# Chapter 1: Introduction

# Terminology

Bit - a binary digit

Byte - a group of 8 bits

Word - an fixed-size group of data

• Used to be 32 bits, is now 64 bits
• Dependent on the register size

Bus - an electrical line that transfers data between two devices
Register - a component that stores data in the CPU

Higher Level Language - Language that's relatively higher in the computer architecture
Lower Level Language - Language that's relatively lower in the computer architecture

CPU - Central Processing Unit

• Invented by von Neumann
• Continuously does the “fetch-decode-execute” cycle

FDE - Fetch, Decode, Execute

• The instructions are “fetched” from RAM
• They're then decoded so the CPU knows what to do
• They're then executed

RAM (aka main memory) - the memory of the computer

# Logarithms and Powers of 2

# Computer Architecture Layers

5 - Problem-Oriented Language Level
↓ Translation (compilation) to assembly
4 - Assembly Language Level
↓ Translation (assembly)
3 - Operating System Language Level
↓ Interpretation (OS)
2 - Instruction Set Language Level
↓ Interpretation (microprograms) or direct execution
1 - Microarchitecture Level
↓ Hardware
0 - Digital Logic Level

# Translation and Interpretation

In order to convert between different language levels, we need to first convert the instructions between different layers.

• This process can usually only go down the layers, not up

Translation is the process of converting a program from one HLL to its equivalent program in an LLL

• Converting an HLL program is done all at once
• After doing so, the HLL program isn't needed anymore

Interpretation is the process of converting an instruction from one HLL to its equivalent set of instructions in an LLL

• Converting an HLL program is done one instruction at a time

# Notes

• The Digital Logic Level is the only hardware level
  • All other levels are software
• Compilation can only go down to assembly level
• Multi-layered computers are a modular system
  • If you want to replace a level, you just need to also replace the conversions between its higher and lower levels
• When compiling and running a program, we end up with six different versions of the same program, each in a different language

# Patterns From Highest to Lowest Level

• Easiest to hardest to understand (for humans)
• Hardest to easiest to understand (for computers)
• Most to least powerful instructions
  • Flexible to rigid
  • Easiest to hardest to execute a high-level instruction
• Least to most amount of instructions
  • Smallest to largest files to execute
• Slowest to fastest levels
  • Most to least conversions to get to the lowest level

# Week 1 Question

# Q4.

Consider a multilevel computer in which all the levels are different. Each level has instructions that are m times as powerful as those of the level below it; that is, one level r instruction can do the work of m level r − 1 instructions. If a level-1 program requires k seconds to run, how long would equivalent programs take at levels 2, 3, and 4, assuming n level r instructions are required to interpret a single r + 1 instruction?

The first step is to define all the variables in the question. From there, we can figure out how to solve it.

Given four variables, r, m, n, k:

r = an arbitrary level in the computer
k = the amount of seconds for a level-1 program to run
m = the power factor of a level's instructions
n = the amount of level r - 1 instructions required to interpret a level-r instruction

• 1 level-r instruction can do the work of m level-(r - 1) instructions
• 1 level-r instruction gets converted into n level-(r - 1) instructions after interpretation

Now, let's calculate how long a level-2 program takes.

A level-1 program takes k seconds to run.

k

A level-2 instruction first has to be converted into level-1 instructions before being executed.

One level-2 instruction gets converted into n level-1 instructions.

n * k

The execution of these instructions then gets sped up by our power factor, m.

\frac{n}{m} * k

One level-2 program will take \frac{n}{m} * k seconds to run.

We can use the same logic to calculate how long a level-3 program takes.

A level-2 program takes \frac{n}{m} * k seconds to run.

\frac{n}{m} * k

A level-3 instruction first has to be converted into n level-2 instructions before being executed.

\frac{n}{m} * n * k

The execution of these instructions then gets sped up by our power factor, m.

\frac{n}{m} * \frac{n}{m} * k

(\frac{n}{m})^2 * k

One level-2 program will take (\frac{n}{m})^2 * k seconds to run.

Notice that, for a level-3 program, the degree of \frac{n}{m} is equal to 2.
For a level-2 program, the degree of \frac{n}{m} is equal to 1.
We can generalize this to say that for a level-r program, the degree of \frac{n}{m} is equal to r - 1.

\frac{n}{m}^{r - 1} * k

This is how we get the formula!