Question

Q3. A computer has 128 MB of main memory organized logically as 32M blocks of 4 bytes each. It has a cache memory of 16 KB. Answer the following questions: a. How many address lines are required to access the main memory? [1 Mark] b. Determine how to split the address (s-r, r, w) for Direct Mapping. [2 Marks] c. Determine how to split the address (s-d, d, w) for 2-way set-associative mapping. [2 Marks]

Answer 1

Main memory size = 128 MB = $\small 2^{27} \;bytes$ as 1 MB =

220 bytes

Block Size = 4 Bytes

Number of blocks in main memory = 32 M =

25

Cache Memory Size = 16 KB =
214 bytes
as 1 KB = $\small 2^{10}\; bytes$

a) Address lines for main memory = $\small \bold{log_2(main \;memory \;size\;(in\;bytes)) = log_2{(2^{27})} = 27 \;bits}$

b)

Number of lines in cache memory = (cache size / block size) = 16 KB / 4 B = 4 K = $\small 2^{12}$

Block index of Main memory (s) = $\small log_2(number\;of\;blocks) = log_2(2^{25}) = 25 \;bits$

word offset (w) = $\small \bold{log_2(block\;size) = log_2(4) = 2 \;bits}$

Line index (r) = $\small \bold{log_2(number\;of\;blocks) = log_2(2^{12}) = 12 \;bits}$

Cache Tag bits (s-r) = 25 - 12 = 13 bits

c)

Number of lines in cache memory = (cache size / block size) = 16 KB / 4 B = 4 K = $\small 2^{12}$

Number of sets in cache memory = (number of blocks / number of ways) = $\small 2^{12}$ / 2 = $\small 2^{11}$

Block index of Main memory (s) = $\small log_2(number\;of\;blocks) = log_2(2^{25}) = 25 \;bits$

word offset (w) = $\small \bold{log_2(block\;size) = log_2(4) = 2 \;bits}$

Set index (d) = $\small \bold{log_2(number\;of\;blocks) = log_2(2^{11}) = 11 \;bits}$

Cache Tag bits (s-d) = 25 - 11 = 14 bits