Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider a non-pipelined machine with 6 stage...
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Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?
Correct answer is '3.11'. Can you explain this answer?
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Consider a non-pipelined machine with 6 stages; the lengths of each st...
Calculation of Cycle Time with Non-Pipelined Machine
Total cycle time = sum of all stages = 20 + 10 + 30 + 25 + 40 + 15 = 140 ns
Calculation of Overhead Added with Pipelining
Calculation of Cycle Time with Pipelined Machine
Total cycle time = sum of all stages = 25 + 15 + 35 + 30 + 45 + 20 = 170 ns
Calculation of Speedup Factor
Speedup factor = Non-pipelined cycle time / Pipelined cycle time = 140 / 170 = 3.11
Explanation of the Answer
Pipelining is a technique used to increase the throughput of a processor by dividing the execution of instructions into multiple stages. In this case, the non-pipelined machine has 6 stages with varying lengths. The total cycle time is the sum of all stages, which is 140 ns. When implementing pipelining, an overhead of 5 ns is added to each stage to account for clock skew and set up. The new cycle time with pipelining is 170 ns, which is longer than the non-pipelined cycle time.
However, the speedup factor is calculated by dividing the non-pipelined cycle time by the pipelined cycle time. In this case, the speedup factor is 3.11, which means the pipelined machine can execute instructions 3.11 times faster than the non-pipelined machine. This is because pipelining allows multiple instructions to be executed simultaneously, which increases the throughput of the processor.
- Stage 1: 20 ns
- Stage 2: 10 ns
- Stage 3: 30 ns
- Stage 4: 25 ns
- Stage 5: 40 ns
- Stage 6: 15 ns
Total cycle time = sum of all stages = 20 + 10 + 30 + 25 + 40 + 15 = 140 ns
Calculation of Overhead Added with Pipelining
- Stage 1: 20 ns + 5 ns overhead = 25 ns
- Stage 2: 10 ns + 5 ns overhead = 15 ns
- Stage 3: 30 ns + 5 ns overhead = 35 ns
- Stage 4: 25 ns + 5 ns overhead = 30 ns
- Stage 5: 40 ns + 5 ns overhead = 45 ns
- Stage 6: 15 ns + 5 ns overhead = 20 ns
Calculation of Cycle Time with Pipelined Machine
- Stage 1: 25 ns
- Stage 2: 15 ns
- Stage 3: 35 ns
- Stage 4: 30 ns
- Stage 5: 45 ns
- Stage 6: 20 ns
Total cycle time = sum of all stages = 25 + 15 + 35 + 30 + 45 + 20 = 170 ns
Calculation of Speedup Factor
Speedup factor = Non-pipelined cycle time / Pipelined cycle time = 140 / 170 = 3.11
Explanation of the Answer
Pipelining is a technique used to increase the throughput of a processor by dividing the execution of instructions into multiple stages. In this case, the non-pipelined machine has 6 stages with varying lengths. The total cycle time is the sum of all stages, which is 140 ns. When implementing pipelining, an overhead of 5 ns is added to each stage to account for clock skew and set up. The new cycle time with pipelining is 170 ns, which is longer than the non-pipelined cycle time.
However, the speedup factor is calculated by dividing the non-pipelined cycle time by the pipelined cycle time. In this case, the speedup factor is 3.11, which means the pipelined machine can execute instructions 3.11 times faster than the non-pipelined machine. This is because pipelining allows multiple instructions to be executed simultaneously, which increases the throughput of the processor.
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Consider a non-pipelined machine with 6 stages; the lengths of each st...
Pipelining: Pipelining is a process of arrangement of hardware elements of the CPU such that its overall performance is increased. Simultaneous execution of more than one instruction takes place in a pipelined processor.
In a Non-Pipelining system, processes like decoding, fetching, execution and writing memory are merged into a single unit or a single step.
For non-pipelining system,
Average execution time = (20+10+30+25+40+15) = 140 ns
For pipelining system,
Average execution time = max[ (20+5), (10+5), (30+5), (25+5), (40+5), (15+5) ]
Cycle (ideal condition) = 45 ns
Speed up = 
Speed up factor
= 140/45
= 140/45
Hence, the correct answer is 3.11.
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Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer?
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Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. 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 Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. 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 Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer?.
Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. 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 Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. 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 Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer?.
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Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer?, a detailed solution for Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer? has been provided alongside types of Consider a non-pipelined machine with 6 stages; the lengths of each stage are 20 ns,10 ns,30 ns,25 ns,40 ns, and 15 ns respectively. Suppose for implementing the pipelining the machine adds 5 ns of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?Correct answer is '3.11'. Can you explain this answer? theory, EduRev gives you an
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