pipelining

Question

We have two designs P 1 and P 2 for a synchronous pipeline processor. P1 has 8 pipeline stages with execution time of 3 nsec, 2 nsec, 4 nsec, 7 nsec, 2 nsec, 5 nsec, 4 nsec and 2 nsec while design P2 has 5 stages each with 6 nsec execution time. How much time can be saved using design P2 over design P1 for executing 400 instructions?

Answer 1

 P1:

k=8 pipeline stages

Execution time (tp)= max(3 nsec, 2 nsec, 4 nsec, 7 nsec, 2 nsec, 5 nsec, 4 nsec and 2 nsec)

tp= 7nsec

total time = (k-1+n)*tp

=(8-1+400)*7 =2849nsec

P2:

k=5

Execution time (tp) = 6 nsec

total time = (5-1+400)*6 =2424nsec

time can be saved using design P2 over design P1 for executing 400 instructions = 2849-2424 =425nsec

Answer 2

$P1$ $ = 8$ stages $ \{ 3 ,2, 4 , 7 , 2 , 5 ,4 ,2 \}$nsec

To operate timingly let $7$ nsec as cycle time.

$P2 \ = 5 $ stages each with $6$ nsec .

Time taken by $P1=$ $1\times 7 \times 8 + 399 \ \times 7 $ $= 2849$ nsec .

Time taken by $P2=$ $1\times 5 \times 6 + 399 \ \times 6 $ $= 2424$ nsec .

Time saved $= 2849-2424=\color{Red}{425 \ nsec}$