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.