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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 ALU 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 between '2.15,2.18'. Can you explain this answer?
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Most Upvoted Answer
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 cl...
Calculation of Clock Cycles:
For non-pipelined processor: 1 instruction = 5 clock cycles
For pipelined processor: 1 instruction = 5 stages = 1 clock cycle
Therefore, 1 clock cycle of pipelined processor = 5 clock cycles of non-pipelined processor
Calculation of Execution Time:
Execution time for non-pipelined processor = number of instructions * 5 clock cycles / 2.5 GHz
Execution time for pipelined processor = number of instructions * 1 clock cycle / 2 GHz
Calculation of Stalls:
Memory instructions: 30% of instructions, 5% of memory instructions cause 50 cycle stalls each
Branch instructions: 10% of instructions, 50% of branch instructions cause 2 cycle stalls each
Calculation of Speedup:
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor
Putting the Numbers:
Let's assume 100 instructions in the program
Memory instructions: 30% of 100 = 30, 5% of 30 = 1.5 instructions cause 50 cycle stalls each
Branch instructions: 10% of 100 = 10, 50% of 10 = 5 instructions cause 2 cycle stalls each
Total number of stalls = 1.5 * 50 + 5 * 2 = 77 cycles
Execution time for non-pipelined processor = 100 * 5 / 2.5 GHz = 0.2 μs
Execution time for pipelined processor = (100 + 77) * 1 / 2 GHz = 0.1885 μs
Speedup = 0.2 / 0.1885 = 1.063
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 2.15 to 2.18.
For non-pipelined processor: 1 instruction = 5 clock cycles
For pipelined processor: 1 instruction = 5 stages = 1 clock cycle
Therefore, 1 clock cycle of pipelined processor = 5 clock cycles of non-pipelined processor
Calculation of Execution Time:
Execution time for non-pipelined processor = number of instructions * 5 clock cycles / 2.5 GHz
Execution time for pipelined processor = number of instructions * 1 clock cycle / 2 GHz
Calculation of Stalls:
Memory instructions: 30% of instructions, 5% of memory instructions cause 50 cycle stalls each
Branch instructions: 10% of instructions, 50% of branch instructions cause 2 cycle stalls each
Calculation of Speedup:
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor
Putting the Numbers:
Let's assume 100 instructions in the program
Memory instructions: 30% of 100 = 30, 5% of 30 = 1.5 instructions cause 50 cycle stalls each
Branch instructions: 10% of 100 = 10, 50% of 10 = 5 instructions cause 2 cycle stalls each
Total number of stalls = 1.5 * 50 + 5 * 2 = 77 cycles
Execution time for non-pipelined processor = 100 * 5 / 2.5 GHz = 0.2 μs
Execution time for pipelined processor = (100 + 77) * 1 / 2 GHz = 0.1885 μs
Speedup = 0.2 / 0.1885 = 1.063
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 2.15 to 2.18.
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Community Answer
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 cl...
Data:
For a non-pipelined processor,
Clock cycles to complete one instruction = 5
Instruction operating frequency = 2.5 GHz
One clock cycle time = 1/ (2.5 GHz) = 0.4 ns
For N number of instructions, clock cycles required = 5N
Time taken to complete 5n clock cycles = 0.4*5n = 2N ns
For a pipelined processor,
Stages of pipeline = 5
Overheads associated = 2 GHz
One clock cycle time = 1/ (2 GHz) = 0.5 ns
For n instructions, clock cycles required
Clock cycles to complete one instruction = 5
Instruction operating frequency = 2.5 GHz
One clock cycle time = 1/ (2.5 GHz) = 0.4 ns
For N number of instructions, clock cycles required = 5N
Time taken to complete 5n clock cycles = 0.4*5n = 2N ns
For a pipelined processor,
Stages of pipeline = 5
Overheads associated = 2 GHz
One clock cycle time = 1/ (2 GHz) = 0.5 ns
For n instructions, clock cycles required
Therefore, time taken by pipelined processor:
0.6N (1) + 0.3N [0.05 (1 + 50) + 0.95 (1)] + 0.1N [0.5 (1 + 2) + 0.5 (1)] cycles
= 1.85N cycles
= 1.85N/2 ns
= 0.925N ns
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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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. 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 ALU 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 between '2.15,2.18'. Can you explain this answer?.
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