Introduction to Operating System - Lecture 3
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The lecture delves into operating system architectures, including Direct Memory Access (DMA), various storage structures, and distinctions between files and directories. It explores operating system designs such as MS-DOS, Unix, and modular systems, highlighting their advantages and disadvantages. The discussion also covers process management, threading, concurrency, deadlocks, binary coding schemes, and basic arithmetic in binary notation.
Introduction to Operating System - Lecture 3
- 1. By Mohamed Gamal Lecture 3: Deep Dive into the Operating System Operating System
- 2. – What’s DMA (Direct Memory Access)? – What’re the different layers of a computer system? – Mention three types of storage structure along with their advantages. – What’re the three basic logic gates found? – What’s the difference between a file and a directory? – What’re the following extensions used for? o .exe, .png, .mp3, .mp4, .c, .jpg Test your knowledge ?
- 3. – Find the one’s complement of the binary number 1001001 – Conversion: • 5 Megabytes to Bytes. • 12 Terabytes to Megabytes. • 1024 Kilobytes to Gigabytes. • 5 Bytes to bits. Test your knowledge ?
- 4. The topics of today’s lecture: Agenda
- 6. – Simple Structure (MS-DOS) – More Complex (UNIX – monolithic) – Layered (Conceptual) – Microkernel (Mach) – Modular (modules) OS Design Structure
- 7. Simple Structure – MS-DOS -MS-DOS is written to provide the most functionality in the least space • Higher performance • Smaller size • Not divided into modules • Interfaces and levels of functionality are not well separated • Very complex to maintain later • Hardware dependent
- 8. More Complex Structure – UNIX
- 9. - Advantages: • Higher performance • Adds some protection - Disadvantages: • Very complex to maintain later • Still Hardware dependent Examples: Windows XP – Monolithic More Complex Structure – UNIX (Cont.)
- 10. Layered Approach (Conceptual) -The operating system is divided into a number of layers (levels), each built on top of lower layers. -The bottom layer (layer 0), is the hardware; the highest (layer N) is the user interface. -With modularity, layers are selected such that each uses functions (operations) and services of only lower-level layers
- 11. - Advantages: • Modularity • Debugging • Modification - Disadvantages: • Layering overhead to the system call Layered Approach (Cont.)
- 13. - Advantages: • Easier to extend • More reliable • More secure - Disadvantages: • Performance overhead of user space to kernel space communication ✓ That’s why Windows XP designed almost as monolithic again! Microkernel System Structure (Cont.)
- 14. - Many modern operating systems implement loadable kernel modules • Uses object-oriented approach • Each core component is separate • Each talks to the others over known interfaces • Each is loadable as needed within the kernel - Similar to layers but with more flexibility (any module can call another module) • Linux, Solaris … etc. Modular System Structure
- 15. Modular System Structure (Cont.)
- 16. Solaris User Interface (GUI)
- 17. Modular System Structure (Cont.) - Advantages: • Resembles a layered system • More flexibility ✓ any module can call any other module • Similar to microkernel approach but more efficient ✓ modules do not need to invoke message passing to communicate
- 18. – Most modern operating systems are actually not one pure model • Hybrid combines multiple approaches to address performance, security, usability needs • Linux and Solaris kernels: ✓ monolithic, plus modular • Windows: ✓ monolithic, plus microkernel Hybrid Systems
- 20. Process Concept - Program: a passive entity stored on disk (executable file – .exe). - Process: a program in execution; process execution must progress in sequential fashion. - Program becomes a process when executable file loaded into memory. - One program can be several processes.
- 21. Process in Memory – Text section • Program instructions – Program counter • Next instruction – Stack • Local variables • Return addresses • Method parameters – Data section • Global variables
- 22. Process State
- 23. Process State (Cont.) – As a process executes, it changes state • new: The process is being created • running: Instructions are being executed (time slice) • Waiting (suspended): The process is waiting for some event to occur • ready: The process is waiting to be assigned to a processor • terminated: The process has finished execution
- 24. Process Control Block (PCB) – Information associated with each process: • Process state (running, waiting/suspended, … etc.) • Program counter (location of next instruction to execute) • CPU registers • CPU scheduling information • Memory-management information • Accounting information (CPU used, clock time elapsed since start, time limits) • I/O status information
- 25. Process Control Block (PCB) Pointer Process state Process number Program counter CPU registers Memory management info I/O status information Accounting Information
- 26. Context Switching – When CPU switches to another process, the system must save the state of the old process and load the saved state for the new process via a context switch. – Context Switching is for Multitasking. – Context-switch time is overhead; the system does no useful work while switching. – The more complex the OS and the PCB, the longer the context switch – Time dependent on hardware support
- 28. Process Example Google Chrome Browser Question: why did they do this?
- 30. Threads Concept - Most modern applications are multithreaded. - Multiple tasks with the application can be implemented by separate threads. - Process creation is heavy-weight. - Thread creation is light-weight. • Only Program Counter is needed - Threads simplify code and increase efficiency + PCB
- 31. Single and Multithreaded Processes
- 32. Benefits of using threads - Responsiveness – may allow continued execution if part of process is blocked • especially important for user interfaces - Resource Sharing – threads share resources of process • easier than shared memory or message passing - Economy – cheaper than process creation • thread switching lower overhead than context switching - Scalability – process can take advantage of multiprocessor architectures
- 33. Concurrency vs. Parallelism - Concurrent execution on single-core system: - Parallelism on a multi-core system:
- 36. - A set of blocked processes each holding a resource and waiting to acquire a resource held by another process in the set. - Example • System has 2 tape drives. • P1 and P2 each hold one tape drive and each needs another one. P1 P2 wait (A); wait(B) wait (B); wait(A) The Deadlock Problem
- 37. One-lane Crossing Example - Traffic only in one direction. - Each section of a bridge can be viewed as a resource. - If a deadlock occurs, it can be resolved if one car backs up (preempt resources and rollback). - Several cars may have to be backed up if a deadlock occurs. - Starvation is possible!
- 38. - Ensure that the system will never enter a deadlock state. - Allow the system to enter a deadlock state and then recover. - Ignore the problem and pretend that deadlocks never occur in the system • used by most operating systems, including UNIX • lets the user handle it • Deadlock prevention is not completely guaranteed and is incredibly hard! Methods for Handling Deadlocks
- 40. - Combination of bits are used to represent: • alphabetic data • numeric data • Symbols • sound and video data - Binary Coding schemes are used for this purpose • EBCDIC (Extended Binary Coded Decimal Interchange Code) • ASCII (American Standard Code for Information Interchange) • Unicode (UTF-8) Binary Coding Schemes
- 43. - Binary System - Octal System - Decimal System - Hexadecimal System › Computers use binary system › Decimal system is used in general › Octal and Hexadecimal systems are also used in computer systems Numbering System Numbering System Base Range Binary Base 2 0 – 1 Octal Base 8 0 – 7 Decimal Base 10 0 – 9 Hexadecimal Base 16 0 – 9, A – F
- 44. Decimal Binary Octal Hexadecimal 0 0000 0 0 1 0001 1 1 2 0010 2 2 3 0011 3 3 4 0100 4 4 5 0101 5 5 6 0110 6 6 7 0111 7 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1101 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F
- 45. - 12510 › 5 x 100 = 5 › 2 x 101 = 20 › 1 x 102 = 100 12510 Decimal Numbering System (Base 10) + +
- 46. - Technique: • Multiply each bit by 2n, where n is the weight of the bit • Add the results – Convert the binary number 110012 to decimal › 24 23 22 21 20 › 1 1 0 0 1 › 16 8 0 0 1 + › 16 + 8 + 0 + 0 + 1 = 25 Binary to Decimal (Base 2) × Result: 2510
- 47. - Technique: • Divide the number repeatedly by 2, while keeping track of the remainders. • Arrange the remainders in reverse order. – Convert the decimal number 2510 to binary Decimal to Binary (Base 10) Number Remainder ÷ 2 25 1 ÷ 2 12 0 ÷ 2 6 0 ÷ 2 3 1 ÷ 2 1 1 ÷ 2 0 — Result: 110012
- 48. - Technique: • Group the binary digits into sets of four. • Convert each group to its corresponding hexadecimal digit. – Convert the binary number 110110112 to hexadecimal › 23 22 21 20 23 22 21 20 › 1 1 0 1 1 0 1 1 › 8+4+0+1 8+0+2+1 › D B Binary to Hexadecimal (Base 2) A: 10 D: 13 B: 11 E: 14 C: 12 F: 15 Result: DB16
- 49. - Technique: • Convert each hexadecimal digit to its binary representation. • Similar to binary to hexadecimal but in a reversed way. – Convert the hexadecimal number DB16 to binary › D B › 23 22 21 20 23 22 21 20 › 8 + 4 + 0 + 1 8 + 0 + 2 + 1 › 1 1 0 1 1 0 1 1 Hexadecimal to Binary (Base 16) B: 11 , D: 13 Result: 110110112
- 50. - Technique: • Sum of the positional values of each hexadecimal digit. – Convert the hexadecimal number DB16 to decimal › D B › 13 x 161 + 11 x 160 › 208 + 11 Hexadecimal to Decimal (Base 16) B: 11 , D: 13 Result: 21910 Note: another way is to first convert the hexadecimal number to binary and then convert the binary number to decimal.
- 51. Decimal to Hexadecimal (Base 10) - Technique: • Divide the number repeatedly by 16, while keeping track of the remainders. • Multiply the quotient by 16 to get the first result • Divide the integer by 16 again … • Arrange the remainders in reverse order. – Convert the decimal number 21910 to hexadecimal Number Remainder ÷ 16 219 0.6875 x 16 = 11 ( B ) ÷ 16 13 0.8125 x 16 = 13 ( D ) ÷ 16 0 — Result: DB16
- 53. Binary Addition 0 0 0 1 0 1 0 1 1 1 1 10 + + + + 1 1 1 1 11 + 1 1 1 10 + 10 1 11 + 1
- 54. Binary Addition (Cont.) – Add the two binary numbers 10112 and 11012 1 1 1 1 – 1111 1 0 1 1 › 1 1 0 1 › 1 1 0 0 0 + Result: 110002 Least Significant bit (LSB) Most Significant bit (MSB) Note: you can verify your answer by converting to decimal. 1110 1310 2410
- 55. Try yourself ! – Add the following binary numbers. 1) 1002 and 12 2) 111102 and 101012 3) 10010012 and 1101102 4) 1000002 and 111112
- 56. 1 Binary Subtraction 0 0 0 1 0 1 0 1 1 1 1 0 – – – – 0 2
- 57. Binary Subtraction (Cont.) – Subtract the two binary numbers 11012 and 10112 0 2 – 1111 › 1 1 0 1 › 1 0 1 1 › 0 0 1 0 – Result: 00102 Note: you can verify your answer by converting to decimal.
- 58. Try yourself ! – Subtract the following binary numbers. 1) 1002 and 12 2) 111102 and 101012 3) 10010012 and 1101102 4) 1000002 and 111112
- 59. Binary Multiplication – Multiply the two binary numbers 10112 and 11012 – 1111 › 0 1 1 › 1 0 1 › 0 1 1 › 0 0 0 0 › 0 1 1 0 0 › 0 1 1 1 1 × Result: 011112 Note: you can verify your answer by converting to decimal.
- 60. Try yourself ! – multiply the following binary numbers. 1) 10112 and 1102 2) 100102 and 1012 3) 11102 and 102 4) 110012 and 112
- 62. – Negative numbers are stored in two's complement notation. – Represents the additive Inverse • If you add the number to its additive inverse, the sum is zero. Two’s Complement Note that 00000001 + 11111111 = 00000000
- 63. – Add the two decimal numbers – 9 and 4. Two’s Complement – Example 9 = 00001001 → Two’s Complement: 11110111 4 = 00000100 11110111 00000100 + 11111001 1 Represents a negative number Most Significant bit (MSB) 1 -128 64 32 16 8 4 2 1 1 1 1 1 1 0 1 1 -128 + 64 + 32 + 16 + 8 + 0 + 2 + 1 = -5
- 64. – CMD – OS – OS Types (VM) – Storage Structure – Caching concept – Difference between files and directories – Viruses types – API Concept – Logic Gates – Fetch-decode Cycle – Multiprocessor/Multicores vs. Multiprogramming – Grid and its derivatives – Real-time systems – LAN vs. WAN Revision
- 65. End of lecture 3 ThankYou!