Question

Consider an unpipelined processor where it takes 8ns to go through the circuits and 0.2ns for the latch overhead. Assume that the Point of Production and Point of Consumption in the unpipelined processor are separated by 4ns. Assume that one-third of all instructions do not introduce a data hazard and two-thirds of all instructions depend on their preceding instruction.
Required:
What is the throughput of the processor (in BIPS) for (i) an unpipelined processor, (ii) a 10-stage pipeline, and (iii) a 20-stage pipeline.

Answer 1

Answer:

i) 125 * 10^9  i

ii) 5.2 ns

iii) 4.8 ns

Explanation:

i) The throughput of the processor for an unpipelined processor

= $\frac{1}{8*10^9} = \frac{10^9}{8 }$  = 125 * 10^9  instructions per set

ii ) for a 10-stage pipeline

1st instruction :

= 0.2 + 8ns/10

= 0.2 + 0.8 =  1ns

2nd instruction

= 0.2 + 1 + 4

= 5.2 ns

iii). for a 20-stage pipeline

1st instruction

= 0.2 + 8ns/20

= 0.2 + 0.4  = 0.6 ns

2nd instruction

= 0.2 + 0.6 + 4

= 4.8 ns