Discrete Mathematics with Graph Theory (3rd Edition) 391

Discrete Mathematics with Graph Theory (3rd Edition) 391 -...

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Section 14.3 389 8 (b) We multiply the capacities by 6, the greatest common divisor of the denominators, and find a maximum flow in the new network, as shown on the left. Dividing by 6, we find a maximum flow in the given network, of value 2, as shown on the right. (Only the flow in each arc is shown.) t 8 t Exercises 14.3 c 20 10 5 l. [BBJ 10 10 15 20 10 20 25,5 10,5 A 10,5 5 5 d 8 20 10 t 20,15 25,15 b B e 15,15 10 10 10 10 10 a 5 20 15 The demand cannot be met; the best that can be done is as shown. Warehouse
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Unformatted text preview: b sends out 20 Klein bottles and the others send out none. Five go to retail outlet d and 15 to retail outlet e. 2. If the arrow on AB is reversed, five more Klein bottles can be sent. They could go from warehouse a to retail outlet e as shown, or they could go from c to e. This is clearly best possible now. A cut right down the middle of the network has capacity 25. A :!-L...
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