Practice Quiz

Exam pattern: 1 small (5-mark) + 2 big problems. Solve on paper first, then reveal. Use the Calculators to check your numbers.

Type A — Small (5-mark)

A1smallCPU clock rate 2.5 GHz. (a) Cycle time? (b) Program needs 5×10⁹ cycles — execution time?

(a)

T = \frac{1}{2.5 \times 10^{9}}

= 0.4 ns.
(b) Time =

\frac{5 \times 10^{9}}{2.5 \times 10^{9}}

= 2 s.

A2small10⁶ instructions, average CPI 3.5, 700 MHz machine. Execution time?

\text{Cycles} = 10^{6} \times 3.5 = 3.5 \times 10^{6} \qquad \text{Time} = \frac{3.5 \times 10^{6}}{700 \times 10^{6}}

5 ms.

A3smallMix: 50% loads (CPI 5), 30% ALU (CPI 1), 20% branches (CPI 2). Average CPI?

0.5 \cdot 5 + 0.3 \cdot 1 + 0.2 \cdot 2 = 2.5 + 0.3 + 0.4

= 3.2.

A4smallX runs a program in 8 s; Y in 12 s. How many times faster is X than Y?

\frac{\text{Time}_Y}{\text{Time}_X} = \frac{12}{8}

= 1.5× faster.

A5smallProgram runs 20 s; new compiler cuts IC to 0.7× but raises CPI 1.2×. New run time?

20 \times 0.7 \times 1.2

= 16.8 s.

B1bigProgram runs 12 s on a 3 GHz machine. A redesign runs it in 8 s but needs 1.4× as many cycles. Required clock rate?

\begin{aligned} \text{Cycles}_{\text{old}} &= 12 \times 3 \times 10^{9} = 36 \times 10^{9} \\ \text{Cycles}_{\text{new}} &= 1.4 \times 36 \times 10^{9} = 50.4 \times 10^{9} \\ \text{Rate} &= \frac{50.4 \times 10^{9}}{8} \end{aligned}

6.3 GHz.

B2bigSame ISA. P: cycle 200 ps, CPI 2.5. Q: cycle 400 ps, CPI 1.0. Which is faster, by how much?

\text{Time}_P = \text{IC} \cdot 2.5 \cdot 200 = 500 \cdot \text{IC} \qquad \text{Time}_Q = \text{IC} \cdot 1.0 \cdot 400 = 400 \cdot \text{IC}

Q is faster, by

500/400 = 1.25\times

B3bigCPI A/B/C = 1/2/4. Seq 1 = (A=5,B=2,C=1); Seq 2 = (A=2,B=2,C=3). Same machine — which is faster, by how much, and each CPI?

Seq 1: IC = 8, cycles =

5 \cdot 1 + 2 \cdot 2 + 1 \cdot 4 = 13

, CPI = 1.625.
Seq 2: IC = 7, cycles =

2 \cdot 1 + 2 \cdot 2 + 3 \cdot 4 = 18

, CPI ≈ 2.57.
Seq 1 is faster by

18/13 \approx 1.38\times

B4big(MIPS pitfall) 2 GHz, CPI A/B/C = 1/2/3. Comp 1 (billions): A=4,B=2,C=1. Comp 2: A=8,B=1,C=1. Compare by MIPS and by execution time.

Comp 1: IC =

7 \times 10^{9}

, cycles =

11 \times 10^{9}

, time = 5.5 s, MIPS ≈ 1273.
Comp 2: IC =

10 \times 10^{9}

, cycles =

13 \times 10^{9}

, time = 6.5 s, MIPS ≈ 1538.
By MIPS Comp 2 wins; by execution time Comp 1 wins. MIPS misleads.

B5bigProgram runs 200 s; memory ops take 120 s. How much faster must memory be for (a) 2× faster, (b) 3× faster overall?

Unaffected = 80 s.

\text{New} = 80 + \frac{120}{x}

(a) target 100 s:

\frac{120}{x} = 20

\Longrightarrow

x = 6.
(b) target 66.67 s < 80 s floor

\Longrightarrow

impossible (max =

200/80 = 2.5\times

B6big(Reverse Amdahl) Vector instructions made 10× faster; program then runs 4× faster overall. What fraction f was vector work?

(1 - f) + \frac{f}{10} = 0.25 \;\Longrightarrow\; 1 - 0.9f = 0.25

f \approx 0.833

(83.3%).

B7bigProgram spends 40% of time on FP; FP sped up 8×. Overall speedup?

\frac{1}{0.6 + \frac{0.4}{8}} = \frac{1}{0.65}

= ≈ 1.538×.

Exam-day checklist
• Watch units (ps/ns/MHz/GHz). • "Same ISA / same program"

\Longrightarrow

IC cancels. • Always compare by execution time. • Amdahl:

\text{New} = \text{Unaffected} + \frac{\text{Affected}}{r}

; if target < unaffected

\Longrightarrow

impossible. • Master equation

\text{CPU time} = \text{IC} \times \text{CPI} \times \frac{1}{\text{clock rate}}

solves almost everything.