Computation I pg 1
Memory Hierarchy, why?• Users want large and fast memories!
SRAM access times are 1 – 10 nsDRAM access times are 20-120 nsDisk access times are 5 to 10 million ns, but it’s bits are very cheap
• Get best of both worlds: fast and large memories:–build a memory hierarchy CPU
Level 1Level 2
Level nSize
Speed
Computation I pg 2
Memory recap• We can build a memory – a logical k × m array of stored bits. Usually m = 8 bits / location
•••
n bits addressk = 2n locations
m bits data / entry
Address Space:number of locations(usually a power of 2)
Addressability:m: number of bits per location(e.g., byte-addressable)
Computation I pg 3
• SRAM:–value is stored with a pair of inverting gates–very fast but takes up more space than DRAM (4 to 6
transistors)
• DRAM:–value is stored as a charge on capacitor (must be
refreshed)–very small but slower than SRAM (factor of 5 to 10)–charge leakes =>
• refresh needed
Memory element: SRAM vs DRAM
Word line
Pass transistor
Capacitor
Bit line
Computation I pg 4
Latest Intel: i7 Ivy Bridge, 22 nm
- Sandy Bridge 32nm -> 22 nm- - incl graphics, USB3, etc.; 3 levels of cache
Computation I pg 5
Exploiting Locality• Locality = principle that makes having a memory hierarchy a
good idea
• If an item is referenced,
temporal locality: it will tend to be referenced again soonspatial locality : nearby items will tend to be referenced soon.
Why does code have locality?
• Our initial focus: two levels (upper, lower)– block: minimum unit of data – hit: data requested is in the upper level– miss: data requested is not in the upper level block
$
lower level
upper level
Computation I pg 6
Cache operationM
emor
y / L
ower
leve
l
Cache / Higher level
block / line
tags data
Computation I pg 7
• Mapping: cache address is memory address modulo the number of blocks in the cache
Direct Mapped Cache
00001 00101 01001 01101 10001 10101 11001 11101
000
Cache
Memory
001
010
011
100
101
110
111
Computation I pg 8
Q:What kind of locality are we taking advantage of in this example?
Direct Mapped Cache
20 10
Byteoffset
Valid Tag DataIndex
012
102110221023
Tag
Index
Hit Data
20 32
31 30 13 12 1 1 2 1 0Address (bit positions)
Computation I pg 9
• This example exploits (also) spatial locality (having larger blocks):
Direct Mapped CacheAddress (showing bit positions)
16 12 Byteoffset
V Tag Data
Hit Data
16 32
4Kentries
16 bits 128 bits
Mux
32 32 32
2
32
Block offsetIndex
Tag
31 16 15 4 32 1 0Address (bit positions)
Computation I pg 10
• Read hits– this is what we want!
• Read misses– stall the CPU, fetch block from memory, deliver to cache, restart
the load instruction
• Write hits:– can replace data in cache and memory (write-through)– write the data only into the cache (write-back the cache later)
• Write misses:– read the entire block into the cache, then write the word
(allocate on write miss)– do not read the cache line; just write to memory (no allocate on
write miss)
Hits vs. Misses
Computation I pg 11
Splitting first level cache• Use split Instruction and Data caches
– Caches can be tuned differently– Avoids dual ported cache
ProgramBlock size in
wordsInstruction miss rate
Data miss rate
Effective combined miss rate
gcc 1 6.1% 2.1% 5.4%4 2.0% 1.7% 1.9%
spice 1 1.2% 1.3% 1.2%4 0.3% 0.6% 0.4%
CPUI$
D$
I&D $ Main Memory
L1 L2
Computation I pg 12
Let’s look at cache&memory performance
Texec = Ncycles • Tcycle = Ninst• CPI • TcyclewithCPI = CPIideal + CPIstallCPIstall = %reads • missrateread • misspenaltyread+
%writes • missratewrite • misspenaltywrite
or:
Texec = (Nnormal-cycles + Nstall-cycles ) • TcyclewithNstall-cycles = Nreads • missrateread • misspenaltyread +
Nwrites • missratewrite • misspenaltywrite (+ Write-buffer stalls )
Computation I pg 13
Performance example (1)• Assume application with:
– Icache missrate 2%– Dcache missrate 4%– Fraction of ld-st instructions = 36%– CPI ideal (i.e. without cache misses) is 2.0– Misspenalty 40 cycles
• Calculate CPI taking misses into account
CPI = 2.0 + CPIstallCPIstall = Instruction-miss cycles + Data-miss cyclesInstruction-miss cycles = Ninstr x 0.02 x 40 = 0.80 NinstrData-miss cycles = Ninstr x %ld-st x 0.04 x 40CPI = 3.36
Slowdown: 1.68 !!
Computation I pg 14
Performance example (2)1. What if ideal processor had CPI = 1.0 (instead of 2.0)
• Slowdown would be 2.36 !
2. What if processor is clocked twice as fast• => penalty becomes 80 cycles
• CPI = 4.75• Speedup = N.CPIa.Tclock / (N.CPIb.Tclock/2) =
3.36 / (4.75/2)• Speedup is not 2, but only 1.41 !!
Computation I pg 15
Improving cache / memory performance
• Ways of improving performance:
–decreasing the miss ratio (avoiding conflicts): associativity
–decreasing the miss penalty: multilevel caches
–Adapting block size: see earlier slides
–Note: there are many more ways to improve memory performance (see e.g. master course 5MD00)
Computation I pg 16
How to reduce CPIstall ?CPIstall = %reads • missrateread • misspenaltyread+
%writes • missratewrite • misspenaltywrite
Reduce missrate: • Larger cache
–Avoids capacity misses–However: a large cache may increase Tcycle
• Larger block (line) size –Exploits spatial locality: see previous lecture
• Associative cache–Avoids conflict misses
Reduce misspenalty: • Add 2nd level of cache
Computation I pg 17
Decreasing miss ratio with associativity
Tag Data Tag Data Tag Data Tag Data Tag Data Tag Data Tag Data Tag Data
Eight-way set associative (fully associative)
Tag Data Tag Data Tag Data Tag Data
Four-way set associative
Set
0
1
Tag Data
One-way set associative(direct mapped)
Block
0
7
1
2
3
4
5
6
Tag Data
Two-way set associative
Set
0
1
2
3
Tag Data
block
2 blocks / set
4 blocks / set
8 blocks / set
Computation I pg 18
An implementation: 4 way associativeAddress
22 8
V TagIndex012
253254255
Data V Tag Data V Tag Data V Tag Data
3222
4-to-1 multiplexor
Hit Data
123891011123031 0
Computation I pg 19
Performance of Associative Caches
0%
3%
6%
9%
12%
15%
Eight-wayFour-wayTwo-wayOne-way
1 KB2 KB4 KB8 KB
Mis
s ra
te
Associativity 16 KB32 KB64 KB128 KB
1 KB
2 KB
8 KB
Computation I pg 20
Further Cache Basics• cache_size = Nsets x Associativity x Block_size• block_address = Byte_address DIV Block_size in bytes
• index size = Block_address MOD Nsets
• Because the block size and the number of sets are (usually) powers of two, DIV and MOD can be performed efficiently
tag index blockoffset
block address
… 2 1 0bit 31 …
Computation I pg 21
Comparing different (1-level) caches (1)• Assume
– Cache of 4K blocks– 4 word block size– 32 bit address
• Direct mapped (associativity=1) : – 16 bytes per block = 2^4– 32 bit address : 32-4=28 bits for index and tag– #sets=#blocks/ associativity : log2 of 4K=12 : 12 for index– Total number of tag bits : (28-12)*4K=64 Kbits
• 2-way associative – #sets=#blocks/associativity : 2K sets– 1 bit less for indexing, 1 bit more for tag– Tag bits : (28-11) * 2 * 2K=68 Kbits
• 4-way associative– #sets=#blocks/associativity : 1K sets– 1 bit less for indexing, 1 bit more for tag– Tag bits : (28-10) * 4 * 1K=72 Kbits
Computation I pg 22
Comparing different (1-level) caches (2)
3 caches consisting of 4 one-word blocks:
• Cache 1 : fully associative• Cache 2 : two-way set associative• Cache 3 : direct mapped
Suppose following sequence of block addresses: 0, 8, 0, 6, 8
Computation I pg 23
Direct MappedBlock address Cache Block
0 0 mod 4=0
6 6 mod 4=2
8 8 mod 4=0
Address of memory block
Hit or miss
Location 0
Location 1
Location 2
Location 3
0 miss Mem[0]
8 miss Mem[8]
0 miss Mem[0]
6 miss Mem[0] Mem[6]
8 miss Mem[8] Mem[6]
Coloured = new entry = miss
Computation I pg 24
2-way Set Associative: 2 sets
Block address Cache Block0 0 mod 2=06 6 mod 2=08 8 mod 2=0
Address of memory block
Hit or miss
SET 0entry 0
SET 0entry 1
SET 1entry 0
SET 1entry 1
0 Miss Mem[0]
8 Miss Mem[0] Mem[8]
0 Hit Mem[0] Mem[8]
6 Miss Mem[0] Mem[6]
8 Miss Mem[8] Mem[6]
LEAST RECENTLY USED BLOCK
(so all in set/location 0)
Computation I pg 25
Fully associative (4 way assoc., 1 set)
Address of memory block
Hit or miss
Block 0 Block 1 Block 2 Block 3
0 Miss Mem[0]
8 Miss Mem[0] Mem[8]
0 Hit Mem[0] Mem[8]
6 Miss Mem[0] Mem[8] Mem[6]
8 Hit Mem[0] Mem[8] Mem[6]
Computation I pg 26
Review: Four Questions for Memory Hierarchy Designers
• Q1: Where can a block be placed in the upper level? (Block placement)
–Fully Associative, Set Associative, Direct Mapped• Q2: How is a block found if it is in the upper level? (Block identification)
–Tag/Block• Q3: Which block should be replaced on a miss? (Block replacement)
–Random, FIFO, LRU• Q4: What happens on a write? (Write strategy)
–Write Back or Write Through (with Write Buffer)
Computation I pg 27
Classifying Misses: the 3 Cs
• The 3 Cs:–Compulsory—First access to a block is always a miss. Also called cold start misses
• misses in infinite cache
–Capacity—Misses resulting from the finite capacity of the cache
• misses in fully associative cache with optimal replacement strategy
–Conflict—Misses occurring because several blocks map to the same set. Also called collision misses
• remaining misses
Computation I pg 28
3 Cs: Compulsory, Capacity, ConflictIn all cases, assume total cache size not changed
What happens if we:1) Change Block Size: Which of 3Cs is obviously affected? compulsory
2) Change Cache Size: Which of 3Cs is obviously affected? capacity misses
3) Introduce higher associativity : Which of 3Cs is obviously affected? conflict misses
Computation I pg 29
Ca che Size (KB)
Mis
s R
ate
per
Ty
pe
0
0 .0 2
0 .0 4
0 .0 6
0 .0 8
0 .1
0 .1 2
0 .1 4
1 2 4 8
16
32
64
12
8
1 -wa y
2 -wa y
4 -wa y
8 -wa y
Ca pa city
Co mpulso ry
3Cs Absolute Miss Rate (SPEC92)
Conflict
Miss rate per type
Computation I pg 30
Second Level Cache (L2)• Most CPUs
– have an L1 cache small enough to match the cycle time (reduce the time to hit the cache)
– have an L2 cache large enough and with sufficient associativity to capture most memory accesses (reduce miss rate)
• L2 Equations, Average Memory Access Time (AMAT):AMAT = Hit TimeL1 + Miss RateL1 x Miss PenaltyL1Miss PenaltyL1 = Hit TimeL2 + Miss RateL2 x Miss PenaltyL2AMAT = Hit TimeL1 + Miss RateL1 x (Hit TimeL2 + Miss RateL2 x Miss PenaltyL2)
• Definitions:– Local miss rate— misses in this cache divided by the total
number of memory accesses to this cache (Miss rateL2)– Global miss rate—misses in this cache divided by the total
number of memory accesses generated by the CPU (Miss RateL1 x Miss RateL2)
Computation I pg 31
Second Level Cache (L2)• Suppose processor with base CPI of 1.0• Clock rate of 500 Mhz• Main memory access time : 200 ns• Miss rate per instruction primary cache : 5%What improvement with second cache having 20ns access time,
reducing miss rate to memory to 2% ?
• Miss penalty : 200 ns/ 2ns per cycle=100 clock cycles• Effective CPI=base CPI+ memory stall per instruction = ?
– 1 level cache : total CPI=1+5%*100=6– 2 level cache : a miss in first level cache is satisfied by second
cache or memory• Access second level cache : 20 ns / 2ns per cycle=10 clock cycles• If miss in second cache, then access memory : in 2% of the cases• Total CPI=1+primary stalls per instruction +secondary stalls per
instruction• Total CPI=1+5%*10+2%*100=3.5
Machine with L2 cache : 6/3.5=1.7 times faster
Computation I pg 32
Second Level Cache
• Global cache miss is similar to single cache miss rate of second level cache provided L2 cache is much bigger than L1.
• Local cache rate is NOT good measure of secondary caches as it is function of L1 cache.
Global cache miss rate should be used.
Computation I pg 33
Second Level Cache
Computation I pg 34
• Make reading multiple words easier by using banks of memory
• It can get a lot more complicated...
How to connect the cache to next level?
CPU
Cache
Bus
Memory
a. One-word-wide memory organization
CPU
Bus
b. Wide memory organization
Memory
Multiplexor
Cache
CPU
Cache
Bus
Memorybank 1
Memorybank 2
Memorybank 3
Memorybank 0
c. Interleaved memory organization