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Physical Database Design Questions and Answers

Answer: A

Correct answer: Nonunique key

A secondary key identifies a set of rows rather than a single one - all customers in a city, all orders on a date - and is used to speed retrieval on a common search condition. It is a candidate for an index precisely because it is not unique, and it plays no part in identifying rows.

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12.

Sequential retrieval on a primary key for sequential file storage has which of the following features?

Answer: A

Correct answer: Very fast

Sequential file storage keeps the rows in primary key order, so reading them in that order is simply reading the file from start to finish - no index lookups and no seeking. That is what the organisation is for, and it is the reason it survives for batch processing despite being poor at everything else.

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13.

The fastest read/write time and most efficient data storage of any disk array type is:

Answer: A

Correct answer: RAID-0.

RAID-0 stripes data across the disks with no redundancy at all, so every byte of capacity holds data and reads and writes are spread over all the spindles. That gives the best speed and the most efficient storage of any array type - and no protection whatever, since the failure of one disk loses the whole array.

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Answer: D

Correct answer: Be careful indexing attributes that have null values.

Nulls complicate an index, since how they are stored and whether they can be searched for varies between products, and a column with many of them may not benefit from indexing at all. The other three rules are each stated backwards: indexes help most on large tables, on columns that do appear in WHERE clauses, and a unique index on the primary key is essential rather than avoidable.

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15.

Some advantages of partitioning include: efficiency, security, and load balancing.

Answer: B

Correct answer: True

All three are among the recognised advantages, along with local optimisation and recovery. Efficiency comes from a query reading only the partition it needs; security from applying different controls to different partitions; and load balancing from placing partitions on separate devices so their accesses do not contend.

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16.

Data-volume and frequency-of-use statistics are not critical inputs to the physical database design process.

Answer: B

Correct answer: False

Data volumes and access frequencies are among the most critical inputs to physical design: they decide which tables are large enough to warrant indexing, which queries run often enough to be worth optimising for, and where the storage and contention will fall. Without them the design decisions are guesses.

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17.

Two basic constructs to link one piece of data with another piece of data: sequential storage and pointers.

Answer: A

Correct answer: True

Every physical structure reduces to one of the two. Sequential storage links records by placing them next to each other, so the next record is found by continuing to read. Pointers link them by holding an address, so the next record can be anywhere. Indexes, hashing and chains are all built from these two ideas.

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18.

What is the best data type definition for Oracle when a field is alphanumeric and has a fixed length?

Answer: A

Correct answer: CHAR

CHAR is fixed width and pads every value to the declared length, which is exactly right for data that genuinely has one length - a two-character state code, a fixed-format reference. For anything varying, the padding is waste, and VARCHAR2 is the correct choice instead.

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19.

Hashing algorithm converts a primary key value into a record address.

Answer: A

Correct answer: True

A hashing algorithm computes an address directly from the key value, so a record is found in one access with no index to traverse. The price is that the records are then in no useful order, so sequential processing by key becomes expensive, and that collisions must be handled when two keys hash to the same address.

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20.

If a denormalization situation exists with a many-to-many or associative binary relationship, which of the following is true?

Answer: B

Correct answer: All fields are stored in two relations.

A many-to-many relationship denormalises into the associative relation and one of the participants, giving two relations rather than three. It cannot be collapsed into one, because doing so would repeat every attribute of one participant for each of its associations - which is the redundancy normalisation exists to prevent.

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