2.3.4 Optimal paths in directed acyclic graphs
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1 .3.4 Optimal paths in directed acyclic graphs Definition: A directed graph G = (N, A) is acyclic if it contains no circuits. A directed acyclic graph is referred to as DAG. circuit Problem Given a directed acyclic G = (N, A) with a cost c ij for each (i, j) A, and nodes s and t, determine a shortest (longest) path from s to t. E. Amaldi Foundations of Operations Research Politecnico di Milano
2 Property The nodes of any directed acyclic graph G can be ordered topologically, that is, indexed so that for each arc (i, j) A we have i < j. Topological order Such a topological order can be exploited in a very efficient Dynamic Programming algorithm to find shortest (or longest) paths in directed acyclic graphs. E. Amaldi Foundations of Operations Research Politecnico di Milano
3 Topological ordering method G = (N, A) is represented via the lists of predecessors - (v) and successors + (v) for each node v - (v ) + (v). Assign the smallest positive integer not yet assigned to a node v N with - (v) =. such a node v exists because G does not contain circuits v. Delete the node v with all its incident arcs. 3. Go back to. until there are nodes in the current subgraph. E. Amaldi Foundations of Operations Research Politecnico di Milano 3
4 Example G E. Amaldi Foundations of Operations Research Politecnico di Milano 4
5 etc... Complexity: O(m) where m= A because each node and each arc is considered at most once. E. Amaldi Foundations of Operations Research Politecnico di Milano 5
6 Example G s t 3 Given G and a topological order of the nodes, find a shortest path from node to node? E. Amaldi Foundations of Operations Research Politecnico di Milano
7 Dynamic Programming for shortest paths in DAGs Any shortest path t : t with at least two arcs can be subdivided into two parts i and (i,t): Definition: For each node i =,..., t i t where i : i is a shortest subpath from s to i. i L(i) = cost of a shortest path from to i. c it Clearly L(t) = min { L() i c it } it, ( t ) where the minimum is taken over all possible predecessors i of t in a path from to t. E. Amaldi Foundations of Operations Research Politecnico di Milano
8 If G is directed and acyclic and the nodes are topologically ordered, the only possible predecessors of t in a shortest path t from to t are those with index i < t and hence: L(t) = min{ L( i) c } i t it Observation: In a graph containing circuits, any node different from t can be a predecessor of t in t! E. Amaldi Foundations of Operations Research Politecnico di Milano
9 For DAGs whose nodes are topologically ordered L(t-),... and L() satisfy the same type of recursive relations: min{ Li () } L(t-) = i( t )... i t c min{ L } L() = () i c = L() + c L() = 0 i which can be solved in the reverse order: L() = 0 L() = L() + c i... min{ L c } L(t) = () i i t it E. Amaldi Foundations of Operations Research Politecnico di Milano
10 Example L() = 0 L() = L() + c = = 4 L(3) = L() + c 3 = 0 + = pred()= pred()= pred(3)= L(4) = min{ L( i) c } = min{0 + 5, 4 + 3, + } = 3 i,,3 i 4 pred(4)=3 G s [.] indicates the predecessor of node i in a shortest path from to i E. Amaldi Foundations of Operations Research Politecnico di Milano 0 t
11 L(5) = min{ Li () c } = min { 0 +, 4 + } = 5 i, min{ Li ( ) c } L() = = min { + } = 3 i 3 i5 i L() = L(5) + c 5 = 5 + = pred(5)= pred()=3 pred()=5 L() = min { Li ( ) c } = min { 3+3, 5+, 3+, + } = 4 G i 4,5,, s i pred()= pred(5)= 3 5 shortest path from to of total cost 4 pred()=5 pred()=5 E. Amaldi Foundations of Operations Research Politecnico di Milano t
12 Complexity DP for shortest paths in DAGs BEGIN Sort the nodes of G topologically; L[] := 0; FOR j:= TO n DO /* in topological order */ L[j] := min {L[i] + c ij : (i,j) - (j), i < j}; pred[j] := v with (v,j) = argmin {L[i] + c ij : (i,j) - (j), i < j}; END-FOR END Topological ordering of the nodes: O(m) with m= A Each node and arc are considered exactly once: O(n+m) Overall complexity: O(m) E. Amaldi Foundations of Operations Research Politecnico di Milano
13 Adaptation of DP for longest paths in DAGs Straightforward adaptation of the Dynamic Programming method to find longest paths in directed acyclic graphs: L(t) = max{ Li ( ) c}... i t it E. Amaldi Foundations of Operations Research Politecnico di Milano 3
14 Proposition: The Dynamic Programming algorithm for finding shortest (longest) paths in DAGs is exact. This is due to: Optimality principle For any shortest (longest) path j : j there exists i < j such that we have the decomposition: i j i where the subpath i : i is of minimum (maximum) length from s to i. c ij E. Amaldi Foundations of Operations Research Politecnico di Milano 4
15 Dynamic Programming (DP) Proposed in 53 by Richard Bellman (0-4) General technique in which an optimal solution, composed of a sequence of elementary decisions, is determined by solving a set of recursive equations. DP is applicable to any sequential decision problem for which the optimality property is satisfied. A wide range of applications: optimal control (rocket launch), equipment maintenance and replacement, selection of inspection points along a production line, E. Amaldi Foundations of Operations Research Politecnico di Milano 5
16 Exercise A company buys a new machine for K. Annual maintenance costs for the next five years: Age (years) Maintenance (K ) Recover (K ) To avoid high maintenance costs of an older machine, the machine can be replaced at the beginning of the nd,3 rd,4 th or 5 th year with a new machine. E. Amaldi Foundations of Operations Research Politecnico di Milano
17 To simplify the model, suppose that the cost of a new machine is constant ( K ). Objective: minimize the net total cost ( price + maintenance recover ) over the 5 year horizon. Age 0 Main. Reco. - If we buy a machine at beginning of st year and sell it at the beginning of nd year, the net cost is: + - = If we buy it at the beginning of st year and sell it at the beginning of 3rd year, the net cost is: = etc... E. Amaldi Foundations of Operations Research Politecnico di Milano
18 Show how the problem of determining an optimal maintenance - replacement policy of minimum total cost (for the next 5 years) can be solved via Dynamic Programming. Hint: Reduce it to a minimum cost path problem in an appropriate acyclic graph. Find all the optimal maintenance-replacement policies. E. Amaldi Foundations of Operations Research Politecnico di Milano
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