Alternative 1:A small D-cache with a hit rate of 94% and a hit access time of 1 cycle (assume that no additional cycles on top of the baseline CPI are added to the execution on a cache hit in this case).Alternative 2: A larger D-cache with a hit rate of 98% and the hit access time of 2 cycles (assume that every memory instruction that hits into the cache adds one additional cycle on top of the baseline CPI). a)[10%] Estimate the CPI metric for both of these designs and determine which of these two designsprovides better performance. Explain your answers!CPI = # Cycles / # InsnLet X = # InsnCPI = # Cycles / XAlternative 1:# Cycles = 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150)CPI= 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150) / X1= X(0.50*2 + 0.50(0.94*2 + 0.06*150) ) / X= 0.50*2 + 0.50(0.94*2 + 0.06*150)= 6.44Alternative 2:# Cycles = 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150)CPI= 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150) / X2= X(0.50*2 + 0.50(0.98*(2+1) + 0.02*150)) / X= 0.50*2 + 0.50(0.98*(2+1) + 0.02*150)= 3.97Alternative 2 has a lower CPI, therefore Alternative 2 provides better performance.
Answer:
def prime_generator(s, e):
for number in range(s, e+1):
if number > 1:
for i in range(2, number):
if (number % i) == 0:
break
else:
print(number)
prime_generator(6,17)
Explanation:
I believe you want to ask the prime numbers between s and e.
- Initialize a for loop that iterates from s to e
- Check if the number is greater than 1. If it is, go inside another for loop that iterates from 2 to that number. If the module of that number to any number in range (from 2 to that number) is equal to 0, this means the number is not a prime number. If the module is not equal to zero, then it is a prime number. Print the number
Answer:
That is called declaring a variable
Explanation:
Answer:
Click anywhere in the dataset
Go to Insert –> Tables –> Pivot Table.
In the Create Pivot Table dialog box, the default options work fine in most of the cases.
click ok
Explanation:
