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CSE 241
Computer Engineering (1) )1 (اتالحاسب هندسة
Lecture #3
Ch. 6Memory System Design
Dr. Tamer Samy GaafarDept. of Computer & Systems Engineering
Characteristics of Memory Systems
1. Location2. Capacity3. Unit of transfer4. Access method5. Performance6. Physical type7. Physical characteristics8. Organization
You want it fast?
• It is possible to build a computer which uses only static RAM
• This would be very fast• This would need no cache• This would cost a very large amount
Design Constraints on Memory
• How much?• Capacity: bigger is better!
• How fast?• Time is money.• Best: keep up with CPU.
• How expensive?• Reasonable compared to other components.
• Trade-off among the three characteristics.• Solution: memory hierarchy.
Memory Hierarchy - Diagram
• Less cost• Higher capacity• Greater access time• Less access frequency
• Registers• L1 Cache• L2 Cache• Main
memory• Disk• Optical• Tape
Locality of Reference
CPU
Main memory
Locality of Reference
• During the course of execution of a program, memory references tend to cluster (for both instructions and data).• e.g., loops, subroutines.• e.g., operations on tables and arrays.
• Over a short period of time, CPU is working with fixed clusters of memory references.
• Over a long period of time, the clusters in use change.
• This principle can be applied across all levels of the memory hierarchy.
Cache Memory – Concept
• Small amount of fast memory.• Sits between normal main memory and CPU.• May be located on CPU chip.• Not usually visible to the programmer or CPU• Volatile, uses semiconductor technology.
• CPU requests contents of memory location.• Check cache for this data.• If present cache hit, get from cache (fast).
• Because of locality of reference, this location, or a close one, is likely to be referenced soon.
• If not present cache miss, read required block from main memory to cache.
• Then deliver from cache to CPU.• Cache includes tags to identify which block of
main memory is in each cache slot.
Cache Memory – Operation
Cache – Read Operation
Miss
Hit
Typical Cache Organization
Cache Memory – Design
1. Mapping function2. Replacement algorithm3. Write policy4. Number of caches5. Addresses6. Size7. Block/line size
Cache Memory – Design
1. Mapping function2. Replacement algorithm3. Write policy4. Number of caches5. Addresses6. Size7. Block/line size
Example (running)
• Main memory:—byte-addressable
Each location is 1-byte long.
—16 MBLength of address (s+w) = log2 16M = 24
—4-byte blocks# of blocks (M) = 16M / 4 = 4M
• Cache:—64 KB—4-bytes lines
– # of lines (m) = 64K / 4 = 16K
Block 0
Block 1
Block 2
Block m-1
. . .
. . .
Block m
Block m+1
Block 2m-1
. . .
. . .
Block M-m
Block M-1
. . .
. . .
Block M-m+1
Block M-m+2
. . .
. . .
. . .
Line 0Line 1Line 2
Line m-1
Main memory
Cache memory
Block m+2
Direct Mapping
Each main memory block maps to only one cache line (B modulo m).i.e. if a block is in cache, it must be in one specific place.
Direct Mapping Cache Line Table
Cache line assigned
Main memory blocks
0 0, m, 2m, 3m, …, 2s-m
1 1,m+1, 2m+1, …, 2s-m+1
m-1 m-1, 2m-1, 3m-1, …, 2s-1
…… ………………………….
2 2,m+2, 2m+2, …, 2s-m+2
• m cache lines.• 2s main memory blocks.
MM Address-Block Number Relationship0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
012345678910111213141516
MM
ad
dre
ss
Block 0
Block 1
Block 2
Block 3
MM Address-Line Number-tag Relationship0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1:::::::::::::::::::::::::::::::::::::::::::::::::::::::::0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1:::::::::::::::::::::::::::::::::::::::::::::::::::::::::0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1:::::::::::::::::::::::::::::::::::::::::::::::::::::::::0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1:::::::::::::::::::::::::::::::::::::::::::::::::::::::::0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
01
::::::::::
655356553665537:::::::::
:13107
113107
213107
3:::::::::
:19660
719660
819660
9:::::::::
:26214
326214
4
MM
ad
dre
ss
16 k blocks
16 k blocks
16 k blocks
16 k blocks
Tag = 0
Tag = 1
Tag = 2
Tag = 3
Direct Mapping
• Address is in two parts.
• Least significant w bits identify unique word.
• Most significant s bits specify one memory block.
• The MSBs are split into a cache line field r and a
tag of s-r (most significant).
Direct MappingAddress Structure
Tag s-r Line or Slot r Word w
8 bits 14 bits 2 bits
• 24-bit address• 2-bit word identifier (4 byte block)• 22-bit block identifier
• 14-bit slot or line• 8-bit tag (=22-14)
• No two blocks that map to the same line have the same Tag field
• Check contents of cache by finding line and checking Tag
Direct Mapping Cache Organization
Direct Mapping Example
• Main memory size = 16 MB • Cache size = 64 kB• Block size = 4 bytes
(16339C)16 =
0001 0110 0011 0011 1001 1100
word
0CE7
LineTag
Direct Mapping Summary
• Address length = (s + w) bits.• Number of addressable units = 2s+w words.• Block size = line size = 2w words.
— w = log2 (# of words in block).
• Number of blocks in MM = M = 2s+ w/2w = 2s.— s = log2 (# of words in MM / # of words in block).
• Number of lines in cache = m = 2r.— r = log2 (# of words in cache / # of words in line).
• Size of tag = (s – r) bits.— (s-r) = log2 (# of words in MM / # of words in cache).
Direct Mapping pros & cons
• Simple.• Inexpensive.• Fixed location for given block.
• If a program accesses 2 blocks that map to the same line repeatedly, cache misses are very high.
Associative Mapping
• A main memory block can load into any line of cache.
• Memory address is interpreted as tag and word.
• Tag uniquely identifies block of memory.• Every line’s tag is examined for a match.• Cache searching gets expensive.
Associative MappingAddress Structure
• 22 bit tag stored with each 32 bit block of data.
• Compare tag field with tag entry in cache to check for hit.
• Least significant 2 bits of address identify which 1-byte word is required from 32 bit data block.
22 bits 2 bits
Tag s Word w
Fully Associative Cache Organization
Associative Mapping Example
• Main memory size = 16 MB • Cache size = 64 kB• Block size = 4 bytes
(16339C)16 =
0001 0110 0011 0011 1001 1100
word
058CE7
Tag
Associative Mapping Summary
• Address length = (s + w) bits.• Number of addressable units = 2s+w words.• Block size = line size = 2w words.
— w = log2 (# of words in block).
• Number of blocks in MM = M = 2s+ w/2w = 2s.— s = log2 (# of words in MM / # of words in block).
• Number of lines in cache = m = undetermined.• Size of tag = s bits.
Set-Associative Mapping
• Cache is divided into a number of sets (v) of equal size.
• Each set contains a number of lines (k).k-way set-associative mapping!
• A block b could map to any line in a set i if and only if i = b modulo v.
Block 0
Block 1
Block 2
Block m-1
. . .
. . .
Block m
Block m+1
Block 2m-1
. . .
. . .
Block M-m
Block M-1
. . .
. . .
Block M-m+1
Block M-m+2
. . .
. . .
. . .
Main memory
Cache memory
Block m+2
Set-Associative Mapping
Set 0
Set 1
Set 2
Set v-1
………….......………….......………….......4-way set associative mapping
Example (running)
• Main memory:—byte-addressable
Each location is 1-byte long.
—16 MBLength of address (s+w) = log2 16M = 24
—4-byte blocks# of blocks (M) = 16M / 4 = 4M
• Cache:—64 KB—4-byte lines
– # of lines (m) = 64K / 4 = 16K
—2-line sets k=2 2-way set-associative– # of sets (v) = 16K / 2 = 8K
Set Associative MappingAddress Structure
• Use set field to determine cache set to look in.
• Compare tag field to see if we have a hit.
9 bits 13 bits 2 bits
Tag s-d Set d Word w
k-Way Set Associative Cache
Two-Way Set Associative Mapping - Example
• Main memory size = 16 MB • Cache size = 64 kB• Block size = 4 bytes• 2-way set associative mapping
Set Associative Mapping Summary
• Address length = (s + w) bits.• Number of addressable units = 2s+w words.• Block size = line size = 2w words.
— w = log2 (# of words in block).
• Number of blocks in MM = M = 2s.— s = log2 (# of words in MM / # of words in block).
• Number of lines in set = k k-way set-associative• Number of sets = v = 2d.• Number of lines in cache = m = k * v = k * 2d.
— d = log2 (# of words in cache / # of words in set).
• Size of tag = (s – d) bits.— (s-d) = log2 (# of blocks in MM / # of sets in cache).
Cache Memory – Design
1. Mapping function2. Replacement algorithm3. Write policy4. Number of caches5. Addresses6. Size7. Block/line size
Replacement Algorithms (1)Direct mapping
• No choice.
• Each block only maps to one line.
• Replace that line.
• Hardware implemented algorithm (speed)
• Least Recently Used (LRU)• e.g. in 2-way set associative: which of the 2 blocks is LRU?
• First In First Out (FIFO)• Replace block that has been in cache longest.
• Least Frequently Used• Replace block which has had fewest hits.
• Random
Replacement Algorithms (2)Associative and Set Associative
Cache Memory – Design
1. Mapping function2. Replacement algorithm3. Write policy4. Number of caches5. Addresses6. Size7. Block/line size
Write Policy – Multiple CPU Organization
• Several CPUs sharing the same MM, each has its own cache memory.
Main Memory
Cache 1
Cache 2
CPU 1
CPU 2
Cache nCPU n
::
Write Policy – Constraints
• Multiple CPUs may have individual caches.
• I/O may access main memory directly.• Must not overwrite a cache block
unless main memory is up to date.
Write through
• All writes go to main memory as well as cache.
• Multiple CPUs can monitor main memory traffic to keep local (to CPU) cache up to date.
• Lots of traffic.
• Slows down writes.
Write back
• Updates initially made in cache only.• Update bit for cache line is set when
update occurs• If block is to be replaced, write to main
memory only if update bit is set.• Other caches get out of sync.• I/O must access main memory through
cache.• N.B. 15% of memory references are writes
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