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 sw sw
45% 20% 20% 15%
1. What is the clock cycle time in a pipelined and non-pipelined processor?
2. What is the total latency of an LW instruction in a pipelined and non-pipelined processor?
3. 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?
4. Assuming there was are no stalls or hazards, what is the utilization of the data memory?
5. Assuming there are no stalls or hazards, what is the utilization of the write-register port of the "Registers" unit?
6. 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

Respuesta :

The clock cycle time in a pipelined and non-pipelined processor will be 350ps and 1250ps respectively.

1. The clock cycle time in a pipelined processor will be the slowest instruction decode which is 350ps. The clock cycle time in a non-pipelined processor will be:

= 250 + 350 + 150 + 300 + 200

= 1250ps

2. The total latency of an LW instruction in a pipelined processor will be:

= 5 × 350 = 1750ps.

The total latency of an LW instruction in a non-pipelined processor will be:

= 250 + 350 + 150 + 300 + 200

= 1250ps

3. Based on the information asked, the stage to be split will be the cycle time. Now, the new cycle time will be based on the longest stage which will be 300ps.

4. It should be noted that the store and load instruction is used for the utilization of memory. The load instruction is 20% of the time while the store instruction is 15% of the time. Therefore, the utilization of data memory will be:

= 20% + 15% = 35%

5. The utilization of the write-register port of the "Registers" unit will be:

= 20% + 45% = 65%

6. The multi cycle execution time will be:

= (5 × 20%) + [4 × (45% + 20% + 15%)]

= (5 × 0.2) + (4 × 0.8)

= 1 + 3.2

= 4.2

The single cycle execution time will be:

= Cycle time non-pipeline / Cycle time pipeline

= 1250/350

= 3.5

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