CHAPTER 2 HARDWARE

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CHAPTER 2HARDWARE

• CGS 3763 - Operating System Concepts
• UCF, Spring 2004

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OVERVIEW

• Binary Hex Data Representation
• Memory Storage Heirarchy
• Addressing
• Primary Secondary Memory
• The disk I/O Mechanism
• Processor Components
• Instruction Execution Cycle
• Program Control Flow The Interrupt Cycle

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BINARY REPRESENTATION

• Binary is the computers native language.
• Binary is a Base 2 numbering system (we represent
  numbers using 2 digits, 0 or 1) wheras our normal
  numbering system is a base 10 system (we
  represent numbers using 10 digits 0-9)
• Example of a typical binary number 10001010
• 8-bit binary number representing the decimal
  number 138
• The computer uses various combinations of 2
  digits to represent letters, numbers, and pixels.
• NOTE At the hardware level binary digits aren't
  really 1s and 0s. They're actually electrical
  impulses in either a (mostly) on state or a
  (mostly) off state.

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BINARY REPRESENTATION (cont).

• With 1 binary digit, we have 2 possible
  combinations
• 0 or 1.
• With 2 binary digits, we have 4 possible
  combinations and are able to represent the
  numbers, 0-3.
• 0 00
• 1 01
• 2 10
• 3 11

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BINARY REPRESENTATION (cont.)

• With 4 binary digit, we have 16 possible
  combinations and are able to represent the
  numbers 0 15.
• 0 0000
• 1 0001
• 2 0010
• 3 0011
• 4 0100
• 5 0101
• 6 0110
• 7 0111
• 8 1000
• 9 1001
• 10 1010
• 11 1011
• 12 1100
• 13 1101
• 14 1110
• 15 1111

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BINARY REPRESENTATION (cont.)

• In general, how many different binary values can
  be represented by a specific number of bits?
• number of different values 2n where n is the
  number of bits .
• Example
• We can represent 256 values using 8 bits
• 256 28
• Q How many bits does it take to represent the
  value 24?
• A. lg2(24) lg2(2n )
• lg2(24) n
• log(24) / log(2) n
• n 4.5 ? 5 bits

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HOW TO READ A BINARY NUMBER

• Example binary to decimal conversion 10001010

127 026 025 024 123 022
121 020 138

• Q Convert the binary number 1011 to decimal?
• A 123 022 121 120 11

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HEX REPRESENTATION

• Most modern architectures are either 32 or 64
  bit.
• If every number we dealt with was 32 or 64 digits
  long, programmers would go nuts!
• The solution is to represent binary numbers in
  hex.
• Hex is a base 16 numbering format which means
  that we represent numbers using 16 digits,
  0-9,A-F
• A hex number is represented with a 0x in front
  of it
• Example 0xA3 or 0xF0 or 0x46

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HEX REPRESENTATION

• Example binary to hex conversion 10001010
• Break binary number into groups of 4 digits
• Treat each 4 digit group as an individual binary
  number.

123 121
123
A
8
0x8A
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HEX CONVERSION (cont.)

• Q Convert 10110011 into hex?
• A
• Break into groups of 4 digits
• 1011 0011
• Treat each 4 digit group as an individual binary
  number.
• 1011 123 022 123 120 0xB
• 0011 023 022 123 120 0x3
• Final Answer 0xB3

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MEMORY

• Memory holds active programs and data.
• Memory write is a destructive operation.
• Memory read is not destructive.
• Memory contents
• Numbers count 0 to infinity...
• In a computer, we can only represent a limited
  subset of them.
• Bit one binary digit (0 or 1).
• Byte eight bits (1 character).
• Word a group of bytes.
• Word size is usually machine dependent.

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MEMORY (cont.)

• Memory can be viewed as a hierarchy based upon
  size .

Words are made up of 1 or more bytes
Bytes are made up of 8 bits
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ADDRESSING MEMORY

• Each byte (or word) is assigned a unique address.
• The bytes (or words) are numbered sequentially.
• Bytes (or words) are the basic addressable units
  of memory.
• Because we cannot represent numbers through
  infinity, there is a maximum size number that can
  be represented on any given machine.
• This number usually corresponds to the maximum
  number of addressable memory units.
• For example, if we had a 16 bit (2 byte) machine,
  we would only be able to address 216 or 65,536
  bytes of memory (because the 16 bits can only
  make 65,536 unique combinations of 0s and 1s).
• Data in memory is stored in binary / hex.

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OVERVIEW TYPES OF MEMORY

• Registers
• Main Memory (RAM ROM)
• Cache
• Magnetic Disk
• Optical Disk
• Magnetic Tape

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VOLATILE VS. NON-VOLATILE STORAGE

• Volatile Storage looses its contents when the
  power to the device is removed
• Examples registers, cache, main memory (RAM)
• Non-volatile Storage retains its contents when
  the power to the device is removed
• Examples magnetic disk, optical disk, magnetic
  tape

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REGISTERS MAIN MEMORY(PRIMARY STORAGE)

• Registers are high speed, temporary storage
  devices used by the CPU to hold control
  information, data, and intermediate results of
  calculations.
• Main Memory (RAM ROM)
• RAM (random access memory) provides moderately
  high speed temporary storage for program code and
  data. It is both readable and writable.
• ROM (read only memory) provides moderately high
  speed permanent storage for program code and data
• Often holds the program that loads the operating
  system (BIOS).

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PRIMARY STORAGE (cont.)

• Main memory and the processor registers are the
  only storage that the CPU can access directly.
• Processor instruction set provides instructions
  that allow a programmer to manipulate the
  contents of memory and registers.
• EXAMPLE
• mov reg, mem
• mov mem, reg
• add reg, reg
• The processor manipulates data stored in memory
  under the control of a program stored in memory.

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CACHE

• Cache is a very fast block of memory that speeds
  up the performance of another device.
• It is used to hold recently-accessed data.
• The computer looks in the cache first to see if
  what it needs is there.
• EXAMPLE
• Main memory is much slower than the CPU registers
• A cache may be placed between the processor and
  main memory to speed up data access.
• If a program needs to load a data value into a
  register using an instruction like mov reg, mem
  the machine first looks to see if it is available
  in the cache before going out to main memory.
• If the data is present it is termed a cache hit
  and will be loaded directly from the cache.
• If the data is not present it is termed a cache
  miss and will be loaded from main memory.

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CACHE RELATIONSHIP WITH THE PROCESSOR MAIN
MEMORY
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PROBLEM WITH PRIMARY STORAGE

• PROBLEM
• Primary memory is both expensive and volatile!
• Because it is expensive, we may not have enough
  memory to store all of the data we need for a
  program.
• Quantity is measured in megabytes.
• Because it is volatile, we have no way to save
  the output of our program.
• SOLUTION
• Secondary Storage
• Provides an extension of main memory.
• We can move data from the secondary storage to
  main memory on an as needed basis.
• Inexpensive so we can have lots of storage!
• Quantity is measured in gigabytes.
• Non-volatile so we can permanently save data.

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SECONDARY STORAGE MAGNETIC DISK

• Magnetic Disks (Hard Disks)
• Magnetic disks rigid metal or glass platters
  covered with magnetic recording material.
• Data is recorded on a series of concentric
  circles on the disk surface called tracks. Tracks
  are subdivided into sectors.
• A sector is the smallest unit of data that can be
  moved between the disk and main memory.
• Storage Capacity 20 - 400 Gigabytes

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I/O Disk Head Mechanism
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CALCULATING DISK CAPACITY

• A disk has 3 platters. There are 3000 tracks on
  the disk (or 1000 tracks on each platter). Each
  track has 32 sectors and each sector has 512
  bytes.
• Q How many cylinders are there?
• A There 3 tracks for every 1 cylinder. Since the
  disk has a total of 3000 tracks, there are 3000/3
  cylinders. The total number of cylinders is
  therefore 1000.

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CALCULATING DISK CAPACITY (cont.)

• A disk has 3 platters. There are 3000 tracks on
  the disk (or 1000 tracks on each platter). Each
  track has 32 sectors and each sector has 512
  bytes.
• Q How many bytes are stored per cylinder?
• A There 32 sectors 512 bytes 16384 bytes per
  track. Since there are 3 tracks per cylinder, a
  cylinder can store 3 16384 49,152 bytes.

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CALCULATING DISK CAPACITY (cont.)

• A disk has 3 platters. There are 3000 tracks on
  the disk (or 1000 tracks on each platter). Each
  track has 32 sectors and each sector has 512
  bytes.
• Q What is the total storage capacity of the
  disk?
• A 1 cylinder can store 49,152 bytes and
  previously determined that there are 1000
  cylinders. Therefore, the total disk capacity is
  49,152 bytes 1000 cylinders 49,152,000 bytes

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Input and Output

• Input
• The act of transferring data into memory from a
  peripheral device.
• Output
• The act of transferring data from memory to a
  peripheral device.
• Input and Output
• Support human interaction with a computer.

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MAGNETIC DISKS I/O MECHANISM

• Steps to read a sector from a hard disk.
• The disk drive is brought up to operating speed.
• The read-write head is positioned over the track
  that holds the data.
• The time required to position the head is called
  the seek time.
• The system waits while the desired sector rotates
  to the read-write head.
• This wait time is called the rotational delay.
• The system reads the data.
• The time required for the system to finish
  reading the data is called the transfer time.
• The disk service time seek time rotational
  delay transfer time

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Reading a sector from disk Step 1
The disk drive is brought up to operating speed.
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Reading a sector from disk Step 2
During seek time, the access mechanism is
positioned over the track that holds the data.
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Reading a sector from disk Step 3
The system waits while the desired sector rotates
to the read/write head (rotational delay).
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CALCULATING DISK SERVICE TIME

• A disk has 3 platters. There are 3000 tracks on
  the disk (or 1000 tracks on each platter). Each
  track has 32 sectors and each sector has 512
  bytes.
• Q Assume the average seek time is 10 ms. The
  rotation speed is 3600 rpm. It additionally takes
  approximately 2 ms to transfer the data contained
  in 1 sector. How long would you expect it to take
  (on average) to read a block of 10 contiguous
  sectors (5120 bytes)?

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CALCULATING DISK SERVICE TIME (cont.)
We want to read 10 consecutive sectors starting
here
Step 2 Calculate the rotational delay for the
desired sector to be positioned over the disk
head.
Step 1 Calculate seek time for the disk Head
to be positioned Over the desired track.
Step 3 Calculate rotational delay for the head
to rotate across all 10 of the sectors we wish to
read.
Step 4 Calculate the time to transfer 10
sectors.
Step 5 disk service time seek time total
rotational delay total transfer time for all 10
consecutive sectors
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CALCULATING DISK SERVICE TIME (cont)

• A The time required to read 10 contiguous
  sectors will be the seek time rotational delay
  transfer time.
• The seek time is given as 10 ms.
• To calculate the rotational delay we must find
  out how long it takes for the disk to make a
  single rotation.
• 3600 rotations / sec 1 rotation / X sec
• 3600x 60
• .0167 sec per rotation
• We expect that on average it will take ½ rotation
  to get to the first sector we wish to read
• The rotational latency to the first sector is
  therefore
• ½ (.0167 sec) .0083 sec 8.3 ms
• Since we wish to read 10 sectors beyond the
  starting point, we also need to calculate the
  rotational latency to the last sector well be
  reading.
• 10 sectors / 32 sectors .0167 sec per
  rotation .3125 .0167 5.2ms
• Lastly, we calculate the transfer time for the 10
  sectors. The time to transfer the data in 1
  sector 2 ms. Therefore, the total transfer time
  is 10 2ms 20 ms.
• Disk service time 10ms (seek time) 8.3ms
  5.2ms (total rotational delay) 20ms
• (total transfer time) 43.5ms

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OTHER STORAGE DEVICES

• Floppy Disks thin circular piece of flexible
  polyester coated with magnetic material
• Storage Capacity 1 MB
• Optical Disks - a flat, circular, plastic disk
  coated with material on which bits may be stored
  in the form of highly reflective areas and
  significantly less reflective areas, from which
  the stored data may be read when illuminated with
  a narrow-beam source, such as a laser diode.
  Usually read only.
• Storage Capacity 700 MB 4 GB
• Examples CD, DVD, CD-R, CD-RW
• Flash Drives a plug-and-play portable storage
  device that uses flash memory. Flash memory
  (sometimes called "flash RAM") is a type of
  constantly-powered nonvolatile memory that can be
  erased and reprogrammed in units of memory called
  blocks.
• Storage Capacity 32 MB 1 GB
• Magnetic Tape - A device, like a tape recorder,
  that reads data from and writes it onto a tape.
• Sequential Access ? slow data access time
• Used primarily for backup
• Storage Capacity gt Magnetic Disk

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THE STORAGE HEIRARCHY

• Storage systems organized in hierarchy according
  to
• Speed
• Cost
• Volatility

VOLATILE
SPEED
NON VOLATILE
COST
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SUMMARY DATA CODE PATH
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WHAT IS A PROGRAM?

• A program can be defined as a series of
  instructions which tells the computer to perform
  one of its basic functions (i.e. add, subtract,
  multiply, compare, ect.).

Each instruction has an operation code and one or
more operands.
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PROCESSOR COMPONENTS
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PROCESSOR COMPONENTS (cont.)

• Instruction Control Unit (ICU) fetches
  instructions from memory
• Arithmetic Logic Unit (ALU) holds or activates
  the computers instruction set and executes
  instructions.
• Registers temporary storage devices that hold
  control information, data, and intermediate
  results.
• Types of registers
• Instruction Counter holds the address of the
  instruction currently being executed.
• Accumulator holds data being used in arithmetic
  calculations (addition, subtraction,
  multiplication, ect.)
• Clock generates precisely timed electronic
  pulses to synchronize the operation of the other
  components.

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TRACING A BASIC MACHINE CYCLE

• Memory holds both program instructions and data.
• The instruction counter points to the first
  instruction to be executed.

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TRACING A BASIC MACHINE CYCLE (cont.)

• The first instruction is fetched from memory and
  stored in the instruction register.
• Note that the instruction counter points to the
  next instruction.

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TRACING A BASIC MACHINE CYCLE (cont.)

• The arithmetic logic unit executes the
  instruction in the instruction register.

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TRACING A BASIC MACHINE CYCLE (cont.)

• The instruction counter points to the next
  instruction.

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TRACING A BASIC MACHINE CYCLE (cont.)

• The next instruction is fetched into the
  instruction register.
• Note that the instruction counter points to the
  next instruction.

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TRACING A BASIC MACHINE CYCLE (cont.)

• The arithmetic and logic unit executes the
  instruction in the instruction register.

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INSTRUCTION / EXECUTION CYCLE

• Modern computers operate using a 2-step cycle
• Fetch
• Get next instruction pointed to by the program
  counter
• Increment the instruction counter
• Decode the instruction and operands
• Execute
• Execute the instruction

An instruction is fetched during I-time and
executed during E-time.
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FLOW OF EXECUTION

• Program control flow is usually sequential (ie.
  instructions are executed in order, one at a
  time).

PROGRAM (sequential execution) 1. mov reg, mem 2.
add reg, reg 3. mov mem, reg 4. test reg, reg

• Non sequential execution may occur because of a
  conditional jump or an interrupt.

PROGRAM (non sequential execution conditional
jmp) 1. mov reg, mem 2. cmp reg, 0 3. jz 8.
//jump to 8. if zero 4. mov reg, mem //otherwise
continue sequentially 5. 6. 7. 8. //execution
resumes here if the test in 2. evaluates to 0 9.
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CONTROL FLOW VISUALIZATION

• Some programs can analyze another program and
  give you a visualization of its control flow.
• Here, IDA Pro generates the control flow graph
  for a function in the Windows kernel.

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FLOW OF EXECUTION (cont.)

• Control flow alterations in response to a
  conditional jump instruction occur in response
  to a change in the internal program state
  (weather a status flag evaluates to true or
  false, for example).
• In contrast, control flow alterations resulting
  from an interrupt occur in response to an
  external event.

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INTERRUPT CYCLE

• When interrupts are introduced, the CPU and the
  operating system driving the system, is
  responsible for the suspension of the program
  currently being run, as well as restoring that
  program at the same point before the interrupt
  was detected.
• Sequence of events the occur following an
  interrupt
• The CPU drops what its doing and saves its state
  (the values contained in all of its registers).
• The CPU determines what type of interrupt
  occurred and looks up the address of its handler
  in a special table called the interrupt vector
  table (IVT).
• Interrupt Vector - a table or array of addresses
  stored in memory which point to each interrupt
  handler.
• Interrupt Handlers - portions of OS program
  specifically written to deal with each interrupt
  type
• It then loads the instruction pointer with the
  interrupt handler extracted from the IVT.
• The CPU executes the code contained at the
  address of the interrupt handler. The interrupt
  handler may disable further interrupts to prevent
  problems with reentrancy.
• The CPU restores the processor registers to their
  state prior to the interrupt and restarts
  execution in the process that was interrupted.

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NOTES ABOUT INTERRUPTS

• Interrupt handlers are provided by the operating
  system.
• What are some uses for interrupts?
• How about gracefully handling an error? The
  interrupt handler detects an error and takes
  control before an errant program can corrupt the
  data of itself or another program.
• How about to detect user input? Consider an input
  device like the keyboard. Most of the time the
  keyboard is idle (no one is typing). Why waste
  CPU cycles constantly checking if there is data
  in the keyboard buffer? Let the keyboard notify
  the OS that there is data available by sending an
  interrupt! When its not receiving interrupts, it
  is free to do other processing.
• What about to manage the CPU resources among
  multiple programs? The timer interrupt allows
  the OS to periodically gain control of the
  processor and give it to another program. This
  sharing of the CPU gives the illusion of many
  programs running at the same time and is the
  basis of the terms, "multi-tasking",
  "multi-processing, and "multiprogramming".
• How about calls to the operating system
  interface? System calls use an interrupt to
  transfer control from an application to the
  operating system in a protected manner.

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