Computer Architecture Assignment: Memory Access and Addressing Modes

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Added on 2024/07/01

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This assignment solution focuses on computer organization and architecture, specifically addressing instruction sets, addressing modes, and memory access. It begins by determining the number of bits required for the opcode based on the given number of operations, calculating the address part for the instruction set, and finding the maximum allowable memory size. The solution then calculates the actual value loaded into the accumulator for different addressing modes, including immediate, direct, indirect, and indexed addressing. Furthermore, it analyzes memory accesses using the MARIE architecture for a given expression and optimizes memory access by utilizing registers. Finally, the assignment provides hexadecimal codes for instructions, constructs a symbol table, and determines the value of the accumulator at program termination.

COMPUTER ORGANISATION AND ARCHITECTURE- ASSIGNMENT 2

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Contents
Q 1:.................................................................................................................................................. 2
Q 2:.................................................................................................................................................. 2
Q 3:.................................................................................................................................................. 3
Q 4:.................................................................................................................................................. 4

Q 1:
As the information is providing in the question
The digital computer contains the memory unit which is 16 bits in every word.
The digital computer uses the instruction set which contains the 122 different types of
operations.
Solution:
a. As the 122 different types of operations used by the instruction set. So 122 instructions
will equal to the 27 therefore, 7 bits will require for the opcode. We will 7 bits will take
because 26 will equal to 64 bits which will smaller to the 122.
b. As the given in the question, 16 bits are used for every word. I am calculating the left
address part for the instruction set which are 16 – 7 = 9 bits.
c. Here, as the calculation is providing the information that 9 bits are left so maximum
allowable memory size is 29
d. One word memory which will accommodate by the digital computer so the largest
unsigned number is 216 – 1.
Q 2:
As the information is providing in the question
1. ADD 1000
2. Ac = 500
3. Content of register = 200
Solution:
Here, I am calculating the actual value that will load in the accumulator for the different types f
the addressing mode.
a. For the immediate addressing
The value at the AC + R1’s content is 1000
Thus value in (AC) accumulator = 1000 + 500
= 1500
b. For the Direct addressing
Value at the 1000’s address is 1400
Thus value in accumulator (AC) = 1400 + 500
= 1900
c. For the Indirect addressing
Value at the 1000’s address is 1400. So for Indirect mode, the value at the address 1400
is 1300

So value in (AC) accumulator = 500 + 1300
= 1800
d. For the Indexed addressing
Value at the AC + address 1200’s content is 1000
So value in (AC) accumulator = 500 + 1000
= 1500
Q 3:
As the information is providing in the question that calculates the memory accesses by using the
MARIE rule. This information is calculating for the expression which is providing in the
question.
According to the code, there will need nine (9) memory accesses so that expression can easily
execute
Code
Load A
ADD B
Store Temp1
Load C
ADD D
Store Temp2
Load Temp1
Sub Temp2
Store Sum
Code
Load R1, A
Load R2, B
ADD R1, R2
Load R3, C
Load R4, D
ADD R3, R4 (this operation doesn’t need memory access)
Sub R1, R4 (this operation doesn’t need memory access)
Store Sum

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As the information is providing in the question that whole architecture contains the 4 registers
such as R1, R2, R3, and R4 therefore only five (5) memory access is needed to execute the
arithmetic operation.
Q 4:
In this question, the code is given in the question. We are performing some question which is
given below.
I. List of the hexadecimal code is given in the table for every instruction.
S. No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Instructions
Load A
ADD B
Store D
Clear
Output
ADDI D
Store B
HALT
HEX 00FC
DEC 14
HEX 0108
HEX 0000
Hexadecimal code
1108
3109
210B
A000
6000
B10B
2109
7000
00FC
000E
0108
0000
II. Here, I am drawing the symbol table on the basis of the label of instructions
S. No.
1.
2.
3.
4.
5.
Location
108
109
10A
10B
100
Symbol
A
B
C
D
Start

III. Here, I am determining the value of the AC when the program will terminate form the
memory. So according to the hexadecimal code, AC will contain the 0108