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in CO and Architecture by Active (4.8k points)
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case size 1024B= 10+tagbits =16 bits without associativity

but given =10+9 =19  

so ans =19-16 =3 bits extra i.e 23 =8 way -associativity

2 Answers

+7 votes
Best answer
Number of address bits = 16.

#Tag bits = 9.

#Offset bits = 3 (block/line size is 8 bytes and byte-addressing assumed).

So, number of set bits = 16 - 9 - 3 = 4.

Number of sets $=2^4=16$.

No. of cache blocks/lines $=\frac{\text{Cache size}}{\text{Block size}} \\=\frac{1024}{8} = 128.$

So, we have 128 cache blocks and 16 sets meaning $\frac{128}{16} = 8$ way associative cache.
by Veteran (431k points)
selected by
How cache memory size and main memory size same here?

Isnot question incomplete?
No, cache size is 1024 bytes and memory size is $2^{16} = 64K$ bytes. Question says cache is "addressed with" - means the physical memory size or virtual memory size depending on the cache addressing policy.
ok, thank u :)
0 votes

Cache size=2^10 Byte

Block Size=2^3 Byte

Format of address split in set associativity

Tag=9bits Set number=4 bits Block Offset=3 bits




number of sets=Lines/2^3=2^4 =4bits

Therefore it is 8 way set associativity.

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