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Shortest Path Networks

# Shortest Path Networks - 8 The Prototype Sequential...

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Unformatted text preview: 8 The Prototype Sequential Decision Process The networks studied in the unstarred sections of this chapter are particu- larly simple. They are ﬁnite and acyclic. A key property of such networks is introduced in terms of Figure 2—2. This network is ﬁnite and acyclic, but the labels have purposely been omitted from its nodes. These nodes can be labeled with the integers 1 through 4 in such a way that each are (i, j) has 1' less than j. Do so. (There is exactly one such labeling.) Labeling Figure 2-2 in this way should convince you that any acyclic network consisting of N nodes can have its nodes labeled with the integers 1 through N so that each are (i, j) has i less than j. For the cognoscenti, We now sketch a labeling procedure that accom- plishes this. Note that at least one node must have no arcs terminating at it (else there would be a cycle). Label this node with the integer I. Then delete it and all arcs emanating from it. Examine the remaining network. Note that it must have at least one node at'which no arcs terminate (for the same reason). : Label that node with the integer 2. Et cetera. FIGURE 2-2. SHORTEST AND LONGEST PATHS The prototype of all dynamic programming problems is to ﬁnd the shortest (or the longest) path through a ﬁnite acyclic network. In acyclic networks, shortest- path and longest—path problems are interchangeable; either can be converted to the other by multiplying the lengths of all the arcs by “I. A shortest-path prob- lem is now worked out in complete detail. Then a longest-path problem is analyzed, with some of the details left to you. Shortest-Route Problem John bicycles to work regularly. Being methodical, he has decided to calculate the fastest of several routes from his home to the ofﬁce. With some effort, he has collected the data summarized in Figure 2—3. Node 1 represents his home, node 9 represents his ofﬁce, and the remaining nodes represent road junctions. Directed arcs represent roads between junctions. The number adjacent to each arc is the travel time in minutes on the road it represents; it takes 15 minutes to travel from junction 4 to junction 7. The arrow is a some— what artiﬁcial device; it represents John’s prior conviction that when bicycling to work he might wish to traverse a road in the direction of the arrow, but Would never wish to do so in the opposing direction. Effectively, all the “streets” are Shortest and Longest Paths 9 FIG U H E 2-3. one-way; a version with two-way streets is described in the chapter’s starred section. Note that the nodes are labeled such that each are (i, j) has i < j. Note also-that the labeling is not unique (e. g., labels 2 and 3 can be interchanged). With a cursory examination of Figure 2-3, you can satisfyyourself that the fastest path from home to oﬂice is (l, 3, 4, 5, 7, 9) and that its travel time is 19 minutes. Having solved John’s problem by inspection, we now do so by dynamic programming. The intent of this is to provide insight into dynamic programming. Let f,r = the minimum travel time from node i to node 9 By deﬁnition, f,. is zero, and you have just observed that fI = 19. Let t‘U denote the travel time along are (t, j), so :47 = 15. Interpret ti} as the travel time of the path from node i to node 9 that ﬁrst traverses are (1', j) and then travels as quickly as possible from node j to node 9. As this is a path from node i to node 9, its travel time must be at least as large as )2. In other ' words, 12 g t” +13, ii 9 But the fastest path from node i to node 9 traverses some are (1', j) ﬁrst and then gets from node j to node 9 as quickly as possible. So some j satisﬁes the displayed inequality as an equality, and ' (2-1) _ f, = min [in +JG}, ia’: 9 The set of those j over which the. right-hand side of equation (2-1) is to be minimized has not been represented explicitly; minimization occurs over those j for which (i, j) is an arc. When i = 4, one minimizes the right-hand side over the values 5, 6, 7, and 8 of j. The case 1' = 9 is not covered by equation (2-1), but we have already observed that ' ' (L2) - f9 = 0 Notice that equation (2—1) speciﬁes 1‘} once f, is known for every j such that (i, j) is an arc. _The nodes in Figure 2~3 have been numbered so that every arc (i, j) has j greater than 1'. Hence, f; can be determined from (2-1) once f} is known for every j greater than 1'. Consequently, (2-1) allows computation of 10 The Prototype Sequential Decision Process f, in decreasing i, starting with i = 8 and ending with 1' = 1. If this is your ﬁrst exposure to dynamic programming, you are advised to ﬁnd pencil and paper, perform the somewhat tedious calculation just described, and compare your results with what follows. (Alternatively, you might do the calculation directly on Figure. 2—3.) fa=10+fg=1Q__ f1: 3+f9= . 7+fs . 7+10 ﬁ—min{15+f9—min{15+ OHIS f5: 4+f5 4+10 . 15+f1_ . 15+ 3_ f4—m1n 7+fa—mm 7+10—14 3+fs 3+15 f3=min{3+f4:min{3+l4=17 4+f,, 4+15 . 12+}; . 12-1-10 2‘ 2 2 0 f2 rnin{6+er mm{6+14 2 __ . 1+fz__ . 1+20ﬁ flam1n{2+f3—m1n{2+l7——l9 Figure 2-4 summarizes the information obtained from this calculation. Travel time f, is recorded above node 1', each are (i, j) attaining the minimum in (2-1) is preserved, and all other arcs are deleted. Note that Figure 2—4 pre- " scribes the fastest path from each node to node 9, not just from node 1 to node 9. It turned out to be easiest to soive John’s problem by embedding it in the more general problem of ﬁnding the shortest path from each node to node 9. Equation (2-1) is the prototype of the equations of dynamic programming. The argument used to justify (2-1) holds whether or not the network’s nodes FIGURE 2-4. Shortest and Longest Paths 11 are labeled so that each arc (1‘ i) has z' < j. It holds for any are lengths, including negative lengths. It even holds for cyclic networks, provided that no cycle has negative length. Although (2—1) holds more generally, the preceding computational method works only for those acyclic networks whose nodes are labeled so that each are (i, j) has i< j. This labeling often arises naturally, as when decision making evolves over time. More general methods of computation are described in this chapter’s starred section. Sequential decision prooesses usually represent decision problems involv- ing trade-offs between immediate costs and future costs. The myopic solution to the shortest-route problem selects for each node i the shortest directed arc (i, j) emanating from it. The myopic solution disregards the node at which an arc terminates, for which reason it may be far from optimal. In fact, the network in Figure 2~3 is designed so that the myopic solution to the shortest-route problem yields the longest path from each node to node 9. When an equation system has exactly one solution, it is said to characterize that solution. The numbers { f1, . . . , f9} satisfy (2-1) and (2-2). Moreover, we have computed these numbers from (2-1) and (2—2), which means that no other numbers satisfy these equations. In other words, (2-1) and (2-2) characterize the set { f1, . . . , f9} of fastest travel times to node 9. [In this, we mean that the functional values are unique, not that the maximizing indices in (2-1) are unique; ties occur when multiple paths have the same length] Shortest- Path Trees The network displayed in Figure 2-4 has one fewer arc than node, and it prescribes exactly one path from each node to node 9. This network is a tree of paths to node 9. The tree’s path from node i to node 9 has length ﬂ, which is the shortest of the lengths of all the paths from node i to node 9. Hence, the network depicted in Figure 2-4 is a tree of shortest paths, or a shortest-path tree. Rather than ﬁnding the shortest path from node 1 to node 9, we have found a tree of shortest paths to node 9 from all other nodes. More generally, a subset of the arcs in a network is called a tree of paths to node j if this are set contains exactly one path from each node i, with i e’: j, to node j. Similarly, a subset of arcs is called a tree of paths from node i if this are set contains exactly one path from node i to each node j, except j = i. No tree can contain a cycle, because then it would have multiple paths. One can show that every tree contains one fewer arc than the number of nodes in the network. When the methods of dynamic programming are applied to network optimization problems, one often gets a tree of shortest paths. Longest-Route Problem One might suspect that only extreme pessimists could be interested in the longest path through a directed acyclic network, but this will turn out to be false. Consider the problem of ﬁnding the longest path from node 1 of Figure DYNAMIC _ PROGRAMMING MODELS AND APPLICATIONS ERIC V. DENARDO School of Organization and Management Yale University El 7—9 (Shortest Route Problem) The network given in Figure 7—7 give diﬁ'erent routes for reaching city B from city A passing through a number c other cities. The lengths of the individual routes are shown on the arrow: It is required to determine the shortest route from A to B. Formulateth problem as a dynamic programming model. Explicitly deﬁne the stage: states, and return function; then ﬁnd the optimal solution. ...
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