© 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+ =