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There is a rather exotic property of certain electronic components that you ll sometimes hear or read about It is called characteristic impedance or surge impedance, and is symbolized Zo It is a specification of an important property of transmission lines It can always be expressed as a positive real number, in ohms

3 5 0 2 20 19 1 10 13 11 0 20

(b) Now relax this to a linear program, and show that there must in fact be an integral optimal solution (as is the case, for example, with maximum ow and bipartite matching)

Now let s get 0 at the extreme left in the second row If we multiply the second row by 3/5, we come up with

Transmission Lines When it is necessary to get energy or signals from one place to another, a transmission line is required These almost always take either of two forms, coaxial or two-wire (also called parallel-wire) Cross-sectional renditions of both types are shown in Fig 15-8 Examples of transmission lines include the ribbon that goes from a television antenna to the receiver, the cable running from a hifi amplifier to the speakers, and the set of wires that carries electricity over the countryside Factors Affecting Zo The Zo of a parallel-wire transmission line depends on the diameter of the wires, on the spacing between the wires, and on the nature of the insulating material separating the wires In general, the Zo

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3 3 0 2 12 19 1 6 13 11 0 20

730 Hall s theorem Returning to the matchmaking scenario of Section 73, suppose we have a bipartite graph with boys on the left and an equal number of girls on the right Hall s theorem says that there is a perfect matching if and only if the following condition holds: any subset S of boys is connected to at least |S| girls Prove this theorem (Hint: The max- ow min-cut theorem should be helpful) 731 Consider the following simple network with edge capacities as shown

3 0 0 2 10 19 1 7 13 11 11 20

either type of line, Zo depends on the conductor diameters and spacing, and on the nature of the dielectric material between the conductors See text for discussion

Now the number 19 in the third row must somehow be made to vanish, which means we must turn it into 0 Let s use the second row to attack it If we multiply the second row through by 19 and the third row through by 10 (combining two moves), we get

(a) Show that, if the Ford-Fulkerson algorithm is run on this graph, a careless choice of updates might cause it to take 1000 iterations Imagine if the capacities were a million instead of 1000! We will now nd a strategy for choosing paths under which the algorithm is guaranteed to terminate in a reasonable number of iterations Consider an arbitrary directed network (G = (V, E), s, t, {ce }) in which we want to nd the maximum ow Assume for simplicity that all edge capacities are at least 1, and de ne the capacity of an s t path to be the smallest capacity of its constituent edges The fattest path from s to t is the path with the most capacity

3 0 0 2 190 190 1 133 130 11 209 200

(b) Show that the fattest s t path in a graph can be computed by a variant of Dijkstra s algorithm (c) Show that the maximum ow in G is the sum of individual ows along at most |E| paths from s to t (d) Now show that if we always increase ow along the fattest path in the residual graph, then the Ford-Fulkerson algorithm will terminate in at most O(|E| log F ) iterations, where F is the size of the maximum ow (Hint: It might help to recall the proof for the greedy set cover algorithm in Section 54) In fact, an even simpler rule nding a path in the residual graph using breadth- rst search guarantees that at most O(|V | |E|) iterations will be needed

Adding the second and third rows and then replacing the third row with the sum, we obtain a matrix in echelon form:

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