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The floating point unit of a processor using a design $D$ takes $2t$ cycles compared to $t$ cycles taken by the fixed point unit. There are two more design suggestions $D_1$ and $D_2$. $D_1$ uses $30\%$ more cycles for fixed point unit but $30\%$ less cycles for floating point unit as compared to design $D$. $D_2$ uses $40\%$ less cycles for fixed point unit but $10\%$ more cycles for floating point unit as compared to design $D$. For a given program which has $80\%$ fixed point operations and $20\%$ floating point operations, which of the following ordering reflects the relative performances of three designs?
($D_i > D_j$ denotes that $D_i$ is faster than $D_j$)

  1. $D_1 > D > D_2$
  2. $D_2 > D > D_1$
  3. $D > D_2 > D_1$
  4. $D > D_1 > D_2$
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answer is a. D1>D>D2

we are asked relative performance , but

D=1.2tcpi

D1=1.32tcpi

D2=.92tcpi

performance we won,t chech throuch CPI , if cpi is greater it means that processor is slow..

 

 

please correct me in case of wrong.
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@Jai

it says Di>Dj denotes that Di is faster than Dj ..

now D=1.2tcpi

D1=1.32tcpi

D2=.92tcpi means D2 takes least cpi hence fastest among these .. so D2>D>D1 is correct,

which is option B

A processor is faster when it takes less amount of time than other processors.

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4

If we assume t=1ns then it will give 

For D = .8*t + .2*2*t = 1.2ns

For D1= .8*1.3*t + .2*2*.7*t = 1.32ns

For D2 = .8*.6*t + .2*2*1.2*t = .96ns

so as the question is about performance so u can see by result 

D2 > D >D1 is the correct answer.

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Aptitude question ?
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2 Answers

50 votes
50 votes
Best answer

(B) is correct.
 

$T = 0.8 \times \text{time taken in fixed point} + 0.2 \times \text{time taken in floating point}$

$D = 0.8 \times t + 0.2 \times 2t = 1.2t$

$D_1 = 0.8\times 1.3 t + 0.2 \times 0.7 \times 2t = 1.04 t + .28t = 1.32t$

$D_2 = 0.8 \times 0.6 t + 0.2 \times 1.1 \times 2t = 0.48t + .44t = 0.92t$
 

So, $D_2$ is the best design for this given program followed by $D$ and then $D_1$. Option B.

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@subhankar das

No, option B is correct.

Di>Dj denotes that Di is faster than Dj 

now  processor A is faster than processor b means processor A takes less amount of time than processor B .

That's why  D2>D>D1 as 

D2 = 0.92 

D = 1.20

D1= 1.32

as D2 is less time taking it is faster among three ..

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cycle and time are inversely proportionl, so more cycles means less time. Answer is A.
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 No cycle and time are directly proportional (obviouslt cause time can be measured in multiple of cycles)

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10 votes
10 votes

Suppose t=1ns, there are total 100 operations, 80 fixed point operations and 20 floating point operations,
Time taken by design D = 80*1ns + 20*2ns = 80+40 =$120 ns$
Time taken by design D1 = 1.3*80 + 0.7*40 = 104 + 28 = $132 ns$
Time taken by design D2 = 0.6*80 + 1.1*40 = 48 + 44 = $92 ns$

Now, we can see that D2 is fastest, D is faster than D1.
Hence (B) is correct option D2 > D > D1

Answer:

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