Discrete Mathematics with Graph Theory (3rd Edition) 387

Discrete Mathematics with Graph Theory (3rd Edition) 387 -...

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Section 14.1 (b) The value of the flow is 5. (c) The capacity of the cut is Cab + Cde + Cdt = 2 + 5 + 4 = 11. (d) The flow can be increased by one unit along sadt and by two units along scdt. (e) The flow is not maximum, as noted in (d). 385 (f) (fab-fba)+(fcb-fbc)+(fde-fed)+(fdt-ftd) = (1-0)+(0-1)+(3-0)+(2-0) = 5. 5. (a) i. [BB] Send one unit through the path sbet. ii. [BB] The flow in (a) has a saturated arc, be. iii. [BB] Here is a maximum flow, of value 6. (Did you get it?) To see that the flow is maximum, con- sider the cutS = {s,a,b}, T = {e,d,e,j,t}. (b) i. Send one unit through seet. ii. Send three units through seet, saturating ee and et. iii. Here is a maximum flow, of value 5. (Did you find it?) a a 3,3 1, ° 1,1 2,0 1,1 c 2,2 To see that this is maximum, consider the cut
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Unformatted text preview: S = {s, a, b, e, d, e, J}, T = {t}. (c) i. Send one unit through selt. ii. Send four units through selt, saturating se and ef. iii. Here is a maximum flow, of value 9. (Did you get it?) s b 2,2 d c 4,4 f To see that this is maximum, consider the cut S = {s, a}, T = {b, e, d, e, I, t}. (d) i. Send one unit through the path set. ii. Send two units through set, saturating et. iii. Here is a maximum flow, of value 10. (Did you get it?) d e To see that the flow is maximum, consider the cut S = {s, a, b, e, d, e}, T = {t}. b 6. [BB] s 1,0 a t c d t f t t 7. [BB] Let {Sl, 7i} be any cut. By Theorem 14.1.4, val(F) ~ cap(Sl, 7i). But val(F) = cap(S, T). So cap( S, T) ~ cap( Sl, 7i). Since {Sl, 7i} was an arbitrary cut, the result is proven....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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