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Consider an unpipelined processor assume that it has a 1 ns clock cycles and that it uses 4 cycles for ALU operation and branches and 5 cycles for memory operations. Assume that the relative frequencies of these operations are 40%, 20% and 40% respectively. Suppose due to clock skew and setup, pipelining the processor adds 0.2 ns of overhead to the clock, ignore any latency impact.
Q. What speedup gain after pipelined the processor? 
  • a)
    3.4 ns
  • b)
    3.8 ns
  • c)
    3.7 ns
  • d)
    1.2 ns
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Consider an unpipelined processor assume that it has a 1 ns clock cycl...
Average instruction time pipelined = 1 + 0.2 = 1.2 ns
Speedup from pipelined
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Consider an unpipelined processor assume that it has a 1 ns clock cycl...
Solution:

Given parameters are:

- Clock cycle time of unpipelined processor, t = 1 ns
- ALU operation and branches take 4 cycles, i.e., 4 ns
- Memory operations take 5 cycles, i.e., 5 ns
- Frequencies of ALU, branch, and memory operations are 40%, 20%, and 40%, respectively.
- Overhead due to clock skew and setup after pipelining, T = 0.2 ns

To find the speedup gain after pipelining the processor, we need to calculate the speed of the unpipelined and pipelined processors.

Speed of unpipelined processor:

The unpipelined processor takes 1 cycle for each instruction. Therefore, the average time taken by an instruction is given by:

Average time per instruction = (40% x 4 ns) + (20% x 4 ns) + (40% x 5 ns) = 4.4 ns

Speed of unpipelined processor = 1 / 4.4 ns = 0.2273 instructions/ns

Speed of pipelined processor:

The pipelined processor has a clock cycle time of t + T = 1.2 ns.

The pipeline stages are:

- Instruction fetch (IF)
- Instruction decode (ID)
- Execute (EX)
- Memory access (MEM)
- Write back (WB)

The time taken by each stage is as follows:

- IF: 1.2 ns
- ID: 1.2 ns
- EX: 1.2 ns - T = 1 ns
- MEM: 1.2 ns - T = 1 ns + 0.2 ns = 1.2 ns
- WB: 1.2 ns

Therefore, the time taken by each instruction depends on the stage it is in:

- Instructions in the IF stage take 1.2 ns
- Instructions in the ID stage take 1.2 ns + 1 cycle time = 2.2 ns
- Instructions in the EX stage take 1.2 ns + 2 cycle times = 5.2 ns
- Instructions in the MEM stage take 1.2 ns + 3 cycle times = 8.2 ns
- Instructions in the WB stage take 1.2 ns + 4 cycle times = 11.2 ns

The throughput of the pipelined processor is given by the inverse of the average time per instruction:

Average time per instruction = (40% x 11.2 ns) + (20% x 5.2 ns) + (40% x 8.2 ns) = 8.24 ns

Speed of pipelined processor = 1 / 8.24 ns = 0.1214 instructions/ns

Speedup gain after pipelining:

Speedup gain = Speed of pipelined processor / Speed of unpipelined processor

Speedup gain = 0.1214 instructions/ns / 0.2273 instructions/ns = 0.534

Therefore, the speedup gain after pipelining the processor is:

Speedup gain = 1 / 0.534 = 1.871

The actual speedup gain is the inverse of the overhead due to pipelining, i.e., 1 / T = 1 / 0.2 ns = 5.
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Consider an unpipelined processor assume that it has a 1 ns clock cycles and that it uses 4 cycles for ALU operation and branches and 5 cycles for memory operations. Assume that the relative frequencies of these operations are 40%, 20% and 40% respectively. Suppose due to clock skew and setup, pipelining the processor adds 0.2 ns of overhead to the clock, ignore any latency impact.Q.What speedup gain after pipelined the processor?a)3.4 nsb)3.8 nsc)3.7 nsd)1.2 nsCorrect answer is option 'C'. Can you explain this answer?
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