Test: Computer Organisation & Architecture - 1 - Computer Science Engineering (CSE) MCQ

Stack-based CPU organization uses zero address instruction.
Key Points

• The computers which use Stack-based CPU Organization are based on a data structure called a stack. 
• It makes use of the Last In First Out (LIFO) access technique, which is the most common in most CPUs.
• The address of the highest member of the stack, known as the Stack pointer, is stored in a register (SP).
• Push and Pop are the two most common operations done on the stack's operators. These two operations are only carried out from one end.
• This instruction contains the opcode only with no address field. It pops the two top data from the stack, subtracting the data, and pushing the result into the stack at the top. 
• PDP-11, Intel’s 8085, and HP 3000 are some examples of stack-organized computers.

The advantages of Stack-based CPU organization –

• Efficient computation of complex arithmetic expressions. 
• Execution of instructions is fast because operand data are stored in consecutive memory locations. 
• The length of instruction is short as they do not have an address field. 

The disadvantages of Stack-based CPU organization

• The size of the program increases.

CPI: Cycles per instruction or clock per instruction is used as a measure of processors' performance, that is, number of CPU cycles required to execute and instruction at low level. One instruction consists of ALU, Load, Store and Branch.

• ALU takes ~30.30% of the total cycles.
• Load takes ~30.30% of the total cycles.
• Store takes ~27.27% of total cycles.
• Branch takes ~12.12% of total cycles.

Concept:

The given data,

Distinct instructions= 300

General-purpose registers = 70

opcode instruction word = 32-bit

Two register operands and an immediate operand.

Each instruction has 32 bits. To support 300 instructions, the opcode must contain 9-bits.

Register operand1 requires 7 bits, since the total registers are 70, Register operand 2 also requires 7 bits.

So, 32 - (9+7+7) = 9 bits are left over for immediate operand.

Hence the correct answer is 9.

Key Points

• The permanent memory of a computer is known as ROM(Read-only memory).
• In computers and other electronic devices, read-only memory (ROM) is a form of non-volatile memory.
• After the memory unit is manufactured, data contained in ROM cannot be electronically changed. Read-only memory, also known as firmware, is useful for storing software that is rarely updated during the life of the device.
• Plug-in cartridges containing ROM can be used to distribute software applications (such as video games) for programmable computers.
• Read-only memory refers to memory that is hard-wired and cannot be modified electronically after manufactures, such as a diode matrix or a mask ROM integrated circuit (IC).

Key Points

• The first instructor of the bootstrap loader program of an operating system is stored in BIOS(Basic Input/Output System).
• The bootstrap loader is a programme that resides on the EPROM, ROM or other non-volatile memory of the machine.
• It is executed by the processor automatically when the device is turned on. To continue installing the computer's operating system, the bootstrap loader reads the boot sector of the hard drives.
• The bootstrap loader first conducts the power-on self-test, also referred to as POST, when the machine is switched on or restarted.
• The bootstrap loader loads the operating system for the machine into memory if the POST is successful and no problems are found.
• It is then possible for the machine to access, load, and operate the operating system.
• In computers that have an EFI (Extensible Firmware Interface), the bootstrap loader has been replaced and is now part of the EFI BIOS.

Key Points

• 1 gigabyte = 2³⁰ bytes = 2¹⁰ × 2²⁰ bytes
• Since 1 megabyte =  220 bytes
• 1 gigabyte =  210 × megabytes
• 1 gigabyte = 1024 megabytes

Therefore option 3 is correct.

• RAM is the fastest to read from and write to than the other kinds of storage in a computer.
  • It is Volatile or temporary memory.
  • Data gets erased when the power supply is off.
  • Faster memory than ROM.
  • It is used in the normal operations of a computer after starting up and loading the operating system.
• CD-ROM for "Compact Disc Read-Only Memory".
  • CD-ROM is a CD that can be read by a computer with an optical drive.
• Floppy Disk is a type of disk storage composed of a disk of the thin and flexible magnetic storage medium.
  • It is used for backup purpose.
  • It is a type of secondary storage.
  • The storage capacity of a common floppy disk is 1.44 MB.
• The hard disk is the main device for storing data in the computer.
  • It uses magnetic technology (nowadays optical technology for storing data).
  • It has a strong coating above the magnetic metallic oxide in order to protect the major components.

Bus: In computer architecture, a bus (related to the Latin “omnibus”, meaning “for all”) is a communication system that transfers data between components inside a computer, or between computers. This expression covers all related hardware components (wire, optical fiber, etc.) and software, including communication protocols.

The types of buses are:

Address Bus:

• It is a unidirectional Bus, which is responsible for only one-way communication.

Data Bus:

• It is a bi-directional bus.
• Data and Instructions pass through this bus to reach the microprocessor.
• This is responsible for the exchange of information between the main memory and microprocessor.

Control Bus:

• The control bus carries control signals partly unidirectional and partly bidirectional
• This tells whether read or write, which operation should be performed that is in which direction the data should flow.

So Utility Bus is not a type of bus used in computer architecture, hence option 4 is the correct answer.

Network of Communication protocol is formed by one master and 247 slaves with a unique address.
Key Points
MODBUS ASCII/RTU is a master-slave communication protocol, able to support up to 247 slaves connected in a bus or a star network.
The protocol uses a simplex connection on a single line. In this way, the communication messages move on a single line in two opposite directions.

Concept:

Option 1: Parallelism is high in the horizontal microprogrammed control unit as compared to a vertical microprogrammed control unit.

True, Parallelism is high in horizontal microprogramming as several operations on different registers can be performed simultaneously.

Option 2: Hardwired control unit is slower compared to the microprogrammed control unit.

False, a Hardwired control unit is faster as compared to the microprogrammed control unit as there won’t be a delay of fetch, decoding, and executing the control instructions in the case of the hardwired control unit.

Option 3: In 2’s complement sum carry flag and overflow are the same.

False, In unsigned numbers, carry out is equivalent to overflow. But in two's complement, carry out tells you nothing about overflow.

Option 4: In 2’s complement sum if the sum of two negative numbers yields a positive result, the sum has overflowed.

True, the Following are the rules for detecting overflow in a two's complement sum:

• If the sum of two positive numbers yields a negative result, the sum has overflowed.
• If the sum of two negative numbers yields a positive result, the sum has overflowed.

Otherwise, the sum has not overflowed.

Hence the correct answer is A and option B.