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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. $D > D_1 > D_2$

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.

@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.

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.

Aptitude question ?

(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.

@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 ..

cycle and time are inversely proportionl, so more cycles means less time. Answer is A.

No cycle and time are directly proportional (obviouslt cause time can be measured in multiple of cycles)

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