Test: Computer Organisation - Grade 11 MCQ

• An address bus is a computer architecture that is used to transfer data between devices that are identified by the physical address which is in binary form.
• Secondary memory in the computer provides ample storage to store a maximum number of files, videos etc. It saves the data permanently.
• The system bus in the computer is divided into the address bus and the data bus.
• The address bus is used to specify a physical address in the memory. A data bus is used to transmit data among components.
• Secondary memory is independent of the address bus. Hence option 4 is correct.

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. 

I/O Device: I/O devices are divided into two categories:

• Block Device: These are the devices with which the device driver communicates by sending the entire block of data. Example - Hard Disk
• Character Device: These are the devices with which the device driver communicates by sending and receiving single characters in either bytes or octets. Example - Serial port, parallel port.

​Device Controller: These are software modules that are plugged into the operating system to handle a particular device. These are not I/O devices but software modules that control I/O devices.

For a multiprocessor system, interconnection network - crossbar switch is an example of a non-blocking network.
A non-blocking network is a type of interconnection network in which concurrent communication can occur between multiple nodes without blocking. The crossbar switch is a non-blocking network because it can connect any input to any output without any blocking. This allows for efficient communication and data transfer between multiple processors, memory modules, and input/output devices in a multiprocessor system.

• A combination of 16 bits is called a word.
• Word "size" refers to the amount of data a CPU's internal data registers can hold and process at one time.
• Computers embedded in appliances and consumer products have word sizes of 8, 16, or 32 bits.

Key Points

• The smallest unit of data in a computer is called Bit.
• A bit has a single binary value which is either 0 or 1.
• A byte is a unit of data that is eight binary digits long.
• The difference between a bit and a byte is 7 bits.
• The difference between a nibble and a byte is 4 bits.

• 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).

• DRAM:
  • It is dynamic random access memory and is widely used as a computer's main memory. Hence, Option 2 is correct.
  • Each DRAM memory cell is made up of a transistor and a capacitor within an integrated circuit, and a data bit is stored in the capacitor.
• DDR-RAM:
  • It stands for Double Data Rate Synchronous Dynamic Random-Access Memory.
  • These are the computer memory that transfers the data twice as fast as regular chips like SDRAM chips because DDR memory can send and receive signals twice per clock cycle as a comparison.
• SRAM:
  • It stands for Static Random Access Memory.
  • It is a form of a semiconductor.
  • It is widely used in microprocessors, general computing applications, and electronic devices.
  • The SRAM is volatile in nature which means the data stored in it gets all wiped out once the power supply is cut.
  • SRAM is comprised of flip-flops. 
• PRAM:
  • In computer science, a parallel random-access machine is a shared-memory abstract machine.
  • As its name indicates, the PRAM is intended as the parallel-computing analogy to the random-access machine.

• CD-ROM, an abbreviation of compact disc read-only memory, type of computer memory in the form of a compact disc that is read by optical means.
• A CD-ROM drive uses a low-power laser beam to read digitized (binary) data that has been encoded in the form of tiny pits on an optical disk.
• The drive then feeds the data to a computer for processing.
• With a storage capacity of 680 megabytes, the CD-ROM found rapid commercial acceptance as an alternative to so-called floppy disks.
• Unlike conventional magnetic storage technologies (e.g., tapes, floppy disks, and hard disks), CDs and CD-ROMs are not recordable hence the tag “read-only.”

• 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.

The term RAM refers to read and write memory, that is, you can both write data into RAM and read data from RAM. Most RAM is volatile, which means that it requires a steady flow of electricity to maintain its contents. Read-only memory (ROM) refers to computer memory on which data has been prerecorded. Once data has been written onto a ROM chip, it cannot be removed and can only be read.

There are three basic units of a computer system:

• Input Unit: Keyboard, Mouse, Scanner, Light pen, Trackball etc.
• Central Processing Unit
• Output Unit: Monitor and Printer
  

Central Processing Unit (CPU): It is the most important unit, where all the processing jobs take place. CPU is the control centre of the computer and hence it is said to be the brain of the computer. CPU has three main components:

• Arithmetic Logic Unit (ALU): It performs all the arithmetic operations like addition (+), multiplication (*), subtraction (-), division (/) on the numerical data directed by the control unit. All the logical operations, like less than (<), greater than (>), equal to (=), not equal to (≠) etc. are also carried out by ALU.
• Control Unit (CU): It controls and coordinates all the operations taking place in the system. It controls the flow of data and information from one unit to the other.
• Registers/Memory Unit (MU): To execute a program, data and instructions need to be stored temporarily. This storage is done in the MU. The data and instructions are retrieved from MU by Control Unit for supplying to ALU as when required by the program.
• Internal bus: To transfer the data

• 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.

Cache memory

• It is a high-speed storage area for temporary storage.
• It is the smaller and fastest memory component in the computer.
• It is used during the reading and writing processes from the disk.
• It acts as a buffer between RAM and the CPU.

Concept
The organization and interconnection of the various components of the computer system are Architecture.

Key Points

• A specification for how a set of software and hardware technology standards combine to build a computer system or platform is known as computer architecture.
• In a nutshell, computer architecture describes how a computer system is built and which technologies it supports.
• Computer architecture is a set of rules and procedures used in computer engineering to explain the functioning, organization, and implementation of computer systems.
• The CPU, memory input/output device, and communication channels that connect them are the primary components in a conventional computer system.

Important Point - In Synchronous Access Organisation, the address lines are common at all the banks.
We know that

• Distance at which address is spaced - 'm'
• Bank Cycle Time - 'T'
• Number of Banks - 'n'

Explanation

• When the distance at which address is spaced is very less than the number of banks, that is, m << n, then the average data access time per word access is - m.(T/n)
• When the distance at which address is spaced is much more than the number of banks, that is, m >> n, then the average data access time per word access is - T
• Combining above two points, we get the average data access time per word access in synchronous organisation as - t = \(\left\{\begin{array}{l} \rm m \cdot \rm\frac{T}{n} \text { for } \rm m<> n \end{array}\right.\)