Answer:
Offset bits: 3-bits
Set number of cache: 12-bits
Tag bits: 7-bits
22-bit physical address
Explanation:
Since the system is 32K so,
=2⁵.2¹⁰
=2¹⁵
As we know that it is 8-way set associative so,
=2¹⁵/2³
=2¹⁵⁻³
=2¹²
2¹² are cache blocks
22-bit physical address
Off-set bits are 3 as they are calulated from 8-way set associative information.
Set number of cache : 12-bits
For tag-bits:
Add off-set bits and cache bits and subtract from the total bits of physical address.
=22 - (12+3)
=22 - 15
=7
Answer:
True
Explanation:
In order to be compliant with the NIST publications, policies must include key security control requirements. One of these key requirements includes certification and accreditation, which is a process that occurs after the system is documented, controls tested, and risk assessment completed. It is required before going live with a major system. Once a system is certified and accredited, responsibility shifts to the owner to operate the system is a true statement.
No it is not. it's quite simple when you get the hang of it
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
