edited by
18,759 views
43 votes
43 votes

In an enhancement of a design of a CPU, the speed of a floating point unit has been increased by $\text{20%}$ and the speed of a fixed point unit has been increased by $\text{10%}$. What is the overall speedup achieved if the ratio of the number of floating point operations to the number of fixed point operations is $2:3$ and the floating point operation used to take twice the time taken by the fixed point operation in the original design?

  1. $1.155$
  2. $1.185$
  3. $1.255$
  4. $1.285$
edited by

11 Answers

Best answer
63 votes
63 votes
SpeedUp $=\dfrac{\text{Original time taken}}{\text{new time taken}}$

Let $x$ be the time for a fixed point operation,

Original time taken $=\dfrac{(3x + 2\times 2x)}{5} =\dfrac{7x}{5}$
New time taken $=\dfrac{\left(\dfrac{3x}{1.1} + \dfrac{4x}{1.2}\right)}{5}=\dfrac{8x}{1.32\times 5}$
So, SpeedUp $=\dfrac{7\times 1.32}{8} =1.155$

Correct Answer: $A$
edited by
69 votes
69 votes

You can tabulate down the given data from question.

  Speed in Original Design Speed in New Design # of Operations given(2:3)
Floating Point (FP) operations 100 120 2x
Fixed Point (XP) operations 200 220 3x

Here, Speed = #operations per unit time (Just like speed = distance/time)

Took 100,200 for FP and XP for easy understanding.

Say 5x is total number of operations, in which 2x FP and 3x XP operations

Speed up = $\frac{Time in Original Design}{Time  in New Design}$

=>$\frac{2x/100 + 3x/200}{2x/120 + 3x/220}$ (Since we know time = #operations/speed)

we get Speed up = 1.155, So Option (A)

Hope this solution was easy to understand

20 votes
20 votes
  • $k$ = no of times one floating point operation takes more time than one fixed point operation.
  • $x:y$ = ratio of no of fixed point operations to that of floating point operations.
  • $a$ = % increase in speed of floating point unit in fraction {i.e. 20% means $a$ = $0.2$}
  • $b$ = % increase in speed of fixed point unit

Then Net speedup 

$\begin{align*} S &= \frac{\text{total time in old design}}{\text{total time in new design}} \\ & = \frac{ky+x}{\frac{ky}{1+a}+\frac{x}{1+b}} \end{align*}$

Putting all these values S = $1.155$.

NOTE: That is not a standard formula. It has been derived , by taking total time  = $T$ and total operations = $N$

9 votes
9 votes
Suppose we have 20 floating instructn and 30 fixed.

Initially Speed of Floating instruction is 20 and fixed is 10,

so total time=20*20+30*10=700

After increasing float speed 20%=speed(float)=24

10% increase in fixs speed=speed(fix)=11

So total time= 24*20+30*11=480+330=810

Speed up=810/700=1.1571
Answer:

Related questions

34 votes
34 votes
3 answers
3