SlideShare ist ein Scribd-Unternehmen logo
1 von 10
Downloaden Sie, um offline zu lesen
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 1
Answer Question No.1 which is compulsory and any five from the rest.
The figure in the right hand margin indicates 2*10=20
1. Answer the following question:
i) How logical address is different from physical address.
Ans: Physical address: Physical address is the actual memory address which denotes a memory
area in your storage device.
Logical address: The operating system or some other programs uses base addresses as a measure
to find addresses. Here base address means starting address of a memory block. According to the
program written the CPU generates a address. This address is added with the base address so that
it forms the physical address. Here in this case the address generated by the CPU is called as
Logical Address. Explain how a ROM is also a RAM.
ii) Explain how a ROM is a RAM?
Ans: RAM - Is memory that can be both written to and read at anytime during processing.
ROM - Is Random Access Memory which can be anything from CD's that have been used that
aren't writable. OR chips that are meant only for storing data PERMANENTLY. Basicly
anything that holds digital data permantently and can no longer be written to.
iii) Define the term memory latency.
Ans: "Latency" in computer terms is how long it takes to get a request back. It's the flip side of the coin
from "bandwidth", which is how fast the data flows.
iv) What is difference between EPROM and EEPROM?
Ans: Erasable Programmable Read Only Memory (EPROM) Chips the information stored in an EPROM
chip can be erased by exposing the chip to strong UV light. EPROM chips are easily recognized by the
small quartz window used for erasure. Once erased the chip can be re-programmed. EPROM is more
expensive to buy per unit cost, but can prove cheaper in the long run for some applications. For example
if PROM was used for firmware that needed upgraded every 6 months or so - it could prove quite
expensive buying new chips!
Electronically Erasable Programmable Read Only Memory: This has the added advantage that the
information stored can be re-written in blocks and hence can be used to store system settings that the
user may want to change periodically. This solid state memory has considerably reduced in price over
recent years and is nowadays commonly used to store system settings such as BIOS settings.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 2
v) How a write through cache differ from write back cache?
Ans: A type of cache with the following feature: when changes are made to cached data, they are not
simultaneously made to the original data as well. Instead, the changed data is marked, and the original
data is updated when the chached data is deallocated. In a write-through cache, by contrast changes
made to cached data are simultaneously made in the original copy. A write-back cache can perform
more quickly than a write-through cache. But in some contexts, differences between cached and original
data could lead to problems, and write-through caches must be used
vi) Define Control word
Ans: The combination of control steps used for the generation of control signals is a control word. A
control word is a word whose individual bits represent the various control signals. In computing, word is
a term for the natural unit of data used by a particular processor design. A word is basically a fixed sized
group of bits that are handled as a unit by the instruction set and/or hardware of the processor. The
number of bits in a word (the word size, word width, or word length) is an important characteristic of a
specific processor design or computer architecture.
vii) Define the term memory access time:
Ans: The time required by a processor to access data or to write data from and to memory chip is
referred as access time.
viii) Differentiate between system software and application software:
Ans: System software performs tasks like transferring data from memory to disk, or rendering text onto
a display device. Specific kinds of system software include loading programs, operating systems, device
drivers, programming tools, compilers, assemblers, linkers, and utility software.
Application software are designed to perform specific data processing or computational tasks for the
user. These programs are specifically designed to meet end-user requirements. (e.g: spreadsheets, word
processors, media players and database applications).
ix) Differentiate between byte address and word addressable computer.
Ans: A byte is a memory unit for storage and a memory chip is full of such bytes. Memory units
are addressable. That is the only way we can use memory. In reality memory is only byte
addressable. It means a binary address always points to a single byte only. A word is just a group
of bytes – 2, 4, 8 depending upon the data bus size of the CPU.
Word no 2 means bytes 4, 5, 6, 7 for 32 bit machine. In reality physical memory is byte
addressable only. So there is trick to handle 'word addressing'.
When MAR is placed on the address bus, its 32 bits do not map onto the 32 address lines 0-31
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 3
respectively. Instead, MAR bit 0 is wired to address bus line 2, MAR bit 1 is wired to address
bus line 3 and so on. The upper 2 bits of MAR are discarded since they are only needed for word
addresses above 2^32 none of which are legal for our 32 bit machine.
2.a) what are the different functional unit present in a digital computer. Briefly explain the function of
each unit.
Ans: FUNCTIONAL UNITS
In order to carry out the operations mentioned in the previous section the computer allocates the
task between its various functional units. The computer system is divided into three separate
units for its operation. They are
1) arithmetic logical unit
2) control unit.
3) central processing unit.
Arithmetic Logical Unit (ALU) Logical Unit
Logical Unit :After you enter data through the input device it is stored in the primary storage
unit. The actual processing of the data and instruction are performed by Arithmetic Logical Unit.
The major operations performed by the ALU are addition, subtraction, multiplication, division,
logic and comparison. Data is transferred to ALU from storage unit when required. After
processing the output is returned back to storage unit for further processing or getting stored.
Control Unit (CU) The next component of computer is the Control Unit, which acts like the
supervisor seeing that things are done in proper fashion. Control Unit is responsible for co
ordinating various operations using time signal. The control unit determines the sequence in
which computer programs and instructions are executed. Things like processing of programs
stored in the main memory, interpretation of the instructions and issuing of signals for other units
of the computer to execute them. It also acts as a switch board operator when several users
access the computer simultaneously. Thereby it coordinates the activities of computer’s
peripheral equipment as they perform the input and output.
Central Processing Unit (CPU) The ALU and the CU of a computer system are jointly known
as the central processing unit. You may call CPU as the brain of any computer system. It is just
like brain that takes all major decisions, makes all sorts of calculations and directs different parts
of the computer functions by activating and controlling the operations.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 4
3.a) how SDRAM is different from DDRSDRAM
Ans: SDRAM has typically less demands on power, is simpler in structure and fast. It's suitable for
simple, everyday electronic devices within larger products - washing machines, stereos, cars, etc. It's
also good for peripheral support in computers but not ideal for the main processors. Essentially, it's
good for quick, simple tasks;
* DDR RAM is more complex and sophisticated, and better-suited for the main processors in computers.
Essentially, it's good for complex tasks with high demands on processing and control.
b)
4. Set Associative mapping scheme combines the simplicity of direct mapping with the flexibility of Fully
Associative mapping. It is more practical than Fully Associative mapping because the associative portion
is limited to just a few slots that make up a set.
In this mapping mechanism, the cache memory is divided into 'v' sets, each consisting of 'n' cache lines.
A block from Main memory is first mapped onto a specific cache set, and then it can be placed anywhere
within that set. This type of mapping has very efficient ratio between implementation and efficiency.
The set is usually chosen by
Cache set number = (Main memory block number) MOD (Number of sets in the cache memory)
If there are 'n' cache lines in a set, the cache placement is called n-way set associative i.e. if there are
two blocks or cache lines per set, then it is a 2-way set associative cache mapping and four blocks or
cache lines per set, then it is a 4-way set associative cache mapping.
Let us assume we have a Main Memory of size 4GB (232
), with each byte directly addressable by a 32-bit
address. We will divide Main memory into blocks of each 32 bytes (25
). Thus there are 128M (i.e. 232
/25
=
227
) blocks in Main memory.
We have a Cache memory of 512KB (i.e. 219
), divided into blocks of each 32 bytes (25
). Thus there are
16K (i.e. 219
/25
= 214
) blocks also known as Cache slots or Cache lines in cache memory. It is clear from
above numbers that there are more Main memory blocks than Cache slots.
5. Illustrate Booth process multiplication with an example.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 5
Example of Booth’s Algorithm
6. Explain with an example how to design the hardwired control unit of a processor.
Ans: Hardwired control is a control mechanism to generate control signals by using appropriate
finite state machine (FSM). Microprogrammed control is a control mechanism to generate
control signals by using a memory called control storage (CS), which contains the control
signals. Although microprogrammed control seems to be advantageous to CISC machines, since
CISC requires systematic development of sophisticated control signals, there is no intrinsic
difference between these 2 control mechanisms.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 6
b) A Micro-programmed Control Unit
As we have seen, the controller causes instructions to be executed by issuing a specific set of
control signals at each beat of the system clock. Each set of control signals issued causes one
basic operation (micro-operation), such as a register transfer, to occur within the data path
section of the computer. In the case of a hard-wired control unit the control matrix is responsible
for sending out the required sequence of signals.
An alternative way of generating the control signals is that of micro-programmed control. In
order to understand this method it is convenient to think of sets of control signals that cause
specific micro-operations to occur as being "microinstructions" that could be stored in a memory.
Each bit of a microinstruction might correspond to one control signal. If the bit is set it means
that the control signal will be active; if cleared the signal will be inactive. Sequences of
microinstructions could be stored in an internal "control" memory. Execution of a machine
language instruction could then be caused by fetching the proper sequence of microinstructions
from the control memory and sending them out to the data path section of the computer. A
sequence of microinstructions that implements an instruction on the external computer is known
as a micro-routine. The instruction set of the computer is thus determined by the set of micro-
routines, the "microprogram," stored in the controller's memory. The control unit of a
microprogram-controlled computer is essentially a computer within a computer.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 7
7.a) Explain the wilkes model of micro programmed control unit.
Ans: Micro-programmed Control
 Use sequences of instructions (see earlier notes) to control complex operations
 Called micro-programming or firmware
• All the control unit does is generate a set of control signals
• Each control signal is on or off
• Represent each control signal by a bit
• Have a control word for each micro-operation
• Have a sequence of control words for each machine code instruction
• Add an address to specify the next micro-instruction, depending on conditions
Micro-instruction Types
• Each micro-instruction specifies single (or few) micro-operations to be performed
— (vertical micro-programming)
• Each micro-instruction specifies many different micro-operations to be performed in parallel
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 8
— (horizontal micro-programming)
Vertical Micro-programming
• Width is narrow
• n control signals encoded into log2 n bits
• Limited ability to express parallelism
• Considerable encoding of control information requires external memory word decoder to
identify the exact control line being manipulated
Horizontal Micro-programming
• Wide memory word
• High degree of parallel operations possible
• Little encoding of control information
8.a) Show the register organization of 8085 microprocessor.
Register organization of 8085 microprocessor.
The Status Flags of the 8080 and 8085 are single bits which indicate the logical conditions that existed as
a result of the execution of the instruction just completed. This allows instructions following to act
accordingly, such as a branch as a result of two values comparing equal. The flags are:
ZERO FLAG: This flag is set to a 1 by the instruction just ending if the A Register contains a result of all
0’s. Besides the obvious mathematical applications, this is useful in determining equality in a compare
operation (a value subtracted from a second value with an answer of 0), or in logical AND or OR
operations where the result left the A Register with no bit set to a 1 (the AND was not satisfied). If any
bits were left set to a 1 in the A Register, the flag will be reset to a 0 condition.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 9
SIGN FLAG: This flag is set to a 1 by the instruction just ending if the leftmost, or highest order, bit of the
A Register is set to a 1. The leftmost bit of a byte in signed arithmetic is the sign bit, and will be 0 if the
value in the lower seven bits is positive, and 1 if the value is negative.
PARITY FLAG: This flag is set to a 1 by the instruction just ending if the A Register is left with an even
number of bits set on, i.e., in even parity. If the number of bits in the A Register is odd, the bit is left off.
This may be useful in I/O operations with serial devices, or anyplace that error checking is to be done.
CARRY FLAG: This flag is set to a 1 by the instruction just ending if a carry out of the leftmost bit
occurred during the execution of the instruction. An example would be the addition of two 8-bit
numbers whose sum was 9 bits long. The 9th bit would be lost, yielding an erroneous answer if the carry
bit was not captured and held by this flag. This flag is also set if a borrow occurred during a subtraction.
AUXILIARY CARRY FLAG: This flag is set to a 1 by the instruction just ending if a carry occurred from bit 3
to bit 4 of the A Register during the instruction’s execution. Because of the relationships of decimal in
pure BCD to hexadecimal coding, it is possible to bring BCD values directly into the A Register and
perform mathematical operations on them. The result, however, will be as if two hex characters are
being processed. If the result must be returned to the program as BCD rather than as hex, the Decimal
Adjust Accumulator (DAA) instruction can make that translation; the Auxiliary Carry Flag is provided to
assist in this operation.
b) Explain the different addressing modes available in 8085 microprocessor.
 Every instruction of a program has to operate on a data.
 The method of specifying the data to be operated by the instruction is called Addressing.
 The 8085 has the following 5 different types of addressing.
1. Immediate Addressing
2. Direct Addressing
3. Register Addressing
4. Register Indirect Addressing
5. Implied Addressing
1. Immediate Addressing:
 In immediate addressing mode, the data is specified in the instruction itself. The data will be a
part of the program instruction.
 EX. MVI B, 3EH - Move the data 3EH given in the instruction to B register; LXI SP, 2700H.
2. Direct Addressing:
 In direct addressing mode, the address of the data is specified in the instruction. The data will be
in memory. In this addressing mode, the program instructions and data can be stored in
different memory.
GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY
FIFT SEMESTER EXAMINATION-2006
COMPUTER ARCHITECTURE & ORGANIZATION
FULL MARK: 70 TIME: - 3HOURS
Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 10
 EX. LDA 1050H - Load the data available in memory location 1050H in to accumulator; SHLD
3000H
3. Register Addressing:
 In register addressing mode, the instruction specifies the name of the register in which the data
is available.
 EX. MOV A, B - Move the content of B register to A register; SPHL; ADD C.
4. Register Indirect Addressing:
 In register indirect addressing mode, the instruction specifies the name of the register in which
the address of the data is available. Here the data will be in memory and the address will be in
the register pair.
 EX. MOV A, M - The memory data addressed by H L pair is moved to A register. LDAX B.
5. Implied Addressing:
 In implied addressing mode, the instruction itself specifies the data to be operated.
 EX. CMA - Complement the content of accumulator; RAL

Weitere ähnliche Inhalte

Was ist angesagt?

Mainmemoryfinal 161019122029
Mainmemoryfinal 161019122029Mainmemoryfinal 161019122029
Mainmemoryfinal 161019122029marangburu42
 
Computer memory book notes
Computer memory book notes Computer memory book notes
Computer memory book notes Navtej Duhoon
 
Mainmemoryfinalprefinal 160927115742
Mainmemoryfinalprefinal 160927115742Mainmemoryfinalprefinal 160927115742
Mainmemoryfinalprefinal 160927115742marangburu42
 
Computer Fundamentals Chapter 07 pam
Computer Fundamentals Chapter  07 pamComputer Fundamentals Chapter  07 pam
Computer Fundamentals Chapter 07 pamSaumya Sahu
 
Typical Embedded System
Typical Embedded SystemTypical Embedded System
Typical Embedded Systemanand hd
 
Computer Fundamentals Chapter 14 os
Computer Fundamentals Chapter 14 osComputer Fundamentals Chapter 14 os
Computer Fundamentals Chapter 14 osSaumya Sahu
 
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأول
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأولشيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأول
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأولأمنية وجدى
 
Chapter 1 - Introduction
Chapter 1 - IntroductionChapter 1 - Introduction
Chapter 1 - IntroductionWayne Jones Jnr
 
multiprocessors and multicomputers
 multiprocessors and multicomputers multiprocessors and multicomputers
multiprocessors and multicomputersPankaj Kumar Jain
 
Computer Fundamentals
Computer FundamentalsComputer Fundamentals
Computer FundamentalsSaumya Sahu
 

Was ist angesagt? (20)

Co question 2010
Co question 2010Co question 2010
Co question 2010
 
Mainmemoryfinal 161019122029
Mainmemoryfinal 161019122029Mainmemoryfinal 161019122029
Mainmemoryfinal 161019122029
 
Computer memory book notes
Computer memory book notes Computer memory book notes
Computer memory book notes
 
Unit 5-lecture-2
Unit 5-lecture-2Unit 5-lecture-2
Unit 5-lecture-2
 
Main memoryfinal
Main memoryfinalMain memoryfinal
Main memoryfinal
 
Mainmemoryfinalprefinal 160927115742
Mainmemoryfinalprefinal 160927115742Mainmemoryfinalprefinal 160927115742
Mainmemoryfinalprefinal 160927115742
 
Computer Fundamentals Chapter 07 pam
Computer Fundamentals Chapter  07 pamComputer Fundamentals Chapter  07 pam
Computer Fundamentals Chapter 07 pam
 
C q 1
C q 1C q 1
C q 1
 
notes2 memory_cpu
notes2 memory_cpunotes2 memory_cpu
notes2 memory_cpu
 
Typical Embedded System
Typical Embedded SystemTypical Embedded System
Typical Embedded System
 
Chapter 3
Chapter 3Chapter 3
Chapter 3
 
Os Question Bank
Os Question BankOs Question Bank
Os Question Bank
 
Computer Fundamentals Chapter 14 os
Computer Fundamentals Chapter 14 osComputer Fundamentals Chapter 14 os
Computer Fundamentals Chapter 14 os
 
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأول
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأولشيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأول
شيتات واجب كمبيوتر للصف الأول الإعدادى الترم الأول
 
Chapter 1 - Introduction
Chapter 1 - IntroductionChapter 1 - Introduction
Chapter 1 - Introduction
 
Memory management
Memory managementMemory management
Memory management
 
multiprocessors and multicomputers
 multiprocessors and multicomputers multiprocessors and multicomputers
multiprocessors and multicomputers
 
Cheap vs High End PC
Cheap vs High End PCCheap vs High End PC
Cheap vs High End PC
 
Unit 5-lecture-1
Unit 5-lecture-1Unit 5-lecture-1
Unit 5-lecture-1
 
Computer Fundamentals
Computer FundamentalsComputer Fundamentals
Computer Fundamentals
 

Andere mochten auch

Lesson plan proforma progrmming in c
Lesson plan proforma progrmming in cLesson plan proforma progrmming in c
Lesson plan proforma progrmming in cSANTOSH RATH
 
Progr in c lesson plan
Progr in c lesson planProgr in c lesson plan
Progr in c lesson planSANTOSH RATH
 
( Becs 2208 ) database management system
( Becs 2208 ) database management system( Becs 2208 ) database management system
( Becs 2208 ) database management systemSANTOSH RATH
 
Btech 2nd ds_2008.ppt
Btech 2nd ds_2008.pptBtech 2nd ds_2008.ppt
Btech 2nd ds_2008.pptSANTOSH RATH
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tcSANTOSH RATH
 
Expected questions for dbms
Expected questions for dbmsExpected questions for dbms
Expected questions for dbmsSANTOSH RATH
 
Module wise format oops questions
Module wise format oops questionsModule wise format oops questions
Module wise format oops questionsSANTOSH RATH
 
Iwt note(module 2)
Iwt note(module 2)Iwt note(module 2)
Iwt note(module 2)SANTOSH RATH
 
Data structure new lab manual
Data structure  new lab manualData structure  new lab manual
Data structure new lab manualSANTOSH RATH
 
Operating system notes
Operating system notesOperating system notes
Operating system notesSANTOSH RATH
 

Andere mochten auch (17)

Structures
StructuresStructures
Structures
 
Lesson plan proforma progrmming in c
Lesson plan proforma progrmming in cLesson plan proforma progrmming in c
Lesson plan proforma progrmming in c
 
Progr1
Progr1Progr1
Progr1
 
2006dbms
2006dbms2006dbms
2006dbms
 
Progr in c lesson plan
Progr in c lesson planProgr in c lesson plan
Progr in c lesson plan
 
( Becs 2208 ) database management system
( Becs 2208 ) database management system( Becs 2208 ) database management system
( Becs 2208 ) database management system
 
Files
FilesFiles
Files
 
2011dbms
2011dbms2011dbms
2011dbms
 
Dma
DmaDma
Dma
 
Btech 2nd ds_2008.ppt
Btech 2nd ds_2008.pptBtech 2nd ds_2008.ppt
Btech 2nd ds_2008.ppt
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tc
 
Array
ArrayArray
Array
 
Expected questions for dbms
Expected questions for dbmsExpected questions for dbms
Expected questions for dbms
 
Module wise format oops questions
Module wise format oops questionsModule wise format oops questions
Module wise format oops questions
 
Iwt note(module 2)
Iwt note(module 2)Iwt note(module 2)
Iwt note(module 2)
 
Data structure new lab manual
Data structure  new lab manualData structure  new lab manual
Data structure new lab manual
 
Operating system notes
Operating system notesOperating system notes
Operating system notes
 

Ähnlich wie GANDH INSTITUTE COMPUTER EXAM

Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)myanddy
 
Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)myanddy
 
Assignment#1 Mapacpac, F M P (Cs3112 Os)
Assignment#1 Mapacpac, F M P  (Cs3112 Os)Assignment#1 Mapacpac, F M P  (Cs3112 Os)
Assignment#1 Mapacpac, F M P (Cs3112 Os)dyandmy
 
Multilevel arch & str org.& mips, 8086, memory
Multilevel arch & str org.& mips, 8086, memoryMultilevel arch & str org.& mips, 8086, memory
Multilevel arch & str org.& mips, 8086, memoryMahesh Kumar Attri
 
Bab 4
Bab 4Bab 4
Bab 4n k
 
System Programming- Unit I
System Programming- Unit ISystem Programming- Unit I
System Programming- Unit ISaranya1702
 
Chapter 8 : Memory
Chapter 8 : MemoryChapter 8 : Memory
Chapter 8 : MemoryAmin Omi
 
CS403: Operating System : Unit I _merged.pdf
CS403: Operating System :  Unit I _merged.pdfCS403: Operating System :  Unit I _merged.pdf
CS403: Operating System : Unit I _merged.pdfAsst.prof M.Gokilavani
 
UNIT 2_ESD.pdf
UNIT 2_ESD.pdfUNIT 2_ESD.pdf
UNIT 2_ESD.pdfSaralaT3
 
Operating Systems - memory management
Operating Systems - memory managementOperating Systems - memory management
Operating Systems - memory managementMukesh Chinta
 
Computer System.ppt
Computer System.pptComputer System.ppt
Computer System.pptjguuhxxxfp
 
LM1 - Computer System Overview, system calls
LM1 - Computer System Overview, system callsLM1 - Computer System Overview, system calls
LM1 - Computer System Overview, system callsmanideepakc
 
please answer these questions number by numberSolution1) An.pdf
please answer these questions number by numberSolution1) An.pdfplease answer these questions number by numberSolution1) An.pdf
please answer these questions number by numberSolution1) An.pdfarishaenterprises12
 
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdfaniyathikitchen
 
Joemary.doc
Joemary.docJoemary.doc
Joemary.doccas123
 
Co question bank LAKSHMAIAH
Co question bank LAKSHMAIAH Co question bank LAKSHMAIAH
Co question bank LAKSHMAIAH veena babu
 

Ähnlich wie GANDH INSTITUTE COMPUTER EXAM (20)

Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)
 
Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)Assignment#1 lograbo, s.f. (cs3112-os)
Assignment#1 lograbo, s.f. (cs3112-os)
 
Assignment#1 Mapacpac, F M P (Cs3112 Os)
Assignment#1 Mapacpac, F M P  (Cs3112 Os)Assignment#1 Mapacpac, F M P  (Cs3112 Os)
Assignment#1 Mapacpac, F M P (Cs3112 Os)
 
Multilevel arch & str org.& mips, 8086, memory
Multilevel arch & str org.& mips, 8086, memoryMultilevel arch & str org.& mips, 8086, memory
Multilevel arch & str org.& mips, 8086, memory
 
Reconfigurable computing
Reconfigurable computingReconfigurable computing
Reconfigurable computing
 
Cao 2012
Cao 2012Cao 2012
Cao 2012
 
Co notes3 sem
Co notes3 semCo notes3 sem
Co notes3 sem
 
Bab 4
Bab 4Bab 4
Bab 4
 
System Programming- Unit I
System Programming- Unit ISystem Programming- Unit I
System Programming- Unit I
 
CS6303 Computer Architecture.pdf
CS6303 Computer Architecture.pdfCS6303 Computer Architecture.pdf
CS6303 Computer Architecture.pdf
 
Chapter 8 : Memory
Chapter 8 : MemoryChapter 8 : Memory
Chapter 8 : Memory
 
CS403: Operating System : Unit I _merged.pdf
CS403: Operating System :  Unit I _merged.pdfCS403: Operating System :  Unit I _merged.pdf
CS403: Operating System : Unit I _merged.pdf
 
UNIT 2_ESD.pdf
UNIT 2_ESD.pdfUNIT 2_ESD.pdf
UNIT 2_ESD.pdf
 
Operating Systems - memory management
Operating Systems - memory managementOperating Systems - memory management
Operating Systems - memory management
 
Computer System.ppt
Computer System.pptComputer System.ppt
Computer System.ppt
 
LM1 - Computer System Overview, system calls
LM1 - Computer System Overview, system callsLM1 - Computer System Overview, system calls
LM1 - Computer System Overview, system calls
 
please answer these questions number by numberSolution1) An.pdf
please answer these questions number by numberSolution1) An.pdfplease answer these questions number by numberSolution1) An.pdf
please answer these questions number by numberSolution1) An.pdf
 
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf
1.CPU INSTRUCTION AND EXECUTION CYCLEThe primary function of the .pdf
 
Joemary.doc
Joemary.docJoemary.doc
Joemary.doc
 
Co question bank LAKSHMAIAH
Co question bank LAKSHMAIAH Co question bank LAKSHMAIAH
Co question bank LAKSHMAIAH
 

Mehr von SANTOSH RATH

Lesson plan proforma database management system
Lesson plan proforma database management systemLesson plan proforma database management system
Lesson plan proforma database management systemSANTOSH RATH
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tcSANTOSH RATH
 
Expected Questions TC
Expected Questions TCExpected Questions TC
Expected Questions TCSANTOSH RATH
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tcSANTOSH RATH
 
Expected questions for dbms
Expected questions for dbmsExpected questions for dbms
Expected questions for dbmsSANTOSH RATH
 
Oops model question
Oops model questionOops model question
Oops model questionSANTOSH RATH
 
System programming note
System programming noteSystem programming note
System programming noteSANTOSH RATH
 
Data structure using c bcse 3102 pcs 1002
Data structure using c bcse 3102 pcs 1002Data structure using c bcse 3102 pcs 1002
Data structure using c bcse 3102 pcs 1002SANTOSH RATH
 
Btech 2nd ds_2007.ppt
Btech 2nd ds_2007.pptBtech 2nd ds_2007.ppt
Btech 2nd ds_2007.pptSANTOSH RATH
 
Btech 2nd ds_2005.ppt
Btech 2nd ds_2005.pptBtech 2nd ds_2005.ppt
Btech 2nd ds_2005.pptSANTOSH RATH
 
Ds lab manual by s.k.rath
Ds lab manual by s.k.rathDs lab manual by s.k.rath
Ds lab manual by s.k.rathSANTOSH RATH
 

Mehr von SANTOSH RATH (18)

Lesson plan proforma database management system
Lesson plan proforma database management systemLesson plan proforma database management system
Lesson plan proforma database management system
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tc
 
Rdbms2010
Rdbms2010Rdbms2010
Rdbms2010
 
Expected Questions TC
Expected Questions TCExpected Questions TC
Expected Questions TC
 
Expected questions tc
Expected questions tcExpected questions tc
Expected questions tc
 
Expected questions for dbms
Expected questions for dbmsExpected questions for dbms
Expected questions for dbms
 
Oops model question
Oops model questionOops model question
Oops model question
 
System programming note
System programming noteSystem programming note
System programming note
 
Os notes
Os notesOs notes
Os notes
 
OS ASSIGNMENT 2
OS ASSIGNMENT 2OS ASSIGNMENT 2
OS ASSIGNMENT 2
 
OS ASSIGNMENT-1
OS ASSIGNMENT-1OS ASSIGNMENT-1
OS ASSIGNMENT-1
 
OS ASSIGNMENT 3
OS ASSIGNMENT 3OS ASSIGNMENT 3
OS ASSIGNMENT 3
 
Ds using c 2009
Ds using c 2009Ds using c 2009
Ds using c 2009
 
Data structure using c bcse 3102 pcs 1002
Data structure using c bcse 3102 pcs 1002Data structure using c bcse 3102 pcs 1002
Data structure using c bcse 3102 pcs 1002
 
Btech 2nd ds_2007.ppt
Btech 2nd ds_2007.pptBtech 2nd ds_2007.ppt
Btech 2nd ds_2007.ppt
 
Btech 2nd ds_2005.ppt
Btech 2nd ds_2005.pptBtech 2nd ds_2005.ppt
Btech 2nd ds_2005.ppt
 
Ds using c 2011
Ds using c 2011Ds using c 2011
Ds using c 2011
 
Ds lab manual by s.k.rath
Ds lab manual by s.k.rathDs lab manual by s.k.rath
Ds lab manual by s.k.rath
 

Kürzlich hochgeladen

OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...
OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...
OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...Soham Mondal
 
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLS
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLSMANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLS
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLSSIVASHANKAR N
 
Coefficient of Thermal Expansion and their Importance.pptx
Coefficient of Thermal Expansion and their Importance.pptxCoefficient of Thermal Expansion and their Importance.pptx
Coefficient of Thermal Expansion and their Importance.pptxAsutosh Ranjan
 
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...ranjana rawat
 
Processing & Properties of Floor and Wall Tiles.pptx
Processing & Properties of Floor and Wall Tiles.pptxProcessing & Properties of Floor and Wall Tiles.pptx
Processing & Properties of Floor and Wall Tiles.pptxpranjaldaimarysona
 
Extrusion Processes and Their Limitations
Extrusion Processes and Their LimitationsExtrusion Processes and Their Limitations
Extrusion Processes and Their Limitations120cr0395
 
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICSHARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICSRajkumarAkumalla
 
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...Christo Ananth
 
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur Escorts
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur EscortsCall Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur Escorts
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur EscortsCall Girls in Nagpur High Profile
 
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Serviceranjana rawat
 
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...ranjana rawat
 
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130Suhani Kapoor
 
Introduction to IEEE STANDARDS and its different types.pptx
Introduction to IEEE STANDARDS and its different types.pptxIntroduction to IEEE STANDARDS and its different types.pptx
Introduction to IEEE STANDARDS and its different types.pptxupamatechverse
 
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur Escorts
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur EscortsHigh Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur Escorts
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur EscortsCall Girls in Nagpur High Profile
 
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...ranjana rawat
 
Software Development Life Cycle By Team Orange (Dept. of Pharmacy)
Software Development Life Cycle By  Team Orange (Dept. of Pharmacy)Software Development Life Cycle By  Team Orange (Dept. of Pharmacy)
Software Development Life Cycle By Team Orange (Dept. of Pharmacy)Suman Mia
 
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).pptssuser5c9d4b1
 
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICS
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICSAPPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICS
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICSKurinjimalarL3
 
Introduction to Multiple Access Protocol.pptx
Introduction to Multiple Access Protocol.pptxIntroduction to Multiple Access Protocol.pptx
Introduction to Multiple Access Protocol.pptxupamatechverse
 

Kürzlich hochgeladen (20)

OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...
OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...
OSVC_Meta-Data based Simulation Automation to overcome Verification Challenge...
 
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLS
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLSMANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLS
MANUFACTURING PROCESS-II UNIT-5 NC MACHINE TOOLS
 
Coefficient of Thermal Expansion and their Importance.pptx
Coefficient of Thermal Expansion and their Importance.pptxCoefficient of Thermal Expansion and their Importance.pptx
Coefficient of Thermal Expansion and their Importance.pptx
 
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(ANJALI) Dange Chowk Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
 
Processing & Properties of Floor and Wall Tiles.pptx
Processing & Properties of Floor and Wall Tiles.pptxProcessing & Properties of Floor and Wall Tiles.pptx
Processing & Properties of Floor and Wall Tiles.pptx
 
Extrusion Processes and Their Limitations
Extrusion Processes and Their LimitationsExtrusion Processes and Their Limitations
Extrusion Processes and Their Limitations
 
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICSHARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
 
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...
Call for Papers - Educational Administration: Theory and Practice, E-ISSN: 21...
 
DJARUM4D - SLOT GACOR ONLINE | SLOT DEMO ONLINE
DJARUM4D - SLOT GACOR ONLINE | SLOT DEMO ONLINEDJARUM4D - SLOT GACOR ONLINE | SLOT DEMO ONLINE
DJARUM4D - SLOT GACOR ONLINE | SLOT DEMO ONLINE
 
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur Escorts
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur EscortsCall Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur Escorts
Call Girls Service Nagpur Tanvi Call 7001035870 Meet With Nagpur Escorts
 
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service
(RIA) Call Girls Bhosari ( 7001035870 ) HI-Fi Pune Escorts Service
 
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
(PRIYA) Rajgurunagar Call Girls Just Call 7001035870 [ Cash on Delivery ] Pun...
 
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130
VIP Call Girls Service Hitech City Hyderabad Call +91-8250192130
 
Introduction to IEEE STANDARDS and its different types.pptx
Introduction to IEEE STANDARDS and its different types.pptxIntroduction to IEEE STANDARDS and its different types.pptx
Introduction to IEEE STANDARDS and its different types.pptx
 
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur Escorts
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur EscortsHigh Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur Escorts
High Profile Call Girls Nagpur Meera Call 7001035870 Meet With Nagpur Escorts
 
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
 
Software Development Life Cycle By Team Orange (Dept. of Pharmacy)
Software Development Life Cycle By  Team Orange (Dept. of Pharmacy)Software Development Life Cycle By  Team Orange (Dept. of Pharmacy)
Software Development Life Cycle By Team Orange (Dept. of Pharmacy)
 
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt
247267395-1-Symmetric-and-distributed-shared-memory-architectures-ppt (1).ppt
 
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICS
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICSAPPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICS
APPLICATIONS-AC/DC DRIVES-OPERATING CHARACTERISTICS
 
Introduction to Multiple Access Protocol.pptx
Introduction to Multiple Access Protocol.pptxIntroduction to Multiple Access Protocol.pptx
Introduction to Multiple Access Protocol.pptx
 

GANDH INSTITUTE COMPUTER EXAM

  • 1. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 1 Answer Question No.1 which is compulsory and any five from the rest. The figure in the right hand margin indicates 2*10=20 1. Answer the following question: i) How logical address is different from physical address. Ans: Physical address: Physical address is the actual memory address which denotes a memory area in your storage device. Logical address: The operating system or some other programs uses base addresses as a measure to find addresses. Here base address means starting address of a memory block. According to the program written the CPU generates a address. This address is added with the base address so that it forms the physical address. Here in this case the address generated by the CPU is called as Logical Address. Explain how a ROM is also a RAM. ii) Explain how a ROM is a RAM? Ans: RAM - Is memory that can be both written to and read at anytime during processing. ROM - Is Random Access Memory which can be anything from CD's that have been used that aren't writable. OR chips that are meant only for storing data PERMANENTLY. Basicly anything that holds digital data permantently and can no longer be written to. iii) Define the term memory latency. Ans: "Latency" in computer terms is how long it takes to get a request back. It's the flip side of the coin from "bandwidth", which is how fast the data flows. iv) What is difference between EPROM and EEPROM? Ans: Erasable Programmable Read Only Memory (EPROM) Chips the information stored in an EPROM chip can be erased by exposing the chip to strong UV light. EPROM chips are easily recognized by the small quartz window used for erasure. Once erased the chip can be re-programmed. EPROM is more expensive to buy per unit cost, but can prove cheaper in the long run for some applications. For example if PROM was used for firmware that needed upgraded every 6 months or so - it could prove quite expensive buying new chips! Electronically Erasable Programmable Read Only Memory: This has the added advantage that the information stored can be re-written in blocks and hence can be used to store system settings that the user may want to change periodically. This solid state memory has considerably reduced in price over recent years and is nowadays commonly used to store system settings such as BIOS settings.
  • 2. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 2 v) How a write through cache differ from write back cache? Ans: A type of cache with the following feature: when changes are made to cached data, they are not simultaneously made to the original data as well. Instead, the changed data is marked, and the original data is updated when the chached data is deallocated. In a write-through cache, by contrast changes made to cached data are simultaneously made in the original copy. A write-back cache can perform more quickly than a write-through cache. But in some contexts, differences between cached and original data could lead to problems, and write-through caches must be used vi) Define Control word Ans: The combination of control steps used for the generation of control signals is a control word. A control word is a word whose individual bits represent the various control signals. In computing, word is a term for the natural unit of data used by a particular processor design. A word is basically a fixed sized group of bits that are handled as a unit by the instruction set and/or hardware of the processor. The number of bits in a word (the word size, word width, or word length) is an important characteristic of a specific processor design or computer architecture. vii) Define the term memory access time: Ans: The time required by a processor to access data or to write data from and to memory chip is referred as access time. viii) Differentiate between system software and application software: Ans: System software performs tasks like transferring data from memory to disk, or rendering text onto a display device. Specific kinds of system software include loading programs, operating systems, device drivers, programming tools, compilers, assemblers, linkers, and utility software. Application software are designed to perform specific data processing or computational tasks for the user. These programs are specifically designed to meet end-user requirements. (e.g: spreadsheets, word processors, media players and database applications). ix) Differentiate between byte address and word addressable computer. Ans: A byte is a memory unit for storage and a memory chip is full of such bytes. Memory units are addressable. That is the only way we can use memory. In reality memory is only byte addressable. It means a binary address always points to a single byte only. A word is just a group of bytes – 2, 4, 8 depending upon the data bus size of the CPU. Word no 2 means bytes 4, 5, 6, 7 for 32 bit machine. In reality physical memory is byte addressable only. So there is trick to handle 'word addressing'. When MAR is placed on the address bus, its 32 bits do not map onto the 32 address lines 0-31
  • 3. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 3 respectively. Instead, MAR bit 0 is wired to address bus line 2, MAR bit 1 is wired to address bus line 3 and so on. The upper 2 bits of MAR are discarded since they are only needed for word addresses above 2^32 none of which are legal for our 32 bit machine. 2.a) what are the different functional unit present in a digital computer. Briefly explain the function of each unit. Ans: FUNCTIONAL UNITS In order to carry out the operations mentioned in the previous section the computer allocates the task between its various functional units. The computer system is divided into three separate units for its operation. They are 1) arithmetic logical unit 2) control unit. 3) central processing unit. Arithmetic Logical Unit (ALU) Logical Unit Logical Unit :After you enter data through the input device it is stored in the primary storage unit. The actual processing of the data and instruction are performed by Arithmetic Logical Unit. The major operations performed by the ALU are addition, subtraction, multiplication, division, logic and comparison. Data is transferred to ALU from storage unit when required. After processing the output is returned back to storage unit for further processing or getting stored. Control Unit (CU) The next component of computer is the Control Unit, which acts like the supervisor seeing that things are done in proper fashion. Control Unit is responsible for co ordinating various operations using time signal. The control unit determines the sequence in which computer programs and instructions are executed. Things like processing of programs stored in the main memory, interpretation of the instructions and issuing of signals for other units of the computer to execute them. It also acts as a switch board operator when several users access the computer simultaneously. Thereby it coordinates the activities of computer’s peripheral equipment as they perform the input and output. Central Processing Unit (CPU) The ALU and the CU of a computer system are jointly known as the central processing unit. You may call CPU as the brain of any computer system. It is just like brain that takes all major decisions, makes all sorts of calculations and directs different parts of the computer functions by activating and controlling the operations.
  • 4. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 4 3.a) how SDRAM is different from DDRSDRAM Ans: SDRAM has typically less demands on power, is simpler in structure and fast. It's suitable for simple, everyday electronic devices within larger products - washing machines, stereos, cars, etc. It's also good for peripheral support in computers but not ideal for the main processors. Essentially, it's good for quick, simple tasks; * DDR RAM is more complex and sophisticated, and better-suited for the main processors in computers. Essentially, it's good for complex tasks with high demands on processing and control. b) 4. Set Associative mapping scheme combines the simplicity of direct mapping with the flexibility of Fully Associative mapping. It is more practical than Fully Associative mapping because the associative portion is limited to just a few slots that make up a set. In this mapping mechanism, the cache memory is divided into 'v' sets, each consisting of 'n' cache lines. A block from Main memory is first mapped onto a specific cache set, and then it can be placed anywhere within that set. This type of mapping has very efficient ratio between implementation and efficiency. The set is usually chosen by Cache set number = (Main memory block number) MOD (Number of sets in the cache memory) If there are 'n' cache lines in a set, the cache placement is called n-way set associative i.e. if there are two blocks or cache lines per set, then it is a 2-way set associative cache mapping and four blocks or cache lines per set, then it is a 4-way set associative cache mapping. Let us assume we have a Main Memory of size 4GB (232 ), with each byte directly addressable by a 32-bit address. We will divide Main memory into blocks of each 32 bytes (25 ). Thus there are 128M (i.e. 232 /25 = 227 ) blocks in Main memory. We have a Cache memory of 512KB (i.e. 219 ), divided into blocks of each 32 bytes (25 ). Thus there are 16K (i.e. 219 /25 = 214 ) blocks also known as Cache slots or Cache lines in cache memory. It is clear from above numbers that there are more Main memory blocks than Cache slots. 5. Illustrate Booth process multiplication with an example.
  • 5. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 5 Example of Booth’s Algorithm 6. Explain with an example how to design the hardwired control unit of a processor. Ans: Hardwired control is a control mechanism to generate control signals by using appropriate finite state machine (FSM). Microprogrammed control is a control mechanism to generate control signals by using a memory called control storage (CS), which contains the control signals. Although microprogrammed control seems to be advantageous to CISC machines, since CISC requires systematic development of sophisticated control signals, there is no intrinsic difference between these 2 control mechanisms.
  • 6. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 6 b) A Micro-programmed Control Unit As we have seen, the controller causes instructions to be executed by issuing a specific set of control signals at each beat of the system clock. Each set of control signals issued causes one basic operation (micro-operation), such as a register transfer, to occur within the data path section of the computer. In the case of a hard-wired control unit the control matrix is responsible for sending out the required sequence of signals. An alternative way of generating the control signals is that of micro-programmed control. In order to understand this method it is convenient to think of sets of control signals that cause specific micro-operations to occur as being "microinstructions" that could be stored in a memory. Each bit of a microinstruction might correspond to one control signal. If the bit is set it means that the control signal will be active; if cleared the signal will be inactive. Sequences of microinstructions could be stored in an internal "control" memory. Execution of a machine language instruction could then be caused by fetching the proper sequence of microinstructions from the control memory and sending them out to the data path section of the computer. A sequence of microinstructions that implements an instruction on the external computer is known as a micro-routine. The instruction set of the computer is thus determined by the set of micro- routines, the "microprogram," stored in the controller's memory. The control unit of a microprogram-controlled computer is essentially a computer within a computer.
  • 7. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 7 7.a) Explain the wilkes model of micro programmed control unit. Ans: Micro-programmed Control  Use sequences of instructions (see earlier notes) to control complex operations  Called micro-programming or firmware • All the control unit does is generate a set of control signals • Each control signal is on or off • Represent each control signal by a bit • Have a control word for each micro-operation • Have a sequence of control words for each machine code instruction • Add an address to specify the next micro-instruction, depending on conditions Micro-instruction Types • Each micro-instruction specifies single (or few) micro-operations to be performed — (vertical micro-programming) • Each micro-instruction specifies many different micro-operations to be performed in parallel
  • 8. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 8 — (horizontal micro-programming) Vertical Micro-programming • Width is narrow • n control signals encoded into log2 n bits • Limited ability to express parallelism • Considerable encoding of control information requires external memory word decoder to identify the exact control line being manipulated Horizontal Micro-programming • Wide memory word • High degree of parallel operations possible • Little encoding of control information 8.a) Show the register organization of 8085 microprocessor. Register organization of 8085 microprocessor. The Status Flags of the 8080 and 8085 are single bits which indicate the logical conditions that existed as a result of the execution of the instruction just completed. This allows instructions following to act accordingly, such as a branch as a result of two values comparing equal. The flags are: ZERO FLAG: This flag is set to a 1 by the instruction just ending if the A Register contains a result of all 0’s. Besides the obvious mathematical applications, this is useful in determining equality in a compare operation (a value subtracted from a second value with an answer of 0), or in logical AND or OR operations where the result left the A Register with no bit set to a 1 (the AND was not satisfied). If any bits were left set to a 1 in the A Register, the flag will be reset to a 0 condition.
  • 9. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 9 SIGN FLAG: This flag is set to a 1 by the instruction just ending if the leftmost, or highest order, bit of the A Register is set to a 1. The leftmost bit of a byte in signed arithmetic is the sign bit, and will be 0 if the value in the lower seven bits is positive, and 1 if the value is negative. PARITY FLAG: This flag is set to a 1 by the instruction just ending if the A Register is left with an even number of bits set on, i.e., in even parity. If the number of bits in the A Register is odd, the bit is left off. This may be useful in I/O operations with serial devices, or anyplace that error checking is to be done. CARRY FLAG: This flag is set to a 1 by the instruction just ending if a carry out of the leftmost bit occurred during the execution of the instruction. An example would be the addition of two 8-bit numbers whose sum was 9 bits long. The 9th bit would be lost, yielding an erroneous answer if the carry bit was not captured and held by this flag. This flag is also set if a borrow occurred during a subtraction. AUXILIARY CARRY FLAG: This flag is set to a 1 by the instruction just ending if a carry occurred from bit 3 to bit 4 of the A Register during the instruction’s execution. Because of the relationships of decimal in pure BCD to hexadecimal coding, it is possible to bring BCD values directly into the A Register and perform mathematical operations on them. The result, however, will be as if two hex characters are being processed. If the result must be returned to the program as BCD rather than as hex, the Decimal Adjust Accumulator (DAA) instruction can make that translation; the Auxiliary Carry Flag is provided to assist in this operation. b) Explain the different addressing modes available in 8085 microprocessor.  Every instruction of a program has to operate on a data.  The method of specifying the data to be operated by the instruction is called Addressing.  The 8085 has the following 5 different types of addressing. 1. Immediate Addressing 2. Direct Addressing 3. Register Addressing 4. Register Indirect Addressing 5. Implied Addressing 1. Immediate Addressing:  In immediate addressing mode, the data is specified in the instruction itself. The data will be a part of the program instruction.  EX. MVI B, 3EH - Move the data 3EH given in the instruction to B register; LXI SP, 2700H. 2. Direct Addressing:  In direct addressing mode, the address of the data is specified in the instruction. The data will be in memory. In this addressing mode, the program instructions and data can be stored in different memory.
  • 10. GANDH INSTITUTE FOR EDUCATION & TECHNOLOGY FIFT SEMESTER EXAMINATION-2006 COMPUTER ARCHITECTURE & ORGANIZATION FULL MARK: 70 TIME: - 3HOURS Prepared by Asst.Prof. Santosh Kumar Rath (CSE DEPARTMENT) Page 10  EX. LDA 1050H - Load the data available in memory location 1050H in to accumulator; SHLD 3000H 3. Register Addressing:  In register addressing mode, the instruction specifies the name of the register in which the data is available.  EX. MOV A, B - Move the content of B register to A register; SPHL; ADD C. 4. Register Indirect Addressing:  In register indirect addressing mode, the instruction specifies the name of the register in which the address of the data is available. Here the data will be in memory and the address will be in the register pair.  EX. MOV A, M - The memory data addressed by H L pair is moved to A register. LDAX B. 5. Implied Addressing:  In implied addressing mode, the instruction itself specifies the data to be operated.  EX. CMA - Complement the content of accumulator; RAL