Question

We begin with a computer implemented in single-cycle implementation. When the stages are split by functionality, the stages do not require exactly the same amount of time. The original machine had a clock cycle time of 7 ns. After the stages were split, the measured times were IF, 1 ns; ID, 1.5 ns; EX, 1 ns; MEM, 2 ns; and WB, 1.5 ns. The pipeline register delay is 0.1 ns.

a. What is the clock cycle time of the 5-stage pipelined machine?
b. If there is a stall every 4 instructions, what is the CPI of the new ?machine?
c. What is the speedup of the pipelined machine over the single- cycle machine?
d. If the pipelined machine had an infinite number of stages, what would its speedup be over the single-cycle machine?

Answer 1

a) First, we need to determine the pipeline stage amounting to the maximum time. In the given case, the maximum time required is 2ns for MEM. In addition, the pipeline register delay=0.1 ns.

Clock cycled time of the pipelined machine= max time+delay

=2ns+0.1 ns

=2.1 ns

b) For any processor, ideal CPI=1. However, since there is a stall after every four instructions, the effective CPI of the new machine is specified by:

`1+(1 / 4)=1.25`

c) The speedup of pipelined machine over the single-cycle machine=avg time per instruction of single cycle/avg time per instruction of pipelined.

Single cycle processor:

CPI=1

Clock period=7 ns

Pipelined processor:

Clock period=2.1 ns

CPI=1.25

Therefore, speedup=
`=7^(*) 1 /\left(2.1^(*) 1.25\right)`

=7/2.625

= 2.67

d) As the number of stages approach infinity, the speedup=k where k is the number of stages in the machine.