In this exercise, we examine how pipelining affects the clock cycle time of the

Question

In this exercise, we examine how pipelining affects the clock cycle time of the processor Problems in this exercise assume that individual stages of the datapath have the following latencies IF ID EX MEM WB 250ps 350ps 150ps 300ps 200ps Also, assume that instructions executed by the processor are broken down as follows alu beq lw SW 45% 20% 20% 15% a. What is the clock cycle time in a pipelined and non-pipelined processor? b. What is the total latency of an LW instruction in a pipelined and non-pipelined processoir c. If we can split one stage of the pipelined datapath into two new stages, each with half the latency of the original stage, which stage would you split and what is the new clock cycle time of the processor? d.Instead of a single-cycle organization, we can use a multi-cycle organization where each instruction takes multiple cycles but one instruction finishes before another is fetched. In this organization, an instruction only goes through stages it actually needs (e.g., ST only takes 4 cycles because it does not need the WB stage). Compare clock cycle times and execution times with single-cycle, multi-cycle, and pipelined organization.

Explanation / Answer

(a)

We can get the Clock cycle time of pipelined processor by slowest stage of the datapath.

We can get the clcok cycle time of the non-pipelined processor by the sum of all the stages.

(b)

Pipelining doesn't affect latency. It only improves throughput i.e, number of instruction per clock cycle time.

Pipelined Proessor:

Non-Piplined Processor:

(c)

Split ID ==> ID1 = ID2 = 175ps.

We know that clock cycle time of pipelined processor is slowest stage in datapath.

IF = 250, ID1=ID2=175, EX=150, MEM=300, WB=200ps.

MEM is the new slowest stage or new longest stage in the datapath.

So, the new Clock cycle time = 300ps.

(d)

Multi-Cycle:

  (0.45 x 4) + (0.20 x 3) + (0.20 x 5) + (0.15 x 4) = 4

Single cycle:

1250 / 350 = 3.57

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