Network Optimization

1. Network Optimization Chapter 3 Shortest Path Problems

2. 3.1 Shortest paths from a single source • In a weighted digraph, a path of minimum weight from vertex v to vertex w is called a shortest path (SP) from v to w, and its length is called the shortest distance (SD) from v to w. • For undirected graph, we can define SP and SD between two vertices. • The shortest path problem can be treated as a transshipment problem.

3. 3.1 Shortest paths from a single source • (a) If we want to find SP and SD from v to w， then: • let v be the only source with a supply of 1 unit; • let w be the only sink with a demand of 1 unit; • let other vertices be intermediate vertices; • let the cost of sending one unit of the commodity from i to j be the weight of the arc (i , j); • we now use the network simplex method to solve this transshipment problem. A 0-1 solution x* will be obtained, and the arcs (i , j) with =1 form a shortest path from v to w.

4. 3.1 Shortest paths from a single source • (b) if we want to find shortest paths from a given vertex v to each of the other n-1 vertices in the digraph, then: • let v be the only source with a supply of n-1 units; • let every other vertex be a sink with a demand of 1 unit; • let the cost of sending one unit of commodity from i to j be the weight of the arc (i , j); • then the shortest path problem is transformed to a transshipment problem, and hence can be solved by the network simplex method.

5. 3.1 Shortest paths from a single source • We study other two algorithms: • Dijkstra’s algorithm to find a SP and the SD from a specified vertex to every other vertex; • Floyd and Warshall algorithm for all-pairs shortest path problem.

6. Main idea about the Dijkstra’s method • Suppose the 5 nearest vertices to v1 are v1,v3,v5,v7 and v9. • Then finding the sixth nearest vertex is easy. • Assume the sixth nearest vertex is v6 and the shortest path is (v1,…v?, v6). • Then v? must be one the 5 nearest vertices. Can you see why?

7. Another important idea!! • Suppose 1356 is the SP from 1 to 6. Then, for sure, 135 is the SP from 1 to 5. Can you see why? • As a result, to save the SP from 1 to 6, I just need to write down 5 and the SP from 1 to 5.

8. 3.1 Shortest paths from a single source • Dijkstra’s algorithm • Let the network G = (V, E), V = {1, 2, …, n}, and the weight of the arc (i , j) be a(i , j). • If there is no arc from i to j (i j), then a(i , j) is taken as a large positive number M. • We want to find the SD and SP from vertex 1 to all other vertices.

9. 3.1 Shortest paths from a single source • In the Dijkstra’s algorithm, each vertex i is assigned a label which is either permanent, or tentative. • The permanent label L(i) of i is the SD from 1 to i; • The tentative label L’(i) of i is an upper bound for the SD from 1 to i. • At each stage of the procedure, V is partitioned to two sets: P and T, whereP is the set of vertices with permanent labels, and T = V \ P is the set of vertices with tentative labels.

10. 3.1 Shortest paths from a single source • We also need to use an index V(i) to record the vertex immediately before i . This index may be updated after each iteration, and when we complete computation, it shows the vertex immediately before vertex i in the shortest path from vertex 1 to i. • Dijkstra’s algorithm • Step 0 (initial step) • Set L(1) = 0. • Set L’(j) = a(1, j) and V(j)=1 for j = 2, 3, …, n. • Set P = {1}, T = {2, 3, …, n}.

11. 3.1 Shortest paths from a single source • Step 1 (Designation of a permanent label) • Find kT such that Declare vertex k to be permanently labeled: • Set T = T - k, P = P + k, L(k) = L’(k). • If T = Ø (i.e. P = V), stop; the computation is completed.

12. 3.1 Shortest paths from a single source • Step 2 (Revision of tentative labels) • Set L’(j) = min {L’(j), L(k) + a(k, j)} for all jT, and if now L’(j) = L(k) + a(k, j), then update V(j) as V(j)=k. • Go to step 1.

13. Informal steps • L’(1)=0, V’(1)=1, P = {1}, T = {2,…} • For all v in T, L(v)=M, V(v)=1, • For any v is still in T and adjacent to v*, Update L’(v), V’(v) • Find v*, s.t. L’(v*)  L(v), v in T. • L(v*) = L’(v*) and V(v*) = V’(v*) • Move v* from T to P.

14. 3.1 Shortest paths from a single source • In each step 1 → step 2 → step 1 iteration, a vertex is moved from T to P. • So, we need to have n-1 iterations to complete computation, and for all j=2,3,…n, the indexes V(2), …, V(n) gives us n-1 arcs which together with all n vertices form a subgraph H of G(V, E).

15. 3.1 Shortest paths from a single source • If there exists a path from vertex 1 to any other vertex, then H must be connected. H is acyclic because if V(k)=i, arc (i, k) is in the shortest path from 1 to k, and i must enter P earlier than k does. • So, H is a spanning tree rooted at vertex1. And H is a shortest distance tree which includes the shortest paths from vertex 1 to other vertices.

16. 3.1 Shortest paths from a single source • Example 3.1 • Obtain the SD from 1 to the remaining vertices in the directed network shown below, using Dijkstra’s algorithm.

17. 3.1 Shortest paths from a single source Iteration 1 Step 1. • P = {1} • L(1)=0 T = {2, 3, 4, 5, 6, 7} L’(2) = 4, L’(3) = 6, L’(4) = 8, L’(5) = L’(6) = L’(7) = M. V(2)=V(3)=V(4)=V(5)=V(6)=V(7)=1. Vertex 2 is assigned a permanent label.

18. 3.1 Shortest paths from a single source Step 2. • P = {1, 2} • L(2) = 4 • Record arc (1,2) T = {3, 4, 5, 6, 7} L’(3) = min {6, L(2) + a(2, 3)}=5 L’(4) = min {8, L(2) + a(2, 4)}=8 L’(5) = min {M, L(2) + a(2, 5)}=11 L’(6) = min {M, L(2) + a(2, 6)}=M L’(7) = min {M, L(2) + a(2, 7)}=M V(3)=2, V(5)=2.

19. 3.1 Shortest paths from a single source Iteration 2 Step1. • P={1,2} • L(1) = 0 • L(2) = 4 T = {3, 4, 5, 6, 7} Min{L’(i) | i in T}= L’(3) Vertex 3 is assigned a permanent label.

20. 3.1 Shortest paths from a single source Step 2 • P = {1, 2, 3} • L(3) = 5 T = {4, 5, 6, 7} L’(4) = min {8, L(3) + a(3, 4)}=7 L’(5) = min {11, L(3) + a(3, 5)}=10 L’(6) = min {M, L(3) + a(3, 6)}=9 L’(7) = min {M, L(3) + a(3, 7)}=M V(4)=3, V(5)=3, V(6)=3.

21. 3.1 Shortest paths from a single source Iteration 3 Step 1. • P = {1, 2, 3} • L(1) = 0 • L(2) = 4 • L(3) = 5 T = {4, 5, 6, 7} Min {L’(i) | i in T} = L’(4) Vertex 4 is assigned a permanent label.

22. 3.1 Shortest paths from a single source • Step 2 • P = {1, 2, 3, 4} • L(4) = 7 T = {5, 6, 7} L’(5) = min {10, L(4) + a(4, 5)}=10 L’(6) = min {9, L(4) + a(4, 6)}=9 L’(7) = min {M, L(4) + a(4, 7)}=M

23. 3.1 Shortest paths from a single source Iteration 4 Step 1 • P = {1, 2, 3, 4} • L(1) = 0, • L(2) = 4, • L(3) = 5, • L(4) = 7 T = {5, 6, 7} Min {L’(i) | i in T} =L’(6) Vertex 6 is assigned a permanent label.

24. 3.1 Shortest paths from a single source Step 2. • P = {1, 2, 3, 4, 6} • L(6) = 9 T = {5, 7} L’(5) = min {10, L(6) + a(6, 5)}=10 L’(7) = min {M, L(6) + a(6, 7)}=17 V(7)=6.

25. 3.1 Shortest paths from a single source Iteration 5 Step 1 • P = {1, 2, 3, 4, 6} • L(1) = 0, • L(2) = 4, • L(3) = 5, • L(4) = 7, • L(6) = 9 T = {5,7} Min {L’(i) | i in T} = L’(5) Vertex 5 is assigned a permanent label.

26. 3.1 Shortest paths from a single source Step 2 • P = {1, 2, 3, 4,6, 5} • L(5) = 10 T = {7} L’(7) = min {17, L(5) + a(5, 7)}=16 V(7)=5

27. 3.1 Shortest paths from a single source Iteration 6 Step 1. • P = {1, 2, 3, 4, 6, 5} • L(1) = 0, L(2) = 4, • L(3) = 5, L(4) = 7, • L(6) = 9, L(5) = 10. T = {7} L’(7) = 16 Vertex 7 gets a permanent label.

28. 3.1 Shortest paths from a single source • Step 2. • P = {1, 2, 3, 4, 6, 5, 7} • L(7) = 16 T is empty

29. 3.1 Shortest paths from a single source • Thus L(1) = 0, L(2) = 4, L(3) = 5, L(4) = 7, L(6) = 9, L(5) = 10 and L(7) = 16, giving the SD from 1 to each vertex. • The indexes V(2)=1, V(3)=2, V(4)=3, V(5)=3, V(6)=3, V(7)=5 show that arcs (1, 2), (2, 3), (3, 4), (3, 5), (3, 6) and (5, 7) constitute a shortest distance tree in the given network as shown below, giving the SP from vertex 1 to every other vertex.

30. 3.1 Shortest paths from a single source • The shortest distance tree: Question: is this a minimum weight spanning tree? No, in MST, (6,5) replaces (3,6)

31. You answer should consist • Sequence of arcs. (according to the order the arcs are moved to P) • Tree (Draw it!!). • Total weight.

32. 3.1 Shortest paths from a single source Theorem 3.1 Dijkstra’s algorithm finds the SD from vertex 1 to every other vertex i (i = 2, …, n). Proof.We prove the theorem by induction on the cardinality of P. We will show that for each P generated in the algorithm, (1)for every i P,L(i) is the SD from 1 to i. (2)for every j T, L’(j) is the length of an SP from 1 to j under the restriction that every intermediate vertex is in P.

33. 3.1 Shortest paths from a single source • First, when |P| = 1, i.e. P = {1}, T = {2, 3, …, n}, the two conclusions hold obviously. • We now show that if conclusions (i) and (ii) are true when |P| = k-1, then they also hold if |P| = k. shore shore

34. 3.1 Shortest paths from a single source • Without loss of generality, assume P = {1, 2, …, k-1}, T = {k, …, n}. • By the assumption, (i)for i P, L(i) is the SD from 1 to i; (ii)for jT, L’(j) is the SD from 1 to j under the restriction that every intermediate vertex is in P.

35. 3.1 Shortest paths from a single source • Also assume that in the current iteration, vertex k moves to P, i.e. • So, L(k) = L’(k), and (△)

36. 3.1 Shortest paths from a single source • We need to show that (i)L(k) is the SD from 1 to k. • If it is not true, let d be the SD from 1 to k. So, d < L(k) = L’(k). As L’(k) is the SD from 1 to k provided that every intermediate vertex of the SP is in P, it means that along any SP from 1 to k, there must be at least one vertex in T. Let v be the first vertex in T along the SP from 1 to k.

37. 3.1 Shortest paths from a single source • Let the SD from 1 to v be d’. Then d’ = L’(v) and d’d < L’(k). • So, L’(v) < L’(k), which contradicts (*). • Therefore, L(k) must be the SD from 1 to k.

38. 3.1 Shortest paths from a single source (ii) We need to show that for each j = {k+1, …, n}, ’(j) is the SD from 1 to j under the restriction that all intermediate vertices are in . • Let be the SD from 1 to j when all intermediate vertices are in . The corresponding SP may have two possibilities:

39. 3.1 Shortest paths from a single source (a) The SP does not go through vertex k. In this case, is the SD from 1 to j under the restriction that every intermediate vertex is in P = {1, …, k-1}. So, = L’(j).

40. 3.1 Shortest paths from a single source (b)The SP includes vertex k. • In this case, k must be the last vertex before j in the SP. If not, the SP arrives at k, then reaches a qP, and at last comes to j. • It means the shortest path from 1 to qmust pass through k, which is impossiblebecause q enters P before k does, i.e. the SP from 1 to q need not to pass k. • Now since k is the last vertex in the SP before reaching j, the SD from 1 to j must be the SD from 1 to k plus a(k, j): L’(k) + a(k, j).

41. 3.1 Shortest paths from a single source • Combining the two cases (a) and (b), = min {L’(j), L’(k) +a(k, j)}. • So, by the formula (△),’(j) = , i.e. ’(j) is the SD from 1 to j under the restriction that every intermediate vertex of SP is in . So, the proof is completed. (△)

42. 3.2 All shortest path algorithm • Let G = (V, A) be a directed network. V = {1, 2, …, n}. • Let auvbe the weight of the arc (u, v). (in the method we allow some negative auv) • We want to calculate the SD from every vertex u to every other vertex v. • We will use The Floyd - Warshall Method

43. The main idea used • Suppose v5 is the second vertex in the shortest path from v1 to v9. • That is, the shortest path from v1 to v9 is (v1, v5, ???, v9). • Besides, suppose (v5, v6, v7, v8, v9) is the shortest path from v5 to v9. • Then, we can assure the shortest path from v1 to v9 is (v1, v5, v6, v7, v8, v9).

44. 3.2 All shortest path algorithm • Let A = [auv] be the n n weight matrix; P = [Puv] be the n n matrix with Puv = v, i.e. • The F&W method is an iterative method which needs to have n iterations. In the j-th iteration, we will have two n n matrices A(j) = [auv(j)] and P(j)=[Puv(j)] (j = 1, …, n). A(0)=A and P(0)=P.

45. 3.2 All shortest path algorithm • In A(j), auv (j) shall be the SD from u to v with intermediate vertices in the vertex set {1, 2, …, j}, and the corresponding element puv(j) in P(j) gives the vertex immediately after u in the path to attain the above SD auv(j).

46. 3.2 All shortest path algorithm • The algorithm is as follows. • Step 0 Let j = 1, A(0) = A and P(0) = P. • Step 1for all (u,v), If auv (j-1) < auj (j-1) + ajv (j-1),then auv (j) = auv (j-1) and puv(j) = puv (j-1),otherwise auv (j) = auj (j-1) + ajv (j-1) and puv (j) = puj (j-1).

47. 3.2 All shortest path algorithm • Step 2If in matrix A(j), one diagonal element is negative, stop, the problem has a negative cycle, and some SD from a vertex to another vertex are unbounded. • If j = n, go to step 3; otherwise let j ← j+1 and return to Step 1.

48. 3.2 All shortest path algorithm • Step 3 Each element auv(n) of the matrix A(n) gives the SD from vertex u to vertex v. • To find the SP from u to v, if puv (n) = j1, then the first arc of the SP is (u, j1). If j1v, we continue by checking pj1v(n). If pj1v(n)=j2, then the second arc is (j1, j2). Repeat the procedure until we reach vertex v. Then the SP is u → j1 → j2 → …. → v.

49. 3.2 All shortest path algorithm • We use an example to explain the algorithm. • Example 3.2 • Obtain the SD matrix and the SP matrix in the directed network as shown below.

50. 3.2 All shortest path algorithm • We begin with the following matrices: