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COMPUTER ARCHITECTURE –CS1251

UNIT I- BASIC STRUCTURE OF COMPUTER

 

 

  1. What are the basic functional units of a computer?

Ans: A computer consists of five functionally independent main parts namely

  1.  
    • Input Unit
    • Memory Unit
    • Arithmetic and logic Unit
    • Output Unit
    • Control Unit

 

  1. Define RAM.

Ans: Memory in which any location can be reached in a short and fixed amount of time after specifying its address is called random access memory.

 

  1. Define memory access time.

Ans: The time required to access one word is called memory access time.

 

  1. What is instruction register (IR) and program counter (PC) used for ?

Ans: The instruction register (IR) holds the instruction that is currently being executed .Its output is available to the control circuits which generate the timing signals that control the various processing elements.

The program counter PC) is used to keep track of the execution of the program. It contains the memory address of the next instruction to be fetched and executed.

 

  1. What do you mean by memory address register(MAR) and memory data register(MDR)?

Ans: The MAR holds the address of the location to be accessed. The MDR contains the data to be written into or read out of the addressed location.

 

  1. What is an interrupt?

Ans: An interrupt is a request from an I/O device for service by the processor. The processor provides the requested service by executing an appropriate interrupt service routine.

 

  1. Explain about Bus.

Ans: Bus is a group of lines that serves as a connecting path for several devices. In addition to the lines that carry the data , the bus must have the lines for address and control purposes.

 

 

  1. What do you mean by multiprogramming or multitasking?

Ans: The operating system manages the concurrent execution of several application programs to make best possible use of computer resources. This pattern of concurrent execution is called multiprogramming or multitasking.

 

  1. Give the basic performance equation.

Ans: The basic performance equation is given as

                                                    T = N ´ S/ R

T = It is the processor time required to execute a program

N= It is the actual number of instruction executions.

S = It is the average number of basic steps neede to execute one machine instruction.

R =  It is the clock rate.

 

  1. Explain the concept of pipelining.

Ans: Pipelining is the means of executing machine instructions concurrently. It is the effective way of organizing concurrent activity in a computer system. It is a process of substantial improvement in the performance by overlapping the execution of successive instructions.

 

  1. What are the two techniques used to increase the clock rate R?

Ans: The two techniques used to increase the clock rate R are:

1. The integrated – circuit (IC) technology can be increased which reduces the time needed to complete a basic step.

2. We can reduce the amount of processing done in one basic step.

 

  1. What is Big – Endian  and Little- Endian representations.

Ans: The Big- endian is used when lower byte addresses are used for the more significant bytes (The leftmost bytes) of the word.

The little-endian is used for the opposite ordering, where the lower byte addresses are used for the less significant bytes ( the rightmost bytes) of the word.

 

  1. What is addressing mode?

Ans: The addressing mode is defined as the different ways in which the location of an operand is specified in an instruction.

 

  1. What are the different types of addressing modes available?

Ans: The different types of addressing modes available are:

  • Immediate addressing mode
  • Register addressing mode
  • Direct or absolute addressing mode
  • Indirect addressing mode
  • Indexed addressing mode
  • Relative addressing mode
  • Autoincrement
  • Autodecrement
  1. What is indirect addressing mode?

Ans: The effective address of the operand is the contents of a register or memory location whose address appears in the instruction

 

  1. What is indexed addressing mode?

Ans: The effective address of the operand is generated by adding a constant value to the contents of a register.

 

  1. Define autoincrement mode of addressing?

Ans: The effective address of the operand is the contents of a register specified in the instruction. After accessing the operand, the contents of this register are automatically incremented to point to the next item in the list.

 

  1. Define autodecrement mode of addressing?

Ans: The contents of a register specified in the instruction are first automatically decremented and are then used as the effective address of the operand.

 

  1. What are condition code flags? What are the commonly used flags?

Ans:The processor has to keep track of the information about the results of various operations for the subsequent conditional branch instructions. This is done by recording required information in individual bits called condition code flags.

Four commonly used flags are:

  • N( Negative )
  • Z(Zero)
  • V(overflow)
  • C(Carry)

 

  1. What do you mean by assembler directives?

Ans: These are the instructions which direct the program to be executed. They have no binary equivalent so they are called pseudo-opcodes. These instructions are used to define symbols, allocate space for variable, generate fixed tables etc.

Examples : END, NAME

 

  1. What is use of loader?

Ans: Loader is utility program which is stored in the memory. When the loader program is executed it performs a sequence of input operations needed to transfer the machine language program from the disk into a specified place in the memory.

 

  1. What do you man by relative addressing mode?

Ans: The effective address is determined by the index mode using the program counter in place of the general purpose register Ri.

 

  1. What is Stack?

Ans: A stack is a list of data elements, usually words or bytes with the accessing restriction that elements can be added or removed at one end of the list only. It follows last in first out (LIFO) mechanism.

24. What is a queue?

Ans: Is a type of datastructure in which the data are stored in and retrieved on a First in first out(FIFO) basis. It grows in the direction of increasing addresses in the memory. New data are added at the back (High-address end) and retrieved from the front (low-address end) of the queue.

 

25. What are the difference between Stack and Queue?

 

STACK

QUEUE

One end of stack is fixed ( the bottom) while the other end rises and falls as data are pushed and popped Both ends of a queue move to higher addresses
Single Pointer is needed to point to the top of the stack Two pointers are needed to keep track of the two ends of the queue

 

 

UNIT II – ARITHMETIC UNIT

 

1.      What is half adder?

Ans: A half adder is a logic circuit with two inputs and two outputs, which adds two bits at a time, producing a sum and a carry.

 

2. What is full adder?

Ans: A full adder is logic circuit with three inputs and two outputs, which adds three bits at a time giving a sum and a carry.

 

3. What is signed binary?

Ans: A system in which the leading bit represents the sign and the remaining bits the magnitude of the number is called signed binary. This is also known as sign magnitude.

 

4. What are the two approaches used to reduce delay in adders?

Ans:1) The first approach is to use the fastest possible electronic technology in implementing the ripple-carry  logic design.

       2) The second approach is to use an augmented logic gate network structure.

 

5. What is a carry look-ahead adder?

Ans: The input carry needed by a stage is directly computed from carry signals obtained from all the preceding stages i-1,i-2,…..0, rather than waiting for normal carries to supply slowly from stage to stage. An adder that uses this principle isa called carry look-ahead adder.

 

6. What are the main features of Booth’s algorithm?

Ans:           1) It handles both positive and negative multipliers uniformly.

2) It achieves some efficiency in the number of addition required when the multiplier has a few large blocks of 1s.

7. How can we speed up the multiplication process?

Ans: There are two techniques to speed up the multiplication process:

1) The first technique guarantees that the maximum number of summands that must be added is n/2 for n-bit operands.

2) The second technique reduces the time needed to add the summands.

 

8. What is bit pair recoding? Give an example.

Ans: Bit pair recoding halves the maximum number of summands. Group the Booth-recoded multiplier bits in pairs and observe the following: The pair (+1 -1) is equivalent to to the pair (0  +1). That is instead of adding -1 times the multiplicand m at shift position i  to +1 ´ M at position i+1, the same result is obtained by adding  +1 ´ M at position i.

 

Eg: 11010 – Bit Pair recoding  value is 0 -1 -2

 

      9.  What are the two methods of achieving the 2’s complement?

  1. Take the 1’s complement of the number and add 1.
  2. Leave all least significant 0’s and the first unchanged and then complement the remaining bits.

     

    10. What is the advantage of using Booth algorithm?

    Ans: 1) It handles both positive and negative multiplier uniformly.

 2) It achieves efficiency in the number of additions required when the multiplier has a few large blocks of 1’s.

             3) The speed gained by skipping 1’s depends on the data.

 

    11. Write the algorithm for restoring division.

    Ans: Do the following for n times:

    1) Shift  A and Q left one binary position.

    2) Subtract M and A and place the answer back in A.

    3) If the sign of A is 1, set q0 to 0 and add M back to A.

                           Where A- Accumulator, M- Divisor, Q- Dividend.

 

   12. Write the algorithm for  non restoring division.

   Ans: Do the following for n times:

 

   Step 1: Do the following for n times:

 

1) If the sign of A is 0 , shift A and Q left one bit position and subtract M from A; otherwise , shift A and Q left and add M to A.

 

2) Now, if the sign of A is 0,set q0 to 1;otherwise , set q0 to0.

 

Step 2: if the sign of A is 1, add M to A.

 

 

13. Give the IEEE standard for floating point numbers for single precision number.

Ans:

   S                       E¢                                                M

 

 

Sign of the           8-bit signed                         23- bit Mantissa fraction

Number               exponent in excess-127

0 – Positive         representations

1 – Negative

                                                value represented= ±  1. M´ 2E¢-127 

 

14. Give the IEEE standard for floating point numbers for double precision number.

Ans:

   S                       E¢                                                M

 

  

Sign of the           11-bit signed                         52- bit Mantissa fraction

Number               exponent in excess-1023

0 – Positive         representations

1 – Negative

                                                value represented= ±  1. M´ 2E¢-1023 

 

 

15. When can you say that a number is normalized?

Ans: When the decimal point is placed to the right of the first (nonzero) significant digit, the number is said to be normalized.

 

16. Explain about the special values in floating point numbers.

Ans: The end values 0 to 255 of the excess-127 exponent E¢ are used to represent special values such as:

a)      When E¢= 0 and the mantissa fraction M is zero the value exact 0 is represented.

b)      When E¢= 255 and M=0, the value ¥ is represented.

c)      When E¢= 0 and M ¹0 , denormal values are represented.

d)      When E¢= 2555 and M¹0, the value represented is called Not a number.

 

17. Write the Add/subtract rule for floating point numbers.

Ans: 1) Choose the number with the smaller exponent and shift its mantissa right a number of steps equal to the difference in exponents.

2) Set the exponent of the result equal to the larger exponent.

3) Perform addition/subtraction on the mantissa and determine the sign of the result

4) Normalize the resulting value, if necessary.

 

18. Write the multiply rule for floating point numbers.

Ans:1) Add the exponent and subtract 127.

2) Multiply the mantissa and determine the sign of the result .

3) Normalize the resulting value , if necessary.

 

19. What is guard bit?

Ans: Although the mantissa of initial operands are limited to 24 bits, it is important to retain extra bits, called as guard bits.

 

20. What are the ways to truncate the guard bits?

Ans: There are several ways to truncate the guard bits:

1) Chooping

2) Von Neumann rounding

3) Rounding

 

21. Define carry save addition(CSA) process.

Ans: Instead of letting the carries ripple along the rows, they can be saved and introduced into the next roe at the correct weighted position. Delay in CSA is less than delay through the ripple carry adder.

 

22. What are generate and propagate function?

Ans: The generate function is given by

                           Gi=xiyi    and

The propagate function is given as

                           Pi=xi+yi.

 

23. What is excess-127 format?

Ans: Instead of the signed exponent E, the value actually stored in the exponent field is and unsigned integer E¢=E+127.This format is called excess-127.

 

24. What is floating point numbers?

Ans: In some cases, the binary point is variable and is automatically adjusted as computation proceeds. In such case, the binary point is said to float and the numbers are called floating point numbers.

 

25. In floating point numbers when so you say that an underflow or overflow has occurred?

Ans: In single precision numbers when an exponent is less than -126 then we say that an underflow has occurred. In single precision numbers when an exponent is less than +127 then we say that an overflow has occurred.

 

 

UNIT III- BASIC PROCESSING UNIT

 

1. What are the basic steps required to execute an instruction by the processor?

Ans: The basic steps required to execute an instruction by the processor are:

1) Fetch the contents of the memory location pointed to by the PC. They are loaded into the IR.

                                    IR¬[[PC]]

2) Assuming that the memory is byte addressable, increment the contents of the PC by 4, that is

                                    PC¬[PC} + 4

3) Carry out the action specified by the instruction in the IR.

 

2. Define datapath in the processor unit.

Ans: The registers , The ALU and the interconnecting bus are collectively referred to as the datapath.

 

3. What is processor clock?

Ans: All operations and data transfers within the processor take place within time periods defined by the processor clock .

 

4. Write down the control sequence for Move (R1), R2.

Ans: The control sequence is :

R1out, MARin,Read

MDRoutE,WMFC

MDRout,R2in

 

5.Define register file .

Ans: A set of general purpose registers are called as register file Each register from register file R0 is individually addressable.

 

6. Draw the hardware organization of two-stage pipeline?

           

 

Instruction Fetch Unit

Execution Unit

 

 

 

 

 

                                         Inter stage buffer

                                                                                          

7.What is the role of cache memory in pipeline?

Ans: The use of cache memory is to solve the memory access problem. When cache is included in the processor the access time to the cache is usually the same time needed to perform other basic operation inside the processor.

 

8.Name the methods for generating the control signals.

Ans: The methods for generating the control signals are:

                        1) Hardwired control

                        2) Microprogrammed control

 

9. Define hardwired control.

Ans: Hard-wired control can be defined as sequential logic circuit that generates specific sequences of control signal in response to externally supplied instruction.

 

10. Define microprogrammed control.

Ans: A microprogrammed control unit is built around a storage unit is called a control store where all the control signals are stored in a program like format. The control store stores a set of microprograms designed to implement the behavior of the given instruction set.

 

11. Differentiate Microprogrammed control from hardwired control.

 

Microprogrammed control

Hardwired control

It is the microprogram in control store that generates control signals.

It is the sequential circuit that generates control signals.

Speed of operation is low, because it involves memory access.

Speed of operation is high.

Changes in control behavior can be implemented easily by modifying the microinstruction in the control store.

Changes in control unit behavior can be implemented only by redesigning the entire unit.

 

12. Define parallelism in microinstruction.

Ans: The ability to represent maximum number of micro operations in a single microinstruction is called parallelism in microinstruction.

 

13. What are the types of microinstructions available?

Ans: 1) Horizontal microinstruction

        2)  Vertical microinstruction

 

14. Differentiate horizontal microinstruction from vertical microinstruction.

Ans:

Horizontal

Vertical

Long Formats

Short Formats

Ability to express a high degree of parallelism

Limited ability to express parallel micro operation

Little encoding of control information

Considerable encoding of control information

 

 

15. What is MFC?

Ans: To accommodate the variability in response time, the processor waits until it receives an indication that the requested read operation has been completed. This is accomplished by a control signal called Memory – Function – Completed.

16. What are the major characteristics of a pipeline?

Ans: The major characteristics of a pipeline are:

a) Pipelining cannot be implemented on a single task, as it works by splitting multiple tasks into a number of subtasks and operating on them simultaneously.

b) The speedup or efficiency achieved by suing a pipeline depends on the number of pipe stages and the number of available tasks that can be subdivided.

c) If the task that can be subdivided has uneven length of execution times, then the speedup of the pipeline is reduced.

d) Though the pipeline architecture does not reduce the time of execution of a single task, it reduces the overall time taken for the entire job to get completed.

 

 

17. What is a pipeline hazard?

Ans: Any condition that causes the pipeline to stall is called hazard. They are also called as stalls or bubbles.

 

17. What are the types of pipeline hazards?

Ans: The various pipeline hazards are:

                        1. Data hazard

                        2. Structural Hazard

                        3. Control Hazard.

 

18. What is data hazard?

Ans: Any condition in which either the source or the destination operands of an instruction are not available at the time expected in the pipeline is called data hazard.

 

19. What is Instruction or control hazard?

Ans: The pipeline may be stalled because of a delay in the availability of an instruction. For example, this may be a result of a miss in the cache, requiring the instruction to be fetched from the main memory. Such hazards are often called control hazards or instruction hazard.

 

20. Define structural hazards.

Ans: This is the situation when two instruction require the use of a given hardware resource at the same time. The most common case in which this hazard may arise is in access to memory.

 

21. What is side effect?

Ans: When a location other than one explicitly named in an instruction as a destination operand is affected, the  instruction is said to have a side effect.

 

22. What do you mean by branch penalty?

Ans: The time lost as a result of a branch instruction is often referred to as branch penalty.

 

 

23. What is branch folding?

Ans: When the instruction fetch unit executes the branch instruction concurrently with the execution of the other instruction, then this technique is called branch folding.

 

24. What do you mean by delayed branching?

Ans: Delayed branching is used to minimize the penalty incurred as a result of conditional branch instruction. The location following the branch instruction is called delay slot. The instructions in the delay slots are always fetched and they are arranged such that they are fully executed whether or not branch is taken. That is  branching takes place one instruction later than where the branch instruction appears in the instruction sequence in the memory hence the name delayed branching.

 

25.  What are the two types of branch prediction techniques available?

Ans: The two types of branch prediction techniques are

                        1) Static branch prediction

                        2) Dynamic branch prediction

 

 

UNIT IV –  MEMORY SYSTEM

 

1. Define Memory Access Time?

Ans: It is the time taken by the memory to supply the contents of a location, from the time, it receives “READ”.

 

2. Define memory cycle time.

Ans: It is defined as the minimum time delay required between the initaiation of two successive memory operations.

 

3. What is RAM?

This storage location can be accessed in any order and access time is independent of the location being accessed

 

4. What is cache memory?

Ans: It is a fast memory that is inserted between the larger slower main memory and the processor. It holds the currently active segments of a program and their data.

 

5. Explain virtual memory.

Ans: The data is to be stored in physical memory locations that have addresses different from those specified by the program. The memory control circuitry translates the address specified by the program into an address that can be used to access the physical memory.

 

6. List the various semiconductors RAMs?

                        i]          Static RAM.

ii]         Dynamic RAM

 

 

7. What do you mean by static memories?

Ans: Memories that consist of circuits capable of retaining their state as long as power is applied are known as static memories.

 

8. Define DRAM’s.

Ans: Atatic Rams are fast but their cost is high so we use dynamic RAMs which do not retain their state indefinitely but here the information are stored in the form of charge on a capacitor

 

9. Define DDR SDRAM.

The double data rate SDRAMs are the faster version of SDRAM. It transfers data on both edges of the clock.

 

10. What is ROM?

ROM is by definition Non Volatile Preprogrammed with information permanently encoded in the chip.

 

11. What is the mapping procedures adopted in the organization of a Cache Memory?  

i)                    Associative mapping.

ii)                   Direct mapping.

            iii)        Set-associative mapping

 

12. Give the format for main memory address using direct mapping f unction for 4096 blocks in main memory and 128 blocks in cache with 16 blocks per cache.

                                       Tag           Block              Word

                                                                                                                      

5 7 4

 

 

 

13. Give the format for main memory address using associative mapping f unction for 4096 blocks in main memory and 128 blocks in cache with 16 blocks per cache.

                               Tag                                     Word

12 4

 

 

14. Give the format for main memory address using set associative mapping f unction for 4096 blocks in main memory and 128 blocks in cache with 16 blocks per cache.

                              

                                  Tag                   Block                 Word

                                                                                              

6 6 4

 

 

 

 

15. Define Hit and Miss?

                        The performance of cache memory is frequently measured in terms of a quantity called hit ratio. When the CPU refers to memory and finds the word in cache, it is said to produce a hit. If the word is not found in cache, then it is in main memory and it counts as a miss.

 

16. Write the formula for the average access time experienced by the processor in a system with two levels of caches.

Ans: The formula for the average access time experienced by the processor in a system with two levels of caches is

t ave= h1C1+(1-h1)h2C2+(1-h1)(1-h2)M

h1= hit rate in the L1 cache.

h2= hit rate in the L2 cache.

C1=time to access information in the L1 cache.

C2= time to access information in the L1 cache.

M= time to access information in the main memory.

 

17. What are the enhancements used in the memory management?

Ans: 1) Write Buffer

        2) Pre fetching

        3) Look- up Cache.

 

18. What do you mean by memory management unit?

Ans: The memory management unit is a hardware unit which translates virtual addresses into physical addresses in the virtual memory techniques.

 

19. Explain main (primary) memory.

Ans: This memory stores programs and data that are active in use. Storage locations in main memory are addressed directly by the CPU’s load and store instructions.

 

20. What do you mean by seek time?

Ans: It is the time required to move the read/write head in the proper track.

 

21. What is disk controller?

Ans: It provides an interface between the disk drive and the bus that connects it to the rest of the computer system.

 

22. What is RAID?

Ans: High performance devices tend to be expensive. So we can achieve very high performance at a reasonable cost by using a number of low-cost devices oerating in parallel. This is called RAID( Redundant array of Inexpensive Disks).

 

23. Define data stripping?

Ans: A single large file is stored in several separate disk units by breaking the file up into a   number of smaller pieces and storing these pieces on different disks. This is called data stripping.

24. How the data is organized in the disk?

Ans: Each surface is divided into concentric tracks and each track is divided into sectors. The set of corresponding tracks on all surfaces of a stack of disks forms a logical cylinder. The data are accessed by using read/write head.

 

25. Define latency time.

Ans: This is the amount of time that elapses after the head is positioned over the correct track until the starting position of the addressed sector passes under the read/write head.

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