Zeal test series

Question

A computer system uses $16-$bit memory addresses.It has a $16KB$ cache organized in a $4-$ way set associative manner with $64 $ bytes per cache block.Assume that size of each memory word is 2B .When a program is executed ,The processor read data sequentially from the following byte addresses : $128,144,2176,2180,128,2176$.All the above addresses shown in decimal values.Assume cache is initially empty .The number of addresses hit in the cache ?

Comments

-first find the total number of set in set associative cache

here given as total cache size =16KB

and cache blcok size =64B

number of block =16*2^10/64  =2^14/2^6  = 2^8=256

number of set = 256 /4= 64(0------------------------------------63)

byte addresses converted into  block number

128/64= 2

144/64=2

2176/64=34

2180/64 =34

128/64 =2

2176/64=34

hence references to block =2,2,34,34,2,34

2 MISS

2 HIT

34 MISS

34 HIT

2 HIT

34 HIT

TOTAL  =4 HIT  2 MISS

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It is like block size = $64$ $B$ so block addresses from $1$ to $64$
as nothing is mentioned about the Type of memory , then consider it as byte addressable.
even if memory is given as word addressable then also answer will remain same
just the block number changes as  $4$,$4$,$68$,$68$,$4$,$68$ for the given byte addresses respectively.

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@Chandrabhan Vishwa 1 @Pranavpurkar Thanks

Answer 1

4 HIT

Comments

Could you please clarify how you got block 2, 2, 34 etc. I don’t understand this.

Edit: I got it. I didn't understand that 128, 144 etc are actually "words" of a block. so if we have 128, it means 128th word (which is 127, 1st word =0th) which should be present in 128/64 = 2nd block (because each block can contain 64words).