Your machine has an L1 cache and memory with the following properties Memory add
ID: 3835013 • Letter: Y
Question
Your machine has an L1 cache and memory with the following properties Memory address space 24 bits Cache block size 16 bytes Cache associativity: direct-mapped Caches size 256 bytes 1. (2 points) Identify the number of tag bits. Show calculation for your answer. 2. (2 points) Identify the number of bits for the cache index. 3. (2 points) Identify the number of bits for the cache size. 4. (15 points) Consider the memory address trace (similar to PA3) and indicate which access is a hit and miss. show work. Memory address HTTIMISS Why is it a hit/miss? 0x000010 0x000014 0x000018 0x 00001 C 0x001010 0x110020 0x110010 0x110014 110024 0x001120Explanation / Answer
Address width: 24bits
Block size: 16 bytes
cache size: 256bytes
Since each block has 16bytes, block offset is 4 bits (answer 3)
To calculate the width of index field, we need to calculate the number of blocks in the cache
so, # of blocks: 256/16 = 16 blocks.
So width of index field is 4bits (answer 2)
So width of tag field is (24 – 8 – 8)bits or 16 bits (answer 1)
4.
Address(0x)
Tag(0x)
Index(0x)
Hit/Miss
Why hit/miss
000010
0000
1
Miss
Cache empty
000014
0000
1
Hit
Spatial locality
000018
0000
1
Hit
Spatial locality
00001C
0000
1
Hit
Spatial locality
001010
0010
1
Miss
Index occupied
110020
1100
2
Miss
Index empty
110010
1100
1
Miss
Index occupied
110014
1100
1
Hit
Spatial locality
110024
1100
2
Hit
Temporal locality
001120
0011
2
Miss
Index occupied
Answer of question 5 and 6 included above.
Address(0x)
Tag(0x)
Index(0x)
Hit/Miss
Why hit/miss
000010
0000
1
Miss
Cache empty
000014
0000
1
Hit
Spatial locality
000018
0000
1
Hit
Spatial locality
00001C
0000
1
Hit
Spatial locality
001010
0010
1
Miss
Index occupied
110020
1100
2
Miss
Index empty
110010
1100
1
Miss
Index occupied
110014
1100
1
Hit
Spatial locality
110024
1100
2
Hit
Temporal locality
001120
0011
2
Miss
Index occupied
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