Question

Consider a program of 15000 instructions executed by a linear pipeline processor with a clock rate of 25 MHZ. Instruction pipeline has 5 stages and one instruction is issued per clock cycle .Calculate speed up ratio, efficiency and throughput of this pipelined processor.

Answer 1

Answer:

here is your answer...

please mark BRAINLIEST

Answer 2

Answer:

Speed up ratio= 4.999 Efficiency= 0.999 and Throughput of this pipelined processor is 24.99 MIPS

Explanation:

Data given:

$Number of instructions n=\mathbf{1 5 , 0 0 0} instructions or tasks.\\clock frequency f=25 \mathrm{MHz} .\\Number of stages in pipeline k=5 stages.\\For the given issued processor.$

1. The Speedup$\left(S_{k}\right) ,$

       $S_{k}=\frac{T_{1}}{T_{k}}=\frac{n k \tau}{k \tau+(n-1) \tau}$    

            $=\frac{n k}{k+(n-1)} \\ =\frac{(15,000)(5)}{5+(15,000-1)} \\ =\frac{75,000}{15,004} =\mathbf{4 . 9 9 9}$

      2. Throughput = $H_K$

       $\\H_{k}=\frac{n f}{k+(n-1)} \quad$  

             $=\frac{(15,000)(25)}{5+(15,000-1)} \\ =\frac{375,000}{15,004} \\ =24.99 MIPS$

         3. Efficiency $E_K$

      $E_{k}=\frac{S_{k}}{k} =\frac{4.999}{5} = 0.998 %$