Difficult to understand the solution provided in the video (travelling salesman problem)

  • #1
WMDhamnekar
MHB
377
28
It is difficult to understand the solution provided in the video to the travelling salesman problem having Hamiltonian circuits with added length constraint.
The travelling salesman problem has been solved in the below given video, but I didn't understand lower bound computed for the following paths.

I also didn't understand the why Path 8,9 10 and 11 have been called 'first tour', 'better tour', 'inferior tour' and 'optimal tour' respectively.

What are the meanings of l=24 for Path 8, l=19 for Path 9, l= 24 for Path 10 and l=16 for Path 11?

Path 4: a,e lb=19,

Path 5:a,b,c lb=16,

Path 6:a,b,d lb=16,

Path 7: a,b,e lb=19,

Path 8:a,b,c,d (e,a) l=24 first tour ,

Path 9: a,b,c,e(d,a) l= 19 (better tour),

Path 10: a,b,d,c (e,a) l=24 (inferior tour),

Path 11: a,b,d,e(c,a) l=16 optimal tour.

Watch the vide here →

Would any member explain me the travelling salesman problem solved in this video?

Can we solve this problem by another method where conditions of the travelling salesman problem is not satisfied?

The condition is the salesman has to visit each city in the tour only once and return to the start city.
 
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  • #2
WMDhamnekar said:
It is difficult to understand the solution provided in the video to the travelling salesman problem having Hamiltonian circuits with added length constraint.
The travelling salesman problem has been solved in the below given video, but I didn't understand lower bound computed for the following paths.
All of them or any in particular or all of the ones below? I'll do a few.
WMDhamnekar said:
I also didn't understand the why Path 8,9 10 and 11 have been called 'first tour', 'better tour', 'inferior tour' and 'optimal tour' respectively.
Traversing that branch of the tree from left to right and calculating the total lengths: The first path is 8. The second path is 9, which is better than path 8. The third path is 10, which is longer than path 9. The final path on that branch is 11, which is optimal for that branch.
WMDhamnekar said:
What are the meanings of l=24 for Path 8, l=19 for Path 9, l= 24 for Path 10 and l=16 for Path 11?
Those are the actual lengths if you add up those path steps.

Lower bounds:
Before any path steps are assumed, the initial lower bound (ILB) is where we get to use the two lowest for every node, a,b,c,d,e. That is ILB=((ac+ab)+(ba+bc)+(ca+ce)+(de+dc)+(ec+ed))/2 = ((1+3)+(3+6)+(1+2)+(3+4)+(2+3))/2 = 14,
WMDhamnekar said:
Path 4: a,e lb=19,
The path (ae) is given, so the ILB values of nodes a and e must be changed. Where possible, replace the larger ILB entries for nodes a and b with ae=8. That gives:
lb=((1+8)+(3+6)+(1+2)+(3+4)+(2+8))/2=19
WMDhamnekar said:
Path 5:a,b,c lb=16,
Now, ab and bc are given. ab was already in the ILB of nodes a and b, so that changes nothing. bc was already in the ILB for node b but not for node c. Change the larger entry of node c to bc=6.
((1+3)+(3+6)+(1+6)+(3+4)+(2+3))/2 = 16

Continue in that way and see if it makes sense for the others.
WMDhamnekar said:
Can we solve this problem by another method where conditions of the travelling salesman problem is not satisfied?
Yes. This is an example of the general subject of integer programming. There has been a large amount of work done on this subject. Entire books have been written on how to solve them.
 

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