CS 325: CS Hardware and Software Organization and Architecture

1. CS 325: CS Hardware and SoftwareOrganization and Architecture Memory Organization

2. Storage/Memory Hierarchy

3. Memory Storage Characteristics • Location • Capacity • Unit of transfer • Access method • Performance • Physical type • Physical characteristics • Organization

4. Memory Storage - Location • CPU • Registers • L1, L2, L3, L4 Cache • Internal • Main Memory (System RAM) • BIOS (EEPROM) • External • Magnetic Disk (HDD) • Non Volatile Solid State (SSD) • Optical • Magnetic Tape

5. Memory Storage - Capacity • Word size • The natural unit of organization • Expected size of most data and instructions • Typically 32 bits or 64 bits • Past: 16 bits • Typical Storage • L1 Cache: 32 – 64 KB per core • L2 Cache: 128 – 512 KB per core • L3 Cache: 2 – 8 MB (shared) • L4 Cache: 0 – 128 MB (video memory) • Main Memory (RAM): 4 – 32 GB (Typical Desktop) • HDD Cache: 16 – 64 MB • SDD: 64 – 512 GB • HDD: 200 – 2000 GB (Inexpensive, but extremely slow) • Optical: • DVD: 4.7 – 17.08 GB • Blu-ray: 25 – 100 GB • Magnetic Tape: 10 – 35 TB per cartridge (uncompressed)

6. Performance – Transfer Rate Example Problem • Assume we have a 32-Mbit SDRAM memory with 8 bits simultaneously read and a cycle time of 250 ns. • How fast can data be moved out of memory? 8b * (1/250ns) = 8b * (4x106/s) = 32 Mbps = 4 MBps

7. Memory Storage – Physical Types • Semiconductor • Cache • Main Memory (RAM) • SSD • Magnetic • HDD • Tape • Optical • CD • DVD • Blu-Ray • Others • Bubble • Hologram

8. Memory Storage – Physical Characteristics • Volatility • Erasable • Power consumption/Heat

9. Memory Storage – Hierarchy List • Registers • L1 Cache • L2 Cache • L3 Cache • Main Memory • Disk Cache • SSD • HDD • Optical • Tape

10. Memory Basics

11. Semiconductor Memory • Random Access Memory (RAM): • All semiconductor memory is random access • Directly accessed by address logic • Read/Write • Volatile • Requires constant power supply • Temporary storage • Static • Holds data • Dynamic • Periodically refreshes charge

12. Static RAM • Bits stored as on/off switches (transistors) • No charges to leak • Does not need refresh circuits • No refreshing needed when powered • Larger per bit • More expensive • Faster • Example: • Cache Memory:

13. Dynamic RAM • Bits stored as charge in capacitors (also uses transistors) • Charges leak from capacitors • Needs refreshing, even when powered • Needs refresh circuits • Smaller per bit • Less expensive • Slower • Asynchronous and Synchronous DRAMs • Example: • Main memory

14. Read Only Memory (ROM) • Permanent storage • Microprogramming • Library subroutines • Systems programs • Function tables

15. Measures of Memory Technology • Density • Latency and cycle time

16. Memory Density • Refers to memory cells per square area of silicon • Usually states as number of bits on standard chip size • Examples: • 1 mb chip • 4 mb chip • Memory cells typically structured in arrays • 1Mb x 1 chip • 256 Kb x 4 chips • Note: higher density chip generates more heat

17. CS 325: CS Hardware and SoftwareOrganization and Architecture Internal Memory

18. Semiconductor Memory Types

19. Flash Memory • Provides block electrical erasure but not byte level • Typical block size 512, 2048, 4096 • High density • One transistor per bit • Fast read speeds, but not as fast as DRAM • Very slow erase speed

20. Error Detection and Correction • Hard Failure • Permanent defect • Caused by • Harsh environmental abuse • Manufacturing defects • Wear • Soft Error • Random, non-destructive • No permanent damage to memory • Caused by • Power supply problems

21. Error Detection and Correction • A single parity bit can be used to detect (most) errors in a word • Parity bit test can fail to detect errors when there is more than one bit error • Hamming codes can be used to detect and correct errors

22. Error Detection and Correction • Bits are occasionally flipped in transmission. • For example: 1101001 is sent, but 0101011 is received. • Adding redundancy can allow us to detect, and possibly correct, some errors of this type. • Simple approach: Repeat each bit • Repeat each bit twice. For bit x, transmit xx. If the receiver gets two different bits, it requests a retransmission. • This is an error detecting code. • Allows for one error to be detected, but is not error correcting since retransmission is necessary • Repeat each bit three times. For each bit x, transmit xxx. • Now the receiver can correct a single error. • Why?

23. Problem with the simple approach • The receiver can detect and correct bit errors if each bit is transmitted three times. • How does this affect performance? • Better approach • Parity check codes • Has the ability to detect odd number of bit flips using a single parity bit.

24. Calculating bit string parity • A bit string has odd parity if the number of 1s in the string is odd. • 100011, 1, 000010 have odd parity • A bit string has even parity if the number if 1s in the string is even. • 01100, 000, 11001001 have even parity • Assume 0 is an even number

25. Parity check code • Assume we are transmitting blocks of k bits. • A block (w) of length (k) is encoded as (wa), where the value of the parity bit (a) is chosen so that (wa) has even parity. • Example: • If w = 10110, we send wa = 101101, which has even parity • With no bit flips in the transmission, the receiver gets the bit string exactly as it was sent by the sender. Bit string has even parity. • If there are an odd number of bit flips in the transmission, the receiver gets a bit string with odd parity. Retransmission is requested. • If there are an even number of bit flips in the transmission, the receiver gets a bit string with even parity. The error(s) go undetected. • Another solution?

26. 2D parity check code • Blocks of bits are organized in rows and columns • m x n matrix • The parity bit of each row is calculated, and appended to the row before it is transmitted • The parity of each column is calculated, and the parity bit of the entire matrix is computed. • These are also transmitted to the receiver • m + n + 1 parity bits are computed • mn + m + n + 1 bits are sent to the receiver • Efficiency becomes greater as block size increases

27. 2D parity check • Example: • Original data: 1100, 1011, 0111, 0101 Row Parity Column Parity Matrix Parity bit • MN + M + N + 1 bits transferred. • 5*5 + 5 + 5 + 1= 36 bits

28. Hamming Code • Linear error detecting/correcting codes invented by Richard Hamming in 1950. • Can detect up to 2 bit errors • Can correct 1 bit errors

29. Hamming Code – Parity bits • Hamming code works by propitiating parity bits throughout a bit string of size (w) • (p) parity bits creates a bit string of size 2m – 1, of which 2m – m – 1 bits can be used for data. • Common Hamming code sizes: • Hamming(3,1), 2 parity bits • Hamming(7,4), 3 parity bits • Hamming(15,11), 4 parity bits • Hamming(31,26), 5 parity bits • Is read as Hamming(total bits, data bits)

30. Hamming Code • Example: • Using Hamming(7,4), create the Hamming codeword for the following 4 bit string: • 0101 • Hamming(7,4) • 7 total bits • 4 data bits • 3 parity bits • Parity bits are always located in the codeword at positions of 2n. • P1 = 20 = 1 • P2 = 21 = 2 • P3 = 22 = 4

31. Hamming Code • Example: • Using Hamming(7,4), create the Hamming codeword for the following 4 bit string: • 0101

32. Hamming Code • Example: • Using Hamming(7,4), create the Hamming codeword for the following 4 bit string: • 0101

33. Hamming Code • Example: • Using Hamming(7,4), create the Hamming codeword for the following 4 bit string: • 0101 • Now, to calculate the parity bits.

34. Hamming Code • To calculate P1: find parity of substring(1,3,5,7)

35. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0

36. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7)

37. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7) • P2 0 0 1 = 1

38. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7) • P2 0 0 1 = 1 • To calculate P3: find the parity of substring(4,5,6,7)

39. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7) • P2 0 0 1 = 1 • To calculate P3: find the parity of substring(4,5,6,7) • P3 1 0 1 = 0

40. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7) • P2 0 0 1 = 1 • To calculate P3: find the parity of substring(4,5,6,7) • P3 1 0 1 = 0 • Now we know P1, P2, and P3.

41. Hamming Code • To calculate P1: find parity of substring(1,3,5,7) • P1 0 1 1 = 0 • To calculate P2: find the parity of substring(2,3,6,7) • P2 0 0 1 = 1 • To calculate P3: find the parity of substring(4,5,6,7) • P3 1 0 1 = 0 • Now we know P1, P2, and P3, and can calculate the codeword: 0 1 0 0 1 0 1

42. CS 325: CS Hardware and SoftwareOrganization and Architecture Cloud Architectures

43. Outline • Introduction • Software as a Service (SaaS) • Platform as a Service (PaaS) • Infrastructure as a Service (IaaS) • Background • Computational Resource Load Balancing

44. Introduction • Scalable resource hosting • Storage • Computational • Software APIs • Applications • Tailored services • Software as a Service (SaaS) • Platform as a Service (PaaS) • Infrastructure as a Service (IaaS) • Billed like a utility • Monthly, depending on usage

45. Introduction • No formal definition! • A set of service oriented architectures, which allow users to access a number of resources in a way that is scalable, elastic, on-demand, and cost-efficient Compute Service Cloud Interface Server Client Compute Service Compute … Other Services Storage Service Compute Client

46. Introduction Software as a service (SaaS) [2-4] Platform as a service (PaaS) [2-4] Infrastructure as a service (IaaS) [2-4] Lowest service level in cloud stack. Provides compute, storage, and networking services using hardware virtualization. Compute Service Cloud Interface Server Compute Service Compute Other Services Storage Service Compute Edmonds, A., S. Johnston, T. Metsch, and G. Mazzaferro Liu, F., J. Tong, J. Mao, R. Bohn, J. Messina, M. Badger, and D. Canonical Group Ltd.

47. Introduction Typical General Purpose Private Cloud Architecture (Eucalyptus [5]) Eucalyptus Systems

48. Types of Clouds • Public Cloud • Marketed based on • Resources offered • Availability • Security • Price • Local Cloud • Cloud architectures tailored to an organization’s needs • Hybrid Cloud • Combination of public and local cloud resources

49. CS 325: CS Hardware and SoftwareOrganization and Architecture Cloud Architecture Background

50. Background • Concept of delivering computing resources through a global network • 1960s • Computer Clusters • 1970s • Grid Computing • 1990s • Cloud: Evolution of Grid and Cluster • 2000s