Direction: Consider a machine with a 2-way set associative data cache of size 64 kbyte and block size 16 byte. The cache is managed using 32 bit virtual addresses and the page size is 4 kbyte. A program to be run on this machine begins as follows.
double ARR [1024] [1024]
int i, j;
/* Initialize array ARR to 0.0 */
for (i = 0; i < 1024; i ++)
for (j = 0; j < 1024; j ++)
ARR [i] [j] = 0.0;
The size of double is 8 byte. Array ARR is located in memory starting at the beginning of virtual page 0 × FF000 and stored in row major order. The cache is initially empty and no pre-fetching is done. The only data memory references made by the program are those to array ARR.
-
The cache hit ratio for this initialization loop is
-
- 0%
- 25%
- 50%
- 75%
- 0%
Correct Option: C
Block size = 16B and one element = 8B, So in one block 2 element will be stored.
For 1024 * 1024 element num of block required = 1024 *1024/2 = 2^19 blocks required. In one block first element will be a miss and second one is hit (since we are transferring two unit at a time).
| ⇒ hit ratio = | ||
| total reference |
| = | = | = 0.5 | ||
| 2^20 | 2 |
= 0. 5 * 100 = 50%
