Performance: clock, cycle time, CPI, execution time, MIPS, speedup, Amdahl's law

0 · Recall Warm-up

Answer from memory first. Answers are in the Answer Vault at the bottom.

• What does the program counter hold? from Lesson 0.1.1
• Which bus is one-way? from Lesson 0.1.1
• What makes a Harvard machine different from a von Neumann machine? from Lesson 0.1.1

1 · The Hook

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Two processors run the same program. One has a 3 GHz clock, the other 2 GHz. The 2 GHz chip finishes first. How? Clock speed is only one of three numbers that decide speed. This lesson gives the formula that every GATE performance question (cache, pipeline, I/O) reduces to.

2 · Predict First

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A program executes 2 × 10⁹ instructions. Processor X runs at 3 GHz and needs on average 3 clock cycles per instruction. Processor Y runs at 2 GHz and needs 1.5 cycles per instruction. Which finishes first, and how long does each take?

Make a guess in your head. The answer opens Section 3.

3 · Core Concept

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Predict First — the answer

X: 2×10⁹ × 3 / (3×10⁹) = 2 s. Y: 2×10⁹ × 1.5 / (2×10⁹) = 1.5 s. Y wins even with the slower clock, because it needs fewer cycles per instruction.

The clock

Every step inside the CPU is timed by a clock: a signal that ticks at a fixed rate.

• Clock rate (frequency) f: ticks per second, in Hz. 1 GHz = 10⁹ Hz.
• Clock cycle time (period) T = 1 / f. 2 GHz → T = 0.5 ns. 500 MHz → T = 2 ns.

CPI

Different instructions need different numbers of cycles. CPI (cycles per instruction) is the average number of cycles per instruction for a program.

With an instruction mix, CPI is a weighted average:

CPI = Σ (fraction of instructions of class i) × (cycles for class i)

Example: 50% ALU (1 cycle), 30% load/store (3 cycles), 20% branch (2 cycles) → CPI = 0.5×1 + 0.3×3 + 0.2×2 = 1.8.

The performance equation

CPU time = IC × CPI × T = IC × CPI / f

where IC is the instruction count (instructions actually executed, not lines of code). Every design trick in this course lowers one of the three factors:

• better instructions (Phase 1) change IC and CPI,
• a faster datapath or deeper pipeline (Phases 2, 5) changes T and CPI,
• caches (Phase 4) cut the cycles lost waiting for memory, which lowers CPI.

MIPS

MIPS (million instructions per second) = f / (CPI × 10⁶). A 2 GHz CPU with CPI 1.6 runs at 2×10⁹ / 1.6 / 10⁶ = 1,250 MIPS. MIPS ignores IC, so a machine with a higher MIPS rating can still be slower if it needs more instructions for the same job.

Speedup and Amdahl's law

Performance = 1 / execution time. Speedup of B over A = time(A) / time(B).

If an improvement speeds up only a fraction F of the original execution time by a factor S, the rest is untouched:

New time = Old time × [(1 − F) + F / S] → Overall speedup = 1 / [(1 − F) + F / S]

This is Amdahl's law. Even with S → ∞ the speedup cannot exceed 1 / (1 − F): the part you did not improve sets the limit.

Example: 40% of the time is floating-point work; a new unit makes it 10× faster. Speedup = 1 / (0.6 + 0.04) = 1 / 0.64 = 1.5625. Limit with an infinitely fast unit: 1 / 0.6 ≈ 1.67.

4 · Visual

CPU time = IC        x   CPI            x   T
           (how many     (cycles each,      (seconds per cycle
           instructions)  on average)        = 1/f)

Amdahl: old time  |------ 1-F ------|--- F ---|
        new time  |------ 1-F ------|-F/S-|        speedup = 1 / ((1-F) + F/S)

5 · Worked Examples

Example 1 Easy

A 2.5 GHz processor. What is its cycle time?

Solution

T = 1 / (2.5×10⁹) s = 0.4 ns.

Example 2 Medium

Instruction mix: 40% ALU (1 cycle), 40% memory (4 cycles), 20% branch (2 cycles). Clock 1 GHz, IC = 5×10⁸. Find CPI and CPU time.

Solution

CPI = 0.4×1 + 0.4×4 + 0.2×2 = 0.4 + 1.6 + 0.4 = 2.4. Time = 5×10⁸ × 2.4 / 10⁹ = 1.2 s.

Example 3 GATE

A program spends 60% of its time in a part that is made 3 times faster. What is the overall speedup (2 decimals)?

Solution

1 / (0.4 + 0.6/3) = 1 / (0.4 + 0.2) = 1 / 0.6 = 1.67.

6 · GATE Traps

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• IC counts executed instructions: a loop body that runs 100 times counts 100 times.
• Cycle time is the reciprocal of frequency: doubling f halves T.
• Amdahl's F is a fraction of the original time, not of the instructions, unless the question says they are the same.
• MIPS can mislead: compare CPU time, never MIPS alone, when instruction counts differ.
• Speedup = old time / new time (bigger than 1 means faster).

7 · Memory Hook

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"Count × Cycles × Clock." CPU time = IC × CPI × T. And for Amdahl: "you can't speed up what you don't touch."

8 · Connect the Dots

Lesson 0.1.1 gave the loop; this lesson measured it: CPU time = IC × CPI × T. Next chapter builds the bit toolkit (binary, powers of two, signed numbers, gates, memory as an array) that every later calculation uses.

9 · Self-Test

GATE format. Try every question before opening the vault.

Q1NAT1 markCPU time

A program executes 4 × 10⁹ instructions with CPI 1.25 on a 2 GHz processor. CPU time in seconds?

Numerical answer: ________

Q2NAT2 marksCPI from mix

Instruction mix: 50% ALU (1 cycle), 20% load (5 cycles), 10% store (4 cycles), 20% branch (3 cycles). What is the CPI?

Numerical answer: ________

Q3NAT1 markMIPS

A 1.5 GHz processor has CPI 2. What is its MIPS rating?

Numerical answer: ________

Q4NAT2 marksAmdahl

An enhancement makes 80% of a program's execution time run 4 times faster. Overall speedup (2 decimals)?

Numerical answer: ________

Q5MCQ2 marksComparing machines

Machine A: 4 GHz, CPI 2.0, runs 10⁹ instructions for a program. Machine B: 3 GHz, CPI 1.2, needs 1.2 × 10⁹ instructions for the same program. Which is true?

• (A) A is faster by a factor of 1.0
• (B) A is faster by a factor 1.2
• (C) B is faster by a factor 1.04
• (D) B is faster by a factor 1.25

10 · Rule Card

• T = 1/f. CPU time = IC × CPI × T = IC × CPI / f.
• CPI of a mix = Σ fraction × cycles.
• MIPS = f / (CPI × 10⁶); compare machines by CPU time, not MIPS.
• Speedup = old time / new time. Amdahl: 1 / ((1−F) + F/S); upper limit 1/(1−F).

11 · Concept Ledger

Unlocked in this lesson. From here on, these terms may be used.

clock rate (frequency)clock cycle timeCPIinstruction mixinstruction count (IC)CPU time = IC × CPI × TMIPSspeedupAmdahl's law

12 · Answer Vault