Assignment Network Fundamental Kfc 2044
description
Transcript of Assignment Network Fundamental Kfc 2044
INSTITUT KEMAHIRAN MARA LUMUT
KFC 2044 NETWORK FUNDAMENTAL
MOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK
950226-10-6035
1401123
EN. MOHD SHUKRI BIN MUHAMAD HUSIN
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 1
ASSIGNMENTKFC 2044 NETWORK FUNDAMENTAL
UNIT 1 COMPUTER NETWORKING FUNDAMENTALS
TABLE OF CONTENTS
NO. TITLE PAGE
1 IP ADDRESS CLASSESCALCULATION
3 – 5
2 PROBLEM 1 6 – 7
3 PROBLEM 2 8 – 9
4 PROBLEM 3 10 – 11
5 PROBLEM 4 12 – 13
6 PROBLEM 5 14 – 15
7 PROBLEM 6 16 – 17
8 SUBNETTINGCALCULATION
18 – 20
9 PROBLEM 1 21 – 23
10 PROBLEM 2 24 – 26
11 PROBLEM 3 27 – 29
12 PROBLEM 4 30 – 32
13 PROBLEM 5 33 – 35
14 PROBLEM 6 36 - 38
15 REFERENCE 39
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IP ADDRESS CLASSES
STEP TO CALCULATION IP ADDRESS CLASSESKFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 3
IP ADDRESS
Task 1 : For a given Ip Address, Determine Network Information.Given :
Host IP Address 172.25.114.250Network Mask 255.255.0.0 (/16)
Find :
Network Address 172.25.0.0Network Broadcast Address 172.25.255.255Total Number Of Host Bits 16Number Of Hosts 65534Step 1 : Translate Host IP Address And Network mask into binary notation:Network Address: 172.25.0.0
17227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 = 17225
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 1 0 0 1
16 + 8 + 1 = 250 0 0 0 0 0 0 00 0 0 0 0 0 0 0
= 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0
= 0Network Broadcast Address : 172.25.255.255
17227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 = 17225
27 26 25 24 23 22 21 20
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128 64 32 16 8 4 2 10 0 0 1 1 0 0 1
16 + 8 + 1 = 25255
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
128 + 64 + 32 + 16 + 8 + 4 + 2 +1 =255255
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.0.0=11111111.11111111.00000000.00000000Host = 2^16 = 65536 – 2 = 65534Hosts = 16Total Number Of Host = 216
65536 – 2 = 65534
Host Ip Address 172.25.114.250Network Mask 255.255.0.0(/16)Network Address 172.25.0.0Network broadcast Address 172.25.255.255Total Number Of Host Bits 16Number Of Hosts 65534
TASK 2 : PROBLEM 1 :Host IP Address 172.30.1.33
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Network mask 255.255.0.0
ANSWERHost IP Address 172.30.1.33Network Mask 255.255.0.0Network Address 172.30.0.0Network Broadcast Address 172.30.255.255Total Address of Host Bits 65534Number of Host 216
Network Address : 172.30.0.0172
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 =17230
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 1 1 1 0
16 + 8 + 4 + 2 = 300
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 00
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 0Network Address = 172.30.0.0Network Broadcast Address : 172.30.255.255
172
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27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 =17230
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 1 1 1 0
16 + 8 + 4 + 2 = 30255
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 +2 + 1 = 255255
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 +2 + 1 = 255
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.0.0=11111111.11111111.00000000.00000000Host = 2^16 = 65536 – 2 = 65534Hosts = 16Total Number Of Host = 216
65536 – 2 = 65534
Problem 2:Host IP Address 172.30.1.33
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Network mask 255.255.255.0
ANSWERHost IP Address 172.30.1.33Network Mask 255.255.255.0Network Address 172.30.1.0Network Broadcast Address 172.30.1.255Total Address of Host Bits 254Number of Host 28
Network Address : 172.30.1.0172
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 =17230
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 1 1 1 0
16 + 8 + 4 + 2 = 301
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 1
= 10
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 0Network Address = 172.30.1.0
Network Broadcast Address : 172.30.255.255
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17227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 =172
3027 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 1 1 1 0
16 + 8 + 4 + 2 = 30
127 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 1
= 1
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 +2 + 1 = 255
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.255.0=11111111.11111111.11111111.00000000Host = 2^8 = 256 – 2 = 256Hosts = 8
Problem 3:
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Host IP Address 192.168.10.234Network mask 255.255.255.0
ANSWERHost IP Address 192.168.10.234Network Mask 255.255.255.0Network Address 191.168.10.0Network Broadcast Address 191.168.10.255Total Address of Host Bits 254Number of Host 28
Network Address : 192.168.10.0 192
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192168
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 =16810
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 1 0 1 0
= 100
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 0Network Address = 192.168.10.0Network Broadcast Address : 192.168.10.255
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19227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192
16827 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 =168
1027 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 1 0 1 0
= 10
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.255.0=11111111.11111111.11111111.00000000Host bits = 8 Total Number of Hosts = 28
256 – 2 = 254
Problem 4:
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Host IP Address 172.17.99.71Network mask 255.255.0.0
ANSWERHost IP Address 172.17.99.71Network Mask 255.255.0.0Network Address 172.17.0.0Network Broadcast Address 172.17.255.255Total Address of Host Bits 65534Number of Host 216
Network Address : 172.17.0.0172
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 = 17217
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 0 0 0 1
16 + 1 = 170
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 00
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 0Network Address = 172.17.0.0Network Broadcast Address : 172.17.255.255
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17227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 1 0 0
128 + 32 + 8 + 4 = 172
1727 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 1 0 0 0 1
16 + 1 = 17
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.0.0=11111111.11111111.00000000.00000000Host = 2^16 = 65536 – 2 = 65534Hosts = 16
Problem 5:
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Host IP Address 192.168.3.219Network mask 255.255.0.0
ANSWERHost IP Address 192.168.3.219Network Mask 255.255.0.0Network Address 192.168.0.0Network Broadcast Address 192.168.255.255Total Address of Host Bits 65534Number of Host 216
Network Address : 192.168.0.0192
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192168
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 = 1680
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 00
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 0 0
= 0Network Address = 192.168.0.0
Network Broadcast Address : 192.168.255.255KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 14
19227 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192
16827 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 = 168
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255
25527 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 1 1 1 1 1 1
= 128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.0.0=11111111.11111111.00000000.00000000Host = 2^16 = 65536 – 2 = 65534Hosts = 16
Problem 6:
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Host IP Address 192.168.3.219Network mask 255.255.255.224
ANSWERHost IP Address 192.168.3.219Network Mask 255.255.0.0Network Address 191.168.3.192Network Broadcast Address 191.168.3.223Total Address of Host Bits 30Number of Host 25
Network Address : 192.168.3.192192
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192168
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 = 1683
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 1 1
2 + 1 = 3192
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192
Network Address = 192.168.3.192
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Network Broadcast Address : 192.168.3.223192
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
128 + 64 = 192
16827 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
128 + 32 + 8 = 168
327 26 25 24 23 22 21 20
128 64 32 16 8 4 2 10 0 0 0 0 0 1 1
2 + 1 = 3
22327 26 25 24 23 22 21 20
128 64 32 16 8 4 2 11 1 0 1 1 1 1 1
128 + 64 +16 + 8 + 4 + 2 + 1 = 223
FORMULA:Host (0) = perlu ditolak (-) dengan 2
255.255.255.224=11111111.11111111.11111111.11100000Host bits = 5Total numbers of hosts = 25
32 – 2 = 30
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Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information.
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SUBNETTING
Given:Host IP Address 172.25.114.250Network Mask 255.255.0.0(/16)Subnet Mask 255.255.255.192(/26)
Find:Number of subnet bits 10Number of subnets 210 = 1024 subnetNumber of Host Bits per Subnet 6Number of usable Hosts per subnet 26 = 64 – 2 = 62 bit per subnetSubnet Address for this IP Address 172.25.114.192IP Address of first host on this Subnet 172.25.114.193IP Address of last host on this Subnet 172.25.114.254Broadcast Address for this Subnet 172.25.114.255
Step 1: Translate host IP Address and subnet mask into binary notation.172 25 114 250
10101100 00011001 01110010 11111010 IP Address11111111 11111111 11111111 11000000 Subnet Mask
10101100 00011001 01110010 11000000Subnet Address
172 25 114 192
Subnet address for this IP address: 172.25.114.192
Determine the bit rangers for subnets and hosts.
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IP ADDRESS172 25 114 250
10101110 00011001 01110010 11111010
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
Total NumberSUBNET MASK = 11111111.11111111.00000000.00000000
= network : 2^8 = 256Number of Host Bits per Subnet
= 16 bitsNumber of Usable Hosts per Subnet
= host : 2^16 = 65536 – 2 = 65534
Problem 1KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 20
IP ADDRESS 172.30.1.33SUBNET MASK 255.255.255.0
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
Total number of Subnet Mask= 11111111.11111111.11111111.00000000= network = 2^8 = 256
Number of Hosts Bits per Subnet = 8 bits
Number of Usable Hosts per Subnet= hosts = 2^8 = 256 – 2 = 254
NETWORK
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK10101110 00011110 00000001 00000000
172 30 1 0 NETWORK = IP ADDRESS + SUBNET MASK172= 10101110 30=00011110 1= 00000001 33= 00100001 11111111 11111111 11111111 00000000 10101110 00011001 00000001 00000000NETWORK = 10101110.00011001. 00000001.00000000
=172.30.1.0
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BROADCAST
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK10101110 00011110 00000001 00000000
172 30 1 0
BROADCAST10101110 00011110 00000001 11111111
172 30 1 255
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK10101110 00011110 00000001 00000000
172 30 1 0
BROADCAST10101110 00011110 00000001 11111111
172 30 1 255
FIRST HOST
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK10101110 00011110 00000001 00000000
172 30 1 0
BROADCAST10101110 00011001 00000001 11111111
172 30 1 255
FIRST HOST172 30 1 1
10101110 00011110 00000001 00000001
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
BROADCAST10101110 00011110 00000001 11111111
172 30 1 255
FIRST HOST172 30 1 1
10101110 00011110 00000001 00000001
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LAST HOST
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK10101110 00011110 00000001 00000000
172 30 1 0
BROADCAST10101110 00011110 00000001 11111111
172 30 1 255
FIRST HOST172 30 1 1
10101110 00011110 00000001 00000001
LAST HOST172 30 1 254
10101110 00011110 00000001 11111110
Last Host = replace the number 0 be the number 1 at 17-31 bits in the Broadcast
FIRST HOST172 30 1 1
10101110 00011110 00000001 00000001
LAST HOST172 30 1 254
10101110 00011110 00000001 11111110
RESULTHOST IP ADDRESS 172.30.1.33SUBNET MASK 255.255.255.0NUMBER OF SUBNET BITS 8 BITSNUMBER OF SUBNET 2^8 = 256NUMBER OF HOST BITS PER SUBNET 256NUMBER OF USABLE HOSTS PER SUBNET 2^8 – 2 = 256 – 2 = 254SUBNET ADDRESS FOR THIS IP ADDRESS 172.30.1.0IP ADDRESS OF FIRST HOST ON THIS SUBNET 172.30.1.1IP ADDRESS OF LAST HOST ON THIS SUBNET 172.30.1.254BROADCAST ADDRESS FOR THIS SUBNET 172.30.1.255
Problem 2
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IP ADDRESS 172.30.1.33SUBNET MASK 255.255.255.252
IP ADDRESS172 30 1 33
10101110 00011001 00000001 00100001
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
Total number of Subnet Mask = 11111111.11111111.11111111.11111100
Network = 2^14 = 16384
Number of Hosts Bits per Subnet = 2 bits
Number of Usable Hosts per Subnet= hosts = 2^2 = 4 – 2 = 2
NETWORK
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK10101110 00011110 00000001 00100000
172 30 1 32NETWORK = IP ADDRESS + SUBNET MASK172= 10101110 30=00011110 1= 00000001 33= 00100001 11111111 11111111 11111111 11111100 10101110 00011001 00000001 00100000NETWORK = 10101110.00011001. 00000001.00100000
=172.30.1.32
FIRST HOST
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IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK10101110 00011110 00000001 00100000
172 30 1 32
FIRST HOST10101110 00011110 00000001 00100001
172 30 1 33
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
NETWORK10101110 00011110 00000001 00100000
172 30 1 32
FIRST HOST10101110 00011110 00000001 00100001
172 30 1 33
LAST HOST
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK10101110 00011110 00000001 00100000
172 30 1 32
LAST HOST10101110 00011110 00000001 00100010
172 30 1 34
Last host = replace a number from 0 to 1 on the host range subnet add except on the last bit remains 0.
NETWORK10101110 00011110 00000001 00100000
172 30 1 32
LAST HOST10101110 00011110 00000001 00100010
172 30 1 34
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BROADCAST
IP ADDRESS172 30 1 33
10101110 00011110 00000001 00100001
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK10101110 00011110 00000001 00100000
172 30 1 32
BROADCAST10101110 00011110 00000001 00100011
172 30 1 35
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK10101110 00011110 00000001 00000000
172 30 1 0
BROADCAST10101110 00011110 00000001 00000011
172 30 1 35
RESULTHOST IP ADDRESS 172.30.1.33SUBNET MASK 255.255.255.252NUMBER OF SUBNET BITS 14 BITSNUMBER OF SUBNET 2^14 = 16384NUMBER OF HOST BITS PER SUBNET 4NUMBER OF USABLE HOSTS PER SUBNET 2^2 – 2 =4 – 2 = 2 host per subnetSUBNET ADDRESS FOR THIS IP ADDRESS 172.30.1.32IP ADDRESS OF FIRST HOST ON THIS SUBNET 172.30.1.33IP ADDRESS OF LAST HOST ON THIS SUBNET 172.30.1.34BROADCAST ADDRESS FOR THIS SUBNET 172.30.1.35
PROBLEM 3
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IP ADDRESS 192.192.10.234SUBNET MASK 255.255.255.0
IP ADDRESS192 192 10 234
11000000 11000000 00001010 11101010
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
Total number of Subnet Mask = 11111111.11111111.11111111.00000000
Network = 2^8 = 256
Number of Hosts Bits per Subnet = 8 bits
Number of Usable Hosts per Subnet= hosts = 2^8 = 256 – 2 = 254
NETWORK
IP ADDRESS192 192 10 234
11000000 11000000 00001010 11101010
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK11000000 11000000 00001010 00000000
192 192 10 0NETWORK = IP ADDRESS + SUBNET MASK192= 11000000 192=11000000 10= 00001010 234= 11101010 11111111 11111111 11111111 11111100 11000000 11000000 00001010 00000000NETWORK = 110000001100000000001010 00000000
=192.192.10.234
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FIRST HOST
IP ADDRESS192 192 10 234
11000000 11000000 00001010 11101010
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
FIRST HOST11000000 11000000 00001010 00000001
192 192 10 1
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
FIRST HOST11000000 11000000 00001010 00000001
192 192 10 1
LAST HOST
IP ADDRESS192 192 10 234
11000000 11000000 00001010 11101010
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
LAST HOST11000000 11000000 00001010 11111110
192 192 10 254
Last host = replace a number from 0 to 1 on the host range subnet add except on the last bit remains 0.
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
LAST HOST11000000 11000000 00001010 11111110
192 192 10 254
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BROADCAST
IP ADDRESS192 192 10 234
11000000 11000000 00001010 11101010
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
BROADCAST11000000 11000000 00001010 11111111
192 192 10 255
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK11000000 11000000 00001010 00000000
192 192 10 0
BROADCAST11000000 11000000 00001010 11111111
192 192 10 255
RESULTHOST IP ADDRESS 192.192.10.234SUBNET MASK 255.255.255.0NUMBER OF SUBNET BITS 8 BITSNUMBER OF SUBNET 2^8 = 256NUMBER OF HOST BITS PER SUBNET 8 BITSNUMBER OF USABLE HOSTS PER SUBNET 2^8 – 2 =256 – 2 = 254 host per subnetSUBNET ADDRESS FOR THIS IP ADDRESS 192.192.10.0IP ADDRESS OF FIRST HOST ON THIS SUBNET 192.192.10.1IP ADDRESS OF LAST HOST ON THIS SUBNET 192.192.10.254BROADCAST ADDRESS FOR THIS SUBNET 192.192.10.255
PROBLEM 4
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IP ADDRESS 172.17.99.71SUBNET MASK 255.255.0.0
IP ADDRESS172 17 99 71
10101100 00010001 01100011 01000111
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
Total number of Subnet Mask = 11111111.11111111.00000000.00000000
Network = 2^0 = 1
Number of Hosts Bits per Subnet = 16 bits
Number of Usable Hosts per Subnet= hosts = 2^16 = 65536 – 2 = 65534
NETWORK
IP ADDRESS172 17 99 71
10101100 00010001 01100011 01000111
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
NETWORK10101100 00010001 00000000 00000000
172 17 0 0NETWORK = IP ADDRESS + SUBNET MASK172= 10101100 17=00010001 99= 01100011 234= 01000111 11111111 11111111 00000000 00000000 10101100 00010001 00000000 00000000NETWORK = 10101100 00010001 00000000 00000000
=172.17.0.0
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 30
FIRST HOST
IP ADDRESS172 17 99 71
10101100 00010001 01100011 01000111
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
FIRST HOST10101100 00010001 00000000 00000001
172 17 0 1
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
FIRST HOST10101100 00010001 00000000 00000001
172 17 0 1
LAST HOST
IP ADDRESS172 17 99 71
10101100 00010001 01100011 01000111
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
LAST HOST10101100 00010001 11111111 11111110
172 17 255 254
Last host = replace a number from 0 to 1 on the host range subnet add except on the last bit remains 0.
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
LAST HOST10101100 00010001 11111111 11111110
172 17 255 254
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 31
BROADCAST
IP ADDRESS172 17 99 71
10101100 00010001 01100011 01000111
SUBNET MASK11111111 111111111 00000000 00000000
255 255 0 0
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
BROADCAST10101100 00010001 11111111 11111111
172 17 255 255
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK10101100 00010001 00000000 00000000
172 17 0 0
BROADCAST10101100 00010001 11111111 11111111
172 17 255 255
RESULTHOST IP ADDRESS 172.17.99.71SUBNET MASK 255.255.0.0NUMBER OF SUBNET BITS 0 BITSNUMBER OF SUBNET 2^0 = 1NUMBER OF HOST BITS PER SUBNET 16 BITSNUMBER OF USABLE HOSTS PER SUBNET 2^16 – 2 =65536 – 2 = 65534 host per
subnetSUBNET ADDRESS FOR THIS IP ADDRESS 172.17.0.0IP ADDRESS OF FIRST HOST ON THIS SUBNET 172.17.0.1IP ADDRESS OF LAST HOST ON THIS SUBNET 172.17.255.254BROADCAST ADDRESS FOR THIS SUBNET 172.17.255.255
PROBLEM 5KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 32
IP ADDRESS 192.168.3.219SUBNET MASK 255.255.255.0
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
Total number of Subnet Mask = 11111111.11111111. 11111111.00000000
Network = 2^8 = 256
Number of Hosts Bits per Subnet = 8 bits
Number of Usable Hosts per Subnet= hosts = 2^8 = 256 – 2 = 254
NETWORK
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK192 168 3 0
11000000 10101000 00000011 00000000NETWORK = IP ADDRESS + SUBNET MASK192= 11000000 168=10101000 3= 00000011 219= 11011011 11111111 11111111 00000000 00000000 11000000 10101000 00000011 00000000NETWORK = 11000000101010000000001100000000
=192.168.3.0
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 33
FIRST HOST
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
FIRST HOST11000000 10101000 00000011 00000001
192 168 3 1
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
FIRST HOST11000000 10101000 00000011 00000001
192 168 3 1
LAST HOST
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
LAST HOST11000000 10101000 00000011 11111110
192 168 3 254
Last host = replace a number from 0 to 1 on the host range subnet add except on the last bit remains 0.
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
LAST HOST11000000 10101000 00000011 11111110
192 168 3 254
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 34
BROADCAST
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 00000000
255 255 255 0
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
BROADCAST11000000 10101000 00000011 11111111
192 168 3 255
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK192 168 3 0
11000000 10101000 00000011 00000000
BROADCAST11000000 10101000 00000011 11111111
192 168 3 255
RESULTHOST IP ADDRESS 192.168.3.219SUBNET MASK 255.255.255.0NUMBER OF SUBNET BITS 8 BITSNUMBER OF SUBNET 2^8 = 256NUMBER OF HOST BITS PER SUBNET 8 BITSNUMBER OF USABLE HOSTS PER SUBNET 2^8 – 2 =256 – 2 = 254 host per subnetSUBNET ADDRESS FOR THIS IP ADDRESS 192.168.3.0IP ADDRESS OF FIRST HOST ON THIS SUBNET 192.168.3.1IP ADDRESS OF LAST HOST ON THIS SUBNET 192.168.3.254BROADCAST ADDRESS FOR THIS SUBNET 192.168.3.255
PROBLEM 6
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 35
IP ADDRESS 192.168.3.219SUBNET MASK 255.255.255.252
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
Total number of Subnet Mask = 11111111.11111111. 11111111.11111100
Network = 2^14 = 16384
Number of Hosts Bits per Subnet = 2 bits
Number of Usable Hosts per Subnet= hosts = 2^2 = 4 – 2 = 2
NETWORK
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK192 168 3 216
11000000 10101000 00000011 11011000NETWORK = IP ADDRESS + SUBNET MASK192= 11000000 168=10101000 3= 00000011 216= 11011000 11111111 11111111 00000000 11111100 11000000 10101000 00000011 11011000NETWORK = 11000000101010000000001111011000
=192.168.3.216
FIRST HOST
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 36
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
FIRST HOST11000000 10101000 00000011 11011001
192 168 3 217
First Host = replace the number 0 be the number 1 at 32 bits in the Broadcast
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
FIRST HOST11000000 10101000 00000011 11011001
192 168 3 217
LAST HOST
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
LAST HOST11000000 10101000 00000011 11011010
192 168 3 218
Last host = replace a number from 0 to 1 on the host range subnet add except on the last bit remains 0.
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
LAST HOST11000000 10101000 00000011 11011010
192 168 3 218
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 37
BROADCAST
IP ADDRESS192 168 3 219
11000000 10101000 00000011 11011011
SUBNET MASK11111111 111111111 11111111 11111100
255 255 255 252
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
BROADCAST11000000 10101000 00000011 11011011
192 168 3 219
BROADCAST = replace the number 0 to number 1 appearing on network address
NETWORK192 168 3 216
11000000 10101000 00000011 11011000
BROADCAST11000000 10101000 00000011 11011011
192 168 3 219
RESULTHOST IP ADDRESS 192.168.3.219SUBNET MASK 255.255.255.252NUMBER OF SUBNET BITS 14 BITSNUMBER OF SUBNET 2^14 = 16384NUMBER OF HOST BITS PER SUBNET 2 BITSNUMBER OF USABLE HOSTS PER SUBNET 2^2 – 2 =4 – 2 = 2 host per subnetSUBNET ADDRESS FOR THIS IP ADDRESS 192.168.3.216IP ADDRESS OF FIRST HOST ON THIS SUBNET 192.168.3.217IP ADDRESS OF LAST HOST ON THIS SUBNET 192.168.3.218BROADCAST ADDRESS FOR THIS SUBNET 192.168.3.219
REFERENCE
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 38
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 39
KFC 2044 NETWORK FUNDAMENTALSMOHAMED FAIRULRASULIZAM BIN ABDUL RAZAK Page 40