Question

Considers pipeline system with 6 segments. Segment delays are 5 ns. 8 ns, 6 ns, 9 ns, 7 ns and 8 ns. Intermediate register delay is 1 ns which is used after each segment. In the pipeline 1000 instructions are executed. Among 1000 instructions 20% are branch instructions each of which incurs 3 pipeline stall cycles. 30% of total 1000 instructions cause resource conflict because of which 1 stall cycles incurred for each instruction. The speed-up of this pipeline as compared to the corresponding non-pipeline system is _______

• A. 3.957
• B. 5.657
• C. 2.257
• D. 4.857

Answer 1

The correct option is C 2.257

option (c)

$\text{Pipeline cycle time}\left(t_p\right)=max(5,8,6,9,7,8)+1=9+1=10ns$

1 instruction execution time in non-pipeline system

$\left(t_a\right)=5+8+6+9+7+8=43ns$

$\text{Pipeline time}=\left[\right(k+n-1)+\text{extra cycles}]t_p$

$=\left[\right(6+1000-1)+(200{}^{\ast }3)+(300{}^{\ast }1\left)\right]10ns$

=1905 * 10ns

=19050 ns

$\text{Non-pipeline time}=n\ast i_n$

=1000 * 43 ns

= 43000ns

$speedup=\frac{43000}{19050}=2.257$