1.7 Main memory(RAM) Addressing

• ⭐️ a RAM that ends in 8Fs or 16Fs, there will be one of two models
• ⭐️ how many bits are we using?
• ⭐️ what is the maximum capacity?
• ⭐️ give one address of RAM in binary, we have data xxx.
• ⭐️ address/positions should be transformed into hexadecimal, data into octal

✅ RAM

storage of open files and applications = processes

• process: open and running file and application

• RAM is divided into address

• addresses are numbered in hexadecimal, read 📌 hexadecimal
• the address at the user area the most top, is 0
• the address the the most bottom is F
• and OS is in secondary memory, but also has a part in RAM that cannot be touched by other programs

Q_1: If the last address of the RAM is FFFFFFFFh, how many bits are there in this address?
A: each F has 4 bits internally, has 4 * 8 = 32 bits
- thus, this RAM is using 32 bits for the address
- thus, this RAM is using 32 bits for addressing(direcionar) the RAM

☑️ Capacity of RAM

• ⭐️ by looking at the last address, we can know the capacity of the RAM

• ⭐️ each F has 4 bits internally

• Following the computing principle, if the RAM has FFFFFFFFh 🟰 32 bits, RAM can have 2^32 combinations of address
• If in each address, I store 1 byte = 8 bits of information
• I can save up to 2^32 * 1 byte in my RAM
• so I can have 2^32 addresses in my RAM
• more or less 4 * 1000 * 1000 * 1000 bytes

1 time 2^10 = 1024 = Kilo
2 times 2^10 = 1024 * 1024 = Mega
3 times 2^10 = 1024 * 1024 * 1024 = Giga
4 times 2^10 = 1024 * 1024 * 1024 * 1024 = Tera
5 times 2^10 = 1024 * 1024 * 1024 * 1024 * 1024 = Petta
6 times 2^10 = 1024 * 1024 * 1024 * 1024 * 1024 * 1024 = Exa

• so, 4 * Giga bytes
• in conclusion, if the last address of the RAM is FFFFFFFFh, this RAM has a total of 4 Giga Bytes

Q: If a person bought 16G of RAM, but can only use 4GB, what is happening?
A: Is is bc the OS is using only 32bits for addressing the RAM
and with 32bits, you can only 4GB of RAM
⭐️ Solution: so you should change your OS to use more bits, OS that uses 64bits for the RAM

• Every OS has 2 versions, version of 32 and version of 64
• and depending on the version, the RAM can be totally useable, or not

📌 Hexadecimal

numbering system that groups bits by 4, starting by the right
for example, 1111 = 15 = F

• 1️⃣ divide bits by grouping them into 4, starting by the right
  • if the group is not complete, complete left with 0
• 2️⃣ then applies the weight rule to each group of 4
• 3️⃣ then changes the weight of 10 into the letter A
• 11 ➡️ B
• 12 ➡️ C
• 13 ➡️ D
• 14 ➡️ E
• 15 ➡️ F
• 4️⃣ add a h

address = 010101111111

1️⃣ divide into groups of 4, from the right
- 1111
- 0111
- 0101

2️⃣ apply weight
- 8421 /  8421 /  8421

- 1111 = 1 + 2 + 4 + 8 = 15
- 0111 = 1 + 2 + 4 = 7
- 0101 = 1 + 4 = 5

3️⃣ change weight into letter
- 15 = F
- 7
- 5

💡 result is 57F

4️⃣ always add h to say this address is hexadecimal
💡 result is 57Fh

address = 1110110

1️⃣ divide into groups of 4, from the right
if group is not complete, add 0 to the left
- 0111
- 0110

2️⃣ apply weight
- 0111 = 7
- 0110 = 6

3️⃣ change weight into letter
nothing to change
- 7
- 6

👉🏻 result is 76

4️⃣ always add h to say this address is hexadecimal
👉🏻 result is 76h

1110
weight = 14
👉🏻 result is Eh

• Sometimes in Windows, the h in hexadecimal is used as 0x

✅ How will the RAM in a OS of 64bits work?

• If we have a RAM of FFFFFFFFFFFFFFFFh(16Fs),
• we have 4 bits * 16 = 64bits for addressing the RAM
• following the computing principle, we can have 2^64 different addresses in the RAM
• if in each of the addresses, if we put 1 byte in each address
• our RAM will be 2^64 = 2^4 * 2^10 * 2^10 * 2^10 * 2^10 * 2^10 * 2^10
• 🟰 16 * 1000 * 1000 * 1000 * 1000 * 1000 * 1000
• 🟰 16 Exa
• So if a RAM of 64bits, we will have 16 Exa bytes of capacity
• This capacity of RAM does not exist yet, bc of limits of
• 👎🏻 temperature(heat)
• and 👎🏻 we do not have technology to divide RAM into 16 trillion addresses

⭐️ Exam Question
Q: If you have 16GB of RAM and it is useable, how many bits will you be using for addressing RAM?
A: You will have 64bits
❓❓❓❓❓
Q: If we have a word width of 1byte per address, how much data can we save on this RAM?
A: 2^64bytes

📌 Computing principle

• Everything in computing is always a power of 2
• there is nothing between for example, 32 and 64

Q: If somebody is selling you 300G of harddisk, is he saying the truth?
A: NO, they are rounding up from 256G, 300G of harddisk does not exist.

Q: If somebody is selling you 2.5Tera, is he saying the truth?
A: NO, maybe they have 2 Tera and 512 Giga.

✅ Word width

Word width: Capacity of each address in RAM

• how much can we save in each address in a RAM

• each adress capacity = Word width

• Q: How could we increase the capacity of a RAM?
• A: save more bytes in each address
• ⭐️ instead of saving one byte per address, save 2, 4, 8, 16...bytes on each address

📌 Octal

numbering system that groups bits by 3, starting by the right

• 1️⃣ create group of 3 bits
• 2️⃣ apply weight 4, 2, 1

original binary: 01010101
1️⃣ create group of 3 bits
- 101
- 010
- 001

2️⃣ apply weight
- 101 = 4 + 1 = 5
- 010 = 2
- 001 = 1

👉🏻 result: 125o
👉🏻 read: one two five octal ⭕️
- reading one hundread twenty five octal ❌ this is reading decimal

• 🛠️ Use of octal
• When we talk about the RAM, we talk about the address in hexadecimal
• However, when we talk about the data size inside the RAM, we use octal
• 👍🏻 we are trying to make numbers shorter, from decimal to octal
• example: In address FA3h(address, hexadecimal), we are saving 125o(data size, octal)

⭐️ Exam question
Q:
In address: 0001110101
we have byte 00011101 saved

Transform this into conventional numbering system for main memory(RAM)

A:
In address 075h, we have data that has size 035o saved inside

• In windows…

Q: If you get a error message in windows saying,
there is an error in 0x 0FA3, where should you be looking?

A: Ox means hexadecimal in windows, so look at address 0FA3

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