This question investigates cache use in different types of cache. Consider a sys
ID: 3830300 • Letter: T
Question
This question investigates cache use in different types of cache. Consider a system of 8Kbytes of byte-addressable main memory partitioned into blocks of 32bytes each. The system has cache of size 512bytes. The main memory blocks are being accessed in the order shown in the tables (the numbers are in decimal). Assuming the cache is empty at the start.
Scenario 3: The system uses 2-way set associative cache and first-in-first-out replacement policy.
How is the memory address to be interpreted?
Address field
Value
Reason
Word field length:
Set field length:
Tag field length:
Show whether each memory block access is a hit or a miss. Then compute the hit ratio.
Memory
block
Address
Tag field
Address
Set field
Address
Word field
Maps to
cache
Hit/Miss
set
block
1
5
21
7
5
1
17
7
17
5
Total
Address field
Value
Reason
Word field length:
Set field length:
Tag field length:
Explanation / Answer
Address field
Value
Reason
Word field length:
5bits
Each block contains 32bytes, and we need Log(32)bit address to access.
Tag field length:
8bits
Memory size is 8kB, so width of address is 13bit. So tag = 13 - 5
Memory
block
Address Tag field
Address Word field
Maps to cache block
Hit/Miss
1
00000001
00000
1
Miss
5
00000101
00000
2
Miss
21
00010100
00000
3
Miss
7
00000111
00000
4
Miss
5
00000101
00000
2
Hit
1
00000001
00000
1
Hit
17
00010001
00000
5
Miss
7
00000111
00000
4
Hit
17
00010001
00000
5
Hit
5
00000101
00000
2
Hit
Total
5/5
Therefore hit ratio is 0.5
NOTE that all Address Word field are marked zero, since only block information is given. We do not have information on which word is being accessed.
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Address field
Value
Reason
Word field length:
5bits
Each block contains 32bytes, and we need Log(32)bit address to access.
Tag field length:
8bits
Memory size is 8kB, so width of address is 13bit. So tag = 13 - 5
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