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COA/Phase 3 · Control Unit

Chapter review LI-3.2 · Item 40 of 87

Microprogrammed control

Read at your own pace. Try the questions before opening the answers.

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ACompressed recap

🧾

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

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