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:
Explanation:
The Hexa-decimal numbers have base 16 and includes numbers:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F
The given steps are performed to convert a decimal number to hexa-decimal number, here to convert decimal number 35 to hexa-decimal number:
- Divide 35 by 16
- Note the remainder, r which is 3 here and quotient which is 2
- Again divide 2 (quotient) by 16 and note the remainder, r' which is 2 and quotient is 0
- We will stop here as the quotient is now 0. Usually division by 16 is repeated until we get quotient = 0
- Now arrange the remainder in reverse order to get the hexa-decimal number as r'r
- The hexa-decimal number is
Answer: like the monitor or the keyboard
Explanation: it is true i even looked it up and it was true
Answer:
import numpy as np
def sample_median(n, P):
return np.median( np.random.choice ( np.arange (1, len(P) + 1 ), n, p = P ) )
print(sample_median(10,[0.1 0.2 0.1 0.3 0.1 0.2]))
print(sample_median(10,[0.1 0.2 0.1 0.3 0.1 0.2]))
print(sample_median(5, [0.3,0.7])
print(sample_median(5, [0.3,0.7])
Explanation:
- Bring in the numpy library to use the median function provided by the numpy library.
- Define the sample_median function that takes in 2 parameters and returns the median with the help of built-in random, choice and arrange functions.
- Call the sample_median function by providing some values to test and then display the results.
Output:
4.5
4.0
2.0
1.0
