NIMCET Organization Of A Computer Previous Year Questions (PYQs)
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Debugger is a program that
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Solution
A debugger is a tool used to test and debug programs. It allows programmers to pause execution at specific points (breakpoints), run code step-by-step, and inspect variables and registers to find logic or runtime errors.
Hence, the correct option is:
✅ Option 3: allows to set breakpoints, execute a segment of program, and display contents of registers.
Dynamic RAM (DRAM) stores each bit of data in a separate capacitor. Due to
leakage, the stored charge tends to dissipate over time and needs to be refreshed
periodically. Consider the following statements:
P: DRAM requires refreshing because it uses capacitors to store bits.
P: DRAM requires refreshing because it uses capacitors to store bits.
Q: SRAM does not require refreshing because it uses flip-flops instead of
capacitors.
R: DRAM is faster than SRAM because it needs less frequent access.
S: DRAM is more suitable for main memory than SRAM due to its density.
R: DRAM is faster than SRAM because it needs less frequent access.
S: DRAM is more suitable for main memory than SRAM due to its density.
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Solution
- P: ✔ True — DRAM cells use capacitors that leak charge and hence must be refreshed periodically.
- Q: ✔ True — SRAM uses flip-flops, so it retains data without refreshing.
- R: ✗ False — DRAM is actually slower than SRAM.
- S: ✔ True — DRAM is denser and cheaper, making it ideal for main memory.
✅ Correct statements: P, Q, and S
In the design of a control unit of a processor, two common approaches are used:
hardware control and microprogrammed control. Consider the following
statements:
- Hardware control units are generally faster but more difficult to modify than microprogrammed control units.
- In a horizontal microprogrammed control unit, each control signal has a separate bit in the control word.
- Vertical microprogramming leads to longer control words but provides greater parallelism.
- Microprogrammed control units are typically easier to implement and modify than hardware control units.
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Solution
- Statement 1 – True: Hardware control units are implemented using combinational logic circuits. They generate control signals directly through hardware, making them faster but difficult to modify since any change requires redesigning the hardware.
- Statement 2 – True: In a horizontal microprogrammed control unit, each control signal is represented by a dedicated bit in the control word. This allows maximum parallelism but results in a wide (long) control word.
- Statement 3 – False: In vertical microprogramming, control words use encoded fields to represent control signals. This makes the word shorter, not longer, but reduces parallelism since only a few signals can be activated at once.
- Statement 4 – True: Microprogrammed control units are easier to implement and modify because control logic is stored in a control memory (microprograms). Modifications can be done by changing microinstructions instead of redesigning hardware.
✅ Correct statements: 1, 2, and 4
Consider a system with a CPU having 6 registers and 32-bit instructions. The
maximum possible size of the main memory is 512 KB (1K = 210). Each instruction
takes two registers and one memory address as operands. Which one of the
following correctly gives the maximum possible distinct instructions that can be
there in the instruction set of the CPU?
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Solution
Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.
Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.
Total operand bits $=19+6=25$.
Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.
Match all items in Group 1 with correct options from those given in Group 2.
| GROUP 1 | GROUP 2 |
| P. Intermediate representation | 1. Activation records |
| Q. Top-down parsing | 2. Code generation |
| R. Runtime environments | 3. Leftmost derivation |
| S. Register allocation | 4. Graph colouring |
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Solution
| Group 1 | Group 2 (Match) | Explanation |
|---|---|---|
| P. Intermediate representation | → 2. Code generation | Intermediate representation (IR) is the form of code produced by the front end of a compiler and used as input to the code generation phase. |
| Q. Top-down parsing | → 3. Leftmost derivation | Top-down parsers construct a parse tree from the root using leftmost derivations. |
| R. Runtime environments | → 1. Activation records | Activation records (stack frames) are part of the runtime environment used to manage function calls and local variables. |
| S. Register allocation | → 4. Graph colouring | Register allocation is commonly implemented using graph-colouring algorithms to assign variables to CPU registers efficiently. |
✅ Final Matching:
P – 2, Q – 3, R – 1, S – 4
Consider a system running under two types of workloads:
(a) CPU-intensive jobs,
(b) I/O-intensive jobs. Which of the following statements about the relative
performance of Interrupt-driven I/O and Programmed I/O is correct?
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Solution
(4) — Interrupt-driven I/O frees the CPU for compute-heavy jobs; under very I/O-intensive loads, polling (programmed I/O) can outperform by avoiding high interrupt overhead.
The time required for fetching and execution of one
simple machine instruction is known as
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Solution
Index register in a digital computer is used for:
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Solution
An index register is mainly used to modify the effective address during instruction execution, especially in array and loop processing.
Consider the program below which uses
six temporary variables a, b, c, d, e, and f.
Assuming that all the above operations take their operands from registers, the minimum number of registers needed to execute this program without spilling is
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Solution
Detailed Register Allocation Analysis
Objective: Determine the minimum number of registers needed to execute the program without spilling.
Live Range Analysis:
- a: Line 1 → 4
- b: Line 6 → 7
- c: Line 3 → 6
- d: Line 8 → 9
- e: Line 7 → 8
- f: Line 5 → 9
Max live variables: 3 (after lines 6 and 7)
✅ Final Answer: 3 registers are required to execute the program without spilling.
Which
of the following components is not a part of an instruction formation in CPU
processing?
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Solution
Instruction Formation in CPU Processing
Question: Which of the following is not a part of instruction formation?
Options:
- Opcode
- Register file
- Source operand
- Destination operand
✅ Correct Answer: Register File
Explanation:
- Opcode, Source operand, Destination operand — all are part of the instruction format.
- Register File — a hardware structure that stores registers, but it is not encoded into the instruction.
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