Question

[5] Assume for a given processor the CPI of arithmetic instructions is 1, the CPI of load/store instructions is 10, and the CPI of branch instructions is 3.

Assume a program has the following instruction breakdowns: 500 million arithmetic instructions, 300 million load/store instructions, 100 million branch instructions.

2.46.1 [5] <§2.19> Suppose that new, more powerful arithmetic instructions are added to the instruction set. On average, through the use of these more powerful arithmetic instructions, we can reduce the number of arithmetic instructions needed to execute a program by 25%, and the cost of increasing the clock cycle time by only 10%. Is this a good design choice? Why?

2.46.2 [5] <§2.19> Suppose that we find a way to double the performance of arithmetic instructions. What is the overall speedup of our machine? What if we find a way to improve the performance of arithmetic instructions by 10 times?

Short answer

2.46.1

Increasing the execution time can’t be a good choice.

2.46.2

By assuming, if the performance of arithmetic instructions is doubled then the overall speedup of that machine will be 2.069.

By assuming, if the performance of arithmetic instructions is increased by 10 times then the speedup will be 1.134.

Step-by-step solution

Define the concept.

2.46.1

$$\begin{gathered} TheCPI=\left(\right(CP{I}_{arithmeticinstructions}\times \frac{breakdow{n}_{arithmeticinstructions}}{breakdow{n}_{all}}) \\ +(CP{I}_{loadorstoreinstructions}\times \frac{breakdow{n}_{loadorstoreinstructions}}{breakdow{n}_{all}}) \\ +(CP{I}_{branchinstructions}\times \frac{breakdow{n}_{branchinstructions}}{breakdow{n}_{all}}\left)\right) \end{gathered}$$

the execution time = the CPI x f x the sum of all these breakdown

2.46.2

The CPI of the arithmetic instructions is 1.

By assuming, the performance of arithmetic instructions is doubled.

So, ( $1\times 0.5$) = 0.5

Speed up =$\frac{CP{I}_{atthestarting}}{CP{I}_{afterincrea\sin gtheperformanceofarithmeticinstructions}}$

Determine the calculation.

Given,

The CPI of the arithmetic instructions is 1.

The CPI of the load or store instructions is 10.

The CPI of the branch instructions is 13.

The breakdown of the arithmetic instructions is 500 million.

The breakdown of the load or store instructions is 300 million.

The breakdown of the branch instructions is 100 million.

The sum of all these breakdown = (500+300+100) = 900 million.

At the start, The CPI

$$\begin{gathered} =10\times \frac{300}{900}+1\times \frac{500}{900}+3\times \frac{100}{900} \\ =\frac{10}{3}+\frac{5}{9}+\frac{1}{3} \\ =\frac{38}{9}=4.22 \end{gathered}$$

Let’s assume, to execute a program the number of arithmetic instructions can be reduced by 25%.

Now, the total arithmetic = 0.75 x 500 =375

At the start, the execution time =$\frac{38}{9}\times f\times 900=3800f$

The sum of these breakdown = (375+300+100) = 775 million.

Now, the CPI $$\begin{gathered} =10\times \frac{300}{775}+1\times \frac{375}{775}+3\times \frac{100}{775} \\ =\frac{3675}{775} \end{gathered}$$

And also by assuming that the cost of increasing the clock cycle time by only 10%

Now, the executive time$$\begin{gathered} =\frac{3675}{775}\times f\times \left(1+0.1\right)\times 775 \\ =\frac{3675}{775}\times f\times 1.1\times 775 \\ =4042.5f \end{gathered}$$

Hence the current execution time > the previous execution time.

Increasing the execution time can’t be a good choice.

2.46.2

By assuming, the performance of arithmetic instructions is doubled.

At the start, the CPI is 4.22.

Now, the CPI $$\begin{gathered} =10\times \frac{3}{9}+0.5\times \frac{5}{9}+3\times \frac{1}{9} \\ =3.944 \end{gathered}$$

Now, the speed up$=\frac{4.22}{3.944}=1.069$

By assuming, the performance of arithmetic instructions is increased by 10 times.

Now, the CPI$$\begin{gathered} =10\times \frac{3}{9}+0.5\times \frac{1}{9}+3\times \frac{1}{9} \\ =\frac{30+0.5+3}{9} \\ =\frac{33.5}{9} \\ =3.72 \\ Now,thespeedup=\frac{4.22}{3.72}=1.134 \end{gathered}$$

Overall speedup = 1.134