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Consider a machine with 4­way set associative data cache of size 32 Kbytes and block size 8 byte. The cache is managed using 32 bit virtual addressed and page size is 5 Kbytes. What is the total size of the tags in the cache directory is _________ (in K bits).
Given answer is 76

I am getting 19.
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Number of lines = $\frac{32 KB}{8 B}$ = 212 lines
Number of sets = $\frac{2^{12}}{4}$ = 210 sets
So 32 bit Virual address will be divided as

19 bits  10 bits 3 bits

Tag size = No of sets * No of lines in each set * Tag bits
Tag size = 210 * 4 * 19 bits
Tag size = 76 K bits

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