The travelling salesman problem 5D › r00 › r... · 2020-04-07 · The travelling salesman...
Transcript of The travelling salesman problem 5D › r00 › r... · 2020-04-07 · The travelling salesman...
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 1
The travelling salesman problem 5D
1 a b
c The better upper bound is 53 since this is lower. 2 a b c The better upper bound is 190 because it is lower. 3 The application to a single vertex has order , so application to entire network has order
Hence the estimate for running time is
4 a b
c
d The better upper bound is 990 because it is lower.
D7B12C8E14A19D = 60
E8C11A13B7D14E = 53
orE8C11D7B13A14E = 53
10 15 40 55 30 70Z X Y V W S Z 220=
X10Z15V40Y45W30S55X = 195
V15Z10X15Y45W30S75V = 190
2n 2 3´ =n n n3200.27 1.25s
12æ ö´ =ç ÷è ø
R150S210T120U180V300W240R = 1200 minutes
U120S150R120V150T180W270U = 990
andU120T150V120R150S240W270U = 1050
V120R150S120U120T180W300V = 990
andV120R150U120S210T180W300V = 1080andV120R150U120T180W240S210V = 1020
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 2
5 a Since , x is the smallest entry in column A, and the nearest neighbour route from A is ADEGCBFA, of total weight 639 + x.
Since , x is the smallest remaining entry in column G after rows B, E, G and D have been
deleted, so the nearest neighbour route from B is BEGDAFCB, of total weight 648 + x. So b 705 miles is an upper bound. In the previous part we have found upper bounds and The second one is better.
126x <
126x <
1419 639 648 2 132 66x x x x= + + + Þ = Þ =
648 714x+ =639 705x+ =