Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
![(1-0.2)*70 s =56s](https://tex.z-dn.net/?f=%20%281-0.2%29%2A70%20s%20%3D56s)
The reduction on this case is ![70-56 s=14s](https://tex.z-dn.net/?f=%2070-56%20s%3D14s)
And since the new total time would be given by ![250-14=236 s](https://tex.z-dn.net/?f=%20250-14%3D236%20s)
b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time ![(1-0.2) *250s =200 s](https://tex.z-dn.net/?f=%20%281-0.2%29%20%2A250s%20%3D200%20s)
The original time for INT operations is calculated as:
![250 = 70+85+40 +t_{INT}](https://tex.z-dn.net/?f=%20250%20%3D%2070%2B85%2B40%20%2Bt_%7BINT%7D)
![t_{INT}=55s](https://tex.z-dn.net/?f=%20t_%7BINT%7D%3D55s)
For this part the only time that was changed is assumed the INT operations so then:
![200 = 70+85+40 \Delta t_{INT}](https://tex.z-dn.net/?f=%20200%20%3D%2070%2B85%2B40%20%5CDelta%20t_%7BINT%7D)
And then: ![\Delta t_{INT}= 200-70-85-40=5 s](https://tex.z-dn.net/?f=%20%5CDelta%20t_%7BINT%7D%3D%20200-70-85-40%3D5%20s)
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 70+85+55=210s, we see that 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 100-20% =80% from the original time
![(1-0.2)*70 s =56s](https://tex.z-dn.net/?f=%20%281-0.2%29%2A70%20s%20%3D56s)
The reduction on this case is ![70-56 s=14s](https://tex.z-dn.net/?f=%2070-56%20s%3D14s)
And since the new total time would be given by ![250-14=236 s](https://tex.z-dn.net/?f=%20250-14%3D236%20s)
Part 2
For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time ![(1-0.2) *250s =200 s](https://tex.z-dn.net/?f=%20%281-0.2%29%20%2A250s%20%3D200%20s)
The original time for INT operations is calculated as:
![250 = 70+85+40 +t_{INT}](https://tex.z-dn.net/?f=%20250%20%3D%2070%2B85%2B40%20%2Bt_%7BINT%7D)
![t_{INT}=55s](https://tex.z-dn.net/?f=%20t_%7BINT%7D%3D55s)
For this part the only time that was changed is assumed the INT operations so then:
![200 = 70+85+40 \Delta t_{INT}](https://tex.z-dn.net/?f=%20200%20%3D%2070%2B85%2B40%20%5CDelta%20t_%7BINT%7D)
And then: ![\Delta t_{INT}= 200-70-85-40=5 s](https://tex.z-dn.net/?f=%20%5CDelta%20t_%7BINT%7D%3D%20200-70-85-40%3D5%20s)
And we can quantify the decrease using the relative change:
of reduction
Part 3
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 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.