University ITC544 Computer Organisation and Architecture Assignment 2

This document presents a comprehensive solution for ITC544 Computer Organisation and Architecture Assignment 2, focusing on MARIE and ISA. The assignment addresses key concepts including opcode bit calculations based on the instruction set, determining the address part of instructions, and calculating maximum memory sizes. It then delves into different addressing modes (immediate, direct, indirect, and indexed) and calculates accumulator values for each. The solution also provides a MARIE program and a register-based assembly program to perform a given arithmetic expression, comparing the memory requirements of each. Finally, it includes an analysis of a provided program, identifying the values in memory locations and registers, and determining the final value in the accumulator after program termination. The assignment showcases a strong understanding of computer architecture principles and assembly language programming.

Running head: ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
Assignment 2: MARIE and ISA
Name of the Student
Name of the University
Author’s Note

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ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
Question 1 answer:
1.1. The number of bits required for the opcode is calculated below:
Number of operation in the instruction set = 122
Operation instruction = 2^7
Thus 7 bits of opcode is needed
1.2. The bits that are left for the instruction address part is calculated below:
Memory unit of the computer = 16 bits /word
Thus the address part = 16 -7
= 9 bits
1.3. The maximum allowable size calculated for the memory is calculated below:
= 2 ^9 bits
1.4. The largest binary number that is unassigned and accommodated in the memory for one word is
calculated as;
2^16 -1
= 16 1s
Question 2 Answer:
For the instruction Add 1000
1.1. Immediate
Value to be added in the accumulator = 1000
Value already in the accumulator = 500
Therefore, the total value in the accumulator = 500+1000
= 1500
1.2. Direct
Effective address = 1000
Value needed to be added in the accumulator = 1400

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ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
Value already in the accumulator = 500
Therefore, the total value in the accumulator = 1400+500
= 1900
1.3. Indirect
Effective address = 1400
Value in the accumulator = 1300
Value already in the accumulator = 500
Therefore, the total value in the accumulator = 1300+500
= 1800
1.4. Indexed
R1 value = 200
Effective address = 1000+200
= 1200
Therefore the value that is in the 1200 is added with the accumulator.
1000 is in the location 1200
Therefore, loaded value in the accumulator = 500+1000
= 1500
Question 3 Answer:
For the expression S = (A+B) – (C+D) the following MARIE program is created:
100 LOAD A
101 ADD B
102 STORE X
103 CLEAR
104 LOAD C
105 ADD D

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ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
106 STORE Y
107 CLEAR
108 LOAD X
109 SUBT Y
10A STORE S
10B HALT
For the register the program developed is given below:
ADD R1, A, B
ADD R2, C, D
SUBT A, R1, R2
Three address are used for storing the result in the first assembly program and a single
memory is needed for storing data in the second program.
Question 4 answer:
a.
Address Hex
100 1108 Start LOAD A
101 3109 ADD B
102 210B STORE D
103 A000 CLEAR
104 F400 OUTPUT
105 B10B ADDI D
106 2014 STORE B
107 7000 HALT
108 0200 A HEX 00FC
109 0014 B DEC 14
10A 0001 C HEX 0108
10B 0000 D HEX 0000
b.

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ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
Symbol Location
A 108
B 109
C 10A
D 10B
Start 100
c.
On the termination of the program the value contained in the 10A is loaded in the accumulator.

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ITC544 COMPUTER ORGANISATION AND ARCHITECTURE
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