Test: Pipelining- 1

Continue with Google

Continue with Facebook

I accept the Terms & Conditions.

Description

This mock test of Test: Pipelining- 1 for Computer Science Engineering (CSE) helps you for every Computer Science Engineering (CSE) entrance exam. This contains 10 Multiple Choice Questions for Computer Science Engineering (CSE) Test: Pipelining- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Pipelining- 1 quiz give you a good mix of easy questions and tough questions. Computer Science Engineering (CSE) students definitely take this Test: Pipelining- 1 exercise for a better result in the exam. You can find other Test: Pipelining- 1 extra questions, long questions & short questions for Computer Science Engineering (CSE) on EduRev as well by searching above.

QUESTION: 1

Pipelining improves CPU performance due to

A.
Reduced memory access time
B.
Increased clock speed
C.
The introduction of parallelism
D.
Additional functional units

Solution:

In pipelining, a number of functional units are employed in sequence to perform a single computation.

QUESTION: 2

An instruction cycle refers to

A.
Fetching an instruction
B.
Clock speed
C.
Fetching, decoding and executing an instruction
D.
Executing an instruction

Solution:

An instruction cycle is the basic operational process of a computer. It is the process by which a computer retrieves a program instruction from its memory, determines what actions the instruction dictates and carries out those actions. Also called as fetch, decode-execute cycle.

QUESTION: 3

Given a 5 stage pipeline with stages taking 1,2, 3, 1, 1 units of time, the clock period of the pipeline is

A.
8
B.
1/8
C.
1/3
D.
3

Solution:

Clock period of pipeline = Maximum delay of stages
= Max ( 1, 2, 3, 1, 1) = 3

QUESTION: 4

Which of following registers processor used for fetch and execute operations?
1. Program counter
2. Instruction register
3. Address register

A.
1 and 3
B.
1 and 2
C.
2 and 3
D.
1, 2 and 3

Solution:

μ-instruction for fetch cycle:

So, PC, IR and address register are used.

QUESTION: 5

A ________ is required to translate such microprogram into executable programs that can be stored in the control memory in microprogramming.

A.
Microassembler
B.
Microcompiler
C.
Microprogrammed CPU
D.
Microprogrammed counter

Solution:

It is the definition of microassembler.

QUESTION: 6

Which of the following statements is false about CISC architectures?

A.
CISC machine instructions may include complex addressing modes, which require many clock cycles to carry out.
B.
CISC control units are typically microprogrammed, allowing the instruction set to be more flexible.
C.
In the CISC instruction set, all arithmetic/logic instructions must be register based.
D.
CISC architectures may perform better in network centric applications than RISC.

Solution:

In RISC instruction set all arithmetic/logic instructions must be register-based but not in CISC.

QUESTION: 7

The register which holds the address of the location to or from which data are to be transferred is known as

A.
Index register
B.
Instruction register
C.
Memory address register
D.
Memory data register

Solution:

MAR is a register that holds the address of location to or from which data are to be transferred.

QUESTION: 8

Consider a case where 4-segment pipeline with a clock cycle time 20 ns in each sub operation to execute 100 tasks. Assume that a non pipeline unit that can perform the same operation. Pipelined system will take how much time to complete the task?

A.
2060 ns
B.
2000 ns
C.
20 ns
D.
1901 ns

Solution:

For pipeline with K segment and n instruction
Time (pipeline ) = ( K + n - i ) t_P
= ( 4 + 100 - 1 ) x 20 nsec
= (4 + 99) x 20 nsec
= (103) x 20 nsec = 2060 nsec

QUESTION: 9

Find out the speed-up ratio between pipelined and non-pipelined system?

A.
3
B.
4
C.
3.88
D.
400

Solution:

The needed for non pipelined processor

Speedup = 8000/2060 = 3.88

QUESTION: 10

Assume that the time required for the eight functional units, which operate in each of the eight cycles, are as follows 5 ns, 8 ns, 6 ns, 10 ns, 15 ns, 12 ns, 6 ns, 8 ns. Assume that pipe lining adds 1 ns of overhead. Find the speedup versus the single cycle data path.

A.
4.67
B.
4.375
C.
4.44
D.
4.285

Solution:

Since the unpipelined machine executes all instructions in a single clock cycle, its average time per instruction is simply clock time. The dock is equal to the sum of the times for each step in the execution.
∴ Average instructions execution time
= 5 + 8 + 6 + 1 0 + 15 + 12 + 6 + 8
= 70
The dock cycle time on the pipelined machine must be the largest time for any stage in the pipeline (15 ns) + the overhead of 1 ns for a total of 16 ns.
∴ Speed from pipelining