医疗运输模型混排

医疗运输模型:下面我们需要考虑在Ebola病毒传播的同时全球药物的运输调配问题。

The prediction of EVD spread is not enough to us to solve the whole problem. So, now we need to build another model for medicine transportation problem over the world.首先需要对问题进行合理化假设简化问题并抽象模型，我们将国家抽象为单个结点，所有的药品运输配送最小单位到结点，假设地球各处的球面半径一致且飞机航行轨迹固定且总完美航行两节点之间最短劣弧长。

假设整个药品运输网络构成一张无向连通图（但并非强连通），假设药品与疫苗只有一种，并且只从一个结点向外运输(应删去)。

假设每个节点的流量均充足不出现拥塞。

假设每个节点对药品有一个固定需求量且不会在短时间（为模型模拟时间长度）变化。

To simplify the problem we make some assumptions here.Use the geographic location of country’s capital instead of the whole country as the destination of our transportation.Assume the Earth is a Uniform sphere there, and the distance between two countries is the minimum flying range, namely minor arc of the two nodes.All transport lines make up an undirected connected network .There is only one country can provide one kind of ,The transport capacity of every single node is sufficient, so there is no jam in the network.The demand of every node should be a constant and will notchange in a certain time period.基于以上假设，我们可以建立图论模型，将整个药品运输网络抽象为无向连通图()G V E=，两个节点之间距离抽象为边权，每个节点的药品需求量抽象为点权。

,那么最优化药品运输问题的实质就是在一张无向图上，求解一个带点权值的最小生成树问题。

经过查阅资料可以了解，带点权值最小生成树问题是NP问题，所以不存在一个多项式算法解决，但由于当前实际情况中节点的数量不多（仅包含疫区和药品生产区），所以算法可以在有效时间内得出结果~Based on all assumptions mentioned, we can make a graph theory model. We abstract the whole medicine transport network as a undirected connected graph G = (V,E). the distance of two nodes is abstracted as edge weight while the amount of the medicine at every node is abstracted as node weight.Then we can optimize the medicine transportation problem. We obtain a minimum spanning tree with vertices weight. This is an NP problem, so there is no such polynomial algorithm answer. We chose 9 nodes all over the world includingLibera，Guinea，Sierra Leone as 3 node stand for affected areas.Egypt , South Africa, and Congo as storage of Africa.The U.S as the storage of America .Germany as the storage of Europe .Japan as the only provider and the storage of Asia.to show our transportation model.对于图的建立，首先需要求解各个节点的边权与生成点权。

我选择了查找各个节点的经纬度资料后自行编写函数来计算劣弧长。

To build our graph, we need to compute the edge weight and node weight of each node. We use the longitude and altitude of each node to calculate the distance with our own function.设A,B 两点的地理坐标（经纬度）分别为()11,x y ，()22,x y 。

以地心为原点O ，赤道平面为xOy 平面，以0度经线圈所在平面为xOz 平面建立空间直角坐标系，则AB 两点的直角坐标分别为Setting the longitudes and altitudes of node 1 and 2 are (), considering the earth ’s center as the origin, the earth equatorial plane as plane XOY, the 0° circle of longitude as plane XOZ, we build an space rectangular coordinate system. So the rectangular coordinates of node 1 and 2 are A(R*CosX1*CosX2，R*SinX1*CosY1，RSinY1)，B （R*CosX2*CosY2，R * SinX2*CosY2，R*SinY2）。

R 为地球半径，取6370KM 。

(R(=6370km) is earth radius).1111122222cos cos ,sin cos ,sin cos cos ,sin cos ,sin ()()A R x y R x y R yB R x y R x y R y其中R 为地球半径，取6370KM 。

AB 两点间的实际距离The actual distance between 1and2 isarccos OA OB d R OA OB ⎛⎫⋅ ⎪= ⎪⋅⎝⎭化简得（化简后公式）Simplify as()121212arccos cos -cos cos sin sin d R x x y y y y =+⎡⎤⎣⎦我们选择Kruskal 算法进行最小生成树的求解，因为图中节点数量相对较少，图的稀疏程度较高，用此算法可以获得更好的时间效率。

（同时编程难度更低）Kruskal 算法其时间复杂度为O （eloge ）。

算法过程：1.将图各边按照权值进行排序2.将图遍历一次，找出权值最小的边，（条件：此次找出的边不能和已加入最小生成树集合的边构成环），若符合条件，则加入最小生成树的集合中。

不符合条件则继续遍历图，寻找下一个最小权值的边。

3.递归重复步骤1，直到找出n-1条边为止（设图有n 个结点，则最小生成树的边数应为n-1条），算法结束。

得到的就是此图的最小生成树。

We chose Kruskal algorithm to obtain the result of the minimum spanning tree. With a small quality of nodes and density of the graph, the algorithm has an excellent time efficiency and low programming difficulty.我们还可以对带点权值最小生成树算法进行下列优化：在实际应用中，疫区国家的边权差值()x y F F δ-很小，所以可取常数,y x y x F F C Max F F ⎧⎫⎪⎪≥⎨⎬⎪⎪⎩⎭；在求解MST 前对于图G 进行预处理，步骤为对于所有边重新赋值给一个新权值，公式为,{}ij ij x x y y x y y x y x x x W W Max F D F D F D F D F D F D '=++++,新的算法流程保持不变~。

We can continue optimize the minimum spanning tree algorithm using the mean follows:Affected countries ‘edge weight δ（Fx - Fy ）is very small, so we can assume it as a constant C; Before solving MST we can pretreat the graph assigning new edge weight to all lines. The equation isThe new algorithm is same as previous algorithm.。