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duration samples. Forecasting procedure of the one period ahead values of the
quantiles of the samples is given in Chapter 3. After defining widely used basic
forecasting techniques, we try to develop a nonparametric method which gives estimates
for  quantiles for  = 0.1, 0.2, 0.3, . . . , 0.9. Estimation and improvement
of model parameters are also examined in this chapter. Applying the defined methods
to given samples, forecast accuracies of the methods are compared. The best
suited method is expected to be chosen by combination. (Conclusion has not been
made yet.)
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Chapter 2
EMPIRICAL ANALYSIS
In this chapter, we examined the statistical properties of Ground Handling Process
Duration samples. Among many data sets, we chose a test set from year 2008 -
Operation Period 2. This set contains ground handling process times of the Boeing
737 − 400 (734) type aircrafts having departed from Amsterdam Schiphol Airport(
AMS), landed in Europe. Our set contains 2725 observations which would
yield good results for testing purposes (see Appendix A - Maximum Likelihood
Estimation).
First, we tried to determine the distribution of the data based on the sub-processes
using Maximum Likelihood Estimation. Then, we followed Least Squares Estimation
method using the quantiles of the data. In order to test the results, we used
Chi-Square Goodness of Fit Test.
For numerical computation purposes the statistical software R has been used.
2.1 DISTRIBUTION ANALYSIS OF THE DATA
In this section, we tried to find a suitable distribution which fits to our data. The
sub-processes are distributed according to either Gamma distribution or Weibull
distribution ([11]Boom, L[2003]). Therefore, we tested if the whole ground handling
process is distributed according to one of those distributions.
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We applied defined maximum likelihood process to the data by using statistical
software R. Clearly, ground handling process takes time greater than 0. However,
the smallest time this process takes is not close to 0 which can be easily explained
by practical reasons. Therefore, if the distribution is gamma, there must be a shift
modification in our data, as Gamma or Weibull distributed random variables start
close to 0. Thus, we used an additional shift parameter for both distributions. First,
the Gamma distribution is tested.
We estimate  = (s, k, ) in  where s is the shift parameter, k is the shape and  is
the scale parameters.
The corresponding R code can be found below:
mlogl < −function(theta, x){sum(−dgamma(x − theta[1], shape = theta[2],
+scale = theta[3], log = TRUE))}
theta.start < −c(1, (mean(x))  2/var(x), var(x)/mean(x))
out < −nlm(mlogl, theta.start, x = x)
theta.hat < −out$estimate
theta.hat
The resulting R output is as follows:
theta.hat
[1] 3.170722 19.604990 1.986347
$minimum
[1] 9744.346
$gradient
[1] − 1.388600e − 05 − 3.699634e − 05 − 4.010511e − 04
$iterations
16
[1]26
Here, ”x” is our random vector of length n = 2725, ”theta = (theta[1], theta[2],
theta[3])” represents our set  = (s, k, ) and ”mlogl” is the log-likelihood function.
We chose ”theta.start = (1, (mean(x))2/var(x), var(x)/mean(x))” as starting
point because sample shape, kn, and sample rate, n, are calculated as:
kn =
μ2n
2n
and n =
2n
μn
where μn is sample mean and 2n
is sample variance. These are the best estimators
one can calculate from the sample.
As seen in the R output, the parameters minimizing log-likelihood function are
 = (s, k, ) = (3.170722, 19.604990, 1.986347).
Now, we have to test how good the Gamma distribution with these parameters fits
to our data.
First of all, a quantile-quantile graph may draw a representative picture for our aim.
y = x − s is used in Figure 2.1 The q-q plot shows that our data does not follow
the estimated Gamma distribution. In order to have more theoretical conclusion,
Chi-square goodness of fit test is applied. Our null hypethesis is that the samle
follows a Gamma distribution with the estimated paameters. Because there is not
a general optimal choice for the bin-width, in this study, we use the following approach:
First, we divide our data into 2n2/5 groups. Then, we check the number
of observations in each bin and rearrange the bins such that each bin has at least 5
observations.
Total number of observations, n, we have is 2725, so 2n2/5 is approximately 47.
Therefore we divide the interval [1.829278, 149.879278] into 47 equal sub-intervals
　
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