Question 1

Database keys and indexing context In database design, which statement best describes a secondary key?

Accepted Answer

Introduction / Context: Keys define how rows are identified and accessed. Beyond the primary key, designers often add additional access paths to speed up lookups or support alternate query predicates. This question checks whether you can distinguish a secondary key from a primary or surrogate key and from storage/physical concerns. Given Data / Assumptions: A table already has a primary key that uniquely identifies rows. Applications run queries filtering on non-PK columns (e.g., email, status, date). Secondary keys may or may not be unique; they frequently are not. Concept / Approach: A secondary key is any non-primary-key attribute (or attribute set) that is indexed to support alternate access to rows. It does not have to be unique. Its purpose is performance and flexible retrieval, not entity identity. Think of searching customers by last_name or orders by order_date: these are classic secondary keys. Step-by-Step Solution: 1) Identify that the “primary key” uniquely identifies each row.2) Recognize that many queries filter on other columns.3) Indexing those columns creates secondary keys, often nonunique.4) Therefore, the best definition is a nonunique key that aids retrieval. Verification / Alternative check: Database texts distinguish: primary key (uniqueness + identity), candidate/surrogate keys (alternative unique identifiers), and secondary keys (additional indexed attributes, not necessarily unique). Why Other Options Are Wrong: Primary key: by definition unique and not “secondary.” “Useful only for denormalization”: secondary keys aid queries regardless of normalization. Tablespace choice: a physical storage concern, not key semantics. “Generated by file system”: keys are logical DB constructs, not filesystem artifacts. Common Pitfalls: Assuming every indexed column must be unique—many secondary keys are not. Confusing candidate keys (unique) with secondary keys (may be nonunique). Final Answer: Nonunique key used for data retrieval and alternative access paths.

Question 2

Choosing data types for robust schemas When selecting a column's data type, which goals should a designer pursue?

Accepted Answer

Introduction / Context: Poor data-type choices propagate pain: wasted space, slow scans, and integrity problems. Picking the right type is foundational for performance and correctness across OLTP and analytics systems. This item tests whether you know the balanced objectives in type selection. Given Data / Assumptions: Columns will store values repeatedly at large scale. Storage, integrity, and representational coverage all matter. Constraints complement but do not replace good type choices. Concept / Approach: Good schema design aims to minimize storage (without losing fidelity), correctly represent the domain (e.g., DATE for dates, NUMERIC for money), and enforce integrity (NOT NULL, CHECK, ENUM/domain constraints). These goals are not mutually exclusive; they reinforce one another when the correct type is used with constraints. Step-by-Step Solution: 1) Map the business domain to native types (e.g., boolean for flags, date for calendar days).2) Choose the smallest type that safely represents values (e.g., SMALLINT vs BIGINT).3) Add integrity constraints to prevent impossible states.4) Conclude that all three goals are desired → “All of the above.” Verification / Alternative check: Vendor guides and normalization practice emphasize domain-correct types with constraints to improve query plans and reduce anomalies, while right-sizing preserves I/O and cache efficiency. Why Other Options Are Wrong: Option E elevates readability above correctness; databases should enforce correctness first. Common Pitfalls: Using overly wide VARCHAR for codes that fit in SMALLINT or fixed CHAR. Storing dates/times as strings, which harms validation and indexing. Final Answer: All of the above.

Question 3

Storage organization terms In file/page storage, the “blocking factor” refers to:

Accepted Answer

Introduction / Context: On-disk layout influences performance. Many DBMSs store data in fixed-size pages or blocks. The blocking factor captures how densely records pack into those pages, directly affecting I/O costs for scans and lookups. This question checks your recall of that definition. Given Data / Assumptions: Data is stored in fixed-size pages/blocks (e.g., 8 KB). Each row has a physical length including overhead. Free space and fragmentation can reduce effective packing. Concept / Approach: Blocking factor = floor(page_size / record_size_with_overhead). Higher values mean more rows per I/O and typically better sequential scan efficiency. Very wide rows reduce the factor and can increase I/O per query. Step-by-Step Solution: 1) Identify page size and average record size.2) Compute how many complete records fit → the blocking factor.3) Recognize its impact on buffer hits and disk reads. Verification / Alternative check: Database storage references and file-organization texts define the term exactly this way; it is not about keys or logical grouping but about physical packing density. Why Other Options Are Wrong: Primary/secondary key groupings describe logical design, not page packing. “Column family” applies to certain NoSQL models and is unrelated to classic blocking factor. Number of indexes has no direct definition link here. Common Pitfalls: Ignoring row overhead (headers, alignment) when estimating packing. Assuming larger pages always help; they can hurt random access. Final Answer: The number of physical records that fit on one page or block.

Question 4

Choosing larger database block sizes Which of the following is NOT a key factor when deciding to switch from a small to a large block/page size?

Accepted Answer

Introduction / Context: DBMSs let you choose a page size (e.g., 4 KB, 8 KB, 16 KB). Larger pages can improve sequential throughput but may increase random-read latency and contention. This question separates signal (row length, contention, access pattern) from noise (column count alone). Given Data / Assumptions: Workloads vary: OLTP point lookups versus analytics scans. Row length and row density shape I/O efficiency. Concurrency and buffer behavior matter for hot pages. Concept / Approach: Decisions should hinge on row size, expected scan vs lookup ratios, and observed contention. The raw count of columns is not meaningful by itself; widths, nullability, and access patterns matter far more than the mere number of attributes. Step-by-Step Solution: 1) Evaluate row size to estimate blocking factor changes.2) Assess workload: large scans like larger blocks; random lookups may prefer smaller ones.3) Measure contention and cache residency for hot objects.4) Conclude that “number of columns” alone is not a deciding factor. Verification / Alternative check: Vendor tuning guides focus on row width, access patterns, and concurrency. Column count is at best a proxy for width and is not decisive independently. Why Other Options Are Wrong: Row length directly affects how many rows fit per page. Contention and random access cost are central to sizing trade-offs. Access pattern is fundamental to page-size benefits. Common Pitfalls: Assuming larger pages are always better—they can hurt cache efficiency for OLTP. Ignoring per-row and per-page overhead when modeling effects. Final Answer: The number of columns in the table, independent of their widths and usage.

Question 5

Denormalization choice for 1-to-1 relationships If a denormalization situation exists with a one-to-one (1:1) binary relationship between two entities, the recommended physical design is to:

Accepted Answer

Introduction / Context: Normalization reduces redundancy; denormalization re-introduces some redundancy or merges entities for performance or simplicity. With a strict 1:1 relationship, a common pragmatic choice is to collapse the entities into a single table to eliminate constant joins and simplify access paths. Given Data / Assumptions: Every row in entity A corresponds to exactly one row in entity B. They are frequently queried together. Security/optional columns do not require strict separation. Concept / Approach: When 1:1 holds and access is co-located, merging eliminates joins and foreign keys. You still enforce constraints with unique keys and CHECKs. If some columns are optional or rarely used, vertical partitioning may be a viable alternative, but the core denormalized design is a single relation. Step-by-Step Solution: 1) Verify true 1:1 cardinality and stable business rules.2) Confirm common access patterns justify merging.3) Merge attributes into one table; preserve unique/NOT NULL constraints.4) Re-assess indexes for typical predicates. Verification / Alternative check: Physical design texts recommend merging 1:1 entities when joins add overhead without adding information content, provided governance needs allow it. Why Other Options Are Wrong: Two or more relations add joins without benefit for strict 1:1. Bridge tables are for many-to-many, not 1:1. Four relations/heap storage are unjustified and harmful to integrity. Common Pitfalls: Merging prematurely when 1:1 later evolves into 1:N. Ignoring column-level security that may require separation. Final Answer: Store all fields in one relation (merge the two logical entities into a single table).

Question 6

Query performance practices Which practice most reliably improves a query's processing time and execution efficiency?

Accepted Answer

Introduction / Context: SQL performance depends on the optimizer’s ability to use indexes, statistics, and selective predicates. Data-type mismatches and non-sargable expressions often prevent index seeks and force scans. This question tests a reliable, general practice for better performance. Given Data / Assumptions: The schema provides appropriate indexes and fresh statistics. Queries can be written in multiple ways. We aim for portable, optimizer-friendly SQL. Concept / Approach: Using compatible types (e.g., comparing INTEGER to INTEGER, DATE to DATE) avoids implicit conversions that block index usage. Writing sargable predicates (Search ARGument ABLE), such as column >= ? and column BETWEEN a AND b, lets the optimizer find efficient plans. Type correctness also prevents needless casts and function-wrapped columns. Step-by-Step Solution: 1) Verify column and parameter types match to avoid implicit casts.2) Use predicates the engine can push to indexes.3) Remove unnecessary nesting and self-joins where simpler joins or CTEs suffice.4) Check statistics so the optimizer estimates cardinalities accurately. Verification / Alternative check: RDBMS tuning guides consistently highlight data-type compatibility and sargability as universal improvements across engines (PostgreSQL, Oracle, SQL Server, MySQL). Why Other Options Are Wrong: Complexity for its own sake usually harms readability and performance. Unnecessary self-joins and nesting add overhead without benefit. Disabling statistics cripples the optimizer, degrading performance. Common Pitfalls: Wrapping indexed columns in functions (e.g., UPPER(col)) making predicates non-sargable. Comparing text to numbers or dates to strings, causing scans. Final Answer: Use compatible and proper data types to enable sargable predicates and good plans.

Question 7

Modeling many-to-many relationships If a denormalization situation involves a many-to-many (associative) binary relationship, the correct baseline structure is to:

Accepted Answer

Introduction / Context: Many-to-many (M:N) relationships cannot be represented directly in a single relational table without redundancy and anomalies. The normalized solution is to introduce an associative (bridge) table that contains foreign keys to each entity. This question verifies that baseline modeling knowledge. Given Data / Assumptions: Two entities, A and B, have M:N cardinality. Each association may carry attributes (e.g., role, quantity, effective dates). We can later consider denormalization for performance. Concept / Approach: The canonical structure is three relations: A, B, and A_B (the bridge) with a composite key (A_id, B_id) or a surrogate, plus foreign keys to enforce referential integrity. Additional attributes of the association reside in the bridge table to avoid duplication across A or B. Step-by-Step Solution: 1) Identify M:N, which cannot be mapped into one FK without duplication.2) Create a bridge table with two FKs (to A and to B).3) Add attributes of the relationship to the bridge as needed.4) Optionally denormalize later based on query patterns. Verification / Alternative check: All relational design references adopt the three-table structure for M:N. Denormalization that collapses tables is an optimization, not the starting point. Why Other Options Are Wrong: One or two tables cannot model M:N without redundancy. Four tables are unnecessary unless there are additional entities. Inferring links from text columns is fragile and violates integrity. Common Pitfalls: Placing relationship attributes onto one side (A or B), creating duplication. Forgetting composite uniqueness on the pair of foreign keys. Final Answer: Store all fields in three relations (two entity tables plus an associative/bridge table).

Question 8

Oracle data type choice for fixed-length text For an alphanumeric column with a fixed length in Oracle, the most appropriate data type is:

Accepted Answer

Introduction / Context: Choosing between CHAR and VARCHAR2 in Oracle depends on whether the text length is fixed or variable. Fixed-length codes (e.g., country_code, status_code) benefit from CHAR for simpler semantics and potential storage/CPU advantages at small sizes. Given Data / Assumptions: Values are alphanumeric and always the same length (e.g., exactly 2 or 3 characters). We are not storing large free-form text. We want predictable storage and comparison behavior. Concept / Approach: Use CHAR(n) for fixed-length fields; Oracle pads to length n for storage/comparison semantics. VARCHAR2(n) is best for variable-length fields. LONG is obsolete and discouraged; CLOB is for large character data; NUMBER is numeric storage only. Step-by-Step Solution: 1) Determine if length is fixed or variable.2) For fixed-length, select CHAR(n).3) For variable-length, prefer VARCHAR2(n).4) Avoid deprecated/unsuitable types (LONG, NUMBER for text). Verification / Alternative check: Oracle documentation advises VARCHAR2 for variable-length character strings and CHAR for fixed-length semantics, with LONG retained only for backward compatibility. Why Other Options Are Wrong: VARCHAR2: variable length, not fixed. LONG: deprecated; use CLOB/VARCHAR2 instead. NUMBER: numeric, not text. CLOB: overkill for short fixed-length codes. Common Pitfalls: Using VARCHAR2 for fixed-length codes causes inconsistent padding/comparisons. Using LONG complicates tooling and indexing. Final Answer: CHAR.

Question 9

Integrity controls supported by DBMSs Which of the following integrity controls may be enforced by a DBMS?

Accepted Answer

Introduction / Context: Modern relational databases provide rich integrity controls so that correctness does not rely solely on application code. Using built-in constraints shrinks bug surface area and improves data quality by design. This question verifies recognition of common DBMS-enforced controls. Given Data / Assumptions: We operate in a relational DBMS with DDL-defined constraints. We can specify defaults, domains/ranges, and NULLability in schema. Constraints are validated on insert/update. Concept / Approach: Defaults fill in values when none are provided; domain constraints limit values via CHECK or enumerated types; NOT NULL ensures presence. These complement primary/foreign keys and unique constraints, forming a comprehensive integrity layer. Step-by-Step Solution: 1) Map each option to a standard DDL feature: DEFAULT, CHECK/ENUM, NOT NULL.2) Confirm all are supported in mainstream DBMSs.3) Therefore, choose “All of the above.” Verification / Alternative check: Documentation for PostgreSQL, Oracle, MySQL, and SQL Server includes DEFAULT, CHECK, and NOT NULL constraints as first-class features. Why Other Options Are Wrong: Option E is false; integrity should be enforced in the database as well as in application logic. Common Pitfalls: Leaving integrity to the app layer only, risking multiple inconsistent writers. Overusing triggers where declarative constraints suffice. Final Answer: All of the above.

Question 10

Disk arrays and performance characteristics Which RAID level offers the fastest read/write performance and the most storage efficiency, but provides no redundancy or fault tolerance?

Accepted Answer

Introduction / Context: RAID configurations trade performance, capacity efficiency, and fault tolerance. Understanding these trade-offs is essential for choosing storage for OLTP logs, data warehouses, or scratch/temp spaces. This question targets the performance-vs-resilience extremes. Given Data / Assumptions: We compare standard RAID levels. “Fastest and most efficient storage” means minimal overhead. No redundancy implies risk of data loss on a single-disk failure. Concept / Approach: RAID-0 stripes data across disks with no parity or mirroring. This maximizes throughput and usable capacity, but a single disk failure loses the entire array. RAID-1 mirrors (50% efficiency), RAID-5/6 add parity (capacity overhead), and RAID-10 mirrors then stripes (performance + redundancy with higher cost). Step-by-Step Solution: 1) Identify the array with no redundancy overhead → RAID-0.2) Confirm it provides best raw throughput due to full striping.3) Note the trade-off: zero fault tolerance. Verification / Alternative check: Systems architecture references list RAID-0 as highest performance/efficiency and highest risk. It is suited to transient data or scratch volumes where loss is acceptable. Why Other Options Are Wrong: RAID-1 mirrors—half capacity, strong redundancy, slower writes. RAID-2/3 are specialized and include parity/ECC overhead. RAID-5 distributes parity, reducing net capacity and write speed. Common Pitfalls: Deploying RAID-0 for critical data without robust backups and replication. Assuming RAID replaces backups; it does not. Final Answer: RAID-0 (striping, no parity).

Question 11

Database indexing best practices in relational systems: Which “rule of thumb” should guide when you choose columns to index for query performance and maintainability?

Accepted Answer

Introduction / Context: Choosing the right columns for indexing is essential in relational database design. The goal is to speed up selective lookups and joins while controlling write overhead and storage. This question tests pragmatic “rules of thumb,” especially around columns that can contain nulls. Given Data / Assumptions: Workload includes OLTP-style queries with WHERE predicates and joins. Some columns may contain null values. We must select the best general guideline among the options. Concept / Approach: Indexes accelerate reads by ordering keys and maintaining pointers, but they add overhead on writes. Columns with many nulls can behave unexpectedly: some engines treat nulls specially in unique indexes; null-heavy columns may have low selectivity; partial or filtered indexes may be preferable. Therefore a sensible rule of thumb is to be cautious when indexing nullable attributes and consider patterns like filtered indexes (WHERE col IS NOT NULL) or enforcing NOT NULL when appropriate. Step-by-Step Solution: Evaluate usefulness: high-selectivity, frequently filtered columns benefit most.Consider null behavior: uniqueness semantics and selectivity can degrade with many nulls.Adopt mitigations: filtered/partial indexes, NOT NULL constraints, or composite keys. Verification / Alternative check: Inspect query plans for predicates involving nullable columns; compare performance with and without a filtered index that excludes nulls. Why Other Options Are Wrong: Indexes more useful on smaller tables: smaller tables are often fully scanned cheaply. Indexes on columns not used in WHERE: provide little benefit. No unique index for primary key: primary keys must be unique; engines auto-create or require a unique index. Common Pitfalls: Indexing low-selectivity columns (e.g., boolean flags) or leaving NULLs rampant in keys used for joins. Final Answer: Be careful indexing attributes that have null values.

Question 12

Oracle data type choice: For an alphanumeric field with variable length, which data type is best to store values efficiently while avoiding unnecessary padding?

Accepted Answer

Introduction / Context: Correct data types ensure storage efficiency and predictable behavior. In Oracle, character columns can be fixed length or variable length; picking the right one avoids wasted space and subtle comparison issues. Given Data / Assumptions: Field is alphanumeric (textual) and varies in length by row. Values should not be padded to a fixed width. Standard Oracle data types are available. Concept / Approach: Use VARCHAR2 for variable-length character data. Oracle stores only the bytes required for the value plus small overhead. CHAR is fixed-length and space-pads values to the declared size, which wastes space and can affect comparisons. LONG is legacy (large unstructured text) and discouraged in favor of CLOB. NUMBER is numeric and not appropriate for alphanumeric text. Step-by-Step Solution: Identify data characteristics: variable-length text.Map to Oracle types: VARCHAR2 handles variable text; CHAR is fixed-length.Select VARCHAR2 to avoid padding and save space. Verification / Alternative check: Compare storage and behavior: inserting 'A' into CHAR(10) pads to 10, whereas VARCHAR2(10) stores 1 character. Why Other Options Are Wrong: CHAR: fixed length; wasteful for varying sizes. LONG: obsolete; limited functions and should be replaced by CLOB. NUMBER: numeric type; cannot store general text. Common Pitfalls: Declaring CHAR for codes that actually vary, leading to trailing-space anomalies in joins and comparisons. Final Answer: VARCHAR2

Question 13

Sequential file organization and primary key scans: When data are stored sequentially by the primary key, how fast is sequential retrieval across that key order typically considered?

Accepted Answer

Introduction / Context: File organization affects access speed. With sequential storage ordered by the primary key, reading records in key order exploits physical locality and minimal repositioning, which is ideal for range scans and reporting. Given Data / Assumptions: Records are stored physically in primary-key order. Access pattern is a forward scan by key (sequential retrieval). Underlying storage favors streaming reads. Concept / Approach: Sequential access minimizes random I/O and leverages read-ahead. Whether in classic sequential files or clustered indexes in databases, a full or range scan in key order is typically very fast compared with scattered random lookups. Step-by-Step Solution: Align storage order with access order (primary key).Use range predicates to read contiguous pages.Benefit from prefetching and large block reads. Verification / Alternative check: Benchmark a range scan versus random key lookups; sequential throughput is markedly higher, especially on spinning media and still favorable on SSDs due to cache coherence. Why Other Options Are Wrong: Moderately fast/Slow/Impractical: contradict established I/O behavior for ordered sequential scans. Common Pitfalls: Assuming random access equals sequential speed; mixing wide rows and excessive page splits can still degrade performance if maintenance is neglected. Final Answer: Very fast

Question 14

Multidimensional database usage: In which environment is a multidimensional database model most commonly used for analytics and aggregated reporting?

Accepted Answer

Introduction / Context: Multidimensional databases (cubes/tabular models) organize measures across dimensions (time, product, geography) to support fast slicing, dicing, and aggregation. Understanding where they fit clarifies modern analytics architectures. Given Data / Assumptions: Goal: rapid summarized analysis over large historical data. Users: business analysts consuming KPIs and trends. Store: centralized analytical repository. Concept / Approach: Data warehouses integrate data from multiple sources, model them in star/snowflake schemas, and often expose OLAP cubes or columnar/tabular models for interactive analytics. While relational systems store detailed records, the multidimensional layer sits on top to accelerate aggregated queries and calculations. Step-by-Step Solution: Identify the primary analytical store: the data warehouse.Recognize cubes/tabular models as common warehouse-serving structures.Conclude the multidimensional model is most associated with data warehouses. Verification / Alternative check: Examine BI stacks: ETL → warehouse (star schema) → OLAP cube → dashboards. Why Other Options Are Wrong: Relational/Hierarchical/Network: general data models; not specifically the multidimensional analytical layer. Common Pitfalls: Confusing the physical warehouse (relational) with the logical multidimensional model presented to users. Final Answer: Data warehouse

Question 15

Table partitioning advantages: Which of the following is a legitimate advantage that partitioning can provide when designing large-scale databases?

Accepted Answer

Introduction / Context: Partitioning splits large tables or indexes into smaller pieces (by range, list, hash) to improve manageability, performance, and availability. Understanding its benefits helps justify design choices in data warehouses and large OLTP systems. Given Data / Assumptions: Very large datasets with distinct segments (e.g., by date or region). DBMS supports native partitioning features. Operational goals include performance, maintenance, and governance. Concept / Approach: Common benefits include faster maintenance (partition switch, truncate), improved query performance via partition pruning, and administrative isolation. Partitioning can also enhance security/governance by isolating sensitive data segments to different storage, filegroups, or access policies, enabling finer-grained controls and easier auditing. Step-by-Step Solution: Map business boundaries (e.g., jurisdictions) to partitions.Apply distinct filegroups or tablespaces with tailored permissions.Leverage pruning to limit scans to authorized partitions. Verification / Alternative check: Review access logs and policies; partitions can be hosted on nodes/storage with restricted access while others remain widely accessible. Why Other Options Are Wrong: Complexity/Extra space/Inconsistent access speed are costs or risks, not advantages. Common Pitfalls: Over-partitioning causing plan instability or administrative overhead; assuming partitioning alone fixes all performance issues without proper indexing. Final Answer: Security

Question 16

In database storage design, is the general guideline that larger block (page) sizes suit OLTP workloads while smaller blocks suit decision support systems (DSS) and data warehousing?

Accepted Answer

Introduction / Context: This item tests your understanding of how database page or block size interacts with different workload patterns: Online Transaction Processing (OLTP) versus Decision Support Systems (DSS) or data warehousing. Choosing an appropriate page size affects I/O efficiency, buffer utilization, and contention. Given Data / Assumptions: Page (block) size determines how much data is read with each physical I/O. OLTP workloads perform many short, highly selective reads and writes of small rows. DSS/warehousing workloads scan large ranges and aggregate many rows. Concept / Approach: Smaller blocks reduce overfetch for point lookups and narrow range scans frequent in OLTP. Larger blocks increase sequential throughput for large scans common in DSS. Thus, the statement in the prompt has the directions reversed. Step-by-Step Solution: Identify OLTP access: selective lookups and frequent updates.Smaller blocks limit unnecessary data movement and reduce latch/lock contention.Identify DSS access: table scans, star-join aggregations, and long-running queries.Larger blocks improve sequential I/O and prefetch efficiency for wide scans. Verification / Alternative check: Vendor guidelines generally recommend smaller blocks for OLTP and larger for analytics, subject to row size and storage engine specifics. Why Other Options Are Wrong: Valid in general reverses the real guideline. Only for in-memory is irrelevant because block size matters primarily for disk I/O. Depends only on indexing ignores workload and I/O path. RAID 10 only is a storage detail, not the determining factor. Common Pitfalls: Assuming one universal page size fits all; ignoring row width and access patterns; forgetting that very large pages may increase buffer churn for point queries. Final Answer: Invalid in general — OLTP benefits from smaller blocks, DSS from larger blocks

Question 17

Are the three newest database architectures “relational, multidimensional, and hierarchical,” or is that characterization inaccurate?

Accepted Answer

Introduction / Context: The question probes your knowledge of database architecture evolution. It contrasts older models (hierarchical, network) with relational and more recent families (document, key–value, wide-column, graph, NewSQL, and cloud-distributed systems). Given Data / Assumptions: “Newest” is interpreted in the modern context of data platforms. Hierarchical databases (e.g., IMS) predate relational systems. Multidimensional storage is an OLAP modeling/physical design, often layered over relational stores. Concept / Approach: Relational is foundational and still dominant, but calling hierarchical and multidimensional “newest” is incorrect. Contemporary innovation centers on distributed, cloud-native, and NoSQL/NewSQL architectures. Step-by-Step Solution: Recognize hierarchical as a legacy model predating relational.Understand that multidimensional is typically an analytical abstraction, not a general-purpose “newest” architecture.Identify modern categories: document, key–value, graph, columnar MPP, lakehouse, NewSQL.Conclude the characterization is invalid. Verification / Alternative check: Survey of current platforms shows widespread adoption of nonrelational, distributed systems alongside relational engines. Why Other Options Are Wrong: Vendor specialization, schema style, or mainframe deployment do not redefine architectural recency. Common Pitfalls: Confusing OLAP “multidimensional” cubes with a standalone, modern general-purpose DB architecture. Final Answer: Invalid — hierarchical is legacy; “newest” includes NoSQL and distributed paradigms

Question 18

Bitmap index concept: is it defined as an index on columns from two or more tables sharing the same value domain, or is that description incorrect?

Accepted Answer

Introduction / Context: This question checks whether you can distinguish a standard bitmap index from related concepts such as a bitmap join index. Bitmap indexes are common in analytic workloads for low-cardinality columns. Given Data / Assumptions: A bitmap index stores a bitmap for each distinct value of a column, indicating row positions where the value occurs. They are typically built on one table's column, enabling fast boolean combinations. A bitmap join index exists in some products and precomputes a join condition, but that is different from the basic definition. Concept / Approach: The prompt’s wording (“columns from two or more tables”) corresponds more to a join index, not to the general definition of a bitmap index. Therefore the statement is invalid. Step-by-Step Solution: Recall: bitmap index = value-to-bitmap mapping for a column in a table.Note: boolean operations on bitmaps accelerate multi-predicate filters.Contrast: bitmap join index pre-joins tables — a specialized feature.Conclude the given description does not match a standard bitmap index. Verification / Alternative check: Vendor docs show separate entries for “bitmap index” and “bitmap join index,” reinforcing the distinction. Why Other Options Are Wrong: Composite keys, foreign-key focus, or columnar storage do not redefine the basic bitmap index concept. Common Pitfalls: Equating all bitmap-based optimizations with the term “bitmap index.” Final Answer: Invalid — bitmap indexes map distinct values to bitmaps in a single table

Question 19

Hashing in physical file organization: does a hashing algorithm convert a primary key value into a record (bucket) address?

Accepted Answer

Introduction / Context: Hash-based file organizations and hash indexes are used to provide near O(1) access by transforming a search key into an address for storage and retrieval. Given Data / Assumptions: A hash function h(key) produces a bucket/page address or slot identifier. Collisions are handled via chaining, open addressing, or overflow areas. The key can be any datatype that the hash function can process (not only numeric). Concept / Approach: The statement accurately reflects primary-key hashing used in static or dynamic hashing schemes to compute a record's home location. Step-by-Step Solution: Define hash function: h(k) → address.Apply to primary key to get target bucket.Handle collisions if multiple keys map to the same address.Retrieve or insert the record at that location. Verification / Alternative check: Standard database textbooks describe hashing as address computation from keys. Why Other Options Are Wrong: Hashing is not mere compression, nor limited to secondary indexes, distributed systems, or numeric keys. Common Pitfalls: Ignoring collision handling and bucket overflow management in design. Final Answer: Valid — hashing maps key values to bucket/record addresses

Question 20

Partitioning advantages: do efficiency, security isolation, and load balancing count among common benefits of partitioning tables/data?

Accepted Answer

Introduction / Context: Partitioning splits large tables or indexes into smaller, more manageable pieces based on range, list, or hash strategies. This question asks whether efficiency, security, and load balancing are legitimate benefits. Given Data / Assumptions: Efficiency: pruning reduces scanned data; maintenance can be targeted. Security/Isolation: partitions can be placed on different tablespaces or have different access controls or retention. Load balancing: partitions can be distributed across storage/compute to spread I/O and parallelize queries. Concept / Approach: These are standard arguments for partitioning, along with manageability, availability, and lifecycle management (rolling windows). Step-by-Step Solution: Define partitioning and common strategies (range/list/hash).Map benefits: pruning → efficiency; per-partition policies → security; distribution → load balancing.Conclude that the statement is valid. Verification / Alternative check: Most major DBMSs document these advantages and provide partition-wise operations for performance. Why Other Options Are Wrong: Benefits are not confined to backups, single nodes, a specific partition method, or indexes alone. Common Pitfalls: Poor partition key choice leading to skew; too many tiny partitions increasing overhead. Final Answer: Valid — these are recognized benefits of partitioning

Physical Database Design Questions