9 MT263 problems 1. For each of the following s, t-networks, use the given initial flow in the Ford-Fulkerson algorithm to determine a maximum s, t-flow and a minimum s, t-cut. (a) a (7,4) b (8,4) (6,0 (4,4) (2,0) (2,0) s > e f 3,0) 1 t (2,0 (2,0), (4,4) 7 (7,4) d (4,0) c Solution: We start with the augmenting path sdct and increase the flow along it by 2. a (7,4) b (6,0 (8,4) (4,4) (2,0 (2,0) s e f 3,0) t (2,2 (2,0). (4,4) (7,6) d (4,2) c Next we take the augmenting path sabt and increase the flow along it by 2. a (7,6) b (8,6) (6,0 (4,4) (2,2) (2,0) s e f 3,0) 1 t (2,2) (2,0), (4.4) 1 (7,6) d (4,2) c Next we take the augmenting path saft and increase the flow along it by 2. a (7,6) b (8,8) (6,2 (4,4) (2,2) (2,0) s > e f 3,2) > t (2,2) (2.0) (4,4) 7 (7,6) d (4,2) c Next we take the augmenting path sebaft and increase the flow along se, af and ft by 1; and decrease the flow along be and ab by 1.
s a (7,5) b (6,3 (8,8) (4,3 (2,2 (2,1) > e f (3,3) > t (2,2 (2,0), (4,4) (7,6) d (4,2) c Next we take the augmenting path sebafdct and increase the flow along the forward arcs by 1; and decrease the flow along the backward arcs by 1. a (7,4) b (8,8) (6,4) (4,2 (2,2) (2,2) s e f (3,3) > t (2,2) (2,1) (4,4) 1 (7,7) d (4,3) c This gives a maximum flow of value 12. The minimum cut is found by looking at the set of vertices reachable by unsaturated s-paths. Since there is only one unsaturated s path (namely the trivial path s), we see that X = {s} defined a minimum cut and it has value 12. (b) a (18,0) b (14,8) (8,8) (14,4) _ (10,10) e s (10,2 (8,8) (20,12) f t (18,18) ~(6,6) (16,6) 7(6,6) (16,12) d c Solution: Take the augmenting path sabft and increase the flow along the forward arcs by 4; and decrease the flow along the backward arcs by 4. a (18,4) b (14,12) (14,0) (8,8) e s (10,2 (8,8) f (10,10) t (20,16) (18,18) ~(6,6) (16,6) 7(6,6) (16,12) d c There are no more augmenting paths, so this is a maximum flow with value 32.
The minimum cut is found by looking at the set of vertices reachable by unsaturated s-paths. This gives X = {s, a, b, d, c, e}, so the arcs in the cut are A(X, V\X) = {(e, f), (a, f), (b, t), (c, t)} minimum cut and it has value 8+ 8 + 6 + 10 = 32. In the figure below the unsaturated s-paths are shown in red, the red vertices are those in X, and the blue in V\X. The edges in the cut are solid and black. a (18,4) b (14,12) (8,8) 14,0) (10,10) e