assignment kfc 2044 network fundamental 2

1

INSTITUT KEMAHIRAN MARA LUMUTCERTIFICATION IN COMPUTER ENGINEERING TECHNOLOGY

(NETWORKING)

NAME: MUHAMMAD FAKHRUL RADZI BIN JAFFER@JAFFAR

IC NO : 950418-08-5371

MATRIX NO: 1401119

SESION : JULY-DISEMBER 2014

SEMESTER: 2

LECTURER: ENCIK MOHD SHUKRI BIN MUHAMMAD HUSIN

KFC 2044 NETWORK FUNDAMENTAL

MUHAMMAD FAKHRUL RADZI BIN JAFFER@JAFFAR

2

NO INDEX PAGE

1 IP ADDRESS CLASSES CALCULATION 3-5

2 PROBLEM 1 6-7

3 PROBLEM 2 8-9

4 PROBLEM 3 10-11

5 PROBELM 4 12-13

6 PROBLEM 5 14-15

7 PROBLEM 6 16-17

8 SUBNETTING CALCUTION 18

9 PROBLEM 1 18-19

10 PROBLEM 2 20-21

11 PROBLEM 3 22-23

12 PROBLEM 4 24-25

13 PROBLEM 5 26-27

14 PROBLEM 6 28-29



3

IP ADDRESS

STEP OF IP ADDRESS CLASSES CALCULATION

TASK 1 : FOR GIVEN IP ADDRESS DETERMINE NETWORK INFORMATION

GIVEN: IP ADDRESS : 172.25.114.250NETWORK MASK : 255.255.0.0(/16)

FIND:NETWORK ADDRESS :172.25.0.0NETWORK BROADCAST :172.25.255.255TOTAL NUMBER OF HOST BITS :66534NUMBER OF HOST : 16

STEP 1

TRANSLATE HOST IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

FORMULA

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1OCTET 1 1 0 1 0 1 1 1 0OCTET 2 0 0 0 1 1 0 0 1OCTET 3 0 1 1 1 0 0 1 0OCTET 4 1 1 1 1 1 0 1 0

BINARY NUMBER

LET



4

OCTET 1 128 + 32 + 8 + 4 = 172OCTET 2 16 +8 + 1 25OCTET 3 64 + 32 + 16 + 2 114OCTET 4 128 + 64 + 32 + 16 + 8+ 2 250

NETWORK MASK

FORMULA27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1OCTET 1 1 1 1 1 1 1 1 1OCTET 2 1 1 1 1 1 1 1 1OCTET 3 0 0 0 0 0 0 0 0OCTET 4 0 0 0 0 0 0 0 0

BINARY NUMBER

LET OCTET 1 128 + 64 + 32 + 16 + 8 + 2 + 1 = 255OCTET 2 128 + 64 + 32 + 16 + 8 + 2 + 1 = 255OCTET 3 0 = 0OCTET 4 0 = 0

STEP 2



5

DETERMINE THE NETWORK ADDRESS

IP ADDRESS172 25 114 250

10101100 00011001 0111001 11111010

NETWORK MASS

225 225 0 0

11111111 11111111 00000000 00000000

NETWORK ADDRESS

172 25 0 0

10101100 00011001 00000000 00000000

STEP 3DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

NETWORK ADDRESS

172 25 0 0

10101100 00011001 00000000 00000000

MASK

255 255 0 0

11111111 11111111 00000000 00000000

BROADCAST

172 25 255 255

10101100 00011001 11111111 11111111

NUMBER OF HOST BITS (n) : 16TOTAL NUMBER OF HOST : 2n - 2 USE THE FORMULA TO CALCULATE



6

= 216 - 2= 65 536 - 2= 65 534

TASK 2 : CHALLENGE

PROBLEM 1

HOST IP ADDRESS 172.30.1.33NETWORK MASK 255.255.0.0NETWORK ADDRESS 172.30.0.0NETWORK BROADCAST ADDRESS 172.30.255.255TOTAL NUMBER OF HOST BITS 65535NUMBER OF HOSTS 16 NETWORK ADDRESS

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1 CALCULATE NETWORK ADDRESS

1 0 1 0 1 1 0 0 128 + 32 + 8 + 4 172

0 0 0 1 1 1 1 0 16 + 8 + 4 + 2 30

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

AND THE ANSWER NETWORK ADDRESS IS 172.30.0.0



7

NETWORK BROADCAST ADDRESS

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1 CALCULATENETWORK

BROADCAST ADDRESS

1 0 1 0 1 1 0 0 128 + 32 + 8 + 4 = 172

0 0 0 1 1 1 1 0 16 + 8 + 4 + 2 + 1 30

1 1 1 1 1 1 1 1128 + 64 + 32 + 16 +

8 + 4 + 2 + 1 255

1 1 1 1 1 1 1 1128 + 64 + 32 + 16 +

8 + 4 + 2 + 1 255

AND THE ANSWER NETWORK BROADCAST ADDRESS IS 172.30.255

USE THE FORMULA N=16

= 2n-2= 216 = 65536 - 2 = 65 534 NUMBER OF HOST BITS = 16

TASK 2 PROBLEM 2



8

HOST IP ADDRESS 172.30.1.33NETWORK MASK 255.255.255.0NETWORK ADDRESS 172.30.1.0NETWORK BROADCAST 172.30.1.255TOTAL NUMBER OF HOST BITS 254NUMBER OF HOSTS 8

NETWORK ADDRESS

27 26 25 24 23 22 21 20


1 0 1 0 1 1 0 0 128 + 32 + 8 + 4

172

0 0 0 1 1 1 1 0 16 + 8 + 4 + 2 30

0 0 0 0 0 0 0 1 1 1

0 0 0 0 0 0 0 0 0 0

THE ANSWER NETWORK ADDRESS IS 172.30.1.0

NETWORK BROADCAST



9

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1 CALCULATE NETWORK BROADCAST

1 0 1 0 1 1 0 0 128 + 32 + 8 + 4 172

0 0 0 1 1 1 1 0 16 + 8 + 4 + 2 30

0 0 0 0 0 0 0 0 1 1

1 1 1 1 1 1 1 1128 + 64 + 32 + 16 + 8 + 4 + 2 +

1255

AND THE ANSWER NETWORK BROADCAST ADDRESS IS 172.30.1.255

USE THE FORMULA N = 8

= 2n - 2= 28 = 256 - 2 = 254

NUMBER OF HOST = 8

TASK 2KFC 2044 NETWORK FUNDAMENTAL


10

PROBLEM 3

HOST IP ADDRESS 192.168.10.234NETWORK MASK 255.255.255.0NETWORK ADDRESS 192.168.10.0NETWORK BROADCAST ADDRESS 192.168.10.255TOTAL NUMBER OF HOST BITS 254NUMBER OF HOSTS 8

NETWORK ADDRESS

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1 CALCULATENETWORK ADDRESS

1 1 0 0 0 0 0 0 128 + 64 192

1 0 1 0 1 0 0 0 128 + 32 + 8 168

0 0 0 0 1 0 1 0 8 + 2 10

0 0 0 0 0 0 0 0 0 0

ANSWER THE NETWORK ADDRESS IS 192.168.10.0

NETWORK BROADCAST



11

27 26 25 24 23 22 21 20


1 1 0 0 0 0 0 0 128 + 64 192

1 0 1 0 1 0 0 0 128 + 32 + 8 168

0 0 0 0 1 0 1 0 8 + 2 10

1 1 1 1 1 1 1 1128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

255

ANSWER THE NETWORK BROADCAST IS 192.168.10.255


= 2n - 2 = 28 = 256 - 2 = 254

NUMBER OF HOST = 8

TASK 2



12

PROBLEM 4

HOST IP ADDRESS 172.17.99.71NETWORK MASK 255.255.0.0NETWORK ADDRESS 172.17.0.0NETWORK BROADCAST ADDRESS 127.17.255.255TOTAL NUMBER OF HOST BITS 65 534NUMBER OF HOSTS 16

NETWORK ADDRESS

27 26 25 24 23 22 21 20


1 0 1 0 1 1 0 0 128 + 32 + 8 + 4

172

0 0 0 1 0 0 0 1 16 + 1 17

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0


NETWORK BROADCAST



13

27 26 25 24 23 22 21 20


1 0 1 0 1 1 0 0 128 + 32 + 8 + 4

172

0 0 0 1 0 0 1 0 16 + 1 17

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

255

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

255

ANSWER THE NETWORK BROADCAST IS 172.17.255


= 2n - 2= 216 = 65536 -2 65534

NUMBER OF HOST BITS = 16

TASK 2



14

PROBLEM 5

HOST IP ADDRESS 192.168.3.219NETWORK MASK 255.255.0.0NETWORK ADDRESS 192.168.0.0NETWORK BROADCAST ADDRESS 192.168.255.255TOTAL NUMBER OF HOST BITS 65 534NUMBER OF HOSTS 16

NETWORK ADDRESS

27 26 25 24 23 22 21 20


BROADCAST

1 1 0 0 0 0 0 0 128 + 64 + 192

1 0 1 0 1 0 0 0 128 + 32 + 8 168

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0




15

NETWORK BROADCAST

27 26 25 24 23 22 21 20


BROADCAST

1 1 0 0 0 0 0 0 128 + 64 + 8 192

1 0 1 0 1 0 0 0 128 + 32 + 8 168

1 1 1 1 1 1 1 1128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

255

1 1 1 1 1 1 1 1128 + 62 +

32 +16 + 8 + 4 + 2 + 1

255


USE THE FORMULA N = 16 = 2n - 2

= 216 = 65 534

NUMBER OF HOST = 16

TASK 2



16

PROBLEM 6

HOST IP ADDRESS 192.168.3.219NETWORK MASK 255.255.255.224NETWORK ADDRESS 192.168.3.192NETWORK BROADCAST ADDRESS 192.168.3.223TOTAL NUMBER OF HOST BITS 30NUMBER OF HOSTS 5

NETWORK ADDRESS

27 26 25 24 23 22 21 20


1 1 0 0 0 0 0 0 128 + 64 192

1 0 1 0 1 0 0 0 128 + 32 + 16 + 8 168

0 0 0 0 0 0 1 1 2 + 1 3

1 1 0 0 0 0 0 0 128 + 64 192


NETWORK BROADCAST



17

27 26 25 24 23 22 21 20


1 0 1 0 0 0 0 0 128 + 32 192

1 0 1 0 1 0 0 0 128 + 32 + 8 168

0 0 0 0 0 0 1 1 2 + 1 3

1 1 0 1 1 1 1 1128 + 64 + 16 + 8 + 4 +

2 + 1223


USE THE FORMULA N= 5

= 2n-2

= 25 = 32 - 2 = 30

NUMBER OF HOST = 5

STEP OF SUBNETING CALCULATION



18

TASK 1 : FOR GIVEN IP ADDRESS AND SUBNET MASK , DETERMINE SUBNET INFORMATION GIVEN :

HOST IP ADDRESS 172.25.114.250 NETWORK MASK 255.255.0.0 (/16)SUBNET MASK 255.255.255.192 (/26)

FIND : PROBLEM 1

NUMBER OF SUBNET BITS 10

NUMBER OF SUBNETS 210=1024 SUBNETS

NUMBER OF HOST BITS PER SUBNET 6 BITS

NUMBER OF USABLE HOST PER SUBNET 26 = 64 -2 = 62 HOSTS PER SUBNET

SUBNET ADDRESS FOR THIS IP ADDRESS 172.25.114.192

IP ADDRESS OF FIRST HOST ON THIS SUBNET 172.25.114.193

IP ADDRESS OF LAST HOST ON THIS SUBNET 172.25.114.254

BROADCAST ADDRESS FOR THIS SUBNET 172.25.114.255

HOST IP 172 30 1 33KFC 2044 NETWORK FUNDAMENTAL


19

ADDRESS

IP ADDRESS 10101100 00011110 00000001 00100001

SUBNET MASK X 11111111 11111111 11111111 11111100

SUBNET ADDRESS

10101100 00011110 00000001

00100000

SUBNET ADDRESS FOR

THIS IP ADDRESS =

172 25 114 192

172.25.114.192

IP ADDREES OF FIRST HOST = 172.30.1.33

192 +

62 HOST

IP ADDRESS OF LAST HOST CALCULATION

= 254IP ADDRESS OF LAST HOST = 172.30.1.34

NETWORK BROADCAST = 172.30.1.255

TASK 1



MULTIPLE

ANSWER

20

PROBLEM 2

HOST IP ADDRESS 172.30.1.33

SUBNET MASK 255.255.255.252

NUMBER OF SUBNET BITS 14

NUMBER OF SUBNETS 214 = 16384

NUMBER OF HOST BITS PER SUBNET 4 BIT

NUMBER OF USABLE HOSTS PER SUBNET 22 =4-2 = 2 HOST PER SUBNET


IP ADDRESS OF FIRST HOST ON THIS SUBNET

172.30.1.33

IP ADDRESS OF LAST HOST ON SUBNET 172.30.1.34




21

MOST IP ADDRESS

172 30 1 33

IP ADDRESS 10101100 00011110 00000001

001000001

SUBNET MASK X 11111111 11111111 11111111 11111100

SUBNET ADDRESS

10101100 00011110 00000001

001000000

SUBNET ADDRESS FOR

THIS IP ADDRESS

172 30 1 32

172.30.1.32

IP ADDRESS OF FIRST HOST= 172.30.1.33

32 +

2 HOSTS

IP ADDRESS OS LAST HOST CALCULATION

= 34IP ADDRESS OF LAST HOST= 172.30.1.34

BROADCAST ADDRESS = 172.30.1.255



MULTIPLE

ANSWER

22

TASK 1

PROBLEM 3


SUBNET MASK 255.255.255.0

NUMBER OF SUBNET BIT 8

NUMBER OF SUBNETS 28 =256


NUMBER OF USEABLE HOSTS PER SUBNET 28 = 256-2 = 254 HOSTS PER SUBNET


IP ADDRESS OF FIRST HOST ON THIS SUBNET 192.192.10.1





23

HOST IP ADDRSS

192 192 10 234

IP ADDRESS 11000000 11000000 00001010 11101010

SUBNET MASK

X 11111111 11111111 11111111 00000000

SUBNET ADDRESS

11000000 11000000 000001010 00000000

SUBNET ADDRESS FOR THIS IP ADDRESS

192 192 10 0

192.192.10.0

IP ADDRESS OF FIRST HOST= 192.192.10.1 0+

254 HOSTS

IP ADDRESS OF LAST CALCULATION

= 254IP ADDRESS OF LAST HOST= 192.192.10.254




MULTIPLE

ANSWER

24

TASK 1

PROBLEM 4



NUMBER OF SUBNETS BITS 0

NUMBER OF SUBNETS 20 = 1

NUMBER OF HOST BITS PER SUBNET16

NUMBER OF USABLE HOSTS PER SUBNET 216 65536 - 2 = 65534 HOSTS PER SUBNET

SUBNETS ADDRESS FOR THIS IP ADDRESS 172.17.0.0


172.17.0.1





25

HOST IP ADDRESS

172 17 99 71

IP ADDRESS 1O1O1100 00010001 0110001101000111

SUBNET MASK X 11111111 11111111 00000000 00000000

SUBNET ADDRESS

10101100 00010001 0000000100000000

SUBNET ADDRESS FOR

THIS IP ADDRESS

172 17 0 0

172.17.0.0

IP ADDRESS OF FIRST = 172.17.0.1

0+

254 HOSTS


= 254IP ADDRESS OF LAST HOST = 172.17.255.254




MULTIPLE

ANSWER

26

TASK 1

PROBLEM 5




NUMBER OF SUBNET 28 = 256


NUMBER OF USABLE MOSTS PER SUBNET 28 = 256 - 2 = 254 HOSTS PER SUBNET



192.168.3.1



HOST IP ADDRESS

192 168 3 219



27

IP ADDRESS 11000000 10101000 0000001111011011

SUBNET MASK X 11111111 1111111 11111111 00000000

SUBNET ADDRESS

11000000 10101000 0000001100000000

SUBNET ADDRESS FOR

THIS IP ADDRESS

192 168 3 0

192.168.3.0

IP ADDRESS OF FIRST HOST = 192.168.3.1 0

+254


= 254IP ADDRESS OF LAST HOST

=192.168.3.254


TASK 1

PROBLEM 6KFC 2044 NETWORK FUNDAMENTAL


MULTIPLE

ANSWER

28




NUMBER OF SUBNET 214 =16384 SUBNET

NUMBER OF HOST BITS PER SUBNET 14

NUMBER OF USABLE HOSTS PER SUBNET 22 =4-2 = 2 HOSTS PER SUBNET


IP ADDRESS OF FIRST HOST THIS SUBNET 192.168.3.217

IP ADDRESS OF LAST HOST THIS SUBNET 192.168.3.218


HOST IP ADDRESS 172 168 3 219

IP ADDRESS 11000000 10101000 00000011 11011011



MULTIPLE

29

SUBNET MASK X 11111111 11111111 11111111 11111100

SUBNET ADDRESS 11000000 10101000 00000011 11011000

SUBNET ADDRESS FOR THIS IP ADDRESS

192 168 3 216

192.168.3.216

IP ADDRESS OF FIRST HOST = 192.168.3.217

216+

2 HOSTS


= 218IP ADDRESS OF LAST HOST =192.168.3.218




ANSWER

assignment kfc 2044 network fundamental 2

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