Computer Organization and Architecture Homework Solution - Course Name

This document presents a solved homework assignment on Computer Organization and Architecture. The solution includes detailed answers to questions involving RAM chip calculations, determining the number of chips needed based on memory requirements, and calculating address bits. The assignment also explores different addressing modes (immediate, direct, indirect, and indexed) with examples, illustrating how values are accessed and manipulated. Furthermore, the solution provides assembly language examples for both 2-address, 1-address, and 0-address machines, demonstrating how the same operation can be performed with different instruction formats. Finally, the document includes a bibliography of relevant academic sources on computer organization and memory allocation.

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

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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer to question number 1
a. We need 256 Kbytes, i.e., 256 x 1024 x 128 bits.
We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.
Number of Ram chips required are (256 * 1024 * 128)/(32 * 1024) = 1024
b. We need 256 Kbytes, i.e., 256 x 1024 x 8 bits.
We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.
Number of Ram chips required are (256 * 1024 * 8)/(32 * 1024) = 64
c. Number of RAM chips required are 1024 = 2^30.
Hence address bits needed for each RAM chip required are 30.
d. The number bits present in the RAM = 1024
Hence the number of memory banks required for this system are: 256/32 = 8
e. Address bits required for all memory = 1024/32 = 32
Answer to question number 2
a. Number of bits for the opcode = 16.
b. Number bits required for the registers = 8
Number of bits required for addresses = 16
c. The largest unsigned binary number: 2^48 -1.

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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer to question number 3
i) Immediate
Value of Add 1000 in immediate addressing is 1000 + 500 = 1500 and the Value is stored in the
accumulator.
ii) Direct
Value of Add 1000 in direct addressing mode is 1400 + 500 =1900.
iii) Indirect
Value of 1000 is 1400. The value of address 1400 is 1300. Hence 1300 + 500 =1800 is loaded
into the accumulator.
iv) Indexed
1000 + R1 = 1200
The value located in 1200 is 1000. Hence the value loaded in the accumulator would be 1000 +
500 = 1500.
Answer to question number 4
2 – addressing machine
MOV R1, B
ADD R1, C
MOV R2, D

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COMPUTER ORGANIZATION AND ARCHITECTURE
SUB R2, E
MUL R1, R2
MOV A, R1
1 – addressing machine
LOAD B
ADD C
STORE T
LOAD D
SUB E
MUL T
STORE A
0 - addressing machine
PUSH B
PUSH C
ADD
PUSH D
PUSH E
SUB

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COMPUTER ORGANIZATION AND ARCHITECTURE
MUL
POP A

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COMPUTER ORGANIZATION AND ARCHITECTURE
Bibliography
Agmon Ben-Yehuda, O., Posener, E., Ben-Yehuda, M., Schuster, A., & Mu'alem, A. (2014,
March). Ginseng: Market-driven memory allocation. In ACM SIGPLAN Notices (Vol. 49,
No. 7, pp. 41-52). ACM.
Berger, E. D., Zorn, B. G., & McKinley, K. S. (2013). OOPSLA 2002: Reconsidering custom
memory allocation. ACM SIGPLAN Notices, 48(4S), 46-57.
Darte, A., Isoard, A., & Yuki, T. (2016, March). Extended lattice-based memory allocation.
In Proceedings of the 25th International Conference on Compiler Construction (pp. 218-
228). ACM.
Tanenbaum, A. S. (2016). Structured computer organization. Pearson Education India.