Abstract:

1. Developing a Mathematical Model of River TravelGavin Stewart and Benjamin Spitz with Charles Stoddard Frontiers of Science Institute, University of Northern Colorado, Greeley, CO 80639 • Abstract: • We consider a problem of boats traveling down a river.  The river has a known length and we know the position of each stop on the river.  Boats on the river travel at a known speed for twelve hours in a day, and must spend between six and eighteen days on the river. To constrain the problem, we consider only schedules where, with the exception of the first and last six day period, an equal number of boats is always on the river.  We show that two types of rivers exist based on the arrangement of stops, and find the largest number of boats which can travel down each type of river in a 180 day season.  • Our Problem: • We examine the case with only boats traveling down the river, where, except for the first and last six days, an equal number of boats are on the river at all times. Next, we classify rivers into two types.  The first type of river we optimize using generating functions, beginning with the case of uniformly distributed stops and generalizing to non-uniformly distributed stops.  Finally, we use graph theory to optimize rivers of the second kind, representing the river as a digraph.   • Two River Types: • There are two restrictions on the number of boats which can finish.  Firstly, boats must take between six and eighteen days to travel down the river, which restricts the number of boats which can finish in a six day period to Y.  Secondly, the geometry of the river can prevent more than a certain number of boats from passing a point.  This restriction, k, is given as the minimum outdegree of a vertex of a digraph representation of the river, where all vertices before the current vertex are over 96 miles from the end of the river.  In the first case, Y ≤ 6k; in the second case, Y > 6k. • References • 2012 MCM problems (2012). In COMAP. Retrieved July 1, 2012, from http://www.comap.com/undergraduate/contests/mcm/contests/ • 2012/problems/ • Aumann, R. J. (2008). Game theory. In The New Palgrave Dictionary of Economics. Retrieved June 29, 2012, from http://www. • dictionaryofeconomics.com/article?result\_number=4\&q=game+theory\&edition=current\&id=pde2008\_G000007\&hh= • \&topicid= • Giordano, F. R., Fox, W. P., Horton, S. B., \& Weir, M. D. (2009). A first course in mathematical modeling (4th ed). Belmont, CA: • Cengage Learning. • Liang, F. (2007, March 22). Chapter 8: Regression Diagnostics: Residuals. Retrieved July 1, 2012, from http://www.stat.tamu.edu/ • ~$fliang/st408ch8.pdf Y ≤ 6k: When Y ≤ 6k, at most Y boats can finish in a six day period.  A schedule reaching this efficiency for rivers with uniformly distributed stops is represented by a generating function as: This schedule forces boats to spend six days on the river (summation of μ(n) over any six day time interval, shows the farthest boat moves to the last stop in six days) and does not move boats a distance greater than 96 miles in a day.  As boats arrive every six day period except the first period, 29Y boats total can travel down the river in a season.  For non-uniformly distributed stops, we redefine μ to obtain This schedule is also legal and sends 29Y boats down the river in a season. We can establish legality by summing μ(n) over a six day time interval, which evaluates to or simply Y. We can also show that boats never travel more than k stops forward in a day. The proof is by cases: either six divides Y, or six does not divide Y. In the first case, While in the second case, Y > 6k: When Y > 6k, at most k boats can finish the river in a day.  We represent the river as a digraph. Then, we divide this digraph into a set of subgraphs, each intersecting at only the beginning and end vertex.  By dividing the river vertices into equivalence classes based on their index modulo k, we can divide the river into k subgraphs, each allowing one boat to finish per day.  No larger set is possible, so k boats can finish in a day. To show that no larger set is possible, we state that since k is the lowest outdegree, we can remove k stops from any river to cut off flow completely (by the definition of outdegree and the river graph). So, if we make k subgraphs, and remove all vertices in those subgraphs from the parent graph, we will always have a parent graph that is not a solvable river, so no more subgraphs can be made, and at most k boats can finish the river each day. Moreover, boats can always finish in 6 days (as they can travel over 96 miles in two days on a solvable river, which is more than one-third the length of the river).  Thus, boats finish on 180-6=174 days of the season, meaning 174k boats finish in a season. Y = 8, k = 1 Y > 6k Discussion: As there are no more stops for boats to fill on the river for rivers where Y≤6k, and boats cannot finish in less time, 29Y is the least upper bound for rivers when Y≤6k. However, for Y>6k, the assumption that an equal number of boats are on the river (except for during the first and last six day period) proves vital in establishing the upper bound, allowing us to utilize flow network techniques to prove only k boats can reach the end of the river in one day. There is no reason to assume the upper bound applies when we remove the requirement of conservation of flow – a greater upper bound may exist for this less constrained problem. Day 1 Day 2 Day 3 1 + 0x + 0x2 … Acknowledgments Charles Stoddard Nicholas Horianopoulos and other FSI Staff Xcel Energy Foundation Kinder Morgan Foundation 1984 Alumni (Brian Kanaga) Spitz Family 1 + 1x + 0x2 … 1 + 1x + 1x2 …