Traveling salesman problem: Game Scheduling Problem Solution: Ant Colony Optimization
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The document describes using ant colony optimization to solve two traveling salesman problems. For the first problem, it finds the shortest route between 4 nodes on a graph. It evaluates the cost of all possible routes and determines that the optimal path is 1-2-4-3-1 with a cost of 80. For the second problem, it schedules games among 5 teams playing 3 types of games. It constructs a weight matrix based on how many games each team participates in. It evaluates all possible paths and determines the optimal schedules are either G1-G2-G3 or G3-G2-G1.
Traveling salesman problem: Game Scheduling Problem Solution: Ant Colony Optimization
- 1. AssignmentNo.1 Findthe Shortesttourfrom the followinggraphusingACO-TSP(antColonyOptimization-Travelling SalesmanProblem) wherethe graphrepresentsasetof nodesanddistance betweeneverypairof nodes.The shortestpossibleroute mustvisit everynode exactlyonce andreturntothe startingpoint. (Assume startnode is1) Solution: Steps: 1. Considernode1asthe startingand endingpoint. 2. Generate all (n-1)!Permutationsof nodes. 3. Calculate costof everypermutationandkeeptrackof minimum costpermutation. 4. Returnthe permutationwithminimumcost. Sl No. Permutations Corresponding cost 1 1-2-3-4-1 95(10+35+30+20) 2 1-2-4-3-1 80(10+25+30+15) 3 1-3-2-4-1 95(15+35+25+20) 4 1-3-4-2-1 80(15+30+25+10) 5 1-4-3-2-1 95(20+30+35+10) 6 1-4-2-3-1 95(20+25+35+15)
- 2. Weightage Matrix: 1 2 3 4 1 0 10 15 20 2 10 0 35 25 3 15 35 0 30 4 20 25 30 0 Let usassume,start node 1, considerasS Iteration No. Cost function Evaluation Cost Path 1 C(4,S) D(4,1) 20 1-4 C(3,S) D(3,1) 15 1-3 C(2,S) D(2,1) 10 1-2 2 C(2,{3}) D(2,3)+C(3,S) 35+15=50 1-3-2 C(2, {4}) D(2,4)+C(4,S) 25+20=45 1-4-2 C(3,{2}) D(3,2)+C(2,S) 35+10=45 1-2-3 C(3,{4}) D(3,4)+C(4,S) 30+20=50 1-4-3 C(4,{2}) D(4,2)+C(2,S) 25+10=35 1-2-4 C(4,{3}) D(4,3)+C(3,S) 30+15=45 1-3-4 3 C(2,{3,4}) MIN(D(2,3) + C(3,{4}), D(2,4)+C(4,{3})) MIN(50+35,45+25) =80 1-3-4-2 C(3,{2,4}) MIN(D(3,2)+C(2,{4}), D(3,4)+C(4,{2})) MIN(45+35,35+30) =65 1-2-4-3 C(4,{2,3}) MIN(D(4,2)+C(2,{3}), D(4,3)+C(3,{2})) MIN(50+25,45+30) =75 1-2-3-4 4 C(1,{2,3,4}) MIN(D(1,2)+C(2,{3,4}) , D(1,3)+C(3,{2,4}), D(1,4)+ C(4,{2,3})) MIN(90,80,95) =80 1-2-4-3-1 SO,THE OPTIMAL PATH 1-2-4-3-1 FOR TOUR HAS LENGTH 80. AssignmentNo.2 There are 5 teams:T1, T2, T3, T4 and T5 ina couple of serieswhere 3typesof games:G1, G2 and G3. NowT1 and T3 bothplaysall three typesof games,T2 playsonlyG1 and G3, T4 plays onlyG2 and G3 andT5 onlyplaysG1 and G3. NowusingAntColonyOptimizationfindoutthe schedulingof gamesamongG1,G2 and G3. Solution: Here create a matrix Cij where ithrepresentsteamandjthrepresentsgame.Now consider1 for playinggame and0 for non-playinggame.
- 3. G1 G2 G3 T1 1 1 1 T2 1 0 1 T3 1 1 1 T4 0 1 1 T5 1 0 1 Now,the weightage Dij representsthe numberof competitorswherethe competitor participatesinboththe gamesI and j. Thisweightage functionformsamatrix whichisknownasweightage matrix. D3x3 G1 G2 G3 G1 0 2 4 G2 2 0 3 G3 4 3 0 SL NO PERMUTATION PATH COST 1 1-2-3-1 9 2 1-3-2-1 9 3 2-1-3-2 9 4 2-3-1-2 9 5 3-1-2-3 9 1 2 3 2 4 3
- 4. 6 3-2-1-3 9 Let us assume start node 1 (S) IterationNo. Cost function Evaluation Cost Path 1 C(2,S) D(2,1) 2 1-2 C(3,S) D(3,1) 4 1-3 2 C(3,{2}) D(3,2)+C(2,S) 3+2=5 1-2-3 C(2,{3}) D(2,3)+C(3,S) 3+4=7 1-3-2 MIN(C(3,{2}),C(2,{3}))=MIN(5,7) =5 SO,PATH 1-2-3 Let us assume start node 2 (S) IterationNo. Cost function Evaluation Cost Path 1 C(1,S) D(1,2) 2 2-1 C(3,S) D(3,2) 3 2-3 2 C(3,{1}) D(3,1)+C(1,S) 4+2=6 2-1-3 C(1,{3}) D(1,3)+C(3,S) 4+3=7 2-3-1 MIN(C(3,{1}),C(1,{3}))=MIN(6,7) =6 SO,PATH 2-1-3 Let us assume start node 3 (S) IterationNo. Cost function Evaluation Cost Path 1 C(2,S) D(2,3) 3 3-2 C(1,S) D(1,3) 4 3-1 2 C(1,{2}) D(1,2)+C(2,S) 2+3=5 3-2-1 C(2,{1}) D(2,1)+C(1,S) 2+4=6 3-1-2 MIN(C(1,{2}),C(2,{1}))=MIN(5,6) =5 SO,PATH 3-2-1 So overall minimizationeitherpathschedule 1-2-3or 3-2-1 So,G1-G2-G3 OR G3-G2-G1.