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 $D_j$)
@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.
(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 ..
usman11 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
