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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