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H0: The data points are independent.
Ha: Not H0
P Value = 0.4883
Thus H0 cannot be rejected.
From the above two i.e. correlation plot and LB test we conclude
that the observations are
independent and hence proceed to fit a distribution.
Fitting a distribution:
From the histogram and the value of coefficient of variation,
which is computed as 0.995, we narrowdown to exponential and
weibull distributions.
The maximum likelihood estimates of the respective parameters
are as follows.
Distribution Shape parameter Scale parameter
We go with exponential distribution because of the error
generated in R due to indeterminacy.
Hence our distribution is exponential with the above
parameter.
Goodness of Fit:
Goodness of fit is tested with Kolmogorov Smirnov test (KS test)
due to the following advantages of
the KS test.
Non necessity of grouping the data Exactly valid for any sample
size n.
C. The service time distribution of each class is given.
Customers of class A and B are to go in to one
queue and class C to another queue.
The minimum number of tellers is to be used. Hence we use one
teller for both the queues. This We
.
Average time a type A spends in the bank:
Time spent by a customer in the bank = time in the queue +
service time.
Accordingly a monitor mA is initialized and the mean is taken.
The same is replicated 50 times
using seeds 100 to 150.
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The 95% confidence interval for the same is (20.05467, 22.64498)
minutes.
Average time a type B spends in the bank:
A monitor mB is initialized and the mean is taken. The same is
replicated 50 times using seeds
100 to 150.
The 95% confidence interval for the same is (11.86976, 12.29478)
minutes
Average time a type C spends in the bank:
A monitor mC is initialized and the mean is taken. The same is
replicated 50 times using seeds
100 to 150.
The 95% confidence interval for the same is (13.61194, 14.18023)
minutes
The above three show that a type A customer spends the most time
in the bank. This might be the
reflection of the lowest priority accorded to him.
Type C spending more time than B in spite of having a higher
priority, may be because of the
increased service time needed by Type C. The mean of C is 4.2
compared to a uniform (3,4)
distribution for the service time needed by B
Average time a type A spends in the queue:
A monitor cA is initializedand mean is taken. A confidence
interval is needed in such a way that
the half-length is less than 5% of the mean. However 50
replications do not yield such a confidence
interval. Hence we iteratively increase the number of
replications. Finally doing 200 replications
from seeds 100 to 300 yields a suitable confidence interval.
The 95% confidence interval for the same is (17.02831, 18.50863)
minutes
Utilization of the teller:
A monitor U is defined. This is assigned to 1 as soon as a
customer enters service and 0 as soon as
he leaves the counter. This is a measure ofthe time when the
teller is occupied. The timeAverage
of U is taken here as this is the area under the function
defined as follows.
B(t) = 1,
