retagged by
1,218 views
2 votes
2 votes
Consider a five stage pipeline having stages IF,ID,OF,EX and WB having stage delays 5,6,14,20 and 8 ns. Buffer delay is 2 ns.(say this is pipeline design 1).

Now in this design, if we break the EX stage into 2 stages having delays 12ns and 8 ns(say this is design 2).

What is the speedup we can get by using design 2 over design 1 for executing 20 instructions ?
retagged by

1 Answer

3 votes
3 votes

@sandeep007734 You forgot to include number of instruction in calculation

Time for executing 20 instruction in 1st design

Tdesign1 = (K + N - 1)T

                = (5 + 20 - 1)*22

                = 528

Now you are dividing EX stage into 2 parts to total stage now becomes 6

Tdesign2 = (K' + N - 1)T'

                = (6 + 20 - 1)*16

                = 400

Speedup = $\frac{Tdesign1}{Tdesign2}$

               = $\frac{528}{400}$

               = 1.32

Related questions