Register

Memory Mapping

• retagged by
1,022 views

1 Answer

Best answer
3 3 votes

Let Tag = x bits, Set = y bits, Block Offset = z bits.

x y z

So, x + y + z = 40.

Cache Size = 512KB = 219B.

Number of cache lines = $\frac{2^{19}}{2^{z}}$ = 2(19-z)

Given the cache is 8 way set associative.

So, number of sets = $\frac{2^{(19-z)}}{2^{3}}$ = 2(16-z)

From the first equation,

x + (16-z) + z = 40

So, x = 24 bits

• selected by
Voting & Accepting an answer helps improve the community
Answer:
← Previous · find the outputNext → · NUMBER SYSTEM FLOATI...
← Previous in category · In COA's sums of hit...Next in category → · Cache Memory
Position:
Show:

Related questions

77 77 votes
9 answers 9 answers
41.3k
41.3k views
Kathleen asked Sep 13, 2014
41,330 views
GATE CSE 1992 | Question: 5-a
The access times of the main memory and the Cache memory, in a computer system, are $500$ n sec and $50$ nsec, respectively. It is estimated that $80\%$ of the main memor...
0 0 votes
1 1 answer
209
209 views
Vishnu__ asked Jan 20
209 views
Practice Type | COA | Memory Mapping |
Tag field size is ‘x’ bits in a 2 way block set associative mapped cache memory. The tag field size (in bits) when the system is upgraded to 8 way block set associative m...
1 1 vote
1 1 answer
385
385 views
js__ asked Jan 25
385 views
Cache mapping
 
0 0 votes
1 answers 1 answer
1.5k
1.5k views
Mrityudoot asked Jun 5, 2023
1,488 views
Cache Mapping
In a 2 level hierarchy, the cache has an access time of 15 ns and the main memory has an access time of 110 ns, the hit rate of the cache is 90%. If the block size of the...