Embed Size (px)
Citation preview
Discussion of exam problems
1
Hint for Homework problem 72: Show how to connect three Full Adders to implement a 2-bit “ripple-carry” adder
2
3
C D A B
X Y
Connect to ground (0)
Z
Chapter 5
Computing Components
Components
Circuits
Gates
Transistors
√
√
√
Yet another layer of
abstraction!
5
5.1 Individual Computer Components
Consider the following ad: Insatavialion 640 Laptop Exceptional Performance and Portability!
It’s just a made-up example!
6
What does all this jargon mean?
• Intel® Core™ 2 Duo (2.66GHz/1066Mhz FSB/6MB cache)
• 4GB Shared Dual Channel DDR2 at 800 MHz• 500 GB SATA Hard Drive at 5400RPM• 15.6” High Definition (1080p) LED Backlit
LCD Display (1366 x 768)• 8X Slot Load DL DVD+/- RW Drive • 14.8”W X 1.2”H X10.1” D, 5.6 lbs.• 512 MB ATI Mobility Radeon Graphics• 85 WHr Lithium Ion Battery• (2) USB 2.0, HDMI, 15-Pin VGA, Ethernet 10/100/1000 IEEE
1394 Firewire, Express Card, Audio line-in, line-out, mic-in
Be patient!You don't knowthem now, but you will getused to them.
Multipliers
7
When referring to computer memory, mega does not mean one million! (but it’s still close to that)
8
Decimal vs. binary multipliers
1000 = 103 1,000,000 = 106 1,000,000,000 = 109
1024 = 210 1024*1024 = 220 1024*1024*1024 = 230
Rule:For memory capacities, the multipliers are binary, for everything else (speed, frequency, pixels, etc.) they are decimal
9
Decimal vs. binary multipliersexamples
What is the meaning of:• 1.5 TB hard-disk • 54 Mbps wireless Ethernet• 6 GB of RAM• 8 Mega-pixel camera• 3.2 GHz CPU
Intel® Core™ 2 Duo (2.66GHz/1066Mhz, FSB/6MB cache)
10
QUIZ: A CPU chip is rated 2.5 GHz.
What is the duration of one clock cycle? Use appropriate units!
11
QUIZ: A CPU chip is rated 2.5 GHz. What is the duration of one clock cycle? Use appropriate units!
Solution: 2.5∙109 Hz = frequency = f == 1/(duration of one CLK cycle) = 1/T
T = 1/f = 1/2.5∙10-9 s = 0.4 ns
12
Quick work for next time:A CPU chip is rated 3.44 GHz.
What is the duration of one clock cycle? Use appropriate units!
13
5.2 Stored-Program Concept
14
The von Neumann (a.k.a. Princeton) architecture is based on two fundamental ideas:1. Instructions and data are the same, so they are stored in the
same circuit (memory)2. Information processing is different from information storage,
so they are performed in different circuits (CPU, memory)
Are there any other architectures?
15
The Harvard architecture is based on two fundamental ideas:1. Instructions and data are not the same, so they are stored in
separate circuits (memories)2. Information processing is different from information storage,
so they are performed in different circuits (CPU, memory)
Image source: http://www.mikroe.com/chapters/view/74/pic-basic-book-chapter-1-world-of-microcontrollers/
Not in text
QUIZ: Which of the diagrams depicts a von Neumann architecture?
16
QUIZ: Which of the diagrams depicts a von Neumann architecture?
17
Main Memory (MM)
18
Remember that in the vN architecture there is only one MM, which stores both data and programs!
19
ComponentsCircuitsGates
Transistors
Which hardware layer does memory belong to?
Image sources: http://www.mypcmobile.com/memory.phphttp://www.imagener.com/top-six-photo-enlargement-tips
Main Memory “Blades”
Main Memory =A collection of cells,each with a uniquephysical addressBoth addresses andcontents are in Binary (or hex)
Cells can be bits, nibbles, bytes, words
QUIZThe address and contents of a memory cell are the following, in hex:
89AB FC
Translate them into decimal and binary.
Addressability = the # of bits in each cell
What is the addressabilityof the memory pictured?
Today, most computers’ memories are byte-addressable
QUIZ1. How many bytes of memory
are in the MM pictured?
2. Express the result using the
appropriate multiplier.
QUIZ
1. How many bits does the
address have in this memory?
2. What is the addressability?
0x789ABCDE 0x789A
EOHF1
QUIZ: Connect each hardware item to the layer it belongs to:
Components
Circuits
Gates
Transistors
• mouse• XOR• Intel 8008• NTE 2996 MOSFET• MUX• motherboard• full adder• SR latch
QUIZ: A CPU’s frequency is 12 KHz.
What is the duration of one clock cycle? Use appropriate units!
26
QUIZ: A CPU’s clock cycle is 55 ns.
What is the frequency?Use appropriate units!
27
28
ALU
ALU
• basic arithmetic operations such as add, subtract, increment, decrement, change sign, multiply, integer division• logical operations such as AND, OR, XOR, NOT
29
The ALU also has a few very fast storage units called registers• The information in a register can be processed
quickly (in one CLK cycle), w/o waiting for a lengthy (~10 ns) memory transfer.
30Figure source: http://www10.edacafe.com/book/
31
The von Neumann architecture
Input/Output devices
Input/Output devices
Input deviceA device through which data and programs fromthe outside world are entered into the computer;
Can you name three?
Output deviceA device through which results stored in thecomputer memory are made available to theoutside world
Can you name three?
32
33
The von Neumann architecture
Control Unit
Control UnitIt is the organizing force in the computerImplements the fetch-execute cycleIncludes two important registers:• Instruction register (IR) →Contains the instruction that is being executed• Program counter (PC) → Contains the address of the next instruction to be executed
ALU + Control Unit = CPU
34
Flow of Information Bus = A set of wires that connect all major units in a computer
35
Data flow through a von Neumann architecture
Flow of Information Bus = A set of wires that connect all major units in a computer
36
Data flow through a von Neumann architectureHow can we tell that it’s a vN architecture?
Flow of Information Bus = A set of wires that connect all major units in a computer
37
Is there another kind of memory?
FSB and BSB
38
Cache
Back-side Bus (BSB)
Front-side Bus (FSB)
Note: Although the CPU can access two memory circuits, this is still considered a vN architecture, b/c the cache is much smaller than the MM.
39
Find the info about bus and cache in the ad!
The Fetch-Execute Cycle
Fetch the next instructionDecode the instructionGet data (if needed)Execute the instruction
Why is it called a cycle?40
Remember: In a vN machine, both instructions and data are stored in the same memory!
The Fetch-Execute Cycle
41
Read carefully pp.133-135 of text!
RAM and ROM
The memory used for the main memory (MM) of a computer is of 2 types: RAM and ROM
• Both can be accessed directly , i.e. in constant time, by providing a memory address.• Both can be read.
• However, when it comes to writing …
RAM and ROM
Random Access Memory (RAM):• can be changed (written)• is volatile
Read Only Memory (ROM):• cannot be changed (written) • is not volatile
Take this with a grain of salt: Today’s ROMs can be written, just not as fast as RAMs.
Secondary Storage
The size of the MM (RAM+ROM) in today’s computers ranges between …
44
The amounts of data we need to store are much larger!
• Example: Youtube video clips ≤ 20 GB
Secondary storage is a.k.a. mass/volume/bulk storage.
Secondary Storage technologies
• Magnetic: drum, tape, floppy disk, hard disk (internal or external)
• Semiconductors: flash drive , SD card• Optical: CD-R (WORM=Write Once Read Many
times), DVD-R, DVD-R DL, Blue-Ray• Emerging: spintronics (MRAM), Ferroelectric RAM, DNA
storage, etc.
45
Other benefits of Secondary Storage
Portability …Reliability (backups) …Modularity (add as you go) …
46
Magnetic TapeThe first truly mass storage device was the magnetic tape drive
Tape drives have amajor problem; canyou spot it?
Figure 5.4 A magnetic tape
The first high-volume auxiliary storage device was the magnetic tape drive
Image sources:http://museum.ipsj.or.jp/en/computer/device/magnetic_tape/0003.htmlhttp://www.computerhistory.org/revolution/memory-storage/8/258/1025http://wodumedia.com/large-hadron-collider-ready-to-restart/
1964
1993
2013
Magnetic Disk - HDD
• Amount of information (bits, bytes) is the same on all tracks• Disc rotates at the same angular velocity no matter which track is being read → same transfer rate on all tracks!• Tracks near center are more densely packed with information
Platter
QUIZ: HDD
A HDD has 512 Bytes/sector, 256 sectors/track, and 10 platters.The diameter of each platter is 100 mm, and each track is 1 mm wide.
Calculate the total capacity of the drive in MB.
Use binary Mega!
Solution
125 MB
A HDD has 512 Bytes/sector, 256 sectors/track, and 10 platters.The diameter of each platter is 100 mm, and each track is 1 mm wide.
Calculate the total capacity of the drive in MB.
Seek timeTime it takes for read/write head to be over right trackLatencyTime it takes for sector to be in position under R/W head
Access time = Seek time + latency
Transfer rate (e.g. 100 MB/s)
52
QUIZ: HDD
The arm of a HDD moves at an average velocity of 40 m/s. The platter diameter is 100 mm.
What are the minimum and maximum possible seek times?Calculate the average seek time.
53
The arm of a HDD moves at an average velocity of 40 m/s. The platter diameter is 100 mm.
What are the minimum and maximum possible seek times?Calculate the average seek time.
54
1.25 ms
Solution
QUIZ: HDD
The spindle of a HDD rotates at 7200 RPM.What are the minimum and maximum possible latencies?Calculate the average latency.
55
The spindle of a HDD rotates at 7200 RPM.What are the minimum and maximum possible latencies?
Calculate the average latency. 4.16 ms
56
Solution
QUIZ: HDD
If the previous 2 quizzes refer to the same HDD, what is its average access time?
57
If the previous 2 quizzes refer to the same HDD, what is its average access time?
4.16 ms + 1.25 ms = 5.41 ms
58
Solution
QUIZ: HDD
A file is on the HDD, on 15 consecutive blocks of the same track. Each block stores 4 KB.
Average seek time = 10 ms, average latency = 5 ms.Once the first block is under the head, data is transferred at a rate
of 50 MB/s.Calculate the total time needed to transfer the file.
59
60
Solution
A file is on the HDD, on 15 consecutive blocks of the same track. Each block stores 4 KB.
Average seek time = 10 ms, average latency = 5 ms.
Once the first block is under the head, data is transferred at a rate of 50 MB/s.
Total time needed to transfer the file: 16.23 ms
61Source: http://nextshark.com/ibm-5mb-hard-drive/
Quick Work for next time
• Read text pages 131-139• Solve exercises 24 - 29
62EoHW2
QUIZ: Computer Components
Explain in your own words the functions of the following components:
• ALU• Control Unit• Input device• Main Memory• Cache
63
What do the acronyms FSB and BSB stand for?
Compare and contrast them.
64
What do the acronyms FSB and BSB stand for?
Compare and contrast them.
65
Magnetic Disks - RemovableFloppy disks (Why "floppy"?)Year when they first became commercially available:1969 (8-inch)1976 (5¼-inch)1982 (3½-inch)
66
1.44 MB87 KB-1.2 MB
80-500 KB
Magnetic Disks - Removable
Zip disks• Iomega, 1994• 100 MB, 250 MB, 750 MB
67
68
Flash memory• IBM 1998• Nonvolatile!• Can be erased and rewritten
– But there’s a rub: Flash Endurance• No moving parts!
Thumb drives
Solid State Drives (SSD)
To do in notebook for next time:Calculate the cost per gigabyte for HDD and SDD
69Source: www.newegg.com October 2013
Optical DisksCD (600 MB)A compact disk that uses a laser to read information stored optically on a plastic disk; data is evenly distributed around track
CD-ROM read-only memoryCD-DA digital audioCD-WORM or CD-R user can write once, read many timesCD-RAM or CD-RW user can both write and read many times
DVD (4.7 GB)Digital Versatile Disk, used for storing audio and videoDL = dual layer → 8.5 GB
Blu-Ray (25 GB)DL = dual layer → 50 GB
70
CD (600 MB)
DVD (4.7 GB)
Blu-Ray (25 GB)
Hints: The average MP3 file is 4 MB.Use binary multipliers!
71
Your turn! How many songs can be stored on each type of optical disk?
CD (600 MB)
DVD (4.7 GB)
Blu-Ray (25 GB)
Hints: The average MP3 file is 4 MB.Use binary multipliers!
72
How many songs can be stored on each type of optical disk?
Solution
Interesting difference between magnetic and optical drives/disks:
– On a magnetic disk, each track has the same amount on data, there is no accounting for length (center or periphery). The limiting track is the innermost one!
– On an optical disk, data is uniformly packed per sector length → Short tracks (near the center) have less datathan long tracks (near the periphery).
– In a HDD, the speed of rotation is constant, but in an optical drive rotation is faster on the inner tracks, resulting in Constant Linear Velocity (CLV)
73
Read and take notes: Touch Screens
74
75
5.3 Embedded Systems
Computers that are dedicated to perform a narrow range of functions as part of a larger system.
“Every computer that doesn’t have a normal display and keyboard”
How many embedded systems do you think are in this room?
76
Embedded systems are everywhere today!• communications• automotive• military• medical• consumer/home• machine control (CNC)• robotics• entertainment
Source: www/ethio-civility.com
Source: http://www.gosphero.com/
77
5.4 Parallel Architectures
Remember that a vN architecture has one main memory and one CPU.There are two ways to evolve into non-vN architectures:1. Multiple main memories (Harward
arch.)2. Multiple CPUs, a.k.a. parallel arch.
78
Parallel Arch.: Synchronous processing
One approach to parallelism is to have multiple processors apply the same program to multiple data sets:
Figure 5.8 Processors in a synchronous computing environment
79
Synchronous processing example:
Image source: http://tatourian.com/2013/09/03/nvidia-gpu-architecture-cuda-programming-environment/
Not in text
GPU = Graphics Processing Unit • It was invented by NVIDIA in 1999
80
Parallel Arch.: Pipelining
Processors are arranged in tandem, with each processor contributing one part to an overall computation
Figure 5.9 Processors in a pipeline
81
Washer-dryer analogyNot in text
82
Train analogyNot in text
Find the # of steps in terms of n and k
83
n k
# of steps is n+k-1
Hint: Concentrate on the last car!
QUIZ Pipelining
A computer pipeline has 3 stages, as shown above.Each stage takes 8 ms to execute, and each instruction must go sequentially through all 3 stages.A program has 5 instructions. Calculate how long it takes to run it:• without pipelining• with pipelining
84
85
Parallel Arch.: Parallel Processor
It’s made of a number of vN processors which communicate through shared memory• Unlike synchronous processing, there is no central control.• Each CPU participates in a distributed control protocol!
Figure 5.10 Shared memory configuration of processors
Quick-work for next time:• Read pp.130-137 of our text and take notes• Answer end-of-chapter questions 30-38, 42-47
86
Homework:39, 40, 41, 48, 49, 54, 56, 63, 64, 66Due Tuesday, Oct.23 at beginning of class
