Using Amdahl's law how do you determine execution time after an improvement? Speeding up a new floating-point unit by 2 slows down data cache accesses by a factor of 2/3 (or a 1.5 slowdown for data caches).  If old FP unit took 20% of program's execution time and data cache accesses took 10% of program's execution time, what is the overall speed up? 

I solved this problem using amdahl's law: 

FeFP = floating point enhanced fraction = .2

FeDC = data cache access enhanced fraction = .1

SeFP = floating point enhanced speedup = 2

SeDC = data cache access enhanced speedup = 2/3

Speedup overall = 1 / (   (1 - FeFP - FeDC)   +   FeFP/SeFP   +    FeDC * SeDC    )

= 1 /  (   (   1 - .2 - .1  ) + .2/2 + (.1) * (2/3)   )
 = 1.154. 

I hope I did this correctly, but I'm confused about the next part asking what percentage of execution time is spent on floating point operations after implementing the new FP unit? 

I know that T[improved ] = T[affected] / improvement factor   + T[unaffected]

But I'm unclear how to use it in the context of this problem.  Would appreciate all / any advice.
Since the FP speedup  slows down  the cache by a factor of $2/3$, the cache is "sped up" by $3/2$. Hence the overall speedup is

$ S = \frac{1}{0.7 + 0.2/2 + 0.1 \times \frac {3}{2}} = 1.05 \equiv 5\%$

The execution times now have the ratio: $0.7:0.1:0.15$ for remaining operations:FP:cache. The percentage time spent on FP now is $\frac{0.1}{0.7+0.1+0.15} = 0.105 \equiv 10.5\%$.