Consider a 6-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is ________.

Let the clock cycle time of each pipeline stage be $\tau$.

1. Non-pipelined execution time per instruction: Since there are 6 perfectly balanced stages,

   $$\text{Time}_{\text{non-pipelined}} = 6 \times \tau$$

2. Pipelined execution time per instruction:

   • Ideal $\text{CPI} = 1$

   • $25\%$ of the instructions incur $2$ stall cycles, so average stalls per instruction:

     $$\text{Stall cycles} = 0.25 \times 2 = 0.5$$

   • Average $\text{CPI}_{\text{pipeline}} = 1 + 0.5 = 1.5$

   • Average time per instruction in pipeline:

     $$\text{Time}_{\text{pipelined}} = \text{CPI}_{\text{pipeline}} \times \tau = 1.5\tau$$

3. Speedup:

   $$\text{Speedup} = \frac{\text{Time}_{\text{non-pipelined}}}{\text{Time}_{\text{pipelined}}} = \frac{6\tau}{1.5\tau} = 4$$

Thus, the speedup achieved is 4.