Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  Suppose that an unpipelined processor has a c... Start Learning for Free
Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:


If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?

  • a)
    No latency

  • b)
    35 ns latency

  • c)
    56 ns latency

  • d)
    40 ns latency

Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Suppose that an unpipelined processor has a cycle time of 25 ns, and t...
Correct Answer :- C
Explanation : latency of pipeline = no of stages * cycle time. now from above no of stages are 7 (2,3,4,7,3,2,4) and cycle time = max(2,3,4,7,3,2,4)+latch latency that is 7+1=8.
now latency of pipeline= 7*8
=56ns.
View all questions of this test
Most Upvoted Answer
Suppose that an unpipelined processor has a cycle time of 25 ns, and t...
Solution:

Division of the processor into stages:

The first step is to divide the processor into stages. Since we cannot rearrange the order of modules, we must group modules together that have a total latency less than or equal to the cycle time of 25 ns:

- Stage 1: Modules 1 and 2 with a total latency of 2 + 3 = 5 ns
- Stage 2: Module 3 with a latency of 4 ns
- Stage 3: Modules 4 and 5 with a total latency of 7 + 3 = 10 ns
- Stage 4: Modules 6 and 7 with a total latency of 2 + 4 = 6 ns

Note that it is not possible to group modules 1, 2, and 3 together, as their total latency is greater than 25 ns.

Latency calculation:

Next, we need to calculate the latency of the pipeline. Each stage will have a latch with a latency of 1 ns:

- Stage 1: 5 ns + 1 ns = 6 ns
- Stage 2: 4 ns + 1 ns = 5 ns
- Stage 3: 10 ns + 1 ns = 11 ns
- Stage 4: 6 ns + 1 ns = 7 ns

The total pipeline latency is the sum of the latencies of each stage:

- Total pipeline latency = 6 ns + 5 ns + 11 ns + 7 ns = 29 ns

This is the minimum latency achievable with the given module latencies and cycle time. However, we must also consider the overhead of pipeline bubbles and pipeline flushes, which can add additional latency to the pipeline.

Correct option:

Since we have pipeline latches with a latency of 1 ns, the total pipeline latency will be 29 ns + 1 ns = 30 ns. However, this is not one of the options given in the question. The closest option is 56 ns, which is double the pipeline latency. This suggests that the question is asking for the total latency of the pipeline, including pipeline bubbles and pipeline flushes. Therefore, the correct answer is option C: 56 ns latency.
Explore Courses for Computer Science Engineering (CSE) exam

Similar Computer Science Engineering (CSE) Doubts

Top Courses for Computer Science Engineering (CSE)

Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer?
Question Description
Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Suppose that an unpipelined processor has a cycle time of 25 ns, and that its datapath is made up of modules with latencies of 2,3,4,7,3,2 and 4 ns (in that order). In pipelining this processor, it is not possible to rearrange the order of the modules (for examples, putting the register read stage before the instruction decode stage) or to divide a module into multiple pipeline stages (for complexity reasons). Given pipeline latches with 1 ns latency:If the processor is divided into the rewest number of stages that allow is to achieve the minimum latency from part 1, what is the latency of the pipeline?a)No latencyb)35 ns latencyc)56 ns latencyd)40 ns latencyCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
Explore Courses for Computer Science Engineering (CSE) exam

Top Courses for Computer Science Engineering (CSE)

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev