Computer Organization
In a certain machine data references constitute 40% of the mix, and that the ideal CPI of the pipelined machine is 1. Assume that the machine with structural hazards has a clock rate that is 1.05 times higher than the clock rate of machine without hazard. What is effective speedup with pipelining without structural hazard?

Options
1. 1.05
2. 1.3
3. 1
4. none of above

Ans: 2

Explanation:
We major the average instruction time on both the machines to access speedup/slowdown.

Avg. instruction time = CPI * clock cycle time

Since there are no stalls/hazards, the average instruction time for ideal machine is the clock cycle times.

The average clock cycle time for the machine with 40% structural hazard is M
= (1+0.4 x 1) x Clock cycle time ideal/1.05 = 1.3 clock cycle time ideal

Machine without structural hazard is 1.3 times faster.