Another example problem:

Question: Computer A runs at 250 picoseconds (ps) per cycle and has CPI 2 cycles/instruction, and computer B runs at 500 ps/cycle and has CPI of 1.2 cycles/instruction. Which is faster and by how much?

Answer: Seconds per instruction of A = 2 cycles/instruction * 250 ps/cycle = 500 ps/instruction Seconds per instruction of B = 1.2 cycles/instruction * 500 ps/cycle = 600 ps/instruction Therefore, A is 600/500 = 1.2 times faster than B

Yet another example problem:

Question: Compiler designer is deciding between two code sequences when compiling to a specific PC architecture. He is given this information:

type of instruction|A|B|C -|-|-|- CPI|1|2|3

code sequence|A|B|C| -|-|-|-|- 1|2|1|2|number of inst. for code seq 1 2|4|1|1|number of inst. for code seq 2

How many total instructions for each code sequence? What is the total number of cycles for each code sequence? What is the CPI for each code sequence?

Answer: Total instructions code seq 1 = 2 instructions + 1 instruction + 2 instructions = 5 instructions Total instructions code seq 2 = 4 instructions + 1 instruction + 1 instruction = 6 instructions Total cycles for code seq 1 = 2 instructions * 1 cycle/instruction + 1 instruction * 2 cycles/instruction + 2 instructions * 3 cycles/instruction = 2 instr. + 2 instr. + 6 instr. = 10 instructions Total cycles for code seq 2 = 4 instructions * 1 cycle/instr + 1 instruction * 2 cycles/instr + 1 instruction * 3 cycles/instr = 9 instructions CPI of code seq 1 = 10 cycles / 5 instructions = 2 cycles/instruction CPI of code seq 2 = 9 cycles / 6 instructions = 1.5 cycles/instruction

Yet another example:

Question: Given the same table above for cycles per each instruction type A, B, C, we are also given the following information about the composition of a program into instruction types:

|A|B|C -|-|-|- proportion|50%|35%|15%

What is the total CPI of this program?

Answer: Total CPI = 0.5 * 1 cycle/instruction + 0.35 * 2 cycles/instruction + 0.15 * 3 cycles/instruction = 0.5 cycles/instruction + 0.7 cycles/instr + 0.45 cycles/instr = 1.65 cycles/instr

Power wall

This section is about the power wall.

The power wall describes the physical limitations that prevent us from making faster computers.

When we increase frequency, more heat is generated so the CPU overheats.

This is an issue because electrical components can get overheated - the transistors in the CPU will begin to misbehave when heat is increased.

A transistor is a small element with an incoming wire, outgoing wire, and one other "switch" wire that determines whether current flows through the incoming and outcoing wire.

There is no sufficiently efficient way to cool CPUs, so we cannot increase frequency.