Question

A computer system has a 64 KB main memory and a 4 KB (data area only) cache. There are 8 bytes=cache line. Determine (1) the number of comparators needed and (2) the size of the tag field, for the following mapping scheme: Direct.

Answer 1

Given:

Main Memory size = 64KB = $2^6$ KB = $2^6*2^{10}$ B = $2^{16}$ B

Cache size = 4KB = $2^2$ KB = $2^2*2^{10}$ B = $2^{12}$ B

Cache line size = Block Size = 8 B = $2^{3}$ B

In Direct mapping,the address is divided into 3 fields. They are Tag bits, Line number, Block off-set represented as:

TAG

LINE NUMBER

BLOCK-OFFSET

From the given data we get,

Total number of bits in address = 16 (since main memory size = $2^{16}$ B)

Number of bits in block off-set = 3 (since block size = $2^{3}$ B)

Number of lines = $\tiny \frac{Cache Size}{Block Size}$ = $\tiny \frac{2^{12}}{2^3}$ = $2^9$

Therefore, number of bits in line number = 9.

Finally, number of tag bits can be calculated as:

(Number of bits in main memory) - (Number of bits in block off-set + Number of bits in line number)

Number of tag bits = 16 - (9 + 3)

Tag bits = 16 - 12 = 4

(1) The number of comparators needed in direct mapping scheme = one

(2) The size of Tag field = $2^4$ B = 16 B

Note:

1 KB = $2^{10}$ B