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.

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.)

12. Microkernel System Structure

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

27. Context Switching (Cont.)

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:

35. Is this Deadlock?

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!