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Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.
Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.
Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.
Correct answer is '2.16'. Can you explain this answer?
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Verified Answer
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 cl...
Non-pipeline:
Clock frequency = 2.5 GHz
Cycle time = 1/2.5 GHz = 0.4 ns
CPI = 5
So, ETnon-pipe = CPI × Cycle time
= 5 × 0.4 ns = 2 ns
Pipeline:
Clock frequency = 2.5 GHz
Cycle time = 1/2GHz =0.5 ns
Number of stalls/instruction = 0.85
Average instruction ETpipe = (1 + Number of stalls/instruction) × Cycle time
= (1 + 0.85) × 0.5 ns = 0.925 ns
Clock frequency = 2.5 GHz
Cycle time = 1/2.5 GHz = 0.4 ns
CPI = 5
So, ETnon-pipe = CPI × Cycle time
= 5 × 0.4 ns = 2 ns
Pipeline:
Clock frequency = 2.5 GHz
Cycle time = 1/2GHz =0.5 ns
Number of stalls/instruction = 0.85
Average instruction ETpipe = (1 + Number of stalls/instruction) × Cycle time
= (1 + 0.85) × 0.5 ns = 0.925 ns
Most Upvoted Answer
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 cl...
Given Information:
- Non-pipelined processor operates at 2.5 GHz.
- It takes 5 clock cycles to complete an instruction.
- Pipelined processor is made with a 5-stage pipeline.
- Overheads associated with pipelining force the pipelined processor to operate at 2 GHz.
- In the given program, 30% are memory instructions, 60% are ALU instructions, and the rest are branch instructions.
- 5% of memory instructions cause stalls of 50 clock cycles each due to cache misses.
- 50% of branch instructions cause stalls of 2 cycles each.
- No stalls are associated with the execution of ALL instructions.
Calculating the Clock Cycles:
To calculate the speedup achieved by the pipelined processor, we need to determine the total number of clock cycles required by both the non-pipelined and pipelined processors for the given program.
Non-Pipelined Processor:
- Operating at 2.5 GHz, each clock cycle takes 1/2.5 GHz = 0.4 ns.
- Each instruction takes 5 clock cycles, so the total execution time for an instruction is 5 * 0.4 ns = 2 ns.
Pipelined Processor:
- Operating at 2 GHz, each clock cycle takes 1/2 GHz = 0.5 ns.
- The pipelined processor has a 5-stage pipeline, so each stage takes 0.5 ns / 5 = 0.1 ns.
- The instruction fetch stage is not pipelined, so it takes 2 ns.
- The remaining stages (decode, execute, memory, writeback) take 0.1 ns each.
- The total execution time for an instruction in the pipelined processor is 2 ns (fetch) + 0.1 ns (decode) + 0.1 ns (execute) + 0.1 ns (memory) + 0.1 ns (writeback) = 2.4 ns.
Calculating the Speedup:
The speedup achieved by the pipelined processor can be calculated as the ratio of the execution time of the non-pipelined processor to the execution time of the pipelined processor.
- Execution time of non-pipelined processor = Number of instructions * Execution time per instruction = 100 * 5 * 0.4 ns = 200 ns.
- Execution time of pipelined processor = Number of instructions * Execution time per instruction = 100 * 2.4 ns = 240 ns.
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor = 200 ns / 240 ns = 0.8333.
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 0.83.
Conclusion:
The speedup achieved by the pipelined processor over the non-pipelined processor is 0.83.
- Non-pipelined processor operates at 2.5 GHz.
- It takes 5 clock cycles to complete an instruction.
- Pipelined processor is made with a 5-stage pipeline.
- Overheads associated with pipelining force the pipelined processor to operate at 2 GHz.
- In the given program, 30% are memory instructions, 60% are ALU instructions, and the rest are branch instructions.
- 5% of memory instructions cause stalls of 50 clock cycles each due to cache misses.
- 50% of branch instructions cause stalls of 2 cycles each.
- No stalls are associated with the execution of ALL instructions.
Calculating the Clock Cycles:
To calculate the speedup achieved by the pipelined processor, we need to determine the total number of clock cycles required by both the non-pipelined and pipelined processors for the given program.
Non-Pipelined Processor:
- Operating at 2.5 GHz, each clock cycle takes 1/2.5 GHz = 0.4 ns.
- Each instruction takes 5 clock cycles, so the total execution time for an instruction is 5 * 0.4 ns = 2 ns.
Pipelined Processor:
- Operating at 2 GHz, each clock cycle takes 1/2 GHz = 0.5 ns.
- The pipelined processor has a 5-stage pipeline, so each stage takes 0.5 ns / 5 = 0.1 ns.
- The instruction fetch stage is not pipelined, so it takes 2 ns.
- The remaining stages (decode, execute, memory, writeback) take 0.1 ns each.
- The total execution time for an instruction in the pipelined processor is 2 ns (fetch) + 0.1 ns (decode) + 0.1 ns (execute) + 0.1 ns (memory) + 0.1 ns (writeback) = 2.4 ns.
Calculating the Speedup:
The speedup achieved by the pipelined processor can be calculated as the ratio of the execution time of the non-pipelined processor to the execution time of the pipelined processor.
- Execution time of non-pipelined processor = Number of instructions * Execution time per instruction = 100 * 5 * 0.4 ns = 200 ns.
- Execution time of pipelined processor = Number of instructions * Execution time per instruction = 100 * 2.4 ns = 240 ns.
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor = 200 ns / 240 ns = 0.8333.
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 0.83.
Conclusion:
The speedup achieved by the pipelined processor over the non-pipelined processor is 0.83.
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Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer?
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Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer?.
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer?.
Solutions for Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor.Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions. 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALL) instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is ________.Correct answer is '2.16'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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