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8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 QCL_1 0 0 0 11 0 0 0 0 0 17 0 0 0 0 0 23 0 0 0 0 0 29 0 0 0 0 34 0 0 0 0 0 1 0 0 10 0 0 0 0 15 0 0 0 0 20 0 0 0 0 25 0 0 0 0 30 0 0 0 0 35 0 0 0 0 2 0 9 0 0 0 13 0 0 0 0 18 0 0 0 22 0 0 0 0 27 0 0 0 31 0 0 0 0 36 0 0 0 3 8 0 0 0 12 0 0 0 16 0 0 0 20 0 0 0 24 0 0 0 28 0 0 0 32 0 0 0 36 0 0 0 1 0 0 0 11 0 0 0 15 0 0 18 0 0 0 22 0 0 25 0 0 0 29 0 0 0 33 0 0 36 0 0 0 3 0 0 10 0 0 13 0 0 0 17 0 0 20 0 0 23 0 0 0 27 0 0 30 0 0 33 0 0 0 37 0 0 2 0 9 0 0 12 0 0 15 0 0 18 0 0 21 0 0 24 0 0 0 28 0 0 31 0 0 34 0 0 37 0 0 1 8 0 0 11 0 0 14 0 0 17 0 0 20 0 0 23 0 0 26 0 0 29 0 0 32 0 34 0 0 37 0 0 1 8 0 0 11 0 13 0 0 16 0 0 19 0 21 0 0 24 0 0 27 0 29 0 0 32 0 0 35 0 37 0 0 1 8 0 10 0 12 0 0 15 0 17 0 0 20 0 0 23 0 25 0 0 28 0 30 0 0 33 0 35 0 0 38 0 2 0 9 0 0 12 0 14 0 0 17 0 19 0 21 0 0 24 0 26 0 28 0 0 31 0 33 0 35 0 0 38 0 2 0 9 0 11 0 13 0 0 16 0 18 0 20 0 22 0 0 25 0 27 0 29 0 31 0 33 0 0 36 0 38 0 3 8 0 0 11 0 13 0 15 0 17 0 19 0 21 0 23 0 25 0 27 0 0 30 0 32 0 34 0 36 0 38 0 4 QCL-1 represents the clustering result, and1,2 3, 4 represent 4 groups of data.
From the graph above, we can get the following clustering result:
3. The 3 step, selecting the most typical camping program.
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Our purpose is to maximize the number of the trips within the minimized number of encounters between different groups. Accordingly we can try to cluster the camping program, select the most typical and find out the optimized program. Measures of selecting:
Step1: Compare two adjacent scales of camping programs (classified by the number of days spent)
For example, compare the programs of 6 and 7 nights, taking 6 nights as a base. The first night of both camping programs are numbered as 5, therefore the difference is 0. We shall calculate the minimum differences between the numbers of the best campsites on each night in the 7 nights program and those in the 6 nights program in proper order. After that the number of total transfer can be obtained by summing all the data together.
Step2: Calculating the repetition rate
In the last calculation, 0 represents that the best campsites of two programs are completely coincided. And a smaller Repetition rate indicates a more minimized distance between the two best campsites. We can calculate the repetition rate of the best campsites of two programs in order to screen the different sorts of clustered data. Step3: Calculating the average number of transfer campsites
Merge the different programs by calculating the average number of transfer campsites. The smaller the number is, the programs are more suitable to be merged and vice versa
Step4: Screening the right camping programs Principles for screening:
Try to select the camping programs from the different sorts obtained by the cluster analysis. In a same sort, the program of a smaller repetition rate is priority.
If the repetition rates have little difference, the program of the minimum number of average transfer campsites is to be chosen.
Under the above circumstance of screening, we can get 4 numbers, 6, 16, 8 and 12. And at the same time, we can find that in the first clustered sort, 6 is the base which requires the least campsites and in that sort existing 5 camping programs. In order to have a more coordinated entire data, we need to pick a program other than 6 in accordance of the principles aforementioned, namely 11.
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Through the cluster analysis and optimal screening, we can now obtain 5 camping programs, as 6, 8, 11, 12, and 16 from the problem of 6 to 18 nights camping. 4.2.3 To find capacity of the river
Prerequisite: starting point and end point of the river is not suppose to camp, and the distance of each camp is equal in between which is 5.6 miles. Number those campsites from the starting point of the river, which are 1-39 Work out the carrying capacity of the river
From the problem itself, we can deduce a reasonable range of Y, and 39 campsites are the fittest. Therefore Y=39 is used in our model.
?Y??Y?Max(??,6)?xi?Min(??,18)?18??6?, then 6?xi?6
So xi=6。
That is to say, we can choose 6 different spots as the accessible campsites of a
camping program. The program is Night_6 trip .Then 39 campsites can be divided to 6 groups .
nnMaxtotal??[180i?1?(xi?1)]??[180i?1?(6?1)]?1050
The carrying capacity of the river, namely the Maximum trips, is also 1050 And then 36 campsites are used among the total 39, so the utilization is
nMaxu??i?1xiY?[180?(xi?1)]180?[180?(6?1)]180?3639?94.95%
Note of the carrying capacity of the river:
The carrying capacity of the river is gained under the situation of maximizing the transport capacity. Now only Night_6 trip is open, which is not suitable for the requirement of varying duration in the title.
4.2.4 Determine Dedicated Pipeline
1. Adjust the camping place .
Adjustment principle: Let the campsite be used the smallest possible. Calculation results show that all the campsites are used fewer than twice.
Make sure that the campsites of Night_6 trip are belong to that of Night_12 trip, so the same as that of Night_8 trip being belong to that of Night_16 trips.
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Note: The campsites of Night_12 trip are totally different from Night_16 trip. 2. Set up pipeline
According to above analysis, the campsites of Night_6 trip are belong to that of Night_12 trip, so the same as that of Night_8 trip being belong to that of Night_16 trip.
Then the camping program for the trips which Night_6 trips and Night_12 trips can be achieved in the same pipeline; similarly, the camping program for the trips which Night_8 trips and Night_16 trips can also be achieved in an pipeline while Night_11 trip should adopt another pipeline. Clustering results:
Those 39 campsites are divided into three kinds of lines:
Line_12: The line is available only for Night_6 trips or Night_12 trips. Line_16: The line is available only for Night_8 trips or Night_16 trips.
Line_11: A free line is available for the trips that camp within 11 nights trips. While there are lots of trips which camp within 11 nights, they can use Line_11 if Line_12 and Line_16 aren't available. Schedule an optimal mix of trips
Inference: Different pipelines, different camp no.
Description: Renumbered Camp site according to the original order of each pipeline as Line_no. As the following picture shows.
Line_12: Only available for the trips which camp 6nights or 12nights.
Scheme1: Only two group of Night_6s are set up in the same days. All the Line_no of the earlier one are even number while the other Night_6 are odd number.
Scheme2: Only one group of Night_12 are set up in the same days. Rest by Line_no.
Important: Too keep the pipeline normal and get a higher utilization, only 7days after, a night_12 can be followed by night_6s and enter the river. Assumption: enough trips of any kinds Utilization of Line_12:
When all the trips are night_6s, we can get the highest utilization. According to the equation mentioned above , what we can get :
