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By: Q. Innostian, M.B.A., M.D.

Co-Director, University of California, Riverside School of Medicine

Further erectile dysfunction medication non prescription order kamagra soft 100 mg fast delivery, when f = k(P impotence 60 years old order kamagra soft 100 mg otc, P) erectile dysfunction treatment honey order generic kamagra soft from india, f is a maximum flow and (P erectile dysfunction treatment natural food generic kamagra soft 100mg with mastercard, P) is an a­z cut of minimum capacity. The first inequality is an equality-the flow from P into P (in the expanded network) equals f -if condition (i) holds; otherwise the flow into P is greater than f. The second inequality is an equality-the flow out of P equals k(P, P)-if condition (ii) holds; otherwise it is less. We first, discuss an intuitive but faulty technique that can sometimes be used as a shortcut in place of the correct algorithm. After the fault in the shortcut is exposed, the correct algorithm can be more easily understood and appreciated. All normal flows can be decomposed into a sum of unit-flow paths from a to z, for short, a­z unit flows (abnormal flows that cannot be so decomposed are discussed in Exercise 23). For example, in a telephone network, the flow from New York to Los Angeles can be decomposed into paths of individual telephone calls. Similarly, flow of oil in a pipeline network can be decomposed into the paths of each individual petroleum molecule. Formally, an a­z unit flow f L along a­z path L is defined as f L (e) = 1 if e is in L and = 0 if e is not in L. An additional unit flow can use only unsaturated edges, edges where the present flow does not equal the capacity. We define the slack s(e) of edge e in flow f by s(e) = k(e) - f (e) If s is the minimum slack among edges in the a­z unit flow f L, then we can put an additional flow along L of s f L = f L + f L + f L + · · · + f L (s times). If in addition, satisfies condition (a)- f (e) k(e), for all e-then f is a valid flow. Example 2: Building a Flow with Flow Paths Let us use the method just outlined to build a maximum a­z flow for the network in Figure 4. Note, as an upper bound, that the value of a flow cannot exceed 10, the sum of the capacities of edges going out of a. The minimum slack on L1 is 3 (at the start, the slack of each edge is just its capacity). The path L3 = a­b­e­z with minimum slack 2 can be used to get the augmenting flow 2 f L 3. The value of the final flow, 9, equals the capacity k(P, P) of this cut, and so by Corollary 2a the flow must be maximum. Let us again choose augmenting a­z flow paths across the top and bottom of the network, now having sizes 5 and 1, respectively. Since edges (a, b) and (c, d), are saturated by these flow paths, the only possible a­z path along unsaturated edges is L5 = a­c­e­z with minimum slack 1 [the minimum occurring in edge (e, z)]. The cut (P0, P 0), where P0 = {a, c, e}, is saturated, and so no more augmenting a­z unit flows exist. We now see that an arbitrary sequence of augmenting a­z unit flows need not inevitably yield a maximum flow. We are also faced with a flow f0 and a saturated a­z cut (P0, P 0) such that f 0 < k(P0, P 0). Thus the 5 units of flow along L use up 10 units of capacity in the cut, whence k(P0, P 0) is 5 units greater than f 0. The reason that the sequence of augmenting flow paths in this example did not lead to a maximum flow can be explained intuitively as follows. Then only 1 unit of flow passing through c can be routed on to e and then along edge (e, z). But much of the flow through c must go to e, since the capacity of (c, d) is only 1. In sum, the initial 5-unit flow along a­b­e­z was a "mistake" because some of the capacity in edge (e, z) should have been "reserved" for flow from c. If we understood where the mistakes were made, we could change some of the flow paths and try a new sequence of augmenting flow paths. Indeed, there may be certain networks in which it is impossible not to make such a mistake, no matter what sequence of flow paths is used. In terms of cuts, we may always end with a saturated a­z cut that one of our flow paths crosses twice. Fortunately, there is a procedure to correct "mistakes" and thereby further increase the flow.