+1 vote
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A computer has a 256 KB, K-way set associative write-back data cache with block size of 32 B. The address sent to the cache controller by the processor is of 32 bits. In addition to the address tag, each cache tag directory contains 2 valid bits and 1 modified bit. If 16 bits are used to address tag. What is the minimum value of K?

A) 6

B) 5

C) 4

D) None of these

edited | 73 views
+1
It should be option C.

Total lines = $\frac{2^{18}}{2^5}= 2^{13}$

Let x is set number.

Now whole address is of 32 bits out of which 16 + x  + 5 (tagbits + set number + offset)

Hence x = 11 i.e $2^{11}$ sets will be there.

Hence k = $\frac{2^{13}}{2^{11}} = 4$
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Yes 4 way set way associative is right..
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Sir, my confusion is that should we always consider that these three bits (2 valid bits and 1 modified bit) are included in address tag ? Since the question is saying "In addition to the address tag, each cache tag directory contains 2 valid bits and 1 modified bit" i thought these three bits are additional to the address tag bits.

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Actually you can't exceed the virtual address range...it's given 32 bits therefore you have to fit extra bits in that range only..
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Got it..sir

This question (https://gateoverflow.in/2192/gate2012-54) is also helpful.