Memory_Unit Cache Main Virtual Associative

1
Memory Organization
Computer Organization Computer Architectures Lab
• Memory Hierarchy
• Main Memory
• Auxiliary Memory
• Associative Memory
• Cache Memory
• Virtual Memory
• Memory Management Hardware
MEMORY ORGANIZATION

2
Memory Organization
MEMORY HIERARCHY
Magnetic
tapes
Magnetic
disks
I/O
processor
CPU
Main
memory
Cache
memory
Auxiliary memory
Register
Cache
Main Memory
Magnetic Disk
Magnetic Tape
Memory Hierarchy is to obtain the highest possible
access speed while minimizing the total cost of the memory system
Memory Hierarchy

3
Memory Organization
MAIN MEMORY
RAM and ROM Chips
Typical RAM chip
Typical ROM chip
Chip select 1
Chip select 2
Read
Write
7-bit address
CS1
CS2
RD
WR
AD 7
128 x 8
RAM
8-bit data bus
CS1 CS2 RD WR
0 0 x x
0 1 x x
1 0 0 0
1 0 0 1
1 0 1 x
1 1 x x
Memory function
Inhibit
Inhibit
Inhibit
Write
Read
Inhibit
State of data bus
High-impedence
High-impedence
High-impedence
Input data to RAM
Output data from RAM
High-impedence
Chip select 1
Chip select 2
9-bit address
CS1
CS2
AD 9
512 x 8
ROM
8-bit data bus
Main Memory

4
Memory Organization
MEMORY ADDRESS MAP
RAM 1
RAM 2
RAM 3
RAM 4
ROM
0000 - 007F
0080 - 00FF
0100 - 017F
0180 - 01FF
0200 - 03FF
Component
Hexa
address
0 0 0 x x x x x x x
0 0 1 x x x x x x x
0 1 0 x x x x x x x
0 1 1 x x x x x x x
1 x x x x x x x x x
10 9 8 7 6 5 4 3 2 1
Address bus
Memory Connection to CPU
- RAM and ROM chips are connected to a CPU
through the data and address buses
- The low-order lines in the address bus select
the byte within the chips and other lines in the
address bus select a particular chip through
its chip select inputs
Address space assignment to each memory chip
Example: 512 bytes RAM and 512 bytes ROM
Main Memory

5
Memory Organization
CONNECTION OF MEMORY TO CPU
Main Memory
}
CS1
CS2
RD
WR
AD7
128 x 8
RAM 1
CS1
CS2
RD
WR
AD7
128 x 8
RAM 2
CS1
CS2
RD
WR
AD7
128 x 8
RAM 3
CS1
CS2
RD
WR
AD7
128 x 8
RAM 4
Decoder
3 2 1 0
WR
RD
9 8 7-1
10
16-11
Address bus
Data bus
CPU
CS1
CS2
512 x 8
ROM
AD9
1- 7
9
8
Data
Data
Data
Data
Data

6
Memory Organization
• Random Access Memory (RAM) is used to store the
programs and data being used by the CPU in real time.
The data on the random access memory can be read,
written, and erased any number of times. RAM is the
hardware element where the data being currently used is
stored. It is a volatile memory.
• Types of RAM:
– Static RAM, or (SRAM) which stores a bit of data using the state of a six
transistor memory cell.
– Dynamic RAM, or (DRAM) which stores a bit data using a pair of
transistor and capacitor which constitute a DRAM memory cell.

7
Memory Organization
• Read Only Memory (ROM) is a type of memory where the
data has been prerecorded. Data stored in ROM is
retained even after the computer is turned off ie, non-
volatile.
• Types of ROM:
– Programmable ROM, where the data is written after the memory chip has
been created. It is non-volatile.
– Erasable Programmable ROM, where the data on this non-volatile
memory chip can be erased by exposing it to high-intensity UV light.
– Electrically Erasable Programmable ROM, where the data on this non-
volatile memory chip can be electrically erased using field electron
emission.
– Mask ROM, in which the data is written during the manufacturing of the
memory chip.

8
Memory Organization
Difference RAM ROM
Data retention
RAM is a volatile memory which could
store the data as long as the power is
supplied.
ROM is a non-volatile memory which
could retain the data even when power
is turned off.
Working type
Data stored in RAM can be retrieved
and altered.
Data stored in ROM can only be read.
Use
Used to store the data that has to be
currently processed by CPU
temporarily.
It stores the instructions required
during bootstrap of the computer.
Speed It is a high-speed memory. It is much slower than the RAM.
CPU Interaction
The CPU can access the data stored
on it.
The CPU can not access the data
stored on it unless the data is stored in
RAM.
Size and Capacity Large size with higher capacity. Small size with less capacity.
Used as/in CPU Cache, Primary memory. Firmware, Micro-controllers
Accessibility The data stored is easily accessible
The data stored is not as easily
accessible as in RAM
Cost Costlier cheaper than RAM.

9
Memory Organization

10
Memory Organization
• How many 32K x 1 RAM chips are needed to
provide a memory capacity of 256K-bytes?
(A) 8
(B) 32
(C) 64
(D) 128

11
Memory Organization
• Answer: (C)
Explanation:
• We need 256 Kbytes, i.e., 256 x 1024 x 8 bits.
We have RAM chips of capacity 32 Kbits = 32 x
1024 bits.
•
(256 * 1024 * 8)/(32 * 1024) = 64

12
Memory Organization
• Number of chips (128 x 8 RAM) needed to provide
a memory capacity of 2048 bytes
(A) 2
(B) 4
(C) 8
(D) 16

13
Memory Organization
Explanation:
chips (128 x 8 RAM) needed to provide a memory of 2048 B = 2048
x 8 / 128 x 8 = 16 x 1

14
Memory Organization

15
Memory Organization
RAM 2048 /256 = 8 chips; 2048 = 211; 256 = 28
• ROM 4096 /1024 = 4 chips; 4096 = 212; 1024 = 210
• Interface 4 × 4 = 16 registers; 16 = 24
• Component Address 16 15 14 13 12 11 10 19 8765
4321
• RAM 0000-O7FF 0 0 0 0 0 3 8
• decoder
• × ←⎯⎯→ ×××× ××××
• ROM 4000-4FFF 0 1 0 0 2 4
• decoder
• × ←⎯⎯→ ×× ×××× ××××
• Interface 8000-800F 1 0 0 0 0 0 0 0 0000 ××××

16
Memory Organization

17
Memory Organization
(a) 8 chips are needed with address lines connected in
parallel.
(b) 16 × 8 = 128 chips. Use 14 address lines (16 k = 214)
10 lines specify the chip address
4 lines are decoded into 16 chip-select inputs.

18
Memory Organization
A RAM chip has a capacity of 1024 words of 8 bits each (1K ×
8). The number of 2 × 4 decoders with enable line needed to
construct a 16K × 16 RAM from 1K × 8 RAM is
(A) 4
(B) 5
(C) 6
(D) 7

19
Memory Organization
Answer: (B)
Explanation:
RAM chip size = 1k ×8[1024 words of 8 bits each]
RAM to construct =16k ×16
Number of chips required = (16k x 16)/ ( 1k x 8) = (16 x
2) [16 chips vertically with each having 2 chips
horizontally]
So to select one chip out of 16 vertical chips, we need 4
x 16 decoder.
Available decoder is 2 x 4 decoder
To be constructed is 4 x 16 decoder
Hence 4 + 1 = 5 decoders are required.

20
Memory Organization
AUXILIARY MEMORY
Information Organization on Magnetic Tapes
EOF
IRG
block 1 block 2
block 3
block 1
block 2
block 3
R1
R2 R3 R4
R5
R6
R1
R3 R2
R5 R4
file i
EOF
Organization of Disk Hardware
Track
Moving Head Disk Fixed Head Disk
Auxiliary Memory

21
Memory Organization
ASSOCIATIVE MEMORY
- Accessed by the content of the data rather than by an address
- Also called Content Addressable Memory (CAM)
Hardware Organization Argument register(A)
Key register (K)
Associative memory
array and logic
m words
n bits per word
Match
register
Input
Read
Write
M
- Compare each word in CAM in parallel with the
content of A(Argument Register)
- If CAM Word[i] = A, M(i) = 1
- Read sequentially accessing CAM for CAM Word(i) for M(i) = 1
- K(Key Register) provides a mask for choosing a
particular field or key in the argument in A
(only those bits in the argument that have 1’s in
their corresponding position of K are compared)
Associative Memory

22
Memory Organization
ORGANIZATION OF CAM
Internal organization of a typical cell Cij
C11
Word 1
Word i
Word m
Bit 1 Bit j Bit n
M1
Mi
Mm
Associative Memory
Aj
R S
Output
Match
logic
Input
Write
Read
Kj
Mi
To
F ij
A1
Aj An
K1
Kj Kn
C1j C1n
Ci1 Cij Cin
Cm1 Cmj Cmn

23
Memory Organization
MATCH LOGIC
Associative Memory
F'i1 F i1
K1 A1
F'i2 F i2
K2 A2
F'in Fin
Kn An
. . . .
Mi

24
Memory Organization
CACHE MEMORY
Locality of Reference
- The references to memory at any given time
interval tend to be confined within a localized areas
- This area contains a set of information and
the membership changes gradually as time goes by
- Temporal Locality
The information which will be used in near future
is likely to be in use already( e.g. Reuse of information in loops)
- Spatial Locality
If a word is accessed, adjacent(near) words are likely accessed soon
(e.g. Related data items (arrays) are usually stored together;
instructions are executed sequentially)
Cache
- The property of Locality of Reference makes the
Cache memory systems work
- Cache is a fast small capacity memory that should hold those information
which are most likely to be accessed
Cache Memory
Main memory
Cache memory
CPU

25
Memory Organization
PERFORMANCE OF CACHE
All the memory accesses are directed first to Cache
If the word is in Cache; Access cache to provide it to CPU
If the word is not in Cache; Bring a block (or a line) including
that word to replace a block now in Cache
- How can we know if the word that is required
is there ?
- If a new block is to replace one of the old blocks,
which one should we choose ?
Memory Access
Performance of Cache Memory System
Hit Ratio - % of memory accesses satisfied by Cache memory system
Te: Effective memory access time in Cache memory system
Tc: Cache access time
Tm: Main memory access time
Te = Tc + (1 - h) Tm
Example: Tc = 0.4 s, Tm = 1.2s, h = 0.85%
Te = 0.4 + (1 - 0.85) * 1.2 = 0.58s
Cache Memory

26
Memory Organization
• Assume that for a certain processor, a read
request takes 50 nanoseconds on a cache miss
and 5 nanoseconds on a cache hit. Suppose while
running a program, it was observed that 80% of the
processor’s read requests result in a cache hit.
The average read access time in nanoseconds
is____________.
(A) 10
(B) 12
(C) 13
(D) 14

27
Memory Organization
• Answer: (D)
Explanation:
• The average read access time in nanoseconds
= 0.8 * 5 + 0.2*50
= 14

28
Memory Organization
Consider a system with 2 level caches. Access times of Level 1
cache, Level 2 cache and main memory are 1 ns, 10ns, and 500 ns,
respectively. The hit rates of Level 1 and Level 2 caches are 0.8
and 0.9, respectively. What is the average access time of the
system ignoring the search time within the cache?
(A) 13.0 ns
(B) 12.8 ns
(C) 12.6 ns
(D) 12.4 ns

29
Memory Organization
Answer: (C)
Explanation: First, the system will look in cache 1. If it is not
found in cache 1, then cache 2 and then further in main
memory (if not in cache 2 also).
The average access time would take into consideration success in
cache 1, failure in cache 1 but success in cache 2, failure in both
the caches and success in main memory.
Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1-H1)(1-H2)*Hm*Tm]
where,
H1 = Hit rate of level 1 cache = 0.8
T1 = Access time for level 1 cache = 1 ns
H2 = Hit rate of level 2 cache = 0.9
T2 = Access time for level 2 cache = 10 ns
Hm = Hit rate of Main Memory = 1
Tm = Access time for Main Memory = 500 ns

30
Memory Organization
So, Average Access Time = ( 0.8 * 1 ) + ( 0.2 * 0.9 * 10 ) + (
0.2 * 0.1 * 1 * 500)
= 0.8 + 1.8 + 10
= 12.6 ns
Thus, C is the correct choice.

31
Memory Organization
MEMORY AND CACHE MAPPING - ASSOCIATIVE MAPPLING -
Associative mapping
Direct mapping
Set-associative mapping
Associative Mapping
Mapping Function
Specification of correspondence between main
memory blocks and cache blocks
- Any block location in Cache can store any block in memory
-> Most flexible
- Mapping Table is implemented in an associative memory
-> Fast, very Expensive
- Mapping Table
Stores both address and the content of the memory word
address (15 bits)
Argument register
Address Data
0 1 0 0 0
0 2 7 7 7
2 2 2 3 5
3 4 5 0
6 7 1 0
1 2 3 4
CAM
Cache Memory

32
Memory Organization
MEMORY AND CACHE MAPPING - DIRECT MAPPING -
Addressing Relationships
Direct Mapping Cache Organization
Memory
address Memory data
00000 1 2 2 0
00777
01000
01777
02000
02777
2 3 4 0
3 4 5 0
4 5 6 0
5 6 7 0
6 7 1 0
Index
address Tag Data
000 0 0 1 2 2 0
0 2 6 7 1 0
777
Cache memory
Tag(6) Index(9)
32K x 12
Main memory
Address = 15 bits
Data = 12 bits
512 x 12
Cache memory
Address = 9 bits
Data = 12 bits
00 000
77 777
000
777
- Each memory block has only one place to load in Cache
- Mapping Table is made of RAM instead of CAM
- n-bit memory address consists of 2 parts; k bits of Index field and
n-k bits of Tag field
- n-bit addresses are used to access main memory
and k-bit Index is used to access the Cache
Cache Memory

33
Memory Organization
DIRECT MAPPING
Direct Mapping with block size of 8 words
Operation
- CPU generates a memory request with (TAG;INDEX)
- Access Cache using INDEX ; (tag; data)
Compare TAG and tag
- If matches -> Hit
Provide Cache[INDEX](data) to CPU
- If not match -> Miss
M[tag;INDEX] <- Cache[INDEX](data)
Cache[INDEX] <- (TAG;M[TAG; INDEX])
CPU <- Cache[INDEX](data)
Index tag data
000 0 1 3 4 5 0
007 0 1 6 5 7 8
010
017
770 0 2
777 0 2 6 7 1 0
Block 0
Block 1
Block 63
Tag Block Word
6 6 3
INDEX
Cache Memory

34
Memory Organization
MEMORY AND CACHE MAPPING - SET ASSOCIATIVE MAPPING -
Set Associative Mapping Cache with set size of two
- Each memory block has a set of locations in the Cache to load
Index Tag Data
000 0 1 3 4 5 0 0 2 5 6 7 0
Tag Data
777 0 2 6 7 1 0 0 0 2 3 4 0
Operation
- CPU generates a memory address(TAG; INDEX)
- Access Cache with INDEX, (Cache word = (tag 0, data 0); (tag 1, data 1))
- Compare TAG and tag 0 and then tag 1
- If tag i = TAG -> Hit, CPU <- data i
- If tag i  TAG -> Miss,
Replace either (tag 0, data 0) or (tag 1, data 1),
Assume (tag 0, data 0) is selected for replacement,
(Why (tag 0, data 0) instead of (tag 1, data 1) ?)
M[tag 0, INDEX] <- Cache[INDEX](data 0)
Cache[INDEX](tag 0, data 0) <- (TAG, M[TAG,INDEX]),
CPU <- Cache[INDEX](data 0)
Cache Memory

35
Memory Organization
BLOCK REPLACEMENT POLICY
Many different block replacement policies are available
LRU(Least Recently Used) is most easy to implement
Cache word = (tag 0, data 0, U0);(tag 1, data 1, U1), Ui = 0 or 1(binary)
Implementation of LRU in the Set Associative Mapping with set size = 2
Modifications
Initially all U0 = U1 = 1
When Hit to (tag 0, data 0, U0), U1 <- 1(least recently used)
(When Hit to (tag 1, data 1, U1), U0 <- 1(least recently used))
When Miss, find the least recently used one(Ui=1)
If U0 = 1, and U1 = 0, then replace (tag 0, data 0)
M[tag 0, INDEX] <- Cache[INDEX](data 0)
Cache[INDEX](tag 0, data 0, U0) <- (TAG,M[TAG,INDEX], 0); U1 <- 1
If U0 = 0, and U1 = 1, then replace (tag 1, data 1)
Similar to above; U0 <- 1
If U0 = U1 = 0, this condition does not exist
If U0 = U1 = 1, Both of them are candidates,
Take arbitrary selection
Cache Memory

36
Memory Organization
CACHE WRITE
Write Through
When writing into memory
If Hit, both Cache and memory is written in parallel
If Miss, Memory is written
For a read miss, missing block may be
overloaded onto a cache block
Memory is always updated
-> Important when CPU and DMA I/O are both executing
Slow, due to the memory access time
Write-Back (Copy-Back)
When writing into memory
If Hit, only Cache is written
If Miss, missing block is brought to Cache and write into Cache
For a read miss, candidate block must be
written back to the memory
Memory is not up-to-date, i.e., the same item in
Cache and memory may have different value
Cache Memory

37
Memory Organization
VIRTUAL MEMORY
Give the programmer the illusion that the system has a very large memory,
even though the computer actually has a relatively small main memory
Address Space(Logical) and Memory Space(Physical)
Address Mapping
Memory Mapping Table for Virtual Address -> Physical Address
virtual address
(logical address) physical address
address space memory space
address generated by programs actual main memory address
Mapping
Virtual address
Virtual
address
register
Memory
mapping
table
Memory table
buffer register
Main memory
address
register
Main
memory
Main memory
buffer register
Physical
Address
Virtual Memory

38
Memory Organization
ADDRESS MAPPING
Organization of memory Mapping Table in a paged system
Address Space and Memory Space are each divided
into fixed size group of words called blocks or pages
1K words group Page 0
Page 1
Page 2
Page 3
Page 4
Page 5
Page 6
Page 7
Block 3
Block 2
Block 1
Block 0
Address space
N = 8K = 213
Memory space
M = 4K = 212
0
000
1
001
1
010
0
011
0
100
1
101
1
110
0
111
1
Block 0
Block 1
Block 2
Block 3
MBR
0 1 0 1 0 1 0 1 0 0 1 1
1 0 1 0 1 0 1 0 1 0 0 1 1
Table
address
Presence
bit
Page no. Line number
Virtual address
Main memory
address register
Memory page table
Main memory
11
00
01
10
01
Virtual Memory

39
Memory Organization
ASSOCIATIVE MEMORY PAGE TABLE
Assume that
Number of Blocks in memory = m
Number of Pages in Virtual Address Space = n
Page Table
- Straight forward design -> n entry table in memory
Inefficient storage space utilization
<- n-m entries of the table is empty
- More efficient method is m-entry Page Table
Page Table made of an Associative Memory
m words; (Page Number:Block Number)
1 0 1 Line number
Page no.
Argument register
1 0 1 0 0
0 0 1 1 1
0 1 0 0 0
1 0 1 0 1
1 1 0 1 0
Key register
Associative memory
Page no.Block no.
Virtual address
Page Fault
Page number cannot be found in the Page Table
Virtual Memory

40
Memory Organization
PAGE FAULT
Processor architecture should provide the ability
to restart any instruction after a page fault.
1. Trap to the OS
2. Save the user registers and program state
3. Determine that the interrupt was a page fault
4. Check that the page reference was legal and
determine the location of the page on the
backing store(disk)
5. Issue a read from the backing store to a free
frame
a. Wait in a queue for this device until serviced
b. Wait for the device seek and/or latency time
c. Begin the transfer of the page to a free frame
6. While waiting, the CPU may be allocated to
some other process
7. Interrupt from the backing store (I/O completed)
8. Save the registers and program state for the other user
9. Determine that the interrupt was from the backing store
10. Correct the page tables (the desired page is now in memory)
11. Wait for the CPU to be allocated to this process again
12. Restore the user registers, program state, and new page table, then
resume the interrupted instruction.
LOAD M
0
Reference
1
OS
trap
2
3 Page is on backing store
free frame
main memory
4
bring in
missing
page
5
reset
page
table
6
restart
instruction
Virtual Memory

41
Memory Organization
PAGE REPLACEMENT
Modified page fault service routine
Decision on which page to displace to make room for
an incoming page when no free frame is available
1. Find the location of the desired page on the backing store
2. Find a free frame
- If there is a free frame, use it
- Otherwise, use a page-replacement algorithm to select a victim frame
- Write the victim page to the backing store
3. Read the desired page into the (newly) free frame
4. Restart the user process
2
f 0 v i
f v
frame
valid/
invalid bit
page table
change to
invalid
4
reset page
table for
new page
victim
1
swap
out
victim
page
3
swap
desired
page in
backing store
physical memory
Virtual Memory

42
Memory Organization
PAGE REPLACEMENT ALGORITHMS
FIFO
0
7
1
7
2 0 3 0 4 2 3 0 3 2 1 2 0 1 7
7 0 1
0 0
7
1
2
0
1
2
3
1
2
3
0
4
3
0
4
2
0
4
2
3
0
2
3
0
1
3
0
1
2
7
1
2
7
0
2
7
0
1
Page frames
Reference string
-
FIFO algorithm selects the page that has been in memory the longest time
Using a queue - every time a page is loaded, its
identification is inserted in the queue
Easy to implement
May result in a frequent page fault
Optimal Replacement (OPT) - Lowest page fault rate of all algorithms
Replace that page which will not be used for the longest period of time
0
7
1
7
2 0 3 0 4 2 3 0 3 2 1 2 0 1 7
7 0 1
0 0
7
1
2
0
1
2
0
3
2
4
3
2
0
3
2
0
1
7
0
1
Page frames
Reference string
Virtual Memory

43
Memory Organization
- OPT is difficult to implement since it requires future knowledge
- LRU uses the recent past as an approximation of near future.
Replace that page which has not been
used for the longest period of time
LRU
0
7
1
7
2 0 3 0 4 2 3 0 3 2 1 2 0 1 7
7 0 1
0 0
7
1
2
0
1
2
0
3
4
0
3
4
0
2
4
3
2
0
3
2
1
3
2
1
0
2
1
0
7
Page frames
Reference string
Virtual Memory
- LRU may require substantial hardware assistance
- The problem is to determine an order for the frames
defined by the time of last use

44
Memory Organization
LRU Approximation
- Reference (or use) bit is used to approximate the LRU
- Turned on when the corresponding page is
referenced after its initial loading
- Additional reference bits may be used
Virtual Memory
4 7 0 7 1 0 1 2 1 2 7 1 2
Reference string
2
1
0
7
4
7
2
1
0
4
LRU Implementation Methods
• Counters
- For each page table entry - time-of-use register
- Incremented for every memory reference
- Page with the smallest value in time-of-use register is replaced
• Stack
- Stack of page numbers
- Whenever a page is referenced its page number is
removed from the stack and pushed on top
- Least recently used page number is at the bottom

45
Memory Organization
MEMORY MANAGEMENT HARDWARE
Basic Functions of MM
- Dynamic Storage Relocation - mapping logical
memory references to physical memory references
- Provision for Sharing common information stored
in memory by different users
- Protection of information against unauthorized access
Segmentation
- A segment is a set of logically related instructions
or data elements associated with a given name
- Variable size
Subroutine
Stack
SQRT
Main
Program
Symbol
Table
User's view of memory
User's view of a program
The user does not think of
memory as a linear array
of words. Rather the user
prefers to view memory as
a collection of variable
sized segments, with no
necessary ordering among
segments.
Memory Management Hardware

46
Memory Organization
SEGMENTATION
- A memory management scheme which supports
user's view of memory
- A logical address space is a collection of segments
- Each segment has a name and a length
- Address specify both the segment name and the
offset within the segment.
- For simplicity of implementations, segments are numbered.
Segmentation Hardware
<
Segment Table
limit base
(s,d)
s
Memory
+
y
n
error
CPU

47
Memory Organization
SEGMENTATION EXAMPLE
Subroutine
Segment 0
Stack
Segment 3
SQRT
Segment 1 Main
Program
Segment 2
Symbol
Table
Segment 4
Segment 0
Segment 3
Segment 2
Segment 4
Segment 1
1400
2400
3200
4300
4700
5700
6300
6700
Segment Table
1000 1400
400 6300
400 4300
1100 3200
1000 4700
limit base
0
1
2
3
4
Logical Address Space

48
Memory Organization
SHARING OF SEGMENTS
Editor
Segment 0
Data 1
Segment 1
Logical Memory
(User 1)
Editor
Segment 0
Data 2
Segment 1
Logical Memory
(User 2)
Editor
43062
Data 1
68348
72773
90003
98556
Data 2
25286 43062
4425 68348
limit base
0
1
Segment Table
(User 1)
25286 43062
8550 90003
limit base
0
1
Segment Table
(User 2)
Physical Memory

49
Memory Organization
SEGMENTED PAGE SYSTEM
Segment Page Word
Segment table Page table
+
Block Word
Logical address
Physical address

50
Memory Organization
IMPLEMENTATION OF PAGE AND SEGMENT TABLES
Implementation of the Page Table
- Hardware registers (if the page table is reasonably small)
- Main memory
Implementation of the Segment Table
Similar to the case of the page table
- Cache memory (TLB: Translation Lookaside Buffer)
- To speedup the effective memory access time,
a special small memory called associative
memory, or cache is used
- Page Table Base Register(PTBR) points to PT
- Two memory accesses are needed to access
a word; one for the page table, one for the word

51
Memory Organization
EXAMPLE
Logical and Physical Addresses
Logical and Physical Memory Address Assignment
Segment Page Word
4 8 8
Block Word
12 8
Physical address format: 4096 blocks of 256 words
each, each word has 32bits
2 x 32
Physical
memory
20
Logical address format: 16 segments of 256 pages
each, each page has 256words
Hexa
address Page number
Page 0
Page 1
Page 2
Page 3
Page 4
60000
60100
60200
60300
60400
604FF
Segment Page Block
6 00 012
6 01 000
6 02 019
6 03 053
6 04 A61
(a) Logical address assignment (b) Segment-page versus
memory block assignment

52
Memory Organization
LOGICAL TO PHYSICAL MEMORY MAPPING
Segment table
0
F
35
6
A3
Page table
00
35 012
36 000
37 019
38 053
39 A61
012
A3
Physical memory
00000
000FF
Block 0
01200
012FF
Block 12
01900
019FF
32-bit word
0197E
Logical address (in hexadecimal)
6 02 7E
Segment and page table mapping
Segment Page Block
6 02 019
6 04 A61
Associative memory mapping

53
Memory Organization
MEMORY PROTECTION
Protection information can be included in the
segment table or segment register of the memory
management hardware
- Format of a typical segment descriptor
- The protection field in a segment descriptor specifies
the Access Rights to the particular segment
- In a segmented-page organization, each entry in the
page table may have its own protection field to
describe the Access Rights of each page
- Access Rights:
Full read and write privileges.
Read only (write protection)
Execute only (program protection)
System only (O.S. Protection)
Base address Length Protection

54
Memory Organization
A Typical Cache and TLB Design
Page
Number
Line
Number
Word in
Line
Virtual Address
Virtual
Address
Real
Address
From translator CPU
Hash
Function
S
Compare Virtual
Addresses
Real Address Data
CPU Memory
A
S
To translator
Compare Addresses
& Select Data
Word Select & Align
Data
Data Out
S = Select
A
Real
Address
To Main Memory
TLB
Cache

55
Memory Organization
Structure of Cache Entry and Cache Set
Cache Entry
Real Address Tag Data Valid
Entry 1 Entry 2    Entry E Replacement status
Cache Set

56
Memory Organization
Cache Operation Flow Chart
Receive Virtual Address
Hash Page Number
Search TLB
In TLB ?
Send Virtual Address
to Translator
Use Page & Segment tables
to Translate Address
Put in TLB
A
Use Line Number
to Select Set
Read Out Address Tags
Compare Addresses
Match ?
Send Real Address
to Main Memory
Receive Line from
Main Memory
Select Correct
Word from Line
Read Out
Store Line
in Cache
Update Replacement
Status in TLB
Update Replacement
Status
Select
Correct
Line
A
yes
yes
no
no

57
Memory Organization
Virtual Address Format - Example
Byte within line
31 21 20 17 12 11 10 4 3 2 1 0
Byte within page
Page number
Byte within
word
Word within
line
Select set
in cache
Select set
in TLB
Line number
Map through
page directory Map through
page table
Virtual Address of Fairchild Clipper

Memory_Unit Cache Main Virtual Associative

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