Worked Examples All 11 examples from the slides, fully solved. Try each before revealing — that's where the real learning happens.
Clock & Cycles WE-1 small Clock frequency = 200 MHz. Find the cycle time.
Show solution T = 1 F = 1 200 × 10 6 = 5 × 10 − 9 s T = \frac{1}{F} = \frac{1}{200 \times 10^{6}} = 5 \times 10^{-9}\,\text{s} T = F 1 = 200 × 1 0 6 1 = 5 × 1 0 − 9 s 5 nanoseconds .
WE-2 big A program runs in 10 s on computer A (4 GHz). Computer B should run it in 6 s, but needs 1.2× as many clock cycles. What clock rate should B target?
Show solution Cycles A = 10 × 4 × 10 9 = 40 × 10 9 Cycles B = 1.2 × 40 × 10 9 = 48 × 10 9 Rate B = 48 × 10 9 6 \begin{aligned} \text{Cycles}_A &= 10 \times 4 \times 10^{9} = 40 \times 10^{9} \\ \text{Cycles}_B &= 1.2 \times 40 \times 10^{9} = 48 \times 10^{9} \\ \text{Rate}_B &= \frac{48 \times 10^{9}}{6} \end{aligned} Cycles A Cycles B Rate B = 10 × 4 × 1 0 9 = 40 × 1 0 9 = 1.2 × 40 × 1 0 9 = 48 × 1 0 9 = 6 48 × 1 0 9 8 GHz .
CPI & Comparing Machines WE-3 big Same ISA. A: cycle 250 ps, CPI 2.0. B: cycle 500 ps, CPI 1.2. Which is faster, by how much?
Show solution Same ISA
⟹ \Longrightarrow ⟹ same IC.
Time A = IC × 2.0 × 250 = 500 ⋅ IC ps Time B = IC × 1.2 × 500 = 600 ⋅ IC ps \begin{aligned} \text{Time}_A &= \text{IC} \times 2.0 \times 250 = 500 \cdot \text{IC}\ \text{ps} \\ \text{Time}_B &= \text{IC} \times 1.2 \times 500 = 600 \cdot \text{IC}\ \text{ps} \end{aligned} Time A Time B = IC × 2.0 × 250 = 500 ⋅ IC ps = IC × 1.2 × 500 = 600 ⋅ IC ps A is faster, by 600 / 500 = 1.2 × 600/500 = 1.2\times 600/500 = 1.2 × WE-4 small 30 ADD (CPI 2) and 10 MUL (CPI 4). Find total cycles and average CPI.
Show solution Cycles = 30 × 2 + 10 × 4 = 100 IC = 40 \text{Cycles} = 30 \times 2 + 10 \times 4 = 100 \qquad \text{IC} = 40 Cycles = 30 × 2 + 10 × 4 = 100 IC = 40 Average CPI =
100 40 \frac{100}{40} 40 100 =
2.5 .
WE-5 big Classes A,B,C have CPI 1,2,3. Seq 1 = (A=2,B=1,C=2); Seq 2 = (A=4,B=1,C=1). Which is faster, by how much, and the CPI of each?
Show solution Seq IC Cycles CPI 1 5 2 ⋅ 1 + 1 ⋅ 2 + 2 ⋅ 3 = 10 2 \cdot 1 + 1 \cdot 2 + 2 \cdot 3 = 10 2 ⋅ 1 + 1 ⋅ 2 + 2 ⋅ 3 = 10 2.0 2 6 4 ⋅ 1 + 1 ⋅ 2 + 1 ⋅ 3 = 9 4 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 9 4 ⋅ 1 + 1 ⋅ 2 + 1 ⋅ 3 = 9 1.5
Seq 2 is faster by 10 / 9 ≈ 1.11 × 10/9 \approx 1.11\times 10/9 ≈ 1.11 × Note: Seq 2 has MORE instructions yet still wins.
WE-6 small A Java app runs 15 s. A new compiler needs 0.6× as many instructions but raises CPI by 1.1×. New run time?
Show solution Time scales with
IC × CPI \text{IC} \times \text{CPI} IC × CPI :
15 × 0.6 × 1.1 15 \times 0.6 \times 1.1 15 × 0.6 × 1.1 =
9.9 s .
The MIPS Pitfall WE-7 big 4 GHz. CPI A/B/C = 1/2/3. Compiler 1 → (A=5,B=1,C=1) billions; Compiler 2 → (A=10,B=1,C=1). Faster by MIPS? By execution time?
Show solution IC Cycles Time MIPS Comp 1 7 × 10 9 7 \times 10^{9} 7 × 1 0 9 10 × 10 9 10 \times 10^{9} 10 × 1 0 9 2.5 s 2800 Comp 2 12 × 10 9 12 \times 10^{9} 12 × 1 0 9 15 × 10 9 15 \times 10^{9} 15 × 1 0 9 3.75 s 3200
By MIPS: Comp 2 (3200) looks faster. By time: Comp 1 (2.5 s) is actually faster. The classic MIPS pitfall — they disagree. Trust execution time.
Amdahl's Law WE-8 small Fraction f of time enhanced by factor r. Prove Speedup = 1/(1 − f + f/r).
Show solution Old =
t t t . Affected =
f ⋅ t f \cdot t f ⋅ t , Unaffected =
( 1 − f ) t (1 - f)\,t ( 1 − f ) t .
New = ( 1 − f ) t + f ⋅ t r Speedup = t New \text{New} = (1 - f)\,t + \frac{f \cdot t}{r} \qquad \text{Speedup} = \frac{t}{\text{New}} New = ( 1 − f ) t + r f ⋅ t Speedup = New t Speedup = 1 1 − f + f r \text{Speedup} = \frac{1}{1 - f + \frac{f}{r}} Speedup = 1 − f + r f 1 ∎
WE-9 big Program runs 100 s; multiply takes 80 s. How much must multiply speed up to make the program (a) 4× faster, (b) 5× faster?
Show solution Unaffected = 20 s.
New time = 20 + 80 x \text{New time} = 20 + \frac{80}{x} New time = 20 + x 80 (a) target = 25 s:
25 = 20 + 80 x 25 = 20 + \frac{80}{x} 25 = 20 + x 80 ⟹ \Longrightarrow ⟹ x = 16 .
(b) target = 20 s:
80 x = 0 \frac{80}{x} = 0 x 80 = 0 ⟹ \Longrightarrow ⟹ impossible (max speedup =
100 / 20 = 5 × 100/20 = 5\times 100/20 = 5 × , only at
x = ∞ x = \infty x = ∞ ).
WE-10 big All FP instructions made 5× faster. Benchmark was 10 s, half spent on FP. Find the overall speedup.
Show solution FP =
5 s → 1 s 5\,\text{s} \to 1\,\text{s} 5 s → 1 s ; other = 5 s. New = 6 s. Speedup =
10 6 \frac{10}{6} 6 10 =
1.667 .
WE-11 big All FP instructions made 5× faster. Program runs 100 s; new hardware gives overall speedup 3. How much of the original time was FP?
Show solution FP =
x x x .
New = 100 − x + x 5 3 = 100 100 − 0.8 x ⟹ 100 − 0.8 x = 33.33 \text{New} = 100 - x + \frac{x}{5} \qquad 3 = \frac{100}{100 - 0.8x} \;\Longrightarrow\; 100 - 0.8x = 33.33 New = 100 − x + 5 x 3 = 100 − 0.8 x 100 ⟹ 100 − 0.8 x = 33.33 x ≈ 83.33 s x \approx 83.33\,\text{s} x ≈ 83.33 s