(Assignment) Kfc 2044 Network Fundamental

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    KFC 2044 NETWORK FUNDAMENTALS | MOHAMED ARIFF BIN MOHAMED AMIZ

    INSTITUT KEMAHIRAN MARA LUMUT

    KFC 2044 NETWORK FUNDAMENTAL

    CERTIFICATE OF ENGINEERING COMPUTER

    CLASS:2A

    LECTURER:

    EN. MOHD SHUKRI BIN MUHAMAD HUSIN

    SUBMITED BY:

    MOHAMED ARIFF BIN MOHAMED AMIZ

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    ASSIGNMENT

    | KFC 2044 NETWORK FUNDAMENTAL |

    UNIT 1:

    COMPUTER NETWORKING FUNDAMENTALS

    NO. TITLE

    1 IP ADDRESS CLASSES CALCULATION

    2 PROBLEM 1

    3 PROBLEM 2

    4 PROBLEM 3

    5 PROBLEM 4

    6 PROBLEM 5

    7 PROBLEM 6

    8 SUBNETTING CALCULATION

    9 PROBLEM 1

    10 PROBLEM 2

    11 PROBLEM 3

    12 PROBLEM 4

    13 PROBLEM 5

    14 PROBLEM 6

    15 REFERENCE

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    IP ADDRESS

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    IP ADDRESS CLASSES

    Calculation for IP address classes

    Task 1: For given IP address, Determine Network Information.

    Given:

    Find:

    Step 1: Translate Host IP address and network mask into binary notation.

    210

    29

    28

    27

    26

    25

    24

    23

    22

    21

    20

    1024 512 256 128 64 32 16 8 4 2 1

    Convert the host IP address and network mask to library:

    172 25 114 250

    IP Address 10101100 00011001 0111001 11111010

    27 26 25 24 23 22 21 20

    1 0 1 0 1 1 0 0

    128 32 8 4

    128 + 32 + 8 + 4 = 172

    27 2

    6 2

    5 2

    4 2

    3 2

    2 2

    1 2

    0

    0 0 0 1 1 0 0 1

    16 8 1

    16 + 8 + 1 = 25

    27 2

    6 2

    5 2

    4 2

    3 2

    2 2

    1 2

    0

    0 1 1 1 0 0 1 0

    64 32 16 2

    64 + 32 + 16 + 2 = 114

    Host IP Address 172.25.114.250

    Network Mask 255.255.0.0 (/16)

    Network Address 172.25.0.0

    Network Broadcast 172.25.255.255

    Total Number of Host Bits 65534

    Number of Host 16

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    27 2

    6 2

    5 2

    4 2

    3 2

    2 2

    1 2

    0

    1 1 1 1 1 0 1 0

    128 64 32 16 8 2

    128 + 64 + 32 + 16 + 8 + 2 = 250

    Step 2: Determine the Network Address.

    172 25 114 250

    IP Address 10101100 00011001 01110010 11111010

    Subnet Mask 11111111 11111111 00000000 00000000

    Network Address 10101100 00011001 00000000 00000000

    172 25 0 0

    Step 3: Determine the broadcast address for the network address

    172 25 0 0

    Network Address 10101100 00011001 00000000 00000000

    Mask 11111111 11111111 00000000 00000000

    Broadcast 10101100 00011001 11111111 11111111

    172 25 255 255

    By counting the number of host bits, we can determine the total number of usable hosts for this

    network.

    Host bits: 16

    Total number of hosts:

    216

    = (2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 65,536

    65,5362 = 65,534 .

    Host IP Address 172.25.114.250

    Network Mask 255.255.0.0 (/16)

    Network Address 172.25.0.0

    Network Broadcast Address 172.25.255.255

    Total Number of Host Bits 65,536

    Number of Hosts 216

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    Task 2: Challenge.

    Problem 1

    Answer and How to solve:

    Network Address 172.30.0.0

    Host IP Address 172.30.1.33

    Network Mask 255.255.0.0Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 172.30.1.33

    Network Mask 255.255.0.0

    Network Address 172.30.0.0

    Network Broadcast Address 172.30.255.255

    Total Number of Host Bits 65,534

    Number of Hosts 216

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 1 1 1 0

    0 0 0 0 0 0 0 0

    0 0 0 0 0 0 0 0

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+8+4+2 = 30

    0 = 0

    0 = 0

    172.30.0.0

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    Network Broadcast Address 172.30.255.255

    Host bits: 16

    Total number of host:

    216

    (2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 65,536

    65,5362 = 65,534

    Problem 2

    Answer and How to solve:

    C 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 1 1 1 0

    1 1 1 1 1 1 1 1

    1 1 1 1 1 1 1 1

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+8+4+2 = 30

    128+64+32+16+8+4+2+1 =

    255

    128+64+32+16+8+4+2+1 =

    255

    172.30.255.255

    Host IP Address 172.30.1.33

    Network Mask 255.255.255.0

    Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 172.30.1.33

    Network Mask 255.255.255.0

    Network Address 172.30.1.0

    Network Broadcast Address 172.30.1.255

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    Network Address 172.30.1.0

    Network Broadcast Address 172.30.1.255

    Host bits: 8

    Total number of host:

    28

    (2x2x2x2x2x2x2x2) = 256

    2562 = 254

    Total Number of Host Bits 254

    Number of Hosts 28

    27

    26

    25

    24

    23

    22

    21

    20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 1 1 1 0

    0 0 0 0 0 0 0 1

    0 0 0 0 0 0 0 0

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+8+4+2 = 30

    1 = 1

    0 = 0

    172.30.1.0

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 1 1 1 0

    0 0 0 0 0 0 0 1

    1 1 1 1 1 1 1 1

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+8+4+2 = 30

    1 = 1

    128+64+32+16+8+4+2+1 =

    255

    172.30.1.255

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    Problem 3

    Answer and How to solve:

    Network Address 192.168.10.0

    Host IP Address 192.168.10.234

    Network Mask 255.255.255.0

    Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 192.168.10.234

    Network Mask 255.255.255.0

    Network Address 192.168.10.0

    Network Broadcast Address 192.168.10.255

    Total Number of Host Bits 254

    Number of Hosts 28

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 1 0 0 0 0 0 0

    1 0 1 0 1 0 0 0

    0 0 0 0 1 0 1 0

    0 0 0 0 0 0 0 0

    BINARY NUMBER

    DECIMAL

    128+64 = 192

    128+32+8 = 168

    8+2 = 10

    0 = 0

    192.168.10.0

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    Network Broadcast Address 192.168.10.255

    Host bits: 8

    Total number of host:

    28

    (2x2x2x2x2x2x2x2) = 256

    2562 = 254

    Problem 4

    Answer and How to solve:

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 1 0 0 0 0 0 0

    1 0 1 0 1 0 0 0

    0 0 0 0 1 0 1 0

    1 1 1 1 1 1 1 1

    BINARY NUMBER

    DECIMAL

    128+64 = 192

    128+32+8 = 168

    8+2 = 10

    128+64+32+16+8+4+2+1 =

    255

    192.168.10.255

    Host IP Address 172.17.99.71

    Network Mask 255.255.0.0

    Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 172.17.99.71

    Network Mask 255.255.0.0

    Network Address 172.17.0.0

    Network Broadcast Address 172.17.255.255

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    Network Address 172.17.0.0

    Network Broadcast Address 172.17.255.255

    Host bits: 16

    Total number of host:

    216

    (2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 65,536

    65,5362 = 65,534

    Total Number of Host Bits 65534

    Number of Hosts 216

    27

    26

    25

    24

    23

    22

    21

    20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 0 0 0 1

    0 0 0 0 0 0 0 0

    0 0 0 0 0 0 0 0

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+1 = 17

    0 = 0

    0 = 0

    172.17.0.0

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 0 1 0 1 1 0 0

    0 0 0 1 0 0 0 1

    1 1 1 1 1 1 1 1

    1 1 1 1 1 1 1 1

    BINARY NUMBER

    DECIMAL

    128+32+8+4 = 172

    16+1 = 17

    128+64+32+16+8+4+2+1 =

    255

    128+64+32+16+8+4+2+1 =

    255

    172.17.255.255

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    Problem 5

    Answer and How to solve:

    Network Address 192.168.0.0

    Host IP Address 192.168.3.219

    Network Mask 255.255.0.0

    Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 192.168.3.219

    Network Mask 255.255.0.0

    Network Address 192.168.0.0

    Network Broadcast Address 192.168.255.255

    Total Number of Host Bits 65534

    Number of Hosts 216

    27

    26

    25

    24

    23

    22

    21

    20

    128 64 32 16 8 4 2 1

    1 1 0 0 0 0 0 0

    1 0 1 0 1 0 0 0

    0 0 0 0 0 0 0 0

    0 0 0 0 0 0 0 0

    BINARY NUMBER

    DECIMAL

    128+64 = 192

    128+32+8 = 168

    0 = 0

    0 = 0

    192.168.0.0

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    Network Broadcast Address 192.168.255.255

    Host bits: 16

    Total number of host:

    216

    (2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 65,536

    65,5362 = 65,534

    Problem 6

    Answer and How to solve:

    27 26 25 24 23 22 21 20

    128 64 32 16 8 4 2 1

    1 1 0 0 0 0 0 0

    1 0 1 0 1 0 0 0

    1 1 1 1 1 1 1 1

    1 1 1 1 1 1 1 1

    BINARY NUMBER

    DECIMAL

    128+64 = 192

    128+32+8 = 168

    128+64+32+16+8+4+2+1 =

    255

    128+64+32+16+8+4+2+1 =

    255

    192.168.255.255

    Host IP Address 192.168.3.219

    Network Mask 255.255.255.224

    Network Address

    Network Broadcast Address

    Total Number of Host Bits

    Number of Hosts

    Host IP Address 192.168.3.219

    Network Mask 255.255.255.224

    Network Address 192.168.3.192

    Network Broadcast Address 192.168.3.223

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    SUBNETTING

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    Problem 2.

    Number of Subnet Bits

    IP Address : 172.30.1.33

    Subnet Mask : 255.255.255.252 = 11111111,11111111,11111111,11111100

    Network Host : 214

    (2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 16384

    16384-2 = 16382

    Number of Host Bits Per Subnet

    214

    =16384

    Number of Usable Host Per Subnet

    16384

    2 = 16382

    IP Address 172.30.1.33 10101100 00011110 00000001 00100001

    Subnet Mask 255.255.255.252 11111111 11111111 11111111 11111100

    Subnet Add. 172.30.1.32 10101100 00011110 00000001 00100000

    IP Address 172.30.1.33 10101100 00011110 00000001 00100001

    Subnet Mask 255.255.255.252 11111111 11111111 11111111 11111100

    Subnet Add. 172.30.1.32 10101100 00011110 00000001 00100000

    First Host 172.30.1.33 10101100 00011110 00000001 00100001Last Host 172.30.1.62 10101100 00011110 00000001 00111110

    Broadcast 172.30.1.63 10101100 00011110 00000001 00111111

    Host IP Address 172.30.1.33

    Subnet Mask 255.255.255.252

    Number of Subnet Bits 14 Bits

    Number of Subnets 2

    14

    = 16384Number of Host Bits per Subnet 14 Bits

    Number of Usable Hosts per Subnet 214

    2 = 163842 = 16382

    Subnet Address for this IP Address 172.30.1.32

    IP Address of First Host on this Subnet 172.30.1.33

    IP Address of Last Host on this Subnet 172.30.1.62

    Broadcast Address for this Subnet 172.30.1.63

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    Problem 3.

    Number of Subnet Bits

    IP Address : 192.192.10.234

    Subnet Mask : 255.255.255.0 = 11111111,11111111,11111111,00000000

    Network Host : 28

    (2x2x2x2x2x2x2x2) = 256

    2562 = 254

    Number of Host Bits Per Subnet

    28= 256

    Number of Usable Host Per Subnet

    256

    2 = 254

    IP Address 192.192.10.234 11000000 11000000 00001010 11101010

    Subnet Mask 255.255.255.0 11111111 11111111 11111111 00000000

    Subnet Add. 192.192.10.0 11000000 11000000 00001010 00000000

    IP Address 192.192.10.234 11000000 11000000 00001010 11101010

    Subnet Mask 255.255.255.0 11111111 11111111 11111111 00000000

    Subnet Add. 192.192.10.0 11000000 11000000 00001010 00000000

    First Host 192.192.10.1 11000000 11000000 00001010 00000001Last Host 192.192.10.254 11000000 11000000 00001010 11111110

    Broadcast 192.192.10.255 11000000 11000000 00001010 11111111

    Host IP Address 192.192.10.234

    Subnet Mask 255.255.255.0

    Number of Subnet Bits 8 Bits

    Number of Subnets 2

    8

    = 256Number of Host Bits per Subnet 8 Bits

    Number of Usable Hosts per Subnet 282 = 2562 = 254

    Subnet Address for this IP Address 192.192.10.0

    IP Address of First Host on this Subnet 192.192.10.1

    IP Address of Last Host on this Subnet 192.192.10.254

    Broadcast Address for this Subnet 192.192.10.255

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    Problem 5.

    Number of Subnet Bits

    IP Address : 192.168.3.219

    Subnet Mask : 255.255.255.0 = 11111111,11111111,11111111,00000000

    Network Host : 28

    (2x2x2x2x2x2x2x2) = 256

    2562 = 254

    Number of Host Bits Per Subnet

    28= 256

    Number of Usable Host Per Subnet

    2562 = 254

    IP Address 192.168.3.219 11000000 10101000 00000011 11011011

    Subnet Mask 255.255.255.0 11111111 11111111 11111111 00000000

    Subnet Add. 192.168.3.0 11000000 10101000 00000011 00000000

    IP Address 192.168.3.219 11000000 10101000 00000011 11011011

    Subnet Mask 255.255.255.0 11111111 11111111 11111111 00000000

    Subnet Add. 192.168.3.0 11000000 10101000 00000011 00000000

    First Host 192.168.3.1 11000000 10101000 00000011 00000001

    Last Host 192.168.3.254 11000000 10101000 00000011 11111110

    Broadcast 192.168.3.255 11000000 10101000 00000011 11111111

    Host IP Address 192.168.3.219

    Subnet Mask 255.255.255.0

    Number of Subnet Bits 8 Bits

    Number of Subnets 28

    = 256

    Number of Host Bits per Subnet 8 Bits

    Number of Usable Hosts per Subnet 282 = 2562 = 254 host

    per subnet

    Subnet Address for this IP Address 192.168.3.0

    IP Address of First Host on this Subnet 192.168.3.1

    IP Address of Last Host on this Subnet 192.168.3.254

    Broadcast Address for this Subnet 192.168.3.255

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    Problem 6.

    Number of Subnet Bits

    IP Address : 192.168.3.219

    Subnet Mask : 255.255.255.252 = 11111111,11111111,11111111,11111100

    Network Host : 214

    (2x2x2x2x2x2x2x2x2x2x2x2x2x2) = 16384

    163842 = 16382

    Number of Host Bits Per Subnet

    22

    = 4

    Number of Usable Host Per Subnet

    22

    2 = 4

    2 = 2 host per subnet

    IP Address 192.168.3.219 11000000 10101000 00000011 11011011

    Subnet Mask 255.255.255.252 11111111 11111111 11111111 11111100

    Subnet Add. 192.168.3.0 11000000 10101000 00000011 11011000

    IP Address 192.168.3.219 11000000 10101000 00000011 11011011

    Subnet Mask 255.255.255.252 11111111 11111111 11111111 11111100

    Subnet Add. 192.168.3.0 11000000 10101000 00000011 11011000

    First Host 192.168.3.1 11000000 10101000 00000011 11011001

    Last Host 192.168.3.254 11000000 10101000 00000011 11011110

    Broadcast 192.168.3.255 11000000 10101000 00000011 11011111

    Host IP Address 192.168.3.219

    Subnet Mask 255.255.255.252

    Number of Subnet Bits 14 Bits

    Number of Subnets 214 = 16384

    Number of Host Bits per Subnet 2 Bits

    Number of Usable Hosts per Subnet 222 = 42 = 2 host per

    subnet

    Subnet Address for this IP Address 192.168.3.216

    IP Address of First Host on this Subnet 192.168.3.217

    IP Address of Last Host on this Subnet 192.168.3.222

    Broadcast Address for this Subnet 192.168.3.223