This blog includes subject like Computer Organization, Microprocessor, Digital Electronics, System Programming

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Thursday, August 15, 2019

COA Unit-1 MCQ - Computer Evolution and Performance


UNIT-1 Computer Evolution and Performance
MCQ For Online Exam
1)    The Von Neumann system was such that                                                   

a)    Different memory for data and program

b)    Same memory for data and program

c)    Program input from user

d)    Separate data memory

Ans: b)
Explanation: In this both program and data is stored on common memory.

2)    The necessary input device is                                                            

a)    Mouse

b)    Keyboard

c)    Scanner

d)    Joystick

Ans: b)
Explanation: Without Keyboard computer system is not started.

3)    Which unit generates the timing signals for data transfer?

a)    ALU

b)    Control Unit

c)    I/O Devices

d)    Memory

Ans: b)
Explanation: CU is responsible for generating timing and control signals.

4)    In Von Neumann system the program is executed in                                 

a)    Concurrently

b)    Randomly

c)    Sequentially

d)    None of these

Ans: c)
Explanation: Most of the processors work with sequential program execution including Harvard's system.

5)                           register is used to store address from where data is to be retrieved.

a)    MAR

b)    PC

c)    IR

d)    DR

Ans: a) MAR
Explanation: The address given by the PC is forwarded to the address bus of the processor through memory address register (MAR).
6)    The Harvard system uses                                                             .

a)    Different memory for program and data

b)    Same memory for program and data

c)    Separate code, stack and data memory.

d)    None of these.

Ans: a)
Explanation: It uses different memory to store instruction and data.

7)    The first Computer ENIAC consist of                                 

a)    Vacuum Tubes

b)    Transistor

c)    Integrated Circuits

d)    VLSI

Ans: a)
Explanation: This is the first generation computer so uses vacuum tubes.

8)    SPEC stands for,
a)    Standard Performance Evaluation Code.
b)    System Processing Enhancing Code.
c)    System Performance Evaluation Corporation.
d)    Standard Processing Enhancement Corporation.

Answer:c
Explanation: SPEC is a corporation started to standardize the evaluation method of a systems performance.
9)      The average number of steps taken to execute the set of instructions can be made to be less than one by following                                    .
a)  ISA
b)  Pipe-lining
c)  Super-scaling
d)  Sequential View Answer
Answer:c
Explanation: The number of steps required to execute a given set of instructions is sufficiently reduced by using super-scaling. In this method a set of instructions are grouped together and are processed.
10)     A processor performing fetch or decoding of different instruction during the execution of another instruction is called                           .
a)    Super-scaling
b)    Pipe-lining
c)    Parallel Computation
d)    None of these 
Answer:b
Explanation: Pipe-lining is the process of improving the performance of the system by processing different instructions at the same time, with only one instruction performing one specific operation.
11)     Two processors A and B have clock frequencies of 700 Mhz and 900 Mhz respectively. Suppose A can execute an instruction with an average of 3 steps and B can execute with an average of 5 steps. For the execution of the same instruction which processor is faster ?
a)    A
b)    B
  C) Both take the same time
d) Insuffient information 
Answer: a
Explanation: The performance of a system can be found out using the Basic performance formula.

12)       The time delay between two successive initiation of memory operation                 .
a)    Memory access time
b)    Memory search time
c)    Memory cycle time
d)    Instruction delay 
Answer:c
Explanation: The time taken to finish one task and to start another.

13)     The ALU makes use of                to store the intermediate results .
a)    Accumulators
b)    Registers
c)    Heap
d)    Stack View Answer
Answer:a
Explanation: The ALU is the computational center of the CPU. It performs all the mathematical and logical operations. In order to perform better it uses some internal memory spaces to store immediate results.
14)     The              format is usually used to store data .
a)    BCD
b)    Decimal
c)    Hecadecimal
d)    Octal View 
Answer:a
Explanation : The data usually used by computers have to be stored and represented in a particular format for ease of use.

15)     The 8-bit encoding format used to store data in a computer is _           .
a)    ASCII
b)    EBCDIC
c)    ANCI
d)    USCII View
Answer:b
Explanation: The data to be stored in the computers have to be encoded in a particular way so as to be provide secure processing of the data.
16)       A source program is usually in                 .
a)  Assembly language
b)  Machine level language
c)  High-level language
d)  Natural language View Answer
Answer:c
Explanation: The program written and before being compiled or assembled is called as a source program.
17)  Which method/s of representation of numbers occupies large amount of memory than others?
a)    Sign-magnitude
b)    1’s compliment
c)    2’s compliment
d)    Both a and b 
Answer:a
Explanation: It takes more memory as one bit used up to store the sign.

18)     Which representation is most efficient to perform arithmetic operations on the numbers ?
a)    Sign-magnitude
b)    1’s compliment
c)    2’S compliment
d)    None of the above View Answer
Answer:c
Explanation: The two’s compliment form is more suitable to perform arithmetic operations as there is no need to involve the sign of the number into consideration.

19)     When we subtract -3 from 2 , the answer in 2’s compliment form is                 . a) 0001
b) 1101
c) 0101
d) 1001 
Answer:c
Explanation: First the 2’s compliment is found and that is added to the number and the overflow is ignored.
20)     The most efficient method followed by computers to multiply two unsigned numbers is
a)  Booth algorithm
b)  Bit pair recording of multipliers
c)  Restoring algorithm
d)  Non restoring algorithm View Answer
Answer: b

21)        The most efficient method followed by computers to multiply two signed numbers is
a)  Booth algorithm
b)  Bit pair recording of multipliers
c)  Restoring algorithm
d)  Non restoring algorithm View Answer
Answer: a

22)        Booth's algorithm will require                         cycles to multiply 25 and -5.

a)  4

b)  5

c)  6

  d)  7 
Answer:c)
Since 25 is in the range of 6-bit representation of 2's complement number, so require 6 cycles.

23)        When 1101 is used to divide 100010010 the remainder is              . 
   a) 101
b)    11
c)    0
d)    1
View Answer Answer:d
24)        IEEE double precision of floating point representation has                bits. 1) 32
  2) 64
  3) 128
  4) None of these.

25) IEEE double precision mantissa has                  bits. 
1) 8
2) 23

3) 11

4) 52

Booth's Algorithm for Multiplication


Booth's Algorithm

  • Booth's algorithm mainly developed to do signed as well as unsigned binary multiplication. 
  • Following flowchart shows the working or Booth's Algorithm:
  • From above flowchart we got mainly four conditions to perform operation on Q0 and Q -1
Sr. No.
Conditions
Operation
Q 0
Q -1
1
0
0
Shift Right A, Q, Q-1
2
0
1
A = A + M & Shift Right A, Q, Q-1
3
1
0
A = A - M & Shift Right A, Q, Q-1
4
1
1
Shift Right A, Q, Q-1

Let us solve the example of signed multiplication using Booth's Algorithm.

Example-1 : 6 * -3

Here, M = 0110, Q = +3 = 0011, 
-3 = 1101 (Take 2's complement of +3 for this first take 1's complement of +3 which is 1100 and add 1 to get 2's complement i. e. 1101 ), 
- M = 1010 (Take 2's complement of M).

In this number of bits used in Q are 4 bits, hence total 4 cycles i. e. value of n = 4.

Cycle No.
A
     Q 
 
              Q0
Q -1
M
Operation
1
0000
1101
0
0110
Initial Value
1010
1101
0
0110
A = A - M
1101
0110
1
0110
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.
2
0011
0110
1
0110
A = A + M
0001
1011
0
0110
Shift Right A, Q, Q -1 (MSB of A is replaced with 0 because of addition operation.
3
1011
1011
0
0110
A = A - M
1101
1101
1
0110
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.
4
1110
MSB
1110
LSB
1
0110
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.

  
Hence 6 * - 3 = - 18
   18 = 1 0 0 1 0
  -18 = 0 1 1 0 1
         +          1
= 1 1 1 0 1 1 1 0  ( Result in A & Q. Result is negative so that MSB bits are 111 indicate sign bit)



Example-2 : Example of unsigned multiplication using Booth's Algorithm. 11 * 5

Here, M = 1011, Q = 5 = 0101, 
-M = 0101 (Take 2's complement of M for this first take 1's complement of M which is 0100 and add 1 to get 2's complement i. e. 0101 ), 

In this number of bits used in Q are 4 bits, hence total 4 cycles i. e. value of n = 4.

Cycle No.
A
     Q 
 
              Q0
Q -1
M
Operation
1
0000
0101
0
1011
Initial Value
0101
0101
0
1011
A = A - M
1010
1010
1
1011
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.
2
0101
1010
1
1011
A = A + M
0010
1101
0
1011
Shift Right A, Q, Q -1 (MSB of A is replaced with 0 because of addition operation.
3
0111
1101
0
1011
A = A - M
1011
1110
1
1011
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.
4
0110

0011
      1110

      0111
1

0
1011

1011
Shift Right A, Q, Q -1 (MSB of A is replaced with 1 because of substraction operation.

  
Hence 11 * 5 = 55
  
55 = 0011 0111 ( Result in A & Q.)