By Richard Bellman, Kenneth L. Cooke
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Extra resources for Algorithms, Graphs, and Computers, Vol. 62
For the sake of a simple illustration let us suppose that there are seven jobs or operations in all, designated J 1 , J 2 , . , J 7 , and consider the graph in Fig. 26. The seven numbered circles represent the basic operations. A directed segment and the number beside it indicate that the operation at the end of the arrow cannot begin until this number of weeks after the commencement of the operation at the beginning of the arrow. Thus, J2 cannot begin until 4 weeks after J l begins, 53 cannot begin until 7 weeks after J1 begins and 2 weeks after J2 begins, and so on.
N , where N is the terminal vertex. Clearly for each i = 1, 2, We leave it to the reader to derive the basic equations fi = max(tij + f i ) j = 1, fi (2) = 0. , N , (3) where the maximum is over all vertices i for which there is a directed edge from i to j . For Fig. 26, these equations are as follows: The values of all the unknowns can be found successively from these equations, since in each equation only previously computed f ’ s are needed. Thus we get f2 = 4, h = 7, $4 = 11, fs = 15, f6 = 21, f7 = 23, and it is also easy to deduce the critical path itself.
To do this, we represent each of the four sections 1, 2, 3 , 4 of the city by a vertex, and each of the bridges by an edge connecting two vertices. The Y 2 Fig. 20 3 Fig. 21 result is shown in Fig. 21. The problem can then be stated in the following language. Is it possible to find a path in the graph which traverses each edge exactly once and returns to its starting point? As pointed out by Euler, it is quite easy to see that there is no such path in Fig. 21. The reason is that it would have to enter each vertex as many times at it leaves i t ; this is true even for the starting vertex.
Algorithms, Graphs, and Computers, Vol. 62 by Richard Bellman, Kenneth L. Cooke