Embarkation Probabilities

This is the probability that a boat is going to embark on a trip during the simulated time period. These probabilities were calculated from the boat interview data discussed in section 3.3.3. Each subject was asked how often they took their boat out, whether they took their boats out on weekends and holidays or weekends, boating hours during the day, and boating season (boating months) during the year. Table 4.1 shows an example of the raw data format.

The data then were normalized to common units. Data were normalized to daily, monthly, and yearly trips. In all cases, trip frequencies were normalized to number of trips per available time units in the given time. For example, if a subject reported that they took their boat out 3 times a month on weekends, usually between the hours of 8 AM and 2 PM, and their boating season was November through May, the trip frequency would be reported in the following units:

$$\frac{Trips}{Time\, units} \times \frac{Time\, units}{Unit\, time} = Trips/Unit\, time$$ (3)

$$\frac{3\, Trips}{month} \times \frac{6\, months}{boating\, year} = 18\, trips /boating\, year$$ (4)

$$\frac{3\,trips}{month} \times \frac{one\,month}{8\,weekend\,days} = \frac{3}{8} trips/weekend\,day$$ (5)

So, everything else being equal, the probable number of trips that that boat would make on that day would be

$$P (embarkation) = \begin{cases} 0 & \text{on a weekday}\ \frac{3}{8} & \text{on a weekend}\end{cases}$$ (6)

The probability of embarkation outside of the boat's season is considered to be 0. This expression gives a uniform probability of a boat embarking on any weekend day during it's boating season. All distributions used in this simulation will be considered to be uniformly distributed unless an empirical justification can be found otherwise in the data. In reality, a uniform distribution may not be the best way to represent the possibility of such an occurrence; information was provided in the boater interviews used in the calibration phase of this study that suggested that boaters are more likely to embark in ``more suitable'' weather. It has also been shown that many boaters only embark on favorable tides (Antonini and Box, 1996). This information could be used to adapt this model to consider various weather scenarios or potential dredging requests, but was not considered in the scope of this study.

This simulation concentrates on daily traffic summaries. However, if one wished to increase the resolution of the simulation to investigate variation in traffic dynamics over the course of the day, one could segment the day into discrete time steps $\Delta t$. Assuming the probability of the boat embarking at any given time t during a weekend day is uniformly distributed across $\Delta t$, the probability of embarkation at time t would be

$$P (embarkation\vert t) = \begin{cases} \frac{3}{8} \times \... ...Delta t} & 0800h < t < 1400h \ 0 & \text{otherwise}\end{cases}$$ (7)

This assumes a uniform probability of embarkation throughout the boating day. In reality, this would probably be better represented with a normal or a Poisson distribution that is skewed towards the middle of the boating day, as a person who boats between 8 AM and 2 PM is not likely to embark at 1:45 PM. This matter would warrant further consideration in situations where one needs to investigate finer temporal variability in traffic patterns.

The units used for normalization of boating data are presented in table 4.2. The conversion units assume 2 boating days per week on weekends and 5 boating days per week on weekdays. It also assumes 30 boating days per month, with 8 weekend and 25 weekday boating days per month.

 

 

unit day week month year

$$\times$$ 2 8

$$\frac{1}{2} \times$$ 4

$$\frac{1}{8} \frac{1}{4}$$ BM

$$\frac{1}{8 \times BM} \frac{1}{4 \times BM} \frac{1}{BM}$$ 5cWeekdays

unit day week month year

$$\times$$ 5 25

$$\frac{1}{5} \times$$ 4

$$\frac{1}{25} \frac{1}{4}$$ BM

$$\frac{1}{25 \times BM} \frac{1}{4 \times BM} \frac{1}{BM}$$ 5cBoth

unit day week month year

$$\times$$ 7 30

$$\frac{1}{7} \times$$ 4

$$\frac{1}{30} \frac{1}{4}$$ BM

$$\frac{1}{30 \times BM} \frac{1}{4 \times BM} \frac{1}{BM}$$

Conversion units used to normalize interview data to common units.
BM = number of Boating Months

A summary of yearly and daily embarkation statistics for the interview population is presented in tables 4.3 and 4.4. If one takes the data in these tables at face value, one could conclude that on any given day in the summer more than 20% of the sailboats and more than 25% of the recreational fishing boats would be out on the water. These are conclusions that would be valid if the data were relatively homogeneous and normally distributed. Investigation of the average annual frequencies of trips (figure 4.2) suggest that, with the exception of a few outliers, the ranges are exponentially distributed. This seems to hold true for daily embarkation probabilities for the various boat classes: figure 4.3 shows daily embarkation distributions for the four most active boat categories on weekdays in the winter. Distributions for summer and weekend statistics show similar distributions, which have a superficial resemblance to exponential distributions; this suggested that exponentially distributed embarkation rates could be used to describe embarkation rates, hence they could be used as a predictor of embarkation probabilities. The following paragraphs describe some of the steps taken to verify this.

An exponential distribution is described in terms of a single parameter $\lambda$, which describes both its mean and standard deviation. The formal description of the probability density curve is

 

$$E (\lambda) = \frac{1}{\lambda} e ^{-\frac{x}{\lambda}}$$ (8)

where the mean $\lambda$ and the variance is $\lambda ^2$. It is usually used to describe processes where events happen at an average rate $\lambda$.

Visual inspection of yearly embarkation rates (figure 4.2), which is related but not identical to daily embarkation rates, suggests that an exponential curve based on the mean is not the best way to describe the embarkation summies. An exponential curve based on the median value gives a more conservative fit than a curve based on the mean; the curve based on the mean tends to overestimate the left hand side of the histogram (where the majority of the observations lie) to accommodate the outliers. However, simulated data based on the mean were more successful at reproducing the empirical data distributions than those based on the median. Generally exponential distributions are described in terms of their means (Pritsker, 1986; Råde and Westergren, 1995), so distribution means were used as a basis of calibration in this study.

   

|t:========:t| 2||lClass 3|cSummer

2||r|Count 1cmean

||-|--|--|| Row 5

Sail 67

Speed 45

Rec. Fish 70

Power Cabin 60

Other Rec. 2

Comm. Fish 5

Other Comm. 2 |b:========:b|

   

|t:========:t| 2||lClass 3|c|Summer

2||r|Count 1cmean 8cTotal

Row 5

Sail 67

Speed 45

Rec. Fish 70

Power Cabin 60

Other Rec 2

Comm. Fish 5

Other Comm. 2 8cWeekend

Row 5

Sail 57

Speed 35

Rec. Fish 58

Power Cabin 50

Other Rec. 2

Comm. Fish 4

Other Comm. 2 8cWeekday

Row 5

Sail 52

Speed 37

Rec. Fish 61

Power Cabin 45

Other Rec. 2

Comm. Fish 5

Other Comm. 2 |b:========:b|

  

Figure 4.2: Distribution of yearly total trips

  

Figure 4.3: Daily embarkation frequencies for most active boat classes.
A: Sail
B: Speed
C: Recreational Fish
D: Power Cabin

One method for evaluating how well an empirical dataset is described by a hypothetical function is the Kolmogorov-Smirnov test, which is described in many textbooks and mathematical handbooks (Råde and Westergren, 1995). The Kolmogorov-Smirnov statistic D_n is calculated as

$$D_n = \max_x \vert S_n (x) - F_0 (x)\vert$$ (9)

where S_n is the empirical distribution, and F₀ is the distribution under the null hypothesis along the same discretized sections, in this case $\lambda e ^{-\frac{x}{\lambda}}$. The statitistic tests if the empirical distribution is different from the theoretical distribution along any of the sections by more than a critical value, and the null hypothesis is rejected if at least one of the sections exceeds that critical value.

Empirical distributions were calculated for each of the four most active boat classes (sail, speed, recreational fish, power cabin) for winter, summer, weekend, and weekday conditions, a total of 16 separate distributions. Relative numbers of boaters who embark within certain ranges of embarkation rates were tabulated. The relative numbers who would embark within these same ranges were according to an exponential curve based on mean embarkation rates were calculated as

$$\int_{x_1} ^{x_2} e ^{-\frac{x}{\lambda}} dx$$ (10)

which is

$$- \lambda \bigl( e ^{-\frac{x_2}{\lambda}} - e^{-\frac{x_1}{\lambda}}\bigr)$$ (11)

 

 

|t:===:t| class summer winter 3cWeekend

Sail - -

$$\diamond$$ Speed -

$$\diamondsuit \diamondsuit$$ Rec. Fish

$$\diamondsuit \diamondsuit$$ Power Cabin 3cWeekday

$$\diamondsuit$$ Sail -

Speed - -

$$\diamondsuit \diamondsuit$$ Rec. Fish

$$\diamondsuit \diamond$$ Power Cabin |b:===:b|

$$\diamondsuit$$ do not reject H₀

- reject H₀

$$\diamond$$ very close

Table 4.5 show the conclusions drawn from the KS tests. Ten of the 16 scenarios were not rejected at the 90% confidence interval, indicating that they are consistent with exponential distributions. The other six distributions were rejected at this significance level. In each case, the rejection came at the lowest descrete range, which includes the boats that have 0 trips per day. It is possible that this is the result of incorrect counting, or that there is a separate process for the case of zero trips that does not fit the exponential curve. Since the non-zero embarkation rates were consistent with an exponential distribution based on the mean value, this was the distribution chosen to model embarkation rates in this simulation.

Inverse transformation of the equation provides the expression with which random numbers from the computer's processor, which are uniformly distributed in [0,1], are mapped to an exponential distribution $E (\lambda)$. The inverse transformation of equation 4.7 is

$$-\lambda \log (X)$$ (12)

where X is a continuous random variable in [0,1], and $\lambda$ is the mean of $E (\lambda)$.