Worked Examples

All 11 examples from the slides, fully solved. Try each before revealing — that's where the real learning happens.

Clock & Cycles

WE-1smallClock frequency = 200 MHz. Find the cycle time.

WE-2bigA 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?

CPI & Comparing Machines

WE-3bigSame ISA. A: cycle 250 ps, CPI 2.0. B: cycle 500 ps, CPI 1.2. Which is faster, by how much?

WE-4small30 ADD (CPI 2) and 10 MUL (CPI 4). Find total cycles and average CPI.

WE-5bigClasses 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?

Seq	IC	Cycles	CPI
1	5	2·1+1·2+2·3 = 10	2.0
2	6	4·1+1·2+1·3 = 9	1.5

WE-6smallA Java app runs 15 s. A new compiler needs 0.6× as many instructions but raises CPI by 1.1×. New run time?

The MIPS Pitfall

WE-7big4 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?

	IC	Cycles	Time	MIPS
Comp 1	7×10⁹	10×10⁹	2.5 s	2800
Comp 2	12×10⁹	15×10⁹	3.75 s	3200

Amdahl's Law

WE-8smallFraction f of time enhanced by factor r. Prove Speedup = 1/(1 − f + f/r).

WE-9bigProgram runs 100 s; multiply takes 80 s. How much must multiply speed up to make the program (a) 4× faster, (b) 5× faster?

WE-10bigAll FP instructions made 5× faster. Benchmark was 10 s, half spent on FP. Find the overall speedup.

WE-11bigAll FP instructions made 5× faster. Program runs 100 s; new hardware gives overall speedup 3. How much of the original time was FP?