This is Set III, the last of our three Computer Organization and Architecture (COA) MCQ sets. It moves from single micro-operations to the whole basic computer: its registers, the instruction cycle, instruction formats, addressing modes, interrupts and the two ways of building a control unit.
The register names and sizes follow Morris Mano's basic computer, which has a 4096-word memory of 16 bits per word. Where a question depends on that design, it says so, so each question can be read on its own.
How to use this set: pick an option for each question. You will see straight away whether it is right, with a short note on why. There are 15 questions.
What this set covers
- Registers of the basic computer: PC, AR, DR, AC, IR, TR
- The fetch, decode and execute cycle
- Instruction formats and opcode sizing
- Direct, indirect, immediate, relative and indexed addressing
- Interrupt cycle
- Hardwired versus microprogrammed control
Computer Organization and Architecture MCQs with answers and explanations
Q1. Which register holds the address of the next instruction to be fetched?
a) Program counter (PC)
b) Address register (AR)
c) Accumulator (AC)
d) Instruction register (IR)
Q2. In Mano's basic computer, memory has 4096 words of 16 bits. Why are PC and AR 12 bits wide while AC, DR and IR are 16 bits wide?
a) PC and AR hold addresses, and 12 bits address 4096 words; the others hold full data words or instructions
b) 16-bit registers can only be used for arithmetic
c) PC and AR store only the opcode of an instruction
d) PC and AR are slower registers, so they are made smaller
Q3. What is the first micro-operation of the fetch phase in the basic computer?
a) AR ← PC
b) AC ← DR
c) PC ← PC + 1
d) IR ← M[AR]
Q4. Which statement correctly describes the second timing step of the fetch, T1: IR ← M[AR], PC ← PC + 1?
a) The instruction is decoded and executed
b) AR is loaded with the address part of IR
c) IR is incremented and PC is read from memory
d) The instruction is read into IR and PC is incremented in the same clock pulse
Q5. In Mano's basic computer, a memory-reference instruction is 16 bits: 1 mode bit (I), 3 opcode bits and 12 address bits. How many memory-reference instructions are possible?
a) 16
b) 8
c) 7
d) 12
Q6. In the basic computer, which register is used for temporary storage during processing and is called TR?
a) Transfer register
b) Trap register
c) Temporary register
d) Timing register
Q7. A computer has a memory of 64K words and uses 24-bit instructions with one mode bit, an opcode and a direct address field. What is the largest number of opcodes it can have?
a) 64
b) 128
c) 16
d) 256
Q8. In which addressing mode is the operand itself part of the instruction?
a) Immediate
b) Register indirect
c) Indirect
d) Direct
Q9. In indirect addressing, the address field of the instruction gives:
a) An offset to be added to the program counter
b) The address of a memory word that holds the address of the operand
c) The number of the register that holds the operand
d) The operand itself
Q10. A two-word instruction is stored at addresses 200 and 201. Its address field is 500 and it uses relative addressing. What is the effective address?
a) 701
b) 702
c) 500
d) 700
Q11. An instruction uses indexed addressing. Its address field is 500 and the index register holds 100. What is the effective address?
a) 500
b) 400
c) 600
d) 100
Q12. In the basic computer, when is the interrupt cycle entered?
a) Whenever the program counter overflows
b) In the middle of decoding an instruction
c) When the interrupt flip-flop R is 1 at the end of an instruction cycle
d) Only when the accumulator becomes zero
Q13. In Mano's basic computer, what does the interrupt cycle do with the return address?
a) Discards it, because the interrupted program restarts from address 0
b) Stores it at memory address 0 and sets PC to 1
c) Copies it into AC
d) Pushes it onto a hardware stack and jumps to address 4096
Q14. Which statement correctly compares hardwired and microprogrammed control?
a) Microprogrammed control cannot implement complex instructions
b) Microprogrammed control is always faster than hardwired control
c) Hardwired control is generally faster; microprogrammed control is easier to change
d) Hardwired control stores control words in a control memory
Q15. In a microprogrammed control unit, what is held in the control memory?
a) The operating system kernel
b) The user's program and data
c) The page table for virtual memory
d) Microinstructions that specify the control signals for each step
Quick revision notes
- Basic computer registers: PC (12), AR (12), DR (16), AC (16), IR (16), TR (16), plus INPR and OUTR (8 bits each).
- Fetch:
T0: AR ← PC,T1: IR ← M[AR], PC ← PC + 1, then decode at T2. - Opcode bits = instruction length minus address bits minus mode bits. Number of opcodes = 2 raised to that.
- Effective address: direct = address field; indirect = M[address]; relative = PC + address; indexed = XR + address; immediate has no memory read for the operand.
- Hardwired control is fast but rigid. Microprogrammed control is flexible and easier to design for large instruction sets.
Frequently asked questions
Why is PC incremented during fetch and not after execution?
Incrementing during fetch uses a clock pulse when PC is otherwise idle, and it means a branch instruction can simply overwrite PC during execution. If the increment came later, it would undo the branch.
Which processors use microprogrammed control today?
Many x86 processors use microcode for complex instructions while handling simple ones with hardwired logic. Most RISC designs such as ARM and RISC-V cores rely mainly on hardwired control.
How should I prepare addressing mode questions for exams?
Learn the effective address formula for each mode and practise with numbers, as in the relative and indexed questions above. Remember that in relative mode PC has already moved past the current instruction.
Comments
Post a Comment