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Figure 3-12 indicates the conditions based on Fig 3-11 By applying the appropriate combinations of probabilities to their costs, we obtain for the cases considered TN = (P d)(P b)(c)+ (P d)(1 P b)(P m)(P c)(r + btt)+ (P d)(1 P b)(P m)(1 P c)(s + r + btt)+ (P d)(1 P b)(1 P m)(s + r + btt)+ (1 P d)(P l)(r + btt)+ (1 P d)(1 P l)(s + r + btt) / case a / / case b / / case c / / case d / / case e / / case f /

24 N 1

you can find the first instance of the characters hre by writing:

The seek terms in cases d and f, marked with a , disappear when the over ow areas are kept on the same cylinders as the corresponding data blocks The probabilities P m and P l can be rewritten in terms of n and o using Eqs 3-35 and 3-36

int index = s1.IndexOf("hre");

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Equation 3-40 was obtained by a detailed case analysis, but as is typical, the result can be made practical by consideration of the relative importance of the terms If we neglect the memory search time, c, for records within the block, if cylinder seeks can be ignored (ie, is large, so that the value of P c 1), and if the over ow areas are on the same cylinders as the data blocks (s = 0, so that cases b and d, as well as e and f combine), then TN 1 Pov n + o Bfr + Pov (r + btt) = (r + btt) Bfr (n + o )Bfr 3-41

This code sets the int variable index to 9, which is the offset of the letters hre in the string s1. Similarly, the LastIndexOf( ) method returns the index of the last occurrence of a string or substring. Though the following code:

We note that the chain length Lc does not a ect this approximation for TN , but over ows still increase the estimate Adding a record to the le will always cause an addition to the over ow chain, either because of push-through or because the new record follows serially a record already in the over ow chain Each insertion requires the reading and rewriting of a predecessor data or over ow block, since a pointer will have to be inserted or changed and also a READ and REWRITE of the over ow block for the pushed or inserted record The probability of using the same over ow block for both is small if records are randomly inserted into a large data le We avoid a detailed case analysis now by using the previous result for TF and making some further assumptions The fetch time for the predecessor is equal to TF , the over ow block is on the same cylinder and requires r +btt to be reached, and each REWRITE will take one revolution TRW = 2r (given the conditions of Eq 2-30), so that 3-42 TI = TF + TRW + r + btt + TRW = TF + 5r + btt

s1.IndexOf("o");

The e ect of the length of the over ow is incorporated into TF Note that the index is not a ected when a record is inserted The cost of insertion is only due to appending the record to the chain and linking it to the predecessor record Alternative issues of chain management, discussed earlier, may be considered An updated record of equal size and identical key can be placed into the place of the previous version of the record, so that the process can be evaluated as a fetch followed by a REWRITE with a cost of TRW : in place 3-43 TU = TF + TRW = TF + 2r

The space available for new records must be managed carefully because the page is not preformatted into slots One way to manage free space is to maintain a pointer (that is, o set from the start of the data area on the page) that indicates the start of the free space area When a new record is too large to t into the remaining free space, we have to move records on the page to reclaim the space freed by records that have been deleted earlier The idea is to ensure that after reorganization, all records appear contiguously, followed by the available free space A subtle point to be noted is that the slot for a deleted record cannot always be removed from the slot directory, because slot numbers are used to identify records by deleting a slot, we change (decrement) the slot number of subsequent slots in the slot directory, and thereby change the rid of records pointed to by subsequent slots The

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