pipeline speed up

Question

consider unpipelined machine with 12 ns clock cycles. It uses 4 cycles for alu operation and branches, whereas 5 cycles for memory operations. Assume that the relative frequencies of these operations are 30% 20% 20%.Pipelining overhead is 1 ns. The speedup of the pipeline is:
2.0  3.1  2.8  3.5

Answer 1

CPI = $\frac{.3*4+.2*4+.2*5}{.3+.2+.2}$ = 4.2 
clock cycle time_non-pipeline = CPI*Clock cycle = 4.2*12 = 50.4

CPI_pipeline = 1 
Clock cycle time = 1*(12+1)  = 13 

S = $\frac{50.4}{13}$ = 3.87