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