Computer Architecture

R Referral /Deferral Portfolio - Section 1

Do all of the following exercises

1 . Perform binary additions of the following binary numbers (show carry digits in the third row

0 1 0 1 1 1 1 0 0 1 0 1 0 1 1 0

0 0 0 1 0 1 1 1 0 0 1 1 0 1 0 0 1 1 1 1

1 1 1

10 1 1 1 0 1 0 1 1 0 0 0 1 0 1 0 0 0 1 0 1 1 1 1

1 0 0 1 1 1 0 1

0 1 1 1 0 1 0 1 1 0 0 0 1 0 1 0

1 1 1 1 1 1 1 1

1 1 1 0 1 0 0 1 0 1 0 0 1 0 0 1 1 1 2 . The following IP (version 4 ) network address is made up of four denary numbers separated by dots . When it is used , it has to be in binary . Convert the four elements separately to 8-bit binary numbers (there may be some leading zeros

193 .61 .107 .23

193

A A /2 (round down ) A mod 2

193 96 1 - LSB

96 48 0

48 24 0

24 12 0

12 6 0

6 3 0

3 1 1

1 0 1 - MSB

193 1100 0001 61

A A /2 (round down ) A mod 2

61 30 1 - LSB

30 15 0

15 7 1

7 3 1

3 1 1

1 0 1 - MSB

61 0011 1101 107

A A /2 (round down ) A mod 2

107 53 1 - LSB

53 26 1

26 13 0

13 6 1

6 3 0

3 1 1

1 0 1 - MSB

107 0110 1011 193

A A /2 (round down ) A mod 2

23 11 1 - LSB

11 5 1

5 2 1

2 1 0

1 0 1 - MSB

193 0001 0111 193 .61 .107 .23 1100 0001 .0011 1101 .0110 1011 .0001 0111

The following IP (version 4 ) network address is shown in binary but when it is recorded it must be shown as four base ten numbers separated by dots . Convert it into this form

p Bit Position Bit 2Bit Position 7 1 - MSB 128 128

6 1 64 64

5 0 32 0

4 0 16 0

3 0 8 0

2 0 4 0

1 0 2 0

0 0 - LSB 1 0

1100 0000 192 br

Bit Position Bit 2Bit Position 7 1 - MSB 128 128

6 0 64 0

5 1 32 32

4 0 16 0

3 1 8 8

2 0 4 0

1 0 2 0

0 0 - LSB 1 0

1010 1000 168 br

Bit Position Bit 2Bit Position 7 0 - MSB 128 0

6 1 64 64

5 1 32 32

4 0 16 0

3 0 8 0

2 0 4 0

1 1 2 2

0 1 - LSB 1 1

0110 0011 99 br

Bit Position Bit 2Bit Position 7 0 - MSB 128 0

6 0 64 0

5 1 32 32

4 0 16 0

3 1 8 8

2 0 4 0

1 1 2 2

0 0 - LSB 1 0

1100 000 42 192 .168 .99 .42

4 . Represent the following numbers as 8-bit 2-s complement binary

-36 -94 23

-36 3610 0010 01002

0010 01002 1101 10111 's complement

1101 10111 's complement 0000 0001 1101 11002 's complement

-36 1101 11002 's complement

-94 9410 0101 11102

0101 11102 1010 00011 's complement

1010 00011 's complement 0000 0001 1010 00102 's complement

--94 1010 00102 's complement

23 2310 0001 01112

Since 23 is a positive number , no further steps are necessary

- 23 0001 01112 's complement

5 . Do the following calculations in 2-s complement binary , use a suitable number of bits

This means that you should convert both numbers to binary , AND THEN do the calculation

38-21 23-54 -43-9

3810 0010 01102 's complement

-2110 0001 01012 1110 10101 's complement 0000 00012 1110 10112 's complement

38 0 0 1 0 0 1 1 0

-21 1 1 1 0 1 0 1 1

1 1

1 1 1 1 0 0 0 1 0 0 0 1

38 - 21 0001 0001

2310 0001 01112 's complement

-5410 0011 01102 1100 10011 's complement 0000 00012 1100 10102 's complement

23 0 0 0 1 0 1 1 1

-54 1 1 0 0 1 0 1 0 1 1 1 1 1 1 1 0 0 0 0 1

23-54 1110 0001

-4310 0010 10112 1101 01001 's complement 0000 00012 1101 01012 's complement

-910 0000 10012 1111 01101 's complement 0000 00012 1111 01112 's complement

-43 1 1 0 1 0 1 0 1

-9 1 1 1 1 0 1 1 1

1 1 1

1 1 1

1 1 1 0 0 1 1 0 0

-43-9

6 . Represent the following binary numbers in octal

0011 10

0011 001 010 011 010 011 12323

00112 123238

10 001 001 101 100 110 11546

102 115468

7 . Convert the following octal numbers to binary

3258 132748

325 011 010 101

3258 1101 01012

13274 001 011 010 111 100

132748 0001 0110 1011 11002

8 . Represent the following binary numbers in hexadecimal

01 001

01 0011 1001 1000 1110 1011 0101 398EA5

012 398EA516

001 11 0010 1110 0001 32E1

0012 32E116

9 . Convert the following hexadecimal numbers to binary

39CD516 48BFA716

39CD5 0011 1001 1100 1101 0101

39CD516

48BFA7 0100 1000 1011 1111 1010 0111

48BFA716 112

10 . Suppose that a simple Real Number representation had the form of a 12-bit 2s complement mantissa (fraction ) and an 8-bit 2s complement exponent (power . What would the following number be in a `normal base-10 representation

1 0 1 1 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 Note : I am assuming IEEE floating point format wherein there is an implied 1 in front of the mantissa . 1 (mantissa ) x 10^exponent

mantissa 02 281610

exponent 2 410

Therefore , the number is 1 .2816 x 104 12 ,81610

11 . The accompanying chart shows the

7-bit ASCII character codes . The row

along the top is the first hex digit of the

code and the row down the side is the

second hex digit such that 4116 is the

code for A (upper case A

Translate the following string of codes

54 : 68 : 65 : 20 : 72 : 61 : 69 : 6E : 20

69 : 6E : 20 : 53 : 50 : 41 : 49 : 4E : 21

And write the result out carefully

Result : The rain in SPAIN

A typical LCD computer monitor screen shows 1024 by 768 pixels and each pixel is represented by three 8-bit integers (for red , green blue intensities . How many kilobytes of data does the image need

Each pixel is represented by 3bytes . Therefore the entire image has (1024 x 768 x 3 ) bytes worth of information . This comes out to be 2 ,359 ,296 bytes bytes in a kilobyte , we get a image

13 . If the image could be downloaded over an ADSL broadband link at 4Mbps , how long would this take

Assuming no overhead losses , the 2 ,304 kb image is a x1024 x 8 ) bits which is 18 ,874 ,368 bits million bits per second gives us 4 .718592 seconds

Referral /Deferral Portfolio - Section 2

Do all of the following exercises . For some you will need to

hand in a CD or floppy disk containing the simulated circuits

1 . Draw the circuit for the following truth tables using simple gates C AB

C A 'B A 'B

2 . Construct the truth tables for the following circuits and identify them

A B C D E

0 0 1 1 1

0 1 1 0 0

1 0 0 1 0

1 1 0 0 0

Circuit is a NOR gate J K L M N

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

Circuit is a NAND Gate

3 . Construct the truth tables for the following two circuits and thus prove they are equivalent . Make sure you include the intermediate nodes in your tables

Using the logic design CAD tool , build a single simulation showing both circuits . Hand in both the drawings (as printouts ) and the simulations on disk

A B C x Out1

A B C y z Out2

0 0 0 0 0

0 0 0 0 0 0

0 0 1 1 0

0 0 1 0 0 0

0 1 0 1 0

0 1 0 0 0 0

0 1 1 1 0

0 1 1 0 0 0

1 0 0 0 0

1 0 0 0 0 0

1 0 1 1 1

1 0 1 0 1 1

1 1 0 1 1

1 1 0 1 0 1

1 1 1 1 1

1 1 1 1 1 1 Note : I used a buffer to create a uniform amount of input to output gate delay for both circuits . Such buffers only affect the timing characteristics of the circuit . They will have no effect on its behaviour

4 . Design and simulate an efficient 2 :4 decoder circuit . This has the following truth table and is used for device selection in a computer system . You should use 3-input gates to simplify your task

A B Sel W X Y Z

0 0 0 0 0 0 0

0 1 0 0 0 0 0

1 0 0 0 0 0 0

1 1 0 0 0 0 0

0 0 1 1 0 0 0

0 1 1 0 1 0 0

1 0 1 0 0 1 0

1 1 1 0 0 0 1 You will need to use the logic design CAD tool with three input switches for A , B and Sel and four output indicator lights for W , X , Y and Z . Hand in both the drawing (as a printout ) and the simulation on disk

Note : I used buffers to create a uniform amount of input to output gate delay . Such buffers only affect the timing characteristics of the circuit . They will have no effect on its behaviour

Referral /Deferral Portfolio - Section 3

Do the following exercise

The following is an exercise in understanding how machine code is broken down into a series of data movements in the processor . You will need to use a simplified Register Transfer Language (RTL ) as a notation for what is going on

An outline structure of the Data Flow of the processor is given together with the initial RTL of a simple instruction . You are asked to make modifications to it in to transform it into other instructions of a slightly more complex kind

The diagram above shows the data flow of a simple processor with its various registers and linkages to the main store . It can perform the following instruction

? MAR send a copy of PC contents to MAR

(increment PC ) PC has 1 added to it so it points to the next instruction

MAR ? store the address of the instruction is sent to the store for a read to occur

store ? MDR the instruction is fetched from store and placed in the MDR

Decode

MDR ? IR instruction is moved to instruction register

(decode ) decoding of opcode occurs

Operand fetch nothing to do here , data (42 ) is part of immediate mode instruction

Execute no data transformations to do

Write

IRADDR ? Acc a simple write to a register

If the instruction was ADD [42] (add to the accumulator the number whose address can be found at location 42 in the store , i .e . indirect mode what changes are necessary ? Fill in a chart of the following kind with your answer . Split the RTL code into phases as in the example

RTL Instruction

Instruction Fetch

PC ( MAR load MAR with PC contents

increment PC increment PC to point to the next instruction

MAR ( store address of the instruction is loaded to store for a read

store ( MDR instruction is loaded from store into the MDR

Instruction Decode

MDR (IR instruction is loaded to the Instruction Register

(decode ) The control circuitry decode the contents of the Instruction register

Operand Fetch

IR[address] ( MAR The lower bits of the IR are sent to the MAR . These lower bits contain the address [42] . Care is taken to sign extend the address with zero bits for its upper bits

MAR ( store The address [42] is sent to the store

store ( MDR The contents of [42] is loaded into the MDR

Acc ( IB The contents of the Accumulator are loaded into the IB register

Execute

IB MDR ( DB The contents of IB and the MDR are added together by the ALU

Write

DB ( Acc The result of the ALU operation is moved into the Accumulator register COMPUTER ARCHITECTURE - 120CT

Referral /Deferral Portfolio - Section 4

Do all of the following exercise

You have been employed to design the computer network facilities for the new headquarters of Easybuild Ltd . The new company 's accommodation comprises one building with 2 floors . There will be 10 offices with 5 desks each with all members of staff having their own PC . In addition there will be one network printer for each office

The company has four main departments : Engineering Design , Marketing Production Management , Accountancy /Personnel

It is envisaged that the Design and the Production Management departments will be located on the same floor

1 . Produce a workable IP addressing scheme for Easybuild Ltd . Justify your decisions including the IP address Class requested from ARIN

Bill of materials

Units Item

50 Workstations

10 Network Printers

10 8 port switches

1 Router Network Plan for Easybuild Ltd . I am assuming that intra-office data communications far outnumber inter-office data communications

We decide to employ a subnetting IP addressing for Easybuild . Subnetting allows us to break down the large Easybuild network into 10 smaller networks , one for each office . From the bill of materials , we see that we are to connect a We are also making 10 subnets , with one subnet per office . With that in mind , the network mask we will use would be 255 .255 .255 .240 and we shall be requesting from ARIN a Class C IPv4 address for Easybuild . Use of 255 .255 .255 .240 gives us 24 possible subnets as we borrowed 4 bits to make the sub network . This also left us with 14 usable hosts per network as it left us 4 bits per host (minus the all-zero and all-one cases Sixteen subnets are enough for our case as we only have 10 offices to consider . Fourteen hosts per subnet is also enough as we only have 6 devices per office . Write a short report for the non-technical management of Easybuild explaining how network printers operate and are linked into the system . Describe their advantages over having a printer to a specific PC in each office

On the surface , network printers do the same thing that home printers do : print onto a sheet of . However , the biggest difference is that home printers connect to a single computer while network printers connect to a network . While a home printer acts as a peripheral to a computer , a network printer acts as a network device

The benefit of connecting a printer to a network lies in the fact that the network printer is used in a different environment than a home printer . In a home setting , there is only one computer that will be using the printer thus no issues arise from connecting the printer directly to the computer . However , that does not apply in an office setting where multiple users may need to use the printer . The home printer would only be able to get printing jobs from the computer it is connected to . This would translate to the expensive option of purchasing a printer for every computer in the office or as an alternative transferring a single printer to each workstation that needs it every time someone needs to print something . One final alternative is to create a dedicated workstation for printing where people could queue up every time they need to print , bringing their s with them . As we can see , the single printer to computer connection is impractical and expensive for an office setting

With this in mind , let us look at the benefits of a network printer . As aforementioned , a network printer is connected to the network . This means that instead of transferring a print job directly from the computer to the printer , a print job is sent by the computer over the network to the printer . More importantly , ANY computer on the network could send print jobs over the network to the printer . This improves productivity as people no longer need to queue up to transfer s to a dedicated printing computer or to physically transfer a printer every time they need to print

Another advantage is administration . Being a network device , the network administrator could allow or disallow certain users from using the printer . The administrator could also monitor the printing use of each person . This helps cut down on cases of employees using the printer for their personal use which will translate to lower costs in terms of ink and supplies

Explain which PC features may need to be different in the Engineering Design department . Why could this be so

Of the four departments , the Engineering Design department is the most likely to have the toughest demands on their computers . Inferring from the name of the department , their computers will most likely need to have bigger screens , a dedicated graphics card , faster processors bigger memory , and larger hard drives . Additionally , the software that the engineers use may dictate their operating system choices which may require the use of multiple operating systems on each machine . Lastly their subnet may need to be separated from the rest of the departments due to a large amount of traffic between computers in the engineering design department

The use of larger screens stem from possible use of CAD software . CAD does not require the use of large screens however productivity when using CAD software is improved with generous screen real estate . A large screen is something that the engineers of Easybuild will be thankful for . Easybuild also has the option of providing their engineers with one large screen or multiple small screens (or even multiple large screens if they so prefer . Additionally , large screens and CAD software will benefit from a dedicated video graphics adapter . While commonly employed by computer gaming aficionados , dedicated video graphics also benefit CAD tools by taking the rendering load off of the main CPU and into the graphics chip . Examples of such CAD and rendering video graphics chips are the Quadro line from Nvidia and the FireGL chips from ATI . Lastly Easybuild should purchase graphics cards that can support multiply monitors if they decide to give their engineers multiple monitors per workstation

Engineering software is also notorious for being computationally intensive whether it is CAD tools or numerical analysis software such as Matlab or Mathematica . As with the case of bigger screens , these kinds of software don 't necessarily require the fastest processor or the largest amount of memory . However , productivity is aided when the workstations use the latest processors and as much memory that is available to them . It will also be future proofing their workstations by buying into the trend of multi-core 64bit processors as software that is optimized for these processors are already starting to penetrate the marketplace

A large hard drive will allow the engineers to store large amounts of data and programs . Additionally , a large hard drive will accommodate multiple operating systems with ease . A large hard drive will also allow the engineers to delete s less often and give them large amounts of space to backup s

A robust network will also facilitate collaboration between the engineers by allowing them to transfer s between workstations with ease . A robust network may also be needed if the engineers may need to communicate with software that is installed on a server such as compilers or object libraries

COMPUTER ARCHITECTURE - 120CT

Referral /Deferral Portfolio - Section 5

This section should consist of evidence of practical activities concerning computer hardware or systems . It can cover any practical or design work or material concerning the application of computer architecture knowledge to a range of computer science activities . Some suggestions are given and though you do NOT need to do all of them , it is possible to mix parts of them or add something entirely different which meets the practical aim of this section

In all cases , a thorough , possibly accompanied by pictures or diagrams , is required

Upgrading my mother 's computer

Sometime late last year , I set out to perform some upgrades on my mother 's 3 year old computer . The truth was I wasn 't supposed to perform an upgrade my help was needed because the computer was not starting up The computer then was running Windows XP on a Duron microprocessor with 128MB of DDR RAM and a 40GB Hard drive . What happened was that the computer would turn on and go through the BIOS but the operating system would not load

It was clear that the processor was still functioning as well as its fan . The computer also beeped only once during start-up which indicated that the memory was not the problem . Additionally , swapping around the stick of memory with another from another computer didn 't fix the problem . This left the hard drive or the software itself as the final culprit . Snooping around on the BIOS , I discovered that the hard disk was not being seen by the BIOS . Additionally , when booting up , the screen kept giving out DISK BOOT ERROR : ABORT RETRY CANCEL ' messages It seemed that the 3 year old hard drive had finally died

I told my mother that purchase of a new hard drive was in . With the purchase of a new hard drive , we also decided to purchase other upgrades for the computer . Together , the items bought were the following

1 x 80GB 7200 RPM Hard drive

1 x DVD - CDRW drive

1 x 256 MB DDR memory

The hard drive was to replace the ailed hard drive . The DVD-CDRW replaced the old 50x CD RW and allowed the viewing of DVD movies on the upgraded computer . The additional 256MB of RAM brought the of the computer to 384MB (only 352MB are available to the processor as the integrated graphics borrows 32MB

Installing the hard drive was easier than I expected . Gone were the days of having to manually set slave and master . The new hard drive had a jumper setting for auto which let the BIOS be in charge of setting which is master and slave . theless , I connected it to the master slot of the IDE ribbon and selected the new hard drive as the master in the BIOS

The new 80GB Seagate hard drive in place

Now that the hard drive was installed , it was time to install Windows XP from the Windows XP CD . During the installation , I partitioned the 80GB hard drive into a 40GB programs partition and a 40GB s partition . I did it so that if I need to reinstall the Operating system , I need not lose all the s during reformatting . With that done , the installation of Windows XP proceeded without errors . I also had to reinstall all the drivers from the peripherals like the printer and modem to the onboard devices like the integrated graphics and sound . All of these drivers were found in the CDs of the peripherals and the CD of the motherboard

At this point , I should also mention the steps taken to protect the insides of the PC from static electricity . I buy an electrostatic wrist ground strap . What I did was to work at the ground floor of the house which had marble flooring and touch the casing chassis of the computer and the floor before working with the computer 's innards . I made all efforts to avoid touching metal leads and contacts and I also avoided rubbing my cotton shirt with the computer . I also worked barefoot which wasn 't much of an issue as I was only working at home . Lastly , before opening the computer I made sure to unplug all connections from the power to the peripherals

When the OS was up and running I then installed the other hardware Installation of the RAM was a cinch , I simply had to line the stick of RAM properly with the slot (it will only go in when oriented in the right direction ) and push it until it clicks . I then rebooted the computer . I was relieved to hear only a single beep instead of the long continuous beep which signalled a memory problem . Immediately , the change was obvious , boot time was reduced from around 3 minutes to around 1 minute . The computer also seems to work faster with simple tasks such as word processing , web surfing and email . The benefits of the RAM upgrade were immediate and noticeable

The installed two sticks of RAM

The RAM installed , I then installed the combo drive . I swapped out the old drive with the new one , connecting it to the power cable and the IDE cable the same way that the old drive was connected . On restart , Windows was able to detect the hardware change and install the drivers . I then installed the CD burning software that came with the combo drive

The new DVD-CDRW drive

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p ?Y ?Computer Organization , Fifth Edition , McGraw-Hill Higher Education , New York

Uffenbeck , J . 2001 , Microcomputers and Microprocessors , Third Edition Pearson Education , Singapore

Patterson , D Hennesy , J . 2004 , Computer Organization and Design : The Hardware /Software Interface , Morgan Kaufman

Peterson , L Davie , B . 2003 , Computer Networks , a Systems Approach Third Edition , Morgan Kaufmann A B C

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