ACompressed recap
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Microprogrammed control: each step's control word stored in control memory; microroutine per instruction; shared fetch routine; µPC; opcode mapping to the start address; End returns to fetch. CM words = fetch + Σ execute steps; CM bits = words × width. Horizontal: one bit per signal (fast, wide). Vertical: field of k mutually exclusive signals = ⌈log₂(k+1)⌉ bits (⌈log₂ k⌉ if always one active). Sequencing: condition select ⌈log₂ #conditions⌉ bits; next address ⌈log₂ CM words⌉ bits; width = control + select + address. Nanoprogramming: N⌈log₂ D⌉ + DW. Hardwired is fast and rigid; microprogrammed is flexible and slower.
BMerged rule card
3.2.1 Microprogrammed control: control memory, control words, microroutines, µPC
- Control word = all signals of one step; microinstruction = stored control word (+ sequencing).
- Microroutine per instruction; fetch routine shared; µPC + starting-address generator; End → fetch.
- CM words = fetch + Σ execute steps; CM bits = words × width.
- Microprogrammed: flexible, slower, CISC-friendly.
3.2.2 Microinstruction formats: horizontal vs vertical, field encoding, control-memory size
- Horizontal: width = number of signals; fast; parallel; big CM.
- Vertical: field of k mutually exclusive signals = ⌈log₂(k+1)⌉ bits (⌈log₂ k⌉ if one is always active).
- Width = Σ fields (+ independent signals) (+ condition select + next address).
- CM bits = words × width.
3.2.3 Microprogram sequencing: next-address field, conditional branches; hardwired vs microprogrammed
- Next microinstruction: µPC + 1; branch to next-address field if condition true; opcode mapping; End → fetch.
- Width = control bits + ⌈log₂ #conditions⌉ + ⌈log₂ CM words⌉; CM words = 2^(next-address bits).
- Nanoprogramming: N·⌈log₂ D⌉ + D·W vs N·W.
- Hardwired: fast, rigid, RISC. Microprogrammed: flexible, slower, CISC.
CInterleaved GATE-style questions
Q1NAT2 marksCM bits
A CPU: fetch 3 microinstructions; 50 instructions with 4 execute microinstructions each; 70 instructions with 6 each. 32-bit microinstructions. Control memory bits?
Numerical answer: ________
Q2NAT1 markHorizontal
A datapath has 96 control signals. Horizontal control-field width?
Numerical answer: ________
Q3NAT2 marksVertical
Mutually exclusive groups of sizes 7, 7, 15 and 3 (idle allowed). Vertical width?
Numerical answer: ________
Q4NAT2 marksField widths
A 24-bit microinstruction has a 12-bit control field and selects one of 4 conditions. How many words can the control memory hold at most?
Numerical answer: ________
Q5MCQ1 markSpeed
Which microinstruction format is fastest for the same datapath?
- (A) All the same
- (B) Vertical
- (C) Horizontal
- (D) Extreme vertical
Q6NAT2 marksWidth
Control field 33 bits; MUX of 7 status bits; control memory of 300 words. Minimum width?
Numerical answer: ________
Q7MSQ2 marksFacts
Which are true?
- (A) The opcode is mapped to the start of its microroutine.
- (B) Each machine instruction has its own fetch microroutine.
- (C) The microprogram is stored in control memory.
- (D) The µPC normally increments after each microinstruction.
Q8NAT2 marksNano
N = 4,096 microinstructions, W = 100 bits, D = 128 distinct words. Bits saved by nanoprogramming?
Numerical answer: ________
DAnswer Vault
Answer Vault
Try the questions first. Open when you are ready to check.
Q1 Answer: 199362 marks
(3 + 200 + 420) × 32 = 623 × 32 = 19,936.
Re-read: Lesson 3.2.1 — Microprogrammed control: control memory, control words, microroutines, µPC
Q2 Answer: 961 mark
One bit per signal.
Re-read: Lesson 3.2.2 — Microinstruction formats: horizontal vs vertical, field encoding, control-memory size
Q3 Answer: 122 marks
⌈log₂ 8⌉ + ⌈log₂ 8⌉ + ⌈log₂ 16⌉ + ⌈log₂ 4⌉ = 3 + 3 + 4 + 2 = 12.
Re-read: Lesson 3.2.2 — Microinstruction formats: horizontal vs vertical, field encoding, control-memory size
Q4 Answer: 10242 marks
24 − 12 − 2 = 10 bits.
Re-read: Lesson 3.2.3 — Microprogram sequencing: next-address field, conditional branches; hardwired vs microprogrammed
Q5 Answer: (C)1 mark
No field decoding.
- (A) is wrong: Decoding adds delay.
- (B) is wrong: Needs decoders.
- (D) is wrong: Needs the most decoding and allows least parallelism.
Re-read: Lesson 3.2.2 — Microinstruction formats: horizontal vs vertical, field encoding, control-memory size
Q6 Answer: 452 marks
33 + 3 + 9 = 45.
Re-read: Lesson 3.2.3 — Microprogram sequencing: next-address field, conditional branches; hardwired vs microprogrammed
Q7 Answer: (A) (C) (D)2 marks
Fetch is shared.
- (B) is wrong: One fetch routine serves all instructions.
Re-read: Lesson 3.2.1 — Microprogrammed control: control memory, control words, microroutines, µPC
Q8 Answer: 3681282 marks
One level: 4,096 × 100 = 409,600. Two levels: 4,096 × 7 + 128 × 100 = 28,672 + 12,800 = 41,472. Saved: 368,128.
Re-read: Lesson 3.2.3 — Microprogram sequencing: next-address field, conditional branches; hardwired vs microprogrammed