Posted: June 26th, 2022
Let’s say that you have 256M*128 memory built with 1024 K*32 RAM Chips. The memory is word-adressable.
How many RAM chips are required?
What number of RAM Chips is there per memory word.
What number of address bits is required for each RAM Chip
What number of banks does this memory support?
What number of address bits is required for all memory?
Digital computers have a 48 bit per word memory unit.
There are 240 operations in the instruction set.
Each instruction has an operation code (opcode), and an address (permitting only one address).
Each instruction is stored in one word.
What number of bits is required for the opcode to work?
How many bits do you need to specify the register’s name?
What number of bits are needed to specify the register?
Is there a binary number larger than unsigned that can be stored in one word of memory?
Take the instruction Add 1000.
The following figure shows memory contents, Accumulator, and R1 Register contents. (For example, memory address 1000 has a value 1400, while memory address 1100 has a value 400, etc.
All numbers are in decimal.
If R1 is assumed to be implied in the index addressing mode, the following addressing modes can be used to determine the actual value loaded into an accumulator (AC).
To implement the expression, write code: A=(B + C), * (D-E) on 1-, 2-, and 0-address machine.
According to programming language practice, the expression must not be modified by computing.
We require 256 Kbytes. That is, 256 x 1024×128 bits.
We have RAM chips with 32 Kbits = 32×1024 bits.
Ram chips are required in the following quantities: (256 * 1024* 128)/(32* 1024) = 1024
We require 256 Kbytes. That is, 256 x 1024×8 bits.
RAM chips with 32 Kbits capacity = 32 x 1024 bit.
Ram chips needed: (256 * 1024 * 8/(32* 1024) = 64
There are 1024 RAM chips, which is equal to 230.
Therefore, there are 30 bits required for each RAM chip.
The RAM has 1024 bits.
This system requires 8 memory banks.
All memory requires address bits = 1024/32 = 32
16 bits = Number of bits in the opcode
Registers: 8 bits
The largest unsigned binary number is 248-1.
Value of 1000 added in immediate addressing = 1000 + 500 = 1500, and the Value is saved in the accumulator.
Direct addressing mode: 1400 + 500 =1900. Value of 1000
1000 = 1400
Address 1400 has a value of 1300.
The accumulator is then loaded with 1300 + 500 = 1800.
1000 is the value found in 1200.
The accumulator value would then be 1000 + 500 = 1500.
2 – Address machine
MOV R1, B
ADD R1, B
MOV R2, D
SUB R2, E
MUL R1, R2
MOV A, R1
1 – Addressing machine
0 – Address machine
Ginseng: Market-driven allocation of memory
ACM SIGPLAN Notes (Vol.
OOPSLA 2002: Custom memory allocation.
ACM SIGPLAN Notifications, 48(4), 46-57.
Extended lattice-based allocation of memory.
Proceedings of the 25th International Conference on Compiler Construction (pp.
Structured computer organization.
Pearson Education India.
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