Gavin's Modeling Project 5

1. The cost for a communication line between two stations is proportional to the length of the line. The cost for conventional minimum spanning trees of a set of stations can often be cut by introduction of ``phantom'' stations and then constructing a new Steiner tree. this device allows costs to be cut by up to 13.4% (=1 - \sqrt(3)/2). Moreover, a network with n stations never requires more than n-2 points to construct the cheapest Steiner tree. Two simple cases are shown in figure 1.
   

   For local networks, it often is necessary to use rectilinear or ``checkerboard'' distances, instead of straight Euclidean lines. distances in this metric are computed as shown in figure 2.
   

   

   Suppose you wish to design a minimum cost spanning tree for a local network for 9 stations. Their rectangular coordinates are: a(0,15), b(5,20), c(16,24), d(20,20), e(33,25), f(23,11), g(35,7), h(25,0), i(10,3). You are restricted to using rectilinear lines. Moreover, all ``phantom'' stations must be located at lattice points (i.e., the coordinates must be integers). The cost for each line is its length.

   1. Find a minimal cost tree for the network.
   2. Suppose each station has a cost d^3/2 w, where d= the degree of the station. If w=1.2, find a minimal cost tree.
   3. Try to generalize this problem.

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2. Some state water-right agencies require from communities data on the rate of water use, in gallons per hour, and the total amount of water used each day. Many communities do not have equipment to measure the flow of water in or out of the municipal tank. Instead, they can measure only the level of water in the tank, within 0.5% accuracy, every hour. More importantly, whenever the level in the tank drops below some minimum level L, a pump fills the tank up to the maximum level H; however, there is no measurement of the pump flow, either. Thus, one cannot readily relate the level in the tank to the amount of water used while the pump is working, which occurs once or twice a day, for a couple of hours each time.

   Estimate the flow out of the tank f(t) at all times, even when the pump is working, and estimate the total amount of water used during the day. Table 1 gives real data, from an actual small town, for one day.

   The table gives the time, in seconds, since the first measurement, and the level of water in the tank, in hundredths of a foot. For example, after 3316 seconds, the depth of the water in the tank reached 31.10 feet. The tank is a vertical circular cylinder, with a height of 40 feet and diameter of 57 feet. Usually, the pump starts filling the tank when the level drops to about 27.00 feet, and the pump stops when the level rises back to about 35.50 feet.

   Time	Level	Time	Level	Time	Level
   0	3175	35932	pump on	68535	2842
   3316	3110	39332	pump on	71854	2767
   6635	3054	39435	3550	75021	2697
   10619	2994	43318	3445	79254	pump on
   13937	2947	46636	3350	82649	pump on
   17921	2892	49953	3260	85968	3475
   21240	2850	53936	3167	89953	3397
   25223	2797	57254	3087	93270	3340
   28543	2752	60574	3012
   32284	2697	64554	2927