculation of the
    upper and lower bounds of the population mean m
    with a probability value called confidence level.
    Typical number of confidence level is 95 or the
    probability P 0.95 (1-a) where a is the
    significance level.

12
t Distribution and Range of m
P Area of the curve between the arrow
Area (1-P)/2
Area (1-P)/2
t?,(1-P)/2
-t?,(1-P)/2
Range of m m - (s/?n) t?,(1-P)/2 ? m ? m
(s/?n) t?,(1-P)/2
t?,(1-P)/2 can be obtained from t distribution
table
13
Number of Replications (Sample size)
  • Use the relation t? (m-m)/(s/?n)
  • Set ? ?? and (m-m)Range/2 (hw or e), Select a
  • From the t distribution table determine t?,(1-a
    )/2
  • Compute n (s t?,(1-a)/2/hw)2
  • This is the number of replications that will
    provide an adequate sample size for meeting the
    desired absolute error (e or hw) and significance
    level a or 1- a.
  • relative error version, use re/(1re)m in
    place of e.

14
Sample Calculations

15
Statistical Issues with Simulation Output
  • Assumptions that must be met regarding the
    sample of observations used to construct the
    confidence interval
  • Observations are independent so that no
    correlation exists between consecutive
    observations.
  • Observations are identically distributed
    throughout the entire duration of the process
    (they are time invariant).
  • Observations are normally distributed.

16
Terminating Simulations
  • A terminating simulation is one in which
    simulation starts at a defined state or time and
    ends when it reaches some other defined state or
    time.
  • 1. Select the initial model state
  • 2. Select a terminating event
  • 3. Determine the number of replications
  • Usually ten replications and then construct a
    confidence interval for the performance measure
    of interest. If you are satisfied with the width
    of the confidence interval, stop and report your
    results otherwise, continue running additional
    replications until the confidence interval is
    reduced to the desired width.

17
Nonterminating Simulations
  • A nonterminating simulation is one which the
    steady state (long-term average) behavior of the
    system is analyzed.
  • 1. Determine and eliminate the initial warm-up
    bias.
  • Figure 9.3. Welch moving average (Table 9.4)
  • 2. Obtain sample observations.
  • By running replications or batch intervals
  • Lag-1 autocorrelation (between -0.20 and 0.20)
  • 3. Determine run length.

18
Nonterminating Simulations
  • Lag-1 autocorrelation (between -0.20 and 0.20)

19
COMPARING SYSTEMS
20
Hypothesis Testing
  • Two strategies identify the one that maximizes
    the throughput of the production
  • Null hypothesis, H0 The value of m1 is not
    significantly different from m2.at the a level of
    significance.
  • H0 m1 m2
  • H1 m1 ? m2 (Alternate hypothesis)

21
Errors in Hypothesis Testing
  • Type I error
  • Rejecting H0 in favor of H1 when in fact H0 is
    true
  • a level of significance is the probability of
    making a Type I error. Typical value 0.05.
  • Type II error
  • Fail to reject H0 in favor of H1 when in fact H1
    is true
  • b is the probability of making a Type II error.
    It increases when a decreases. (Caution Dont
    make a too small.)

22
Confidence Interval method
  • Equivalent to conducting a two-tailed test of
    hypothesis
  • P(m1-m2)-hw(m1-m2)(m1-m2)hw 1-a
  • If the two populations means are the same, then
    m1-m2 0, which is the null hypothesis.
  • If the confidence interval includes zero, we
    fail to reject H0. Conclusion The value of m1
    is not significantly different than the value of
    m2 at the a level of significance.

23
Comparing two alternative system designs
  • Recall assumptions Observations are independent
    and normally distributed.
  • Two common methods for constructing a confidence
    interval for evaluating hypotheses
  • 1.Welch Confidence Interval method (modified
    two-sample-t confidence interval)
  • 2. Paired-t Confidence Interval method

24
Welch C.I. method
  • Observations from each population be normally
    distributed and independent within a population
    and between populations
  • Does not require the number of samples from one
    population (n1) equal the number of samples from
    another population (n2)
  • Does not require that the two populations have
    equal variances

25
Welch C.I. method continued..
  • P(m1-m2)-hw(m1-m2)(m1-m2)hw 1-a
  • Welch c.i is given as above with
  • Where
  • df is estimated by

26
Paired-t C.I.method
  • Observations from each population be normally
    distributed and independent within a population
    but this method does not require the observations
    between populations be independent
  • Does require the number of samples from one
    population (n1) equal the number of samples from
    another population (n2)
  • Does not require that the two populations have
    equal variances

27
Paired-t C.I. method continued..
  • Pm1-2-hw(m1-2)m1-2hw 1-a
  • Find sample mean and sample S.D. for the paired
    sample (difference in sample values)
  • Paired-t c.i is given as above with
  • H0 m1-20
  • H1 m1-20

28
Comparing more than two alternative system designs
  • 1.The Bonferroni Approach (for comparing 3 to 5
    designs)
  • 2. Analysis of Variance (ANOVA) in conjunction
    with multiple comparison test

29
Bonferroni Approach
  • Construct a series of confidence intervals to
    compare all system designs to each other (all
    pairwise comparisons)
  • For three designs (1-2), (2-3), (1-3)
  • K designs C.Is K(K-1)/2
  • Significance level ai a/ CIs (not necessary
    that all individual ones are to be equal) but S
    ai a
  • H0 m1 m2 m3 m
  • H1 m1 ? m2 or m1 ? m3 or m2 ? m3

30
Advanced Statistical Models (ANOVA and multiple
comparison test)
  • Analysis of Variance (ANOVA) in conjunction with
    multiple comparison test
  • H0 m1 m2 m3 mK m for K alternative
    systems
  • H1 m1 ? m2 or m1 ? m3 or mi ? mj for at
    least one pair
  • Number of factor levels number of alternative
    system designs K
  • Number of observations for each factor level n
  • Total number of observations N nK

31
Analysis of Variance (ANOVA test)
  • Analysis of Variance (ANOVA) Single Factor
  • F calc tested against F critical
  • Do the Excel sheet
  • 3 strategies, 10 observations each
  • DFT 2, DFE 27
  • Find SS (sum of squares), SSE, SST
  • MSE, MST, Fcalc MST/MSE
  • F critical

32
Multiple Comparison Test
  • Do this after confirming from ANOVA that at
    least one strategy performs differently than the
    other strategies.
  • Fishers Least Significant Difference Test
  • If m1-m2 gt LSD (a) then m1 and m2 are
    significantly different at the a level of
    significance.
  • LSD (a) t (dfe, a/2) sqrt (2 MSE/n)
  • Do the calculations on Excel sheet
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