Computer Organization and Architecture - Desklib

This article discusses various concepts related to Computer Organization and Architecture. It covers topics such as memory size, addressing modes, and programming. The article provides answers to questions related to opcode, word size, and memory size. It also explains the different addressing modes and provides a sample program for each. The article concludes with a sample program and its corresponding symbol table.

Running head: COMPUTER ORGANIZATION AND ARCHITECTURE
Computer Organization and Architecture
Name of the Student:
Name of the University:
Author Note

1
COMPUTER ORGANIZATION AND ARCHITECTURE
Answer to question number 1
a. 122 instructions imply that 2^7, hence 7 bits are required for the opcode.
b. The number of bits in a word is 16. Hence the address part is 16 – 7 = 9 bits.
c. Maximum allowable size of the memory is 2^9 bits.
d. 2 ^16 - 1 or 16 ones is the highest number to be allocated in the memory.
Answer to question number 2
1. Immediate Addressing mode
Value added into the accumulator is 1000.
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1000 +500 =1500.
2. Direct Addressing mode
Effective address: 1000
Value added into the accumulator is 1400.
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1400 +500 = 1900.
3. Indirect Addressing mode
The effective address is the value in 1000 which is 1400.
Therefore, the value added into the accumulator is 1300
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1300 +500 = 1800.
4. Indexed Addressing mode
The value of R1 is 200.

2
COMPUTER ORGANIZATION AND ARCHITECTURE
The effective address is 1000 + 200 = 1200. Hence the value in 1200 location would
be added to the accumulator.
Value in location 1200 is 1000. Hence the value loaded into the accumulator is 1000 +
500 = 1500.
Answer to question number 3
The following MARIE program can be utilized to realise the expression S = (A+B)-
(C+D).
100 LOAD A
101 ADD B
102 STORE X
103 CLEAR
104 LOAD C
105 ADD D
106 STORE Y
107 CLEAR
108 LOAD X
109 SUBT Y
10A STORE S
10B HALT
Now for the Register memories the following program is to be used:
ADD R1, A, B

3
COMPUTER ORGANIZATION AND ARCHITECTURE
ADD R2, C, D
SUBT A, R1, R2
Hence for the first program the system would be required to store the results in three
addresses and for the second program would require only a single memory unit for storing the
data.
Answer to question number 4
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.
Symbol Location

4
COMPUTER ORGANIZATION AND ARCHITECTURE
A 108
B 109
C 10A
D 10B
Start 100
c.
When the program terminates the accumulator would store the value 10A in hex.