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Note how similar this process is to looking up a topic in the index of a book. Each page of the index is labeled with a word or letter that represents the topics listed on that page. The page labels are analogous to the keys in the internal nodes of the search tree. The actual page number listed next to the topic in the book s index is analogous to the disk address of file name that leads you to the actual data. The last step of the search process is searching through that page in the book, or through that file on the disk. This analogy is closer if the book s index itself had an index. Each internal level of the multiway tree corresponds to another index level. Algorithm 12.2 Inserting into a B-Tree To insert a record with key k using a B-tree index of order m: 1. If the tree is empty, create a root node with two dummy leaves, insert k there, and return true (indicating that the insertion was successful). 2. Let x be the root. 3. Repeat steps 4 6 until x is a leaf node. 4. Apply the binary search to node x for the key k i , where k i 1 < k k i (regarding k 0 = and k m = ).

Execute hPrb10_12.CIRi of Problem 10.12 and plot v2 V 3 . From the plot, determine the peak-to-peak ripple voltage v2 . Ans: v2 0:383 V

The four are trademarks, trade secrets, patents, and copyrights Trademarks and Service Marks Trademarks are the symbols, names, and pictures that companies use to identify their companies and products Service marks are essentially the same thing, but they identify a service, such as insurance, rather than a tangible product The Kodak logo, for instance, is a trademark So is the Kleenex brand name The name Novell Online Training Provider is a service mark of the Novell Corporation Trademarks are granted by the government, and they can have a very long life A trademark granted by the US Patent and Trademark Office has a term of 10 years, and it can be renewed indefinitely for 10 year terms The application is relatively inexpensive, and can cost as little as a few hundred dollars.

Theorem 1411 The quick sort runs in O(n lgn) time in the best case The best case is when the sequence values are uniformly randomly distributed so that each call to the quick partition algorithm will result in balanced split of the sequence In that case, each recursive call to the quick sort algorithm divides the sequence into two subsequences of nearly equal length As with the binary search and the merge sort (Algorithm 145 on page 261), this repeated subdivision takes lgn steps to get down to size 1 subsequences, as illustrated in the diagram in Figure 142 on page 262 So there are O(lgn) calls made to the quick partition algorithm which runs in O(n) time, so the total running time for the quick sort algorithm is O(n lgn) Theorem 1412 The quick sort runs in O(n2) time in the worst case.

Determine the smallest value of inductance that could have been used for the buck-boost converter of Problem 10.15 and the inductor current remain continuous. Ans: L Lc 24 H

The worst case is when the sequence is already sorted (or sorted in reverse order) In that case, the quick partition algorithm will always select the last element (or the first element, if the sequence is sorted in reverse order), resulting in the most unbalanced split possible: One piece has n 2 elements, and the other piece has 1 element Repeated division of this type will occur O(n) times before both pieces get down to size 1 So there are O(n) calls made to the quick partition algorithm which runs in O(n) time, so the total running time for the quick sort algorithm is O(n2) Note that in the worst case, the quick sort reverts to the selection sort (Algorithm 142 on page 257) because each call to quick partition amounts to selecting the largest element from the subsequence passed to it So actually, Theorem 14.

0 The actual gain GV for the buck-boost converter with inherent inductor resistance was determined in 0 Problem 10.14. Determine the duty cycle D Dp for which GV has a maximum value.