Capital University of Science and Technology
Department of Computer Science
CS 2523 – Computer Organization and Assembly Language
QUIZ NO. 1: Basics of Computer Organization and Assembly Language
Section# 3
CLO: 1. Define concepts in the design of microprocessor as state machine and designing its
data path and its controller.
Semester: Fall 24 Max Marks: 10
Instructor: Ms. Tayyaba Zaheer Date: October 09, 2024 Max Time: 10 Minutes
Name: Reg. No.
Question No.1 [02 Marks]
Please choose the best possible option:
1. (01 mark) Assembly Language programs are written using:
a) Hex code
b) ASCII code
c) Mnemonics
d) None of the mentioned
Solution: c
2. (01 mark) CPU has built-in ability to execute a particular set of machine instructions, called as
__________
a) Registers
b) Instruction Set
c) Sequence Set
d) None of the mentioned
Solution: b
Explanation: An instruction is any task which is to be performed by the processor. Instructions are stored
in the register. Instruction set is the set of machine instructions.
Question No.2 [03 Marks]
Write the following in the correct order of 5 levels of programmer’s view of a computer: [You can only
write numbers in order like ii, iii, vi…]
i. Microarchitecture
ii. Application Programs High-Level Language
iii. Assembly Language
iv. Digital Logic
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v. Instruction Set Architecture
vi. Operating System
Solution:
ii, iii, vi, v, i, iv
ii. Application Programs High-Level Language,
iii. Assembly Language,
vi. Operating System,
v. Instruction Set Architecture,
i. Microarchitecture
iv. Digital Logic
Question No. 3 [05 Marks]
Elaborate 5 sub-operations of the control unit in the given scenario of Instruction Cycle:
Solution:
1. Fetch: Instruction at address 102 would be fetched from memory. Because PC has address of the
instruction i.e. 102.
2. Decode: Instruction would be decoded by the control unit as per the opcode i.e. load (opcode
means load data from memory to register). Control unit would set the controls as per the opcode
of the instruction.
3. Fetch Operands: This sub operation would not be executed for this particular instruction as
memory reading is involved in the given instruction.
4. Execute: This sub operation would not be executed for this particular instruction as there is
nothing to be executed by ALU.
5. Store: This sub operation would store back to memory in this instruction i.e. M[501] = 11
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