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Consider the unpipelined machine with 10 ns clock cycles. It uses four cycles for ALU operations and branches, whereas five cycles for memory operations. Assume that the relative frequencies of there operations are 40%, 20%, 40% respectively. Suppose that due to clock skew and setup, pipelining the machine adds 1 ns overhead to the clock. How much speed up in the instruction execution rate will we gain from a pipeline?

  • a)
    5 times

  • b)
    4 times

  • c)
    8 times

  • d)
    4.5 times

Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the unpipelined machine with 10 ns clock cycles. It uses four...


Total clock cycle



Under non pipeline time = 4.4 x (10) = 44

Under pipeline time

= 10 + overhead = 10 + 1 = 11

So, speedup  = 44/11 = 4 times
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Most Upvoted Answer
Consider the unpipelined machine with 10 ns clock cycles. It uses four...
Given information:
- Unpipelined machine with 10 ns clock cycles
- 4 cycles for ALU operations and branches
- 5 cycles for memory operations
- Relative frequencies: 40% for ALU, 20% for branches, and 40% for memory
- Pipelining adds 1 ns overhead to the clock

To calculate the speedup gained from pipelining, we need to compute the original execution time and the pipelined execution time.

Original execution time:
- 40% of instructions take 4 cycles, which is 40% * 4 = 1.6 cycles on average
- 20% of instructions take 4 cycles, which is 20% * 4 = 0.8 cycles on average
- 40% of instructions take 5 cycles, which is 40% * 5 = 2 cycles on average
- Total cycles per instruction = 1.6 + 0.8 + 2 = 4.4 cycles
- Total execution time for one instruction = 4.4 cycles * 10 ns/cycle = 44 ns

Pipelined execution time:
- Assuming a 5-stage pipeline, each stage takes 2 ns (1 ns overhead + 1 ns for work)
- Ideal speedup = 10 ns / 2 ns per stage = 5 times
- However, due to pipeline hazards (e.g., data dependencies, control hazards), the pipeline may stall and not achieve ideal speedup
- Assuming a 90% pipeline efficiency (i.e., 90% of the time, all pipeline stages are busy and not stalled), the effective speedup is 1 / (1 - 0.9) = 10 times
- Total execution time for one instruction = 1 pipeline stage * 2 ns/stage + 4 previous pipeline stages * 1 ns/stage = 6 ns

Speedup gained from pipelining:
- Original execution time = 44 ns
- Pipelined execution time = 6 ns
- Speedup = original execution time / pipelined execution time = 44 ns / 6 ns = 7.33 times (approximately 8 times)
- Therefore, the correct answer is option (B) 8 times.
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Consider the unpipelined machine with 10 ns clock cycles. It uses four cycles for ALU operations and branches, whereas five cycles for memory operations. Assume that the relative frequencies of there operations are 40%, 20%, 40% respectively. Suppose that due to clock skew and setup, pipelining the machine adds 1 ns overhead to the clock. How much speed up in the instruction execution rate will we gain from a pipeline?a)5 timesb)4 timesc)8 timesd)4.5 timesCorrect answer is option 'B'. Can you explain this answer?
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