Another pitfall cited is expecting to improve the overall performance

Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means ${\displaystyle 100-20\mathrm{\%}=80\mathrm{\%}}$ from the original time
${\displaystyle (1-0.2)\cdot 70s=56s}$
The reduction on this case is ${\displaystyle 70-56s=14s}$
And since the new total time would be given by ${\displaystyle 250-14=236s}$
b) For this case the total time is reduced 20% so that means that the new total time would be ${\displaystyle (1-0.2)=0.8}$ times the original total time ${\displaystyle (1-0.2)\cdot 250s=200s}$
The original time for INT operations is calculated as:
${\displaystyle 250=70+85+40+t_{INT}}$
${\displaystyle t_{INT}=55s}$
For this part the only time that was changed is assumed the INT operations so then:
${\displaystyle 200=70+85+40\mathrm{△}{t}_{INT}}$
And then: ${\displaystyle \mathrm{△}{t}_{INT}=200-70-85-40=5s}$
c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add ${\displaystyle 70+85+55=210s}$, we see that ${\displaystyle 210>205}$ so then we cannot reduce the total time 20% just reducing the branch intructions.
Explanation: From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.
Part 1 For this case the new time to run the FP operation would be reduced 20% so that means ${\displaystyle 100-20\mathrm{\%}=80\mathrm{\%}}$ from the original time
${\displaystyle (1-0.2)\cdot 70s=56s}$
The reduction on this case is ${\displaystyle 70-56s=14s}$
And since the new total time would be given by ${\displaystyle 250-14=236s}$
Part 2
For this case the total time is reduced 20% so that means that the new total time would be ${\displaystyle (1-0.2)=0.8}$ times teh original total time ${\displaystyle (1-0.2)\cdot 250s=200s}$
The original time for INT operations is calculated as:
${\displaystyle 250=70+85+40+t_{INT}}$
${\displaystyle t_{INT}=55s}$
For this part the only time that was changed is assumed the INT operations so then:
${\displaystyle 200=70+85+40\mathrm{△}{t}_{INT}}$
And then: ${\displaystyle \mathrm{△}{t}_{INT}=200-70-85-40=5s}$
And we can quantify the decrease using the relative change:
${\displaystyle \mathrm{\%}Change=\frac{5s}{55s}\cdot 100=9.09\mathrm{\%}}$ of reduction
Part 3
A reduction of the total time implies that the total time would be 205s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55s, so then if we add ${\displaystyle 70+85+55=210s}$, we see that ${\displaystyle 210>205}$ so then we cannot reduce the total time 20% just reducing the branch intructions.

Answer:
1) 236 s
2) 91%
3) no
Explanation:
solution:
The new time required to run FP operations is

0.8  * 70 s = 56 s
So, the new time required to run the program is
250 - (70 - 56) = 236 s (because we reduced the time of execution by 70 - 56 = 14 seconds).
The new total time of execution is

0.8 * 250 = 200s
If we assume that we changed only the time needed to execute INT operations, the new time needed to execute INT operations is
200 - 70 - 85 - 40 = 5 s
Since the old time was 250 - 70 - 85 - 40 = 55 seconds, and
$\frac{5}{55}=0.09$
this is the decrease of a whooping 91%!
Let's assume that we completely avoid using branch operations. Then the time of execution is
$55+70+85=205$
This is the decrease of 18%, since $\frac{205}{250}=0.82$ This means that we cannot decrease the total time by 20% by just decreasing the time of branch operations.