Answer:
The answer is The Cache Sets (S) = 32, Tag bits (t)=24, Set index bits(s) = 5 and Block offset bits (b) = 3
Explanation:
Solution
Given Data:
Physical address = 32 bit (memory address)
Cache size = 1024 bytes
Block size = 8 bytes
Now
It is a 4 way set associative mapping, so the set size becomes 4 blocks.
Thus
Number of blocks = cache size/block size
=1024/8
=128
The number of blocks = 128
=2^7
The number of sets = No of blocks/set size
=128/4
= 32
Hence the number of sets = 32
←Block ←number→
Tag → Set number→Block offset
←32 bit→
Now, =
The block offset = Log₂ (block size)
=Log₂⁸ = Log₂^2^3 =3
Then
Set number pc nothing but set index number
Set number = Log₂ (sets) = log₂³² =5
The remaining bits are tag bits.
Thus
Tag bits = Memory -Address Bits- (Block offset bits + set number bits)
= 32 - (3+5)
=32-8
=24
So,
Tag bits = 24
Therefore
The Cache Sets = 32
Tag bits =24
Set index bits = 5
Block offset bits = 3
Note: ←32 bits→
Tag 24 → Set index 5→Block offset 3
Answer:
See explaination
Explanation:
Square bracket notation is useful when working with objects.
When working with bracket notation, make sure the property identifiers is a String. They can include any characters, including spaces. Variables may also be used as long as the variable resolves to a String.
Below is a representation of the bracket for the question;
a)
racer[0] = 22;
b)
img[1][0] = 89;
c)
img[15][2] = 64;
d)
rgb_img[5][0][1] = 320;
PPTP is what uses chap-secrets. So I imagine that's it
Answer:
(Probably) unknown
Explanation:
First off, you have to get rblx to agree to re-open a deleted, which I can imagine would take a herculean effort and most likely a stream of emails and phone calls. If they happen to archive deleted accounts, they may be able to recover it, but this is done by humans, so there is no definitive way to find out how long it would take a human to do this. If deleted accounts aren't archived, then you're outta luck because they won't be able to get it back.
