Lower bound algorithm

1 Start from A. Delete vertex A and all edges meeting at A. A B C D 4 5 E Find the length of the minimum spanning tree for BCDE using Prim’s algorithm. 3 Add up the MST and the two shortest edges attached to A. Lower bound = MST + two shortest edges from A = (1+2+3) + (3+4) = 6+7 = 13 Notice that the lower bound does not represent a tour. The algorithm can be started from any vertex.

Lower bound algorithm Why there is no tour shorter than a lower bound? Let’s consider a network with 5 vertices. Unless AB and AE are the shortest edges from A, and BCDE is the MST, the tour ABCDE is longer than the lower bound, obtained by deleting vertex A. A A B B C C D D E E Any real tour (e.g. ABCDEA) can be split into two bits: 2 edges from vertex A, say, and the edges, connecting the rest of the vertices.

A salesman whose office is in town B wants to visit four other towns A,C,D and E and get back to his office in B as quick as possible. B A E D C ABCDE A-4583 B4-621 C56-43 D824-7 E3137- The table shows the times, in hours, it takes to travel between each pair of towns. Let’s find a lower bound for the optimal tour. Lower Bound Algorithm on table

ABCDE A-4583 B4-621 C56-43 D824-7 E3137- Lower bound Algorithm on table Find the lower bound by deleting vertex A. Delete 2143 MST+(two shortest edges from A) = 6+7 = 13 Note two shortest edges from A and then delete row A and column A. B A E 4 3 Apply Prim’s algorithm on the rest of the table. B E 1 D 2 3 C Lower Bound =13 The length, T, of the optimal tour is either equal to or greater than the lower bound,i.e. 13 ≤ T BE-1 BD-2 EC-3

A salesman whose office is in town B wants to visit four other towns A,C,D and E and get back to his office in B as quick as possible. B A E D C ABCDE A-4583 B4-621 C56-43 D824-7 E3137- The shortest possible time, T, for a tour lies between 13 and 15 hours. Travelling Salesperson Problem summary Lower Bound =13 Upper Bound=15 13 ≤ T ≤ 15 Suggested tour BDCAEB will take 15 hours