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8/11/2019 assignment kfc 2044 - networking fundamental
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INSTITUT KEMAHIRAN MARA LUMUT
ASSIGNMENT
KFC 2044
NETWORK FUNDAMENTAL
TOPIC
COMPUTER NETWORKING FUNDAMENTALS
CERTIFICATE IN COMPUTER ENGINEERING TECHNOLOGY
(NETWORKING)
NAME : NURUL ISWANI BINTI AHMAD SHAWARI
STUDENT ID : 1401114
NO. IC : 940525-08-6426
CLASS : SKN 2A
SESSION : JULDIS 2014
LECTURER : EN. MOHD SHUKRI BIN MUHAMAD HUSIN
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IP ADDRESS CLASES
CALCULATION OF IP ADDRESS CLASSES
TASK 1 : For a given IP Address, determine network information
Given IP address > 172.25.114.250
Network Mask > 255.255.0.0 (/16)
Find Network Address > 172.25.0.0
Network Broadcast Address > 172.25.255.255
Total Number of Host Bit > 66534
Number of Host > 16
STEP 1
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
IP ADDRESS
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 172 25 114 250
FORMULA
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 0 1 0 1 1 0 0
OCTET 2 0 0 0 1 1 0 0 1
OCTET 3 0 1 1 1 0 0 1 0
OCTET 4 1 1 1 1 1 0 1 0
BINARY NUMBER
Let
= 172= 25
= 114
= 250
OCTET 1 128 + 32 + 8 + 4OCTET 2 16 + 8 + 1
OCTET 3 64 + 32 + 16 + 2
OCTET 4 128 + 64 + 32 + 16 + 8 + 2
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NETWORK MASK
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 0 0
FORMULA
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 0 0 0 0 0 0 0 0
OCTET 4 0 0 0 0 0 0 0 0
BINARY NUMBER
Let
= 255
= 255
= 0
= 0
STEP 2
DETERMINE THE NETWORK ADDRESS
172 . 25 . 0 . 0
172 25 114 250
IP ADDRESS 10101100 00011001 01110010 11111010
SUBNET MASK 11111111 11111111 00000000 00000000
255 255 0 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
10101100 00011001 01110010 11111010
x 11111111 11111111 00000000 00000000
________________________________________
10101100 00011001 00000000 00000000
________________________________________
So
172 255 0 0
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 0
OCTET 4 0
NETWORK
ADDRESS10101100 00011001 00000000 00000000
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STEP 3
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
172 . 25 . 255 . 255
172 25 0 0
NETWORK ADD. 10101100 00011001 00000000 00000000
SUBNET MASK 11111111 11111111 00000000 00000000
BROADCAST ADD. 10101100 00011001 11111111 11111111
172 25 255 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 16
Total Number of Host Bits : 2n2 use this formula to calculate
= 2162
= 65 5362
= 65 534
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TASK
PROBLEM
Host IP Address 172.30.1.33Network Mask 255.255.0.0 ( /16 )
Network Address 172.30.0.0
Network Broadcast Address 172.30.255.255
Total Number of Host Bits 65534
Number of Host 16
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 172 30 1 33
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 0 1 0 1 1 0 0
OCTET 2 0 0 0 1 1 1 1 0
OCTET 3 0 0 0 0 0 0 0 1
OCTET 4 0 0 1 0 0 0 0 1
Let
= 172
= 30
= 1
= 33
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 0 0
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 0 0 0 0 0 0 0 0
OCTET 4 0 0 0 0 0 0 0 0
Let
= 255
= 255
= 0
= 0
OCTET 1 128 + 32 + 8 + 4
OCTET 2 16 + 8 + 4 + 2
OCTET 3 1
OCTET 4 32 + 1
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 0
OCTET 4 0
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DETERMINE THE NETWORK ADDRESS
172 . 30 . 0 . 0
172 30 0 33
IP ADDRESS 10101100 00011110 00000000 00100001
SUBNET MASK 11111111 11111111 00000000 00000000
255 255 0 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
10101100 00011110 00000000 00100001
x 11111111 11111111 00000000 00000000
________________________________________
10101100 00011110 00000000 00000000
________________________________________
So
172 30 0 0
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
172 . 30 . 255 . 255
172 30 0 0
NETWORK ADD. 10101100 00011001 00000000 00000000
SUBNET MASK 11111111 11111111 00000000 00000000
BROADCAST ADD. 10101100 00011001 11111111 11111111
172 30 255 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 16
Total Number of Host Bits : 2n2 use this formula to calculate
= 216
2
= 65 5362
= 65 534
NETWORK
ADDRESS10101100 00011110 00000000 00000000
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TASK
PROBLEM 2
Host IP Address 172.30.1.33Network Mask 255.255.255.0 ( /24 )
Network Address 172.30.1.0
Network Broadcast Address 172.30.1.255
Total Number of Host Bits 254
Number of Host 8
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 172 30 1 33
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 0 1 0 1 1 0 0
OCTET 2 0 0 0 1 1 1 1 0
OCTET 3 0 0 0 0 0 0 0 1
OCTET 4 0 0 1 0 0 0 0 1
Let
= 172
= 30
= 1
= 33
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 255 0
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 1 1 1 1 1 1 1 1
OCTET 4 0 0 0 0 0 0 0 0
Let
= 255
= 255
= 255
= 0
OCTET 1 128 + 32 + 8 + 4
OCTET 2 16 + 8 + 4 + 2
OCTET 3 1
OCTET 4 32 + 1
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 4 0
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DETERMINE THE NETWORK ADDRESS
172 . 30 . 1 . 0
172 30 1 33
IP ADDRESS 10101100 00011110 00000001 00100001
SUBNET MASK 11111111 11111111 11111111 00000000
255 255 255 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
10101100 00011001 00000001 00100001
x 11111111 11111111 11111111 00000000
________________________________________
10101100 00011001 00000001 00000000
________________________________________
So
172 30 1 0
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
172 . 30 . 1 . 255
172 30 1 0
NETWORK ADD. 10101100 00011001 00000001 00000000
SUBNET MASK 11111111 11111111 11111111 00000000BROADCAST ADD. 10101100 00011001 00000001 11111111
172 30 1 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 8
Total Number of Host Bits : 2n2 use this formula to calculate
= 282
= 2562= 254
NETWORK
ADDRESS10101100 00011001 00000001 00000000
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TASK
PROBLEM 3
Host IP Address 192.168.10.234Network Mask 255.255.255.0 ( /24 )
Network Address 192.168.10.0
Network Broadcast Address 192.168.10.255
Total Number of Host Bits 254
Number of Host 8
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 192 168 10 234
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 0 0 0 0 0 0
OCTET 2 1 0 1 0 1 0 0 0
OCTET 3 0 0 0 0 1 0 1 0
OCTET 4 1 1 1 0 1 0 1 0
Let
= 192
= 168
= 10
= 234
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 255 0
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 1 1 1 1 1 1 1 1
OCTET 4 0 0 0 0 0 0 0 0
Let
= 255
= 255
= 255
= 0
OCTET 1 128 + 64
OCTET 2 128 + 32 + 8
OCTET 3 8 + 2
OCTET 4 128 + 64 + 32 + 8 + 2
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 4 0
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DETERMINE THE NETWORK ADDRESS
192 . 168 . 10 . 0
192 168 10 234
IP ADDRESS 11000000 10101000 00001010 11101010
SUBNET MASK 11111111 11111111 11111111 00000000
255 255 255 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
11000000 10101000 00001010 11101010
x 11111111 11111111 11111111 00000000
________________________________________
11000000 10101000 00001010 00000000
________________________________________
So
192 168 10 0
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
192 . 168 . 10 . 255
192 168 10 0
NETWORK ADD. 11000000 10101000 00001010 00000000
SUBNET MASK 11111111 11111111 11111111 00000000
BROADCAST ADD. 11000000 10101000 00001010 11111111
192 168 10 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 8
Total Number of Host Bits : 2n2 use this formula to calculate
= 282
= 2562
= 254
NETWORK
ADDRESS
11000000 10101000 00001010 00000000
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TASK
PROBLEM 4
Host IP Address 172.17.99.71Network Mask 255.255.0.0 ( /16 )
Network Address 172.17.0.0
Network Broadcast Address 172.17.255.255
Total Number of Host Bits 65534
Number of Host 16
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 172 17 99 71
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 0 1 0 1 1 0 0
OCTET 2 0 0 0 1 0 0 0 1
OCTET 3 0 1 1 0 0 0 1 1
OCTET 4 0 1 0 0 0 1 1 1
Let
= 172
= 17
= 99
= 71
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 0 0
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 0 0 0 0 0 0 0 0
OCTET 4 0 0 0 0 0 0 0 0
Let
= 255
= 255
= 0
= 0
OCTET 1 128 + 32 + 8 + 4
OCTET 2 16 + 1
OCTET 3 64 + 32 + 2 + 1
OCTET 4 64 + 4 + 2 + 1
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 0
OCTET 4 0
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DETERMINE THE NETWORK ADDRESS
172 . 17 . 0 . 0
172 17 99 71
IP ADDRESS 10101100 00010001 01100011 01000111
SUBNET MASK 11111111 11111111 00000000 00000000
255 255 0 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
10101100 00010001 01100011 01000111
x 11111111 11111111 00000000 00000000
________________________________________
10101100 00010001 00000000 00000000
________________________________________
So
172 17 0 0
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
172 . 17 . 255 . 255
172 17 0 0
NETWORK ADD. 10101100 00010001 00000000 00000000
SUBNET MASK 11111111 11111111 00000000 00000000
BROADCAST ADD. 10101100 00010001 11111111 11111111
172 17 255 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 16
Total Number of Host Bits : 2n2 use this formula to calculate
= 216
2
= 655362
= 65534
NETWORK
ADDRESS10101100 00010001 00000000 00000000
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TASK
PROBLEM 5
Host IP Address 192.168.3.219Network Mask 255.255.0.0 ( /16 )
Network Address 192.168.0.0
Network Broadcast Address 192.168.255.255
Total Number of Host Bits 65534
Number of Host 16
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 172 17 99 71
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 0 1 0 1 1 0 0
OCTET 2 0 0 0 1 0 0 0 1
OCTET 3 0 1 1 0 0 0 1 1
OCTET 4 0 1 0 0 0 1 1 1
Let
= 172
= 17
= 99
= 71
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 0 0
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 0 0 0 0 0 0 0 0
OCTET 4 0 0 0 0 0 0 0 0
Let
= 255
= 255
= 0
= 0
OCTET 1 128 + 32 + 8 + 4
OCTET 2 16 + 1
OCTET 3 64 + 32 + 2 + 1
OCTET 4 64 + 4 + 2 + 1
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 0
OCTET 4 0
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DETERMINE THE NETWORK ADDRESS
172 . 17 . 0 . 0
172 17 99 71
IP ADDRESS 10101100 00010001 01100011 01000111
SUBNET MASK 11111111 11111111 00000000 00000000
255 255 0 0
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
10101100 00010001 01100011 01000111
x 11111111 11111111 00000000 00000000
________________________________________
10101100 00010001 00000000 00000000
________________________________________
So
172 17 0 0
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
172 . 17 . 255 . 255
172 17 0 0
NETWORK ADD. 10101100 00010001 00000000 00000000
SUBNET MASK 11111111 11111111 00000000 00000000
BROADCAST ADD. 10101100 00010001 11111111 11111111
172 17 255 255
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 16
Total Number of Host Bits : 2n2 use this formula to calculate
= 216
2
= 655362
= 65534
NETWORK
ADDRESS10101100 00010001 00000000 00000000
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TASK
PROBLEM 6
Host IP Address 192.168.3.219Network Mask 255.255.255.224 ( /27 )
Network Address 192.168.3.192
Network Broadcast Address 192.168.3.223
Total Number of Host Bits 30
Number of Host 5
TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION
OCTET 1 OCTET 2 OCTET 3 OCTET 4
IP ADDRESS 192 168 3 219
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 0 0 0 0 0 0
OCTET 2 1 0 1 0 1 0 0 0
OCTET 3 0 0 0 0 0 0 1 1
OCTET 4 1 1 0 1 1 0 1 1
Let
= 192
= 168
= 3
= 219
OCTET 1 OCTET 2 OCTET 3 OCTET 4
NETWORK MASK 255 255 255 224
27 2
6 2
5 2
4 2
3 2
2 2
1 2
0
128 64 32 16 8 4 2 1
OCTET 1 1 1 1 1 1 1 1 1
OCTET 2 1 1 1 1 1 1 1 1
OCTET 3 1 1 1 1 1 1 1 1
OCTET 4 1 1 1 0 0 0 0 0
Let
= 255
= 255
= 255
= 224
OCTET 1 128 + 64
OCTET 2 128 + 32 + 8
OCTET 3 2 + 1
OCTET 4 128 + 64 + 16 + 8 + 2 + 1
OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
OCTET 4 128 + 64 + 32
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DETERMINE THE NETWORK ADDRESS
192 . 168 . 3 . 192
192 168 3 219
IP ADDRESS 11000000 10101000 00000011 11011011
SUBNET MASK 11111111 11111111 11111111 11100000
255 255 255 224
Let the binary notation of IP Address times with binary notation of subnet mask
( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )
11000000 10101000 00000011 11011011
x 11111111 11111111 11111111 11100000
________________________________________
11000000 10101000 00000011 11000000
________________________________________
So
192 168 3 192
DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS
192 . 168 . 3 . 223
192 168 3 192
NETWORK ADD. 11000000 10101000 00000011 11000000
SUBNET MASK 11111111 11111111 11111111 11100000BROADCAST ADD. 11000000 10101000 00000011 11011111
192 168 3 223
Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask
= 5
Total Number of Host Bits : 2n2 use this formula to calculate
= 252
= 322= 30
NETWORK
ADDRESS
11000000 10101000 00000011 11000000
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STEP OF SUBNETING CALCULATION
TASK 1 : For a 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
Number of Subnet Bits 10
Number of Subnets 210
=1 024 subnets
Number of Host Bits per Subnet 6 bit
Number of Useable Host per Subnet 26
= 642 =62 hosts per subnet
Subnet Address for this IP Address 172.25.114.192
IP Address of First Host on this Subnet 172.25.114.193IP Address of Last Host on this Subnet 172.25.114.254
Broadcast Address for this Subnet 172.25.114.255
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
172 25 114 250
IP ADDRESS 10101100 00011001 01110010 11111010SUBNET MASK 11111111 11111111 11111111 11000000
255 255 255 192
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
10101100 00011001 01110010 11111010
X 11111111 11111111 11111111 11000000
_____________________________________________
10101100 00011001 01110010 11000000
_____________________________________________
Subnet Address > 172.25.114.192
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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
172 . 25 . 114 . 192
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
192
11000000
SUBNET ADDRESS 172
10101100
25
00011001
114
01110010
192
11000000
BROADCAST
ADDRESS
10101100
172
00011001
25
01110010
114
11111111
255
To find IP Address of first Host, the subnet address is add with 1
172 . 25 . 114 . 192
+ 1
________________
172 . 25 . 114 . 193 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
172 . 25 . 114 . 255
- 1
________________
172 . 25 . 114 . 254 IP Address of Last Host
________________
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DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS10101100 00011001 01110010 11 111010
SUBNET MASK 11111111 11111111 11111111 11 000000
SUBNET ADDRESS10101100 00011001 01110010 11 000000
subneting host
counting counting
range range
FIRST HOST 10101100
172
00011001
25
01110010
114
11 000001
193
LAST HOST 10101100
172
00011001
25
01110010
114
11 111110
254BROADCAST 10101100
172
00011001
25
01110010
114
11 111111
255
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
172.25.114.250
255.255.255.255
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
172.25.114.250
255.255.255.255 = 11111111
11000000 (host)
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Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 11000000 = 10 ones of subnet bits
2n2
= 210
2
= 10242
= 1022 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 11000000 = 6 zeros of host bits
2n2
= 262
= 642
= 62 useable host addresses
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TASK 2
CHALLENGE
PROBLEM
Given
Host IP Address 172.30.1.33
Subnet Mask 255.255.255.0
Find
Number of Subnet Bits 8
Number of Subnets 28=256 subnets
Number of Host Bits per Subnet 8 bit
Number of Useable Host per Subnet 28
= 2562 =254 hosts per subnetSubnet Address for this IP Address 172.30.1.0
IP Address of First Host on this Subnet 172.30.1.1
IP Address of Last Host on this Subnet 172.30.1.254
Broadcast Address for this Subnet 172.30.1.255
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
172 30 1 33
IP ADDRESS 10101100 00011110 00000001 00100001
SUBNET MASK 11111111 11111111 11111111 00000000
255 255 255 0
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
10101100 00011110 00000001 00100001
X 11111111 11111111 11111111 00000000
_____________________________________________
10101100 00011110 00000001 00000000
_____________________________________________
Subnet Address > 172.30.1.0
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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
172 . 30 . 1 . 192
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
0
00000000
SUBNET ADDRESS 172
10101100
30
00011110
1
00000001
0
00000000
BROADCAST
ADDRESS
10101100
172
00011110
30
00000001
1
11111111
255
To find IP Address of first Host, the subnet address is add with 1
172 . 30 . 1 . 0
+ 1
________________
172 . 30 . 1 . 1 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
172 . 30 . 1 . 255
- 1
________________
172 . 30 . 1 . 254 IP Address of Last Host
________________
8/11/2019 assignment kfc 2044 - networking fundamental
23/44
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS10101100 00011110 00000001 00 100001
SUBNET MASK 11111111 11111111 11111111 00 000000
SUBNET ADDRESS10101100 00011110 00000001 00 000000
subneting host
counting counting
range range
FIRST HOST 10101100
172
00011110
30
00000001
1
00 000001
1
LAST HOST 10101100
172
00011110
30
00000001
1
11 111110
254BROADCAST 10101100
172
00011110
30
00000001
1
11 111111
255
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
172.30.1.33
255.255.255.0
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
172.30.1.33
255.255.255.0 = 11111111
00000000 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
24/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 00000000 = 8 ones of subnet bits
2n2
= 282
= 2562
= 254 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 00000000 = 8 zeros of host bits
2n2
= 282
= 2562
= 254 useable host addresses
8/11/2019 assignment kfc 2044 - networking fundamental
25/44
TASK 2
CHALLENGE
PROBLEM 2
Given
Host IP Address 172.30.1.33
Subnet Mask 255.255.255.252
Find
Number of Subnet Bits 14
Number of Subnets 214
=16834 subnets
Number of Host Bits per Subnet 2 bit
Number of Useable Host per Subnet 22 = 4 2 =2 hosts per subnet
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 172.30.1.34
Broadcast Address for this Subnet 172.30.1.35
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
172 30 1 33
IP ADDRESS 10101100 00011110 00000001 00100001
SUBNET MASK 11111111 11111111 11111111 11111100
255 255 255 252
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
10101100 00011110 00000001 00100001
X 11111111 11111111 11111111 11111100
_____________________________________________
10101100 00011110 00000001 00100000
_____________________________________________
Subnet Address > 172.30.1.32
8/11/2019 assignment kfc 2044 - networking fundamental
26/44
DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
172 . 30 . 1 . 32
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
252
11000000
SUBNET ADDRESS 172
10101100
30
00011110
1
00000001
32
00100000
BROADCAST
ADDRESS
10101100
172
00011110
30
00000001
1
00100011
35
To find IP Address of first Host, the subnet address is add with 1
172 . 30 . 1 . 32
+ 1
________________
172 . 30 . 1 . 33 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
172 . 30 . 1 . 35
- 1
________________
172 . 30 . 1 . 34 IP Address of Last Host
________________
8/11/2019 assignment kfc 2044 - networking fundamental
27/44
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS10101100 00011110 00000001 00 100001
SUBNET MASK 11111111 11111111 11111111 11 111100
SUBNET ADDRESS10101100 00011110 00000001 00 100000
subneting host
counting counting
range range
FIRST HOST 10101100
172
00011110
30
00000001
1
00 100001
33
LAST HOST 10101100
172
00011110
30
00000001
1
00 100010
34BROADCAST 10101100
172
00011110
30
00000001
1
00 100011
35
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
172.30.1.33
255.255.255.252
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
172.30.1.33
255.255.255.252 = 11111111
11111100 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
28/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 11111100 = 14 ones of subnet bits
2n2
= 214
2
= 163842
= 16382 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 11111100 = 2 zeros of host bits
2n2
= 222
= 42
= 2 useable host addresses
8/11/2019 assignment kfc 2044 - networking fundamental
29/44
TASK 2
CHALLENGE
PROBLEM 3
Given
Host IP Address 192.192.10.234
Subnet Mask 255.255.255.0
Find
Number of Subnet Bits 8
Number of Subnets 28=256 subnets
Number of Host Bits per Subnet 8 bit
Number of Useable Host per Subnet 28 = 2562 =254 hosts per subnet
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
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
192 192 10 234
IP ADDRESS 11000000 11000000 00001010 11101010
SUBNET MASK 11111111 11111111 11111111 00000000
255 255 255 0
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
11000000 11000000 00001010 11101010
X 11111111 11111111 11111111 00000000
_____________________________________________
11000000 11000000 00001010 00000000
_____________________________________________
Subnet Address > 192.192.10.0
8/11/2019 assignment kfc 2044 - networking fundamental
30/44
DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
192 . 192 . 10 . 0
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
0
00000000
SUBNET ADDRESS 192
11000000
192
11000000
10
00001010
0
00000000
BROADCAST
ADDRESS
11000000
192
11000000
192
00001010
10
11111111
255
To find IP Address of first Host, the subnet address is add with 1
192 . 192 . 10 . 0
+ 1
________________
192 . 192 . 10 . 1 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
192 . 192 . 10 . 255
- 1
________________
192 . 192 . 10 . 254 IP Address of Last Host
________________
8/11/2019 assignment kfc 2044 - networking fundamental
31/44
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS11000000 11000000 00001010 11 101010
SUBNET MASK 11111111 11111111 11111111 00 000000
SUBNET ADDRESS11000000 11000000 00001010 00 000000
subneting host
counting counting
range range
FIRST HOST 11000000
192
11000000
192
00001010
10
00 000001
1
LAST HOST 11000000
192
11000000
192
00001010
10
11 111110
254BROADCAST 11000000
192
11000000
192
00001010
10
11 111111
255
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
192.192.10.234
255.255.255.0
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
172.25.10.234
255.255.255.0 = 11111111
00000000 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
32/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 00000000 = 8 ones of subnet bits
2n2
= 282
= 2562
= 254 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 00000000 = 8 zeros of host bits
2n2
= 282
= 2562
= 254 useable host addresses
8/11/2019 assignment kfc 2044 - networking fundamental
33/44
TASK 2
CHALLENGE
PROBLEM 4
Given
Host IP Address 172.17.99.71
Subnet Mask 255.255.0.0
Find
Number of Subnet Bits 0
Number of Subnets 20=1 subnets
Number of Host Bits per Subnet 16 bit
Number of Useable Host per Subnet 216
= 6553 -2 =65534 hosts per subnetSubnet Address for this IP Address 172.17.0.0
IP Address of First Host on this Subnet 172.17.0.1
IP Address of Last Host on this Subnet 172.17.255.254
Broadcast Address for this Subnet 172.17.255.255
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
172 17 99 71
IP ADDRESS 10101100 00010001 01100011 01000111
SUBNET MASK 11111111 11111111 00000000 00000000
255 255 0 0
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
10101100 00010001 01100011 01000111
X 11111111 11111111 00000000 00000000
_____________________________________________
10101100 00010001 00000000 00000000
_____________________________________________
Subnet Address > 172.17.0.0
8/11/2019 assignment kfc 2044 - networking fundamental
34/44
DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
172 . 17 . 0 . 0
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
0
00000000
0
00000000
SUBNET ADDRESS 172
10101100
17
00010001
0
00000000
0
00000000
BROADCAST
ADDRESS
10101100
172
00010001
17
11111111
255
11111111
255
To find IP Address of first Host, the subnet address is add with 1
172 . 17 . 0 . 0
+ 1
________________
172 . 17 . 0 . 1 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
172 . 17 . 255 . 255
- 1
________________
172 . 17 . 255 . 254 IP Address of Last Host
________________
8/11/2019 assignment kfc 2044 - networking fundamental
35/44
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS10101100 00010001 011100011 11 111010
SUBNET MASK 11111111 11111111 11111111 11 000000
SUBNET ADDRESS10101100 00010001 00000000 00 000000
subneting host
counting counting
range range
FIRST HOST 10101100
172
00010001
17
00000000
0
00 000001
1
LAST HOST 10101100
172
00010001
17
11111111
255
11 111110
254BROADCAST 10101100
172
00010001
17
11111111
255
11 111111
255
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
172.17.99.71
255.255.0.0
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
172.17.99.71
255.255.0.0 = 00000000
00000000 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
36/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
00000000 00000000 = 0 ones of subnet bits
2n
= 20
= 1 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
00000000 00000000 = 16 zeros of host bits
2n2
= 216
2
= 655362
= 65534 useable host addresses
8/11/2019 assignment kfc 2044 - networking fundamental
37/44
TASK 2
CHALLENGE
PROBLEM 5
Given
Host IP Address 192.168.3.219
Subnet Mask 255.255.255.0
Find
Number of Subnet Bits 8
Number of Subnets 28=256 subnets
Number of Host Bits per Subnet 8 bit
Number of Useable Host per Subnet 28
= 2562 =254 hosts per subnet
Subnet Address for this IP Address 192.168.3.0IP 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
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
192 168 3 219
IP ADDRESS 11000000 10101000 00000011 11011011
SUBNET MASK 11111111 11111111 11111111 00000000
255 255 255 0
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
11000000 10101000 00000011 11011011
X 11111111 11111111 11111111 00000000
_____________________________________________
11000000 10101000 00000011 00000000
_____________________________________________
Subnet Address > 192.168.3.0
DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
8/11/2019 assignment kfc 2044 - networking fundamental
38/44
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
192 . 168 . 3 . 0
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
0
00000000
SUBNET ADDRESS 192
11000000
168
10101000
3
00000011
0
00000000
BROADCAST
ADDRESS
11000000
192
10101000
168
00000011
3
11111111
255
To find IP Address of first Host, the subnet address is add with 1
192 . 168 . 3 . 0+ 1
________________
192 . 168 . 3 . 1 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
192 . 168 . 3 . 255
- 1
________________
192 . 168 . 3 . 254 IP Address of Last Host
________________
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
8/11/2019 assignment kfc 2044 - networking fundamental
39/44
M.D S.D
IP ADDRESS11000000 10101000 00000011 11 011011
SUBNET MASK11111111 11111111 11111111 00 000000
SUBNET ADDRESS 10101100 10101000 00000011 00 000000
subneting host
counting counting
range range
FIRST HOST 11000000
192
10101000
168
00000011
3
00 000001
1
LAST HOST 11000000
192
10101000
168
00000011
3
11 111110
254
BROADCAST 11000000
192
10101000
168
00000011
3
11 111111
255
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address192.168.3.219
255.255.255.0
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
192.168.3.219
255.255.255.0 = 11111111
00000000 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
40/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 00000000 = 8 ones of subnet bits
2n2
= 282
= 2562
= 254 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 00000000 = 8 zeros of host bits
2n2
= 282
= 2562
= 254 useable host addresses
8/11/2019 assignment kfc 2044 - networking fundamental
41/44
TASK 2
CHALLENGE
PROBLEM 6
Given
Host IP Address 192.168.3.219
Subnet Mask 255.255.255.252
Find
Number of Subnet Bits 14
Number of Subnets 214
=16384 subnets
Number of Host Bits per Subnet 2 bit
Number of Useable Host per Subnet 22 = 4 2 =2 hosts 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 192.168.3.218
Broadcast Address for this Subnet 192.168.3.219
TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION
192 168 3 219
IP ADDRESS 11000000 10101000 00000011 11011011
SUBNET MASK 11111111 11111111 11111111 11111100
255 255 255 252
DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS
Binary notation of IP address X Binary notation of subnet mask
( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )
11000000 10101000 00000011 11011011
X 11111111 11111111 11111111 11111100
_____________________________________________
11000000 10101000 00000011 11011000
_____________________________________________
Subnet Address > 192.168.3.216
8/11/2019 assignment kfc 2044 - networking fundamental
42/44
DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION
AND WHICH CONTAIN HOST INFORMATION
Subnet Address
Network . Network . Host . Host
192 . 168 . 3 . 216
DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST
To find broadcast address, use subnet mask and subnet address
SUBNET MASK 255
11111111
255
11111111
255
11111111
252
11111100
SUBNET ADDRESS 192
11000000
168
10101000
3
00000011
216
11011000
BROADCAST
ADDRESS
11000000
192
10101000
168
00000011
3
11011011
219
To find IP Address of first Host, the subnet address is add with 1
192 . 168 . 3 . 216
+ 1
________________
192 . 168 . 3 . 217 IP Address of First Host
________________
To find IP Address of last Host, the broadcast address is subtract with 1
192 . 168 . 3 . 219
- 1
________________
192 . 168 . 3 . 218 IP Address of Last Host
________________
8/11/2019 assignment kfc 2044 - networking fundamental
43/44
DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS
ON THIS SUBNET
M.D S.D
IP ADDRESS11000000 10101000 00000011 11 011011
SUBNET MASK 11111111 11111111 11111111 11 111100
SUBNET ADDRESS11000000 10101000 00000011 11 011000
subneting host
counting counting
range range
FIRST HOST 11000000
192
10101000
168
00000011
3
11 011001
217
LAST HOST 11000000
192
10101000
168
00000011
3
11 011010
218BROADCAST 11000000
192
10101000
168
00000011
3
11 011011
219
DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS
Firstly, use the first octet of the IP address to determine the class of address and to determine which
octet are available for host.
Network.network.host.host
IP Address
192.168.3.219
255.255.255.252
Subnet Mask
Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask to
determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set
to one, proceed to count the number of host bits.
192.168.3.219
255.255.255.252 = 11111111
11111100 (host)
8/11/2019 assignment kfc 2044 - networking fundamental
44/44
Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to
count the number of subnets.
11111111 11111100 = 14 ones of subnet bits
2n2
= 214
2
= 163842
= 16382 useable subnet
Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to
calculate the number of host bits.
11111111 11111100 = 2 zeros of host bits
2n2
= 222
= 42
= 2 useable host addresses
Recommended