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# UNC-Chapel Hill GEOG 192 - Lecture 14 Network Analysis (5)

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Lecture 14 Network Analysis (5)14-2 TSP Example14-2 TSP Example (Cont.)Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 1614-3 TSP Applications14-4 Shipment ProblemSlide 19Slide 2014-4 Shipment Problem (Cont.)Slide 22Slide 23Slide 24Slide 2519/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC11Lecture 14 Network Analysis (5)14-1 Intro to Traveling Salesman ProblemTraveling Salesman Problem (TSP) is an important and practical application of network analysis.-To find the optimal route (minimize total impedance cost) for trips that involve multiple stops.-Requires a salesman to travel from an origin node to several sites of known location.-Data include: origin, multiple destinations, cost factor.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC2214-2 TSP Example14-2 TSP ExampleA publishing company delivers magazines to six book stores located at nodes A through F in the below figure. These locations are connected by a freeway network with distances labeled.553010154020101515ABCDFE19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC3314-2 TSP Example (Cont.)14-2 TSP Example (Cont.)First step is to construct a cost matrix among the locations.Each iteration requires four steps:(1)a row minimum operation(2)a column minimum operation(3)penalty assignment, and(4)path identification19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC4414-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C D E FMinA - 10 30 - 55 - 10B 10 - 15 20 - 40 10C 30 15 - 15 - - 15D - 20 15 - - 10 10E 55 - - - - 15 15F - 40 - 10 15 - 10Iteration 1 – Cost matrix 119/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC5514-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C D E FA - 0 20 - 45 -B 0 - 5 10 - 30C 15 0 - 0 - -D - 10 5 - - 0E 40 - - - - 0F - 30 - 0 5 -min 0 0 5 0 5 0Iteration 1 – Cost matrix 219/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC6614-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C D E FA - 0 15 - 40 -B 0 - 0 10 - 30C 15 0 - 0 - -D - 10 0 - - 0E 40 - - - - 0F - 30 - 0 0 -Iteration 1 – Cost matrix 319/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC7714-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C D E FA - 0(15)15 - 40 -B 0(15)- 0(0)10 - 30C 15 0(0)- 0(0)- -D - 10 0(0)- - 0(0)E 40 - - - - 0(40)F - 30 - 0(0)0(40)-Penalty matrix of iteration 1For all 0-value cell, penalty value equals to the sum of the minimum row value and the minimum column value. The maximum penalty value - link EF is selected in this iteration. Each iteration selects a link to be included.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC8814-2 TSP Example (Cont.)14-2 TSP Example (Cont.)At the end of each iteration, the cost matrix must be revised by:(1) Eliminating the row and column of the location pair corresponding to the selected link, and (2) Revising to infinity the cost of the link opposite the currently selected link in order to rule out the possibility of a return trip on the same link.A B C D E FA - 0(15)15 - 40 -B 0(15)- 0(0)10 - 30C 15 0(0)- 0(0)- -D - 10 0(0)- - 0(0)E 40 - - - - 0(40)F - 30 - 0(0)0(40)-A B C D EA - 0 15 - 40B 0 - 0 10 -C 15 0 - 0 -D - 10 0 - -F - 30 - 0 -Cost Matrix 419/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC9914-2 TSP Example (Cont.)14-2 TSP Example (Cont.)Penalty matrix of iteration 2A B C D EA - 0(0)15 - 0(-)B 0(15)- 0(0)10 -C 15 0(0)- 0(0)-D - 10 0(10)- -F - 30 - 0(30)-This iteration selects the A-E link according to the penalty matrix. Adding this link in a route sequence of A-E-F. (From iteration 1, we get E-F.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC101014-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C DB 0 - 0 10C 15 0 - 0D - 10 0 -F - 30 - 0Cost matrix 5, a result from penalty matrix of iteration 2.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC111114-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B C DB 0(15)- 0(0)10C 15 0(10)- 0(0)D - 10 0(10)-F - 30 - 0(30)Penalty assignment of Iteration 3.In iteration 3, the candidate of maximum penalty is the F-D link. This link is therefore included in the route: AEF + FD=> AEFD19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC121214-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B CB 0 - 0C 15 0 -D - 10 0Cost matrix 6, a result from penalty matrix of iteration 3. F row and D column have been eliminated from Penalty assignment matrix of iteration 3. (FD link has been selected in the previous step.)19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC131314-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A B CB 0(15)- 0(0)C 15 0(25)-D - 10 0(10)Penalty assignment matrix for iteration 4.In iteration 4, the candidate of maximum penalty is the C-B link. This link is therefore included in the route: AEFD + CB.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC141414-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A CB 0 0D - 0Cost matrix 7, a result from penalty matrix of iteration 4. C row and B column have been eliminated from Penalty assignment matrix of iteration 4. (Note: CB link has been selected in the previous step.)19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC151514-2 TSP Example (Cont.)14-2 TSP Example (Cont.)A CB 0(-)0(0)D - 0(-)Penalty assignment matrix for iteration 5.In iteration 5, the candidates of maximum penalty are BA link and DC link. So we have two solutions: (1) AEFD + CB + BA => AEFD + CBA => CBAEFD, or(2) AEFD + CB + DC => AEFDC + CB => AEFDCB19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC161614-2 TSP Example (Cont.)14-2 TSP Example (Cont.)553010154020101515ABCDFECost(CBAEFD) = 15+10+55+15+10 = 105Cost(AEFDCB) = 55+15+10+15+15 = 110Both solutions are identical to each other. (Why?)19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC171714-3 TSP Applications14-3 TSP Applications(1) Constructions of delivery networks. Suppose you have multiple origins and receiving locations. (TSP) – page 244(2) Visit a predefined sequence of stops and then return to origin. (shortest path) – page245(3) Combined applications. Predefined stops + receiving locations – Shortest path and TSP.19/1/1319/1/13Jun Liang, Geography @ UNCJun Liang, Geography @ UNC181814-4 Shipment Problem14-4 Shipment ProblemShipment problems deal with optimizing the transportation of goods or people from multiple origins to multiple

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