Well thats false. They are only notes, not extra details.
Answer:
a) the average CPI for machine M1 = 1.6
the average CPI for machine M2 = 2.5
b) M1 implementation is faster.
c) the clock cycles required for both processors.52.6*10^6.
Explanation:
(a)
The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4
= 1.6
The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4
= 2.5
(b)
The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6
Given 80MHz = 80 * 10^6
The average MIPS ratings for M1 = 80 x 10^6 / 1.6 x 10^6
= 50
Given 100MHz = 100 * 10^6
The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6
= 40
c)
Machine M2 has a smaller MIPS rating
Changing instruction set A from 2 to 1
The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)
and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.
Answer:
def insSort(arr):
ct=0;
for i in range(1, len(arr)):
key = arr[i]
j = i-1
while j >=0 and key < arr[j] :
arr[j+1] = arr[j]
j -= 1
ct=ct+1;
arr[j+1] = key
return arr,ct;
print(insSort([2,1]))
Output of the program is also attached.
It is called a Cooperative program
A Cooperative program refers to a combination of both academic study and vocational activities in one curriculum of education. The purpose of this program is to provide the students with both knowledge in theory and practical skills that make them more prepared in the real world.
