ব্যাংক আইটি অফিসার প্রশ্ন ও সমাধান

Correct answer: (c) Block-level striping with distributed parity disks

RAID 5 stripes data across the disks in blocks, and in each stripe one block holds parity (the XOR of that stripe's data blocks). The parity block rotates from disk to disk, so parity is distributed over all members instead of living on one dedicated disk; this removes the parity-disk bottleneck on writes. If any one disk fails, each lost block is rebuilt as the XOR of the surviving blocks of its stripe.

Minimum 3 disks; usable capacity = (N − 1) × C. Example: 4 disks × 2 TB → (4 − 1) × 2 TB = 6 TB usable.

(The option says "distributed parity disks"; strictly the parity is spread across all the disks rather than stored on separate parity disks, but the meaning is clear.)

Why the other options are wrong

• (a) Bit-level striping with a dedicated parity disk: this is the older RAID 3 idea (bit/byte-interleaved data with one parity disk), not RAID 5.
• (b) Block-level striping with a dedicated parity disk: this is RAID 4; every write must update its single parity disk, the bottleneck RAID 5 removes by rotating parity.
• (d) Mirroring without parity: this is RAID 1.

Tip: RAID 3 = byte/bit + dedicated parity; RAID 4 = block + dedicated parity; RAID 5 = block + distributed parity; RAID 6 = block + double distributed parity.

Sources: Hardware RAID configuration on the IBM Power platform (IBM Developer); MIT 6.033 lecture slides (Spring 2017): reliability via replication and RAID; RAID (CS 4410, Summer 2015) (Cornell University)

Correct answer: (b) Can get stuck in local optima

Hill climbing is a greedy local search: from the current state it moves to a better neighbour and stops as soon as no neighbour is better. That stopping point may be only a local optimum, better than everything around it but worse than the global optimum. Because the method never backtracks or accepts a worse move, it has no way out once it is there. Plateaus (flat areas) and ridges cause the same kind of stall.

Why the other options are wrong

• (a) High memory requirement: hill climbing keeps only the current state and its neighbours under comparison, so it needs very little memory; that is its main advantage.
• (c) Expands all nodes: it follows a single path and never revisits alternatives; expanding every node is what exhaustive searches such as BFS do.
• (d) Requires heuristic functions: it does need an evaluation function to compare neighbours, but every informed search needs one; it is a requirement, not the method's main weakness.

Tip: Hill climbing: little memory, but stuck at local optima, plateaus and ridges; remedies are random restarts and simulated annealing.

Sources: NIST DADS: local optimum; Hill climbing (Wikipedia)

Correct answer: (b) Switch

Layer 2 of the OSI model is the data link layer, which deals with frames and MAC addresses. A (basic) switch reads the destination MAC address of each frame, looks it up in its MAC address (CAM) table and sends the frame only out of the matching port. That makes it the classic Layer 2 device, together with the bridge.

Why the other options are wrong

• (a) Router: Works at Layer 3 (network layer) and forwards packets by IP address.
• (c) Hub: Works at Layer 1 (physical layer); it repeats electrical signals to all ports without reading any address.
• (d) Modem: Converts between digital data and analog signals on the line, a physical-layer function.

Tip: Hub = L1, switch/bridge = L2 (MAC), router = L3 (IP). A "Layer 3 switch" is a switch with routing added.

Sources: Troubleshoot LAN Switching Environments - Cisco; RFC 1812: Requirements for IP Version 4 Routers

Correct answer: (b) (7B.6)₁₆

Taking (123.4) as a decimal number, convert the two parts separately.

Integer part (divide by 16, read remainders bottom-up):
123 ÷ 16 = 7 remainder 11 (B); 7 ÷ 16 = 0 remainder 7 → 7B₁₆ (check: 7×16 + 11 = 123)

Fraction part (multiply by 16, read integer parts top-down):
0.4 × 16 = 6.4 → 6; 0.4 × 16 = 6.4 → 6; … the 6 repeats for ever

So 123.4₁₀ = 7B.666…₁₆, which is 7B.6₁₆ to one hex place (rounding also gives 6, because the remainder 0.4 is below one half). Check: 7B.6₁₆ = 123 + 6/16 = 123.375.

Why the other options are wrong

• (a) (7B.4)₁₆: It copies the decimal digit 4 into the fraction; 0.4₁₀ is not 0.4₁₆ (0.4₁₆ = 4/16 = 0.25).
• (c) (123.6)₁₆: It copies the decimal integer digits; 123₁₆ = 291₁₀, not 123.
• (d) (7C.6)₁₆: The integer part is off by one: 7C₁₆ = 124.

Exam note: The stem does not print the base of (123.4); it must be decimal, the only reading that matches any option (as octal, 123.4₈ = 83.5₁₀ = 53.8₁₆). The exact hex value never ends (7B.666…₁₆); option (b) gives it to one hex place.

Tip: Integer part: divide by 16, read remainders upward. Fraction: multiply by 16, read integer parts downward.

Correct answer: (d) All of the above

For a coil of N turns, cross-sectional area A and length l, the self-inductance is L = μN²A ÷ l, where μ = μ₀μᵣ is the permeability of the core material. So inductance:

• rises with the square of the number of turns (double N → 4 × L);
• rises in proportion to the core permeability (a magnetic core of relative permeability μᵣ multiplies L by μᵣ compared with the same coil in air);
• rises in proportion to the cross-sectional area;
• falls as the coil is made longer for the same number of turns.

Since (a), (b) and (c) are all genuine factors, "All of the above" is the complete answer.

Why the other options are wrong

• (a) Number of turns in the coil: True (L ∝ N²), but it is only one of the factors.
• (b) Permeability of the core material: True (L ∝ μ), but incomplete on its own.
• (c) Cross-sectional area of the coil: True (L ∝ A), but incomplete on its own.

Tip: L = μN²A ÷ l: more turns, a magnetic core or a fatter coil raise L; a longer coil lowers it.

Source: What Is Inductance? – Keysight Oscilloscope Glossary

Correct answer: (d) Using global variables

Encapsulation keeps an object's data together with the methods that work on it, and hides that data so outside code can reach it only through those methods. A global variable does the opposite: it belongs to no object, and any function anywhere in the program can read or change it directly, with no checks. It gives neither bundling nor hiding.

Why the other options are wrong

• (a) Using private access specifier: private limits a member to its own class. This is the main tool for data hiding.
• (b) Using classes in object-oriented programming: a class puts fields and the methods that operate on them into one unit. That is the bundling half of encapsulation.
• (c) Using getter and setter methods: they are the controlled path to private fields and can validate values (for example, reject a negative age). That is controlled access.

Tip: encapsulation = bundle + hide + controlled access. Anything that lets every part of the program change data freely breaks it. (Java has no true global variables; public static fields come closest and weaken encapsulation in the same way.)

Source: What Is an Object? (The Java Tutorials, Oracle)

Correct answer: (c) FROM

A query is written SELECT … FROM … WHERE … ORDER BY, but it is evaluated in a different logical order. The engine first needs the rows to work on, so the FROM clause (with its joins) is processed first. Then WHERE filters rows, GROUP BY and HAVING form and filter groups, SELECT computes the output columns, and ORDER BY sorts the result:

FROM → WHERE → GROUP BY → HAVING → SELECT → DISTINCT → ORDER BY

This is the logical (binding) order. The optimiser may choose a different physical plan, but the result must be the same as if this order were followed.

Why the other options are wrong

• (a) WHERE: It comes second; it filters the rows that FROM has produced, so it cannot run before them.
• (b) SELECT: It is evaluated after WHERE, GROUP BY and HAVING. That is why a column alias defined in SELECT cannot be used in WHERE in standard SQL.
• (d) ORDER BY: It comes last (only TOP/LIMIT/OFFSET follow it), which is why ORDER BY can use a SELECT alias.

Tip: Logical order starts with FROM and ends with ORDER BY, whatever the written order.

Sources: SELECT (Transact-SQL) - Microsoft Learn; SELECT - PostgreSQL Documentation; Problems with Column Aliases - MySQL 8.4 Reference Manual

Correct answer: (c) Heap

A priority queue must always hand out the item with the highest (or lowest) priority. A binary heap is a complete binary tree in which every parent's key is at least as extreme as its children's, so the top-priority item sits at the root.

• Find max/min: read the root → O(1)
• Insert: add at the last position and sift up along the height ⌊log₂ n⌋ → O(log n)
• Delete max/min: move the last item to the root and sift down → O(log n)

Why the other options are wrong

• (a) Array: Unsorted: insert O(1) but finding/removing the top item O(n). Sorted: removal is cheap but each insert costs O(n) shifting.
• (b) Linked List: Same trade-off — one of insert or delete-top is O(n).
• (d) Stack: LIFO — it returns the newest item, ignoring priority.

Tip: Priority queue ⇒ heap: O(log n) insert and delete, O(1) peek.

Sources: priority queue — NIST Dictionary of Algorithms and Data Structures; binary heap — NIST Dictionary of Algorithms and Data Structures

Correct answer: (b) RSA

A digital signature needs an asymmetric algorithm: the signer uses a private key to generate the signature over a hash of the message, and anyone verifies it with the matching public key. RSA is one of the signature algorithms approved in the U.S. Digital Signature Standard (FIPS 186-5), alongside ECDSA and EdDSA.

Why the other options are wrong

• (a) AES: a symmetric block cipher; both sides share the same key, so it cannot prove which of them produced a message.
• (c) DES: also a symmetric block cipher, with the same shared-key limitation.
• (d) Diffie-Hellman: a key-agreement (key-establishment) method used to set up a shared secret; it neither encrypts messages nor produces signatures.

Tip: Signature algorithms: RSA, DSA, ECDSA, EdDSA. Diffie–Hellman = key exchange.

Source: FIPS 186-5: Digital Signature Standard (DSS); RFC 4949: Internet Security Glossary, Version 2

Correct answer: (c) It is inflexible and not suitable for changing requirements

The waterfall model runs its phases strictly in order: one phase cannot start until the previous one is finished, and the requirements are expected to be completely specified at the end of the requirements phase. A requirement that changes later forces rework of design, code and tests already completed, and working software appears only near the end. That rigidity is its best-known drawback.

Why the other options are wrong

• (a) It is difficult to manage: the opposite; clear phases, milestones and deliverables make it one of the easiest models to plan and track.
• (b) It requires too many resources: resource use is not its defining problem; it needs no more staff than other models for the same scope.
• (d) It lacks proper documentation: the waterfall model is document-driven; every phase ends with a document.

Tip: waterfall = stable, well-understood requirements; changing requirements → iterative or Agile models.

Source: SWEBOK Guide V4.0a (IEEE Computer Society)