Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
FOR ANYONE THAT WANTS TO STUDY AND CHILL SCREENSHOT BEFORE KATLE DELEATES THIS
meeting id: 932-8097-2909
password: z99LtX
Explanation:
Answer:
public static boolean isReverse(int [ ]a, int [ ]b ){
for (int i=0;i<a.length;i++)
{
if(!(a[i] == b[a.length-i-1]))
return false;
}
return true;
}
Explanation:
Using a for loop, we go through the elements of the first array. The if comapres and checks if any of the values are not the same as the appropriate value on the other array, if it is so, then it is not a reverse, and we return false. else we return true.
Answer:
mylist = [ ]
for i in range(7):
mylist.append (int(input("Enter the number of bugs for each day ")))
print(mylist)
print("The highest number of bugs is ")
print(max((mylist)))
Explanation:
- Create and Initialize an empty (mylist)
- Using the .append method, we request and add the bugs for each day into the list
- print out the list
- Use the max function to find the highest number of bugs in the list and print it out
