A non-pipelined instruction execution unit that operates at 1.6 GHz clock…
2026
A non-pipelined instruction execution unit that operates at 1.6 GHz clock takes an average of 5 clock cycles to complete the execution of an instruction. To improve the performance, the system was pipelined with a goal of achieving an average throughput of one instruction per clock cycle. However, it could operate only at 1.2 GHz due to pipeline overheads. While executing a program in the pipelined design, 30% of instructions encountered a stall of 2 cycles due to pipeline hazards. The speed-up obtained by the pipelined design over the non-pipelined one for this program is ___________
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Correct answer: 2.3 to 2.4
Step 1: Calculate execution time for non-pipelined design. Time = Cycles × (1 / Frequency) = 5 × (1 / 1.6 GHz).
Step 2: Calculate execution time for pipelined design. Effective Cycles = 1 + (0.30 × 2) = 1.6. Time = 1.6 × (1 / 1.2 GHz).
Step 3: Calculate Speed-up. Speed-up = Time_non-pipelined / Time_pipelined = (5/1.6) / (1.6/1.2) = 2.34375.