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The value of the a­z flow f (e) symptoms uti in women discount betahistine american express, denoted f medicine ball core exercises betahistine 16mg on-line, equals the sum of the flow in edges coming out of a treatment for depression discount 16 mg betahistine free shipping, or equivalently by Theorem 1 medicine qid cheap betahistine 16 mg amex, the flow into z. One upper bound is the sum of the capacities of the edges leaving a, since f = eOut(a) f (e) eOut(a) k(e) Similarly, the sum of the capacities of the edges entering z is an upper bound for f. Intuitively, f is bounded by the sum of the capacities of any set of edges that cut all flow from a to z-that is, the edges in an a­z cut. We define the capacity k(P, P) of the cut (P, P) to be k(P, P) = e(P,P) k(e) the capacity of the a­z cut (P, P), where P = {a, b, c}, in Figure 4. Theo r e m 2 For any a­z flow f and any a­z cut (P, P) in a network N, f k(P, P). Proof Informally, f should equal the total flow from P to P [which is bounded by k(P, P)], but we cannot prove an equality of this type using condition (b) without first modifying N. We cannot use (b) in N because condition (b) requires that P not contain the source a. Expand the network N by adding a new source vertex a with an edge e = (a, a) of immense capacity. Assign a flow value of f to e, yielding a valid a ­z flow in the expanded network (see Figure 4. Observe that new source a is playing a role similar to the source a we added to the multiple-source network in Example 1. It says that the flow into P, which is at least f (other flow could come into P along edges from P), equals the flow out of P. Thus f e(P,P) f (e) = by(b) e(P,P) f (e) e(P,P) k(e) = k(P, P) Since Theorem 2 says that the value of an a-z flow can never exceed the capacity of any a-z cut, it follows that if the value of some a-z flow f equals the capacity of some a-z cut, then f has to be a flow of the maximum possible value. We shall show that for any flow network, we can always construct an a-z flow whose value equals the capacity of some a-z cut. To see what properties a flow must have to make the inequality in Theorem 2 an equality, we need to look carefully at the two inequalities in expression (*) that was used to prove Theorem 2. Coroll ary 2 a For any a­z flow f and any a­z cut (P, P) in a network N, f = k(P, P) if and only if (i) For each edge e (P, P), f (e) = 0. Further, when f = k(P, P), f is a maximum flow and (P, 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.

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If a special fairing is required to get the wires treatment yeast purchase 16mg betahistine with amex, for example treatment deep vein thrombosis 16mg betahistine, into the model or if they are taped to the mounting strut symptoms 5 dpo purchase betahistine 16mg amex, the tares for these items must be evaluated medicine you can take while pregnant order betahistine 16 mg with amex. The easiest method is to make a set of pitch and yaw runs with the same model configuration with and without the wires. However, if the model struts enter the model fuselage, the trunnion, which usually is on the balance moment center and the tunnel centerline, is not at the centerline of the model. When the model is inverted, the exposed portion of the image strut and the distance from its fairing to the model must be exactly the same as the upright model. This often requires the length of the image fairing exposed to the tunnel flow to be either longer or shorter than the standard, or lower, fairing. Thus, it generally is a good policy to make the image fairing longer than the normal fairing and let it extend through the tunnel ceiling. In the three-run method of taking tare and interference there is a possibility of air flowing in and out of the gap between the mounting strut and the fairing if the standard struts do not have a seal at their tops. To get the correct tare and interference, the image system must duplicate exactly the geometry in this region. For a support system that uses two supports on the wing and a tail support, an alternate method is sometimes used. The reason for this alternate treatment is that the length of the tail support varies as angle of attack is changed. This factor so complicates the dummy arrangements that a system is usually employed that does not require a complete dummy tail support. The procedure is as follows: Consider the second method of evaluating the tare and interference. When the image system is brought down to the inverted model, a short support is added to the then upper surface of the model where the tail support would attach in a normal run. For angles of attack other than that corresponding to minimum length of exposed tail support, the drag of the extra exposed tail support length must be evaluated and subtracted. A rear-support windshield that moves with the rear support to keep a constant amount of strut in the airstream could be employed as long as the added interference of the moving shield is evaluated by a moving-shield dummy setup. The evaluation of the tare, interference, and alignment of a wing-alone test follows the procedure outlined above, except that further complication is introduced by the presence of a sting that must be added to the wing to connect it to the rear strut of the support system. The tare and interference caused by the sting may be found by adding a second sting during the image tests. This complication is needed to account for the interference of the strut on the sting as follows: When the wing is held at a high angle of attack, there will be an obtuse angle below the sting. When the wing is inverted and held at a high angle relative to the wind, there will be an acute angle below the sting, for the rear strut will then be extended to its full length. To eliminate this difference between the normal and inverted tests, the support strut is extended above the sting attachment point, so that the sum of the angles between the sting and the support is always 180". Note that although the angles between the sting and the rear support vary with the angle of attack, the image sting is always at right angles to its short rear-support strut. Furthermore, the image rear-support strut does not remain vertical but changes its angle with the wing. The error incurred by failing to have the sting image system simulate the exact interference and rear-strut angle is believed to be negligible. Tare and interference for the tail strut alone may be evaluated, as shown in Figure 7. The setup for determining the tare and interference for a fork support is shown in Figure 7. Here the model is supported externally and a small clearance is left where the struts come into the wing. This measures the drag of the fork plus the effect of the wing on the fork but not the effect of the fork on the wing, which experience has shown to be small. Hence, the tares are used as approximate tares where the desire to save tunnel time precludes taking the usual tare and interference runs. The wing in this particular test was small, and the corrections for tare, interference, and alignment are correspondingly large, but the variation of the corrections is typical. A large amount of interference may arise from air that bleeds through the windshields that surround the support struts to protect them from the windstream. These struts frequently attach at points of low pressure on the model, and if the shield is brought close to the model, a considerable flow may be induced that will run along the model. This flow may stall the entire underside of the model and produce results that are not only wrong but also unsteady and difficult to evaluate.

The assumption of integer capacities and flows is not restrictive-that is medicine 44291 buy betahistine 16 mg visa, the units we count could be thousandths of an ounce rather than barrels medications of the same type are known as cheap betahistine 16mg overnight delivery. The requirement that there be a single supply vertex treatment vitiligo discount betahistine 16 mg fast delivery, the source a medicine xarelto buy generic betahistine 16 mg, and a single demand vertex, the sink z, also turns out not to be a restriction. The reason is that we can recast network problems with multiple sources and multiple sinks into a single-source, single-sink network, as illustrated in the following example. Example 1: Flow Networks with Supplies and Demands Consider the network of solid edges in Figure 4. Vertex b can supply up to 60 units of flow, and vertices c and d can each supply 40 units. Can we b 60 50 a 40 c 50 40 d 30 g 40 30 30 20 30 f 30 e 20 30 h 50 30 i 40 20 30 40 40 j z 30 Figure 4. The sources could be oil refineries and the sinks oil-truck distribution centers, or the sources factories and the sinks warehouses. We model this network problem with a standard one-source, one-sink network as follows. Add a new source vertex a and edges (a, b), (a, c), and (a, d) having capacities 60, 40, and 40, respectively. Next make h, i, and j regular nonsink vertices and add a new sink vertex z and edges (h, z), (i, z), and (j, z), having capacities 50, 40, and 40, respectively. A flow satisfying the original demands is equivalent to a flow in the new network that saturates the edges coming into z, that is, a flow of value 130. Let (P, P) denote the set of all edges (x, y) with vertex x P and y P) (where P denotes the complement of P). The a­z cuts in a network are important, because all flow from a to z must cross each a­z cut. The combined capacity of the edges in any a­z cut is thus an upper bound on how much flow can get from a to z. Let f (e) be an a­z flow in the network N and let P be a subset of vertices in N not containing a or z. Summing together the conservation-of-flow equations in condition (b) for each x P, we obtain f (e) = xP eIn(x) xP eOut(x) f (e) Certain f (e) terms occur on both sides of the preceding equality: namely, edges from one vertex in P to another vertex in P. After eliminating such f (e) from both sides, the left side becomes e(P,P) f (e) and the right side e(P,P) f (e). Thus we have (b) For each vertex subset P not containing a or z, f (e) = Ї e(P,P) e(P,P) f (e) That is, the flow into P equals the flow out of P. Remember that the only flow into P from {a, z} must be from a, since condition (c) forbids flow from z. Then (P, P) consists of edges going into z (from P) and (P, P) consists of edges going out of a (to P). Thus flow out of a = e(P,P) f (e) = by(b) e(P,P) f (e) = flow into z the flow equality still holds if there is flow in an edge (a, z). 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.

Suppose symptoms rabies 16mg betahistine for sale, for example treatment math definition discount betahistine 16mg otc, that there is a "universe" of 100 students in a math course and there are 30 students who are not math majors in the course-that is treatment xanax withdrawal purchase 16 mg betahistine free shipping, N (A) = 30 treatment bee sting 16 mg betahistine for sale, where A is the set of math majors. Then the number N(A) of math majors in the course is N (A) = N - N (A) = 100 - 30 = 70. Let the universe U be all students in a school, let F be the set of students taking French, and let L be the set of students taking Latin. We want formulas for the number of students taking French or Latin N (F L) and the number taking neither language N (F L) in terms of N, N(F), N(L), and N (F L). Note that N (F L) is not simply N (F) + N (L), because N (F) + N (L) counts each student taking both languages two times. Subtracting N (F L) from N (F) + N (L) corrects the double counting of students taking two languages. It is called an inclusion­exclusion formula because first we include the whole set, N; then we exclude (subtract) the single sets F and L; and then we include (add) the 2-set intersection F L. With more sets, this alternating inclusion and exclusion process will continue several rounds. Example 1: Students Taking Neither Language If a school has 100 students with 50 students taking French, 40 students taking Latin, and 20 students taking both languages, how many students take no language? Before using formula (2) to solve this problem, let us consider a more direct approach and see why it fails. If an 8 or 9 were chosen for the first digit, there will be eight choices for the last digit, while if neither 8 nor 9 were chosen for the first digit, F F L L Figure 8. Let F be the set of all arrangements with a 0 or 1 in the first digit, and let L be the set of all arrangements with an 8 or 9 in the last digit. Then the number of arrangements with first digit greater than 1 and the last digit less than 8 is N (F L). The inclusion­ exclusion formula tells us how to put together the answers to the complementary-set calculations solve the original problem. The sets in the inclusion­exclusion formula are always defined so that the final answer (the left side of (2)) is the number of items that are in none of the sets. Thus, we need to define sets that represent the complements of the original constraints we are given. For example, in Example 2, the set of arrangements where the first digit is not greater than 1-the complement of the given constraint on the first digit-is defined to be F, so that the original constraint of the first digit being greater than 1 is now F. We can correct this first formula by adding the number of students taking two languages. Thus we propose the formula N (F L G) = N - N (F) - N (L) - N (G) +N (F L) + N (L G) + N (F G)? A student taking no language is counted once (by the term N)-such students are exactly the ones we want to count. The challenge is to make sure that all other students are counted a net of 0 times. The students taking one language are counted once by N and subtracted once by the term -[N (F) + N (L) + N (G)], for a net count of 0. The students taking two languages are counted once by N, subtracted twice by -[N (F) + N (L) + N (G)] (since they are in exactly two of the three sets), and then added once by the term +[N (F L) + N (L G) + N (F G)] (since they are in exactly one of the three pairwise intersections), for a net count of 0. They are counted once by N, then subtracted three times by the sum of the three sets (since they are in all three sets), then added three times by the pairwise intersections (since they are in all three of these subsets). Then we must correct formula (3) by subtracting N (F L G) to make the net count of students with all three languages 0: N (F L G) = N - [N (F) + N (L) + N (G)] + [N (F L) + N (L G) + N (F G)] - N (F L G) (4) For general sets A1, A2, A3, we rewrite (4) as N (A1 A2 A3) = N - i N (Ai) + ij N (Ai A j) - N (A1 A2 A3) (5) where the sums are understood to run over all possible i and all i,j pairs, respectively. Example 3: Students Taking None of Three Languages If a school has 100 students with 40 taking French, 40 taking Latin, and 40 taking German, 20 students are taking any given pair of languages, and 10 students are taking all three languages, then how many students are taking no language? Here N = 100, N (F) = N (L) = N (G) = 40, N (F L) = N (L G) = N (F G) = 20, and N (F L G) = 10. Then by (4), the number of students taking no language is N (F L G) = 100 - (40 + 40 + 40) + (20 + 20 + 20) - 10 = 30. Example 4: Relatively Prime Numbers How many positive integers 70 are relatively prime to 70? Then we want to count the number of integers 70 that do not have 2 or 5 or 7 as divisors.

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