1. Give brief and clear answers to the following:

(a) Explain how non-volatile RAM can help speed up disk writes

(b) Explain the main advantage(s) and disadvantage(s) of keeping the data ?le sorted.

(c) 17.28 parts (a) and (b)

(d) 18.3

(e) 18.7

(f) 18.8

(g) Since indexes speed up searches, why wouldn't the DBMS automatically create an index for every column of a table?

(h) Explain the main disadvantage(s) of static hashing.

(i) Brie?y describe one situation in which ordered ?les may be better than static hashing.

(j) Explain the main advantage B-trees have over a multilevel index of the type shown in Figure 18.6.

2. In this problem, you have to calculate the number of disk accesses required in the worst case for di?erent index organizations. The data base keeps track of di?erent pieces of information for a student: name (25 bytes), idnumber (15 byes), gpa (10 bytes) and address (90 bytes). There are ?ve hundred thousand students (it is a large university!). Each block can store 2000 bytes of information. A block pointer requires 10 bytes. The data ?le is sorted by the primary key, the student name. The idnumber is a secondary key. How many disk accesses are required if:

(a) access is through student name and there is a primary index on student name (similar to the index in Figure 18.1). You can assume that the index is small enough to be kept in RAM, and is already in RAM.

(b) access is through student name and there is a primary index on student name (similar to the index in Figure 18.1) which also has to be kept on disk.

(c) access is through student name and there is no index

(d) access is through student id and there is a secondary index on the student id (similar to the index in Figure 18.4) which also has to be kept on disk.

(e) access is through student name and there is a 2-level primary index on student name (similar to the index in Figure 18.6) in which the second level of the index is in memory, while the ?rst level of the index is on disk. This is somewhat di?erent from the way the Figure 18.6 index is organized in the sense that we have only 2 levels; the second level already in RAM, and the ?rst level on the disk.