Task 1 - Nur Athirah Binti Ghazaly
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Transcript of Task 1 - Nur Athirah Binti Ghazaly
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7/24/2019 Task 1 - Nur Athirah Binti Ghazaly
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PROBLEM 1
Host IP address 172.30.1.33
Network Mask 255.255.0.0
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 172.30.1.33 -> 10101100.00011110.00000001.00100001
By using the repeated division-by-2 method , you can change a decimal into binary
2 172 0
2 86 0
2 43 1
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
2 30 0
2 15 1
2 7 1
2 3 1
2 1 1
0
0
0
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2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 0 0
0
0
0
0
0
0
0
2 0 0
0
0
0
0
0
0
0
Inverse Number of Network Mask -> 00000000.00000000.11111111.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
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Ip address 10101100 00011110 00000001 00100001
Subnet mask 11111111 11111111 00000000 00000000
Network address 10101100 00011110 00000000 00000000
172 30 0 0
Network address 10101100 00011110 00000000 00000000
mask 11111111 11111111 00000000 00000000
Broadcast 10101100 00011110 11111111 11111111
172 30 255 255
Total number of network mask is 8 + 8 = 16
n =16
216 =65 536
216 - 2 =65 534
PROBLEM 2
Host IP address 172.30.1.33
Network Mask 255.255.255.0
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 172.30.1.33 -> 10101100.00011110.00000001.00100001
By using the repeated division-by-2 method , you can change a decimal into binary
2 172 0
2 86 0
2 43 1
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
FORMULA :2n - 2
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2 30 0
2 15 1
2 7 1
2 3 1
2 1 1
0
0
0
2 1 1
0
0
0
0
0
0
0
2 33 1
2 16 0
2 8 0
2 4 0
2 2 0
2 1 1
0
0
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2 0 0
0
0
0
0
0
0
0
Inverse Number of Network Mask -> 00000000.00000000.00000000.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
Ip address 10101100 00011110 00000001 00100001
Subnet mask 11111111 11111111 11111111 00000000
Network address 10101100 00011110 00000001 00000000
172 30 1 0
Network address 10101100 00011110 00000001 00000000
mask 11111111 11111111 11111111 00000000
Broadcast 10101100 00011110 00000001 11111111
172 30 1 255
Total number of network mask is 8
n =8
216 =256
216 - 2 =254
FORMULA :2n - 2
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PROBLEM 3
Host IP address 192.168.10.234
Network Mask 255.255.255.0
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 192.168.10.234 -> 11000000.10101000.00001010.11101010
By using the repeated division-by-2 method , you can change a decimal into binary
2 192 0
2 96 0
2 48 0
2 24 0
2 12 0
2 6 0
2 3 1
2 1 1
2 168 0
2 84 0
2 42 0
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
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2 10 0
2 5 1
2 2 0
2 1 1
0
0
0
0
2 234 0
2 117 1
2 58 0
2 29 1
2 14 0
2 7 1
2 3 1
2 1 1
Network Mask, 255.255.255.0 -> 11111111.11111111.11111111.00000000
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
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2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 0 0
0
0
00
0
0
0
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Inverse Number of Network Mask -> 00000000.00000000.00000000.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
Ip address 11000000 10101000 00001010 11101010
Subnet mask 11111111 11111111 11111111 00000000
Network address 11000000 10101000 00001010 00000000
192 168 10 0
Network address 11000000 10101000 00001010 00000000
mask 11111111 11111111 11111111 00000000
Broadcast 11000000 10101000 00001010 11111111
192 168 10 255
Total number of network mask is 8
n =8
216 =256
216 - 2 =254
FORMULA :2n - 2
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PROBLEM 4
Host IP address 172.17.99.71
Network Mask 255.255.0.0
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 172.17.99.71-> 10101000.00010001.01100011.01000111
By using the repeated division-by-2 method , you can change a decimal into binary
2 172 0
2 86 0
2 43 1
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
2 17 1
2 8 0
2 4 0
2 2 0
2 1 1
0
0
0
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2 99 1
2 49 1
2 24 0
2 12 0
2 6 0
2 3 1
2 1 1
0
2 71 1
2 35 1
2 17 1
2 8 0
2 4 0
2 2 0
2 1 1
0
Network Mask, 255.255.0.0 -> 11111111.11111111.00000000.00000000
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
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2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 0 0
0
0
0
0
0
0
0
2 0 0
0
0
0
0
0
0
0
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Inverse Number of Network Mask -> 00000000.00000000.00000000.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
Ip address 10101000 00010001 01100011 01000111
Subnet mask 11111111 11111111 00000000 00000000
Network address 10101000 00010001 00000000 00000000
172 17 0 0
Network address 10101000 00010001 00000000 00000000
mask 11111111 11111111 00000000 00000000
Broadcast 10101000 00010001 11111111 11111111
172 17 255 255
Total number of network mask is 8 + 8 = 16
n =16
216 =65 536
216 - 2 =65 534
FORMULA :2n - 2
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PROBLEM 5
Host IP address 192.168.3.219
Network Mask 255.255.0.0
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 192.168.3.219-> 11000000.10101000.00000011.11011011
By using the repeated division-by-2 method , you can change a decimal into binary
2 192 0
2 96 0
2 48 0
2 24 0
2 12 0
2 6 0
2 3 1
2 1 1
2 168 0
2 84 0
2 42 0
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
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2 3 1
2 1 1
0
0
0
0
0
0
2 219 1
2 109 1
2 54 0
2 27 1
2 13 1
2 6 0
2 3 1
2 1 1
Network Mask, 255.255.0.0 -> 11111111.11111111.00000000.00000000
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
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2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 0 0
0
0
0
0
0
0
0
2 0 0
0
0
0
0
0
0
0
Inverse Number of Network Mask -> 00000000.00000000.00000000.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
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Ip address 11000000 10101000 00000011 11011011
Subnet mask 11111111 11111111 00000000 00000000
Network address 11000000 10101000 00000000 00000000
192 168 0 0
Network address 11000000 10101000 00000000 00000000
mask 11111111 11111111 00000000 00000000
Broadcast 11000000 10101000 11111111 11111111
192 168 255 255
Total number of network mask is 8 + 8 = 16
n =16
216 =65 536
216 - 2 =65 534
PROBLEM 6
Host IP address 192.168.3.219
Network Mask 255.255.255.224
Network Address
Change IP Address and Network Mask info Binary Number
IP Address , 192.168.3.219-> 11000000.10101000.00000011.11011011
By using the repeated division-by-2 method , you can change a decimal into binary
2 192 0
2 96 0
2 48 0
2 24 0
2 12 0
2 6 0
2 3 1
2 1 1
FORMULA :2n - 2
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2 168 0
2 84 0
2 42 0
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
2 3 1
2 1 1
0
0
0
0
0
0
2 219 1
2 109 1
2 54 0
2 27 1
2 13 1
2 6 0
2 3 1
2 1 1
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Network Mask, 255.255.255.224-> 11111111.11111111.11111111.11100000
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
2 255 1
2 127 1
2 63 1
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
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2 224 0
2 112 0
2 56 0
2 28 0
2 14 0
2 7 1
2 3 1
2 1 1
Inverse Number of Network Mask -> 00000000.00000000.00000000.11111111
And then do Logical And Gate between 2 binary number , whereas;
1 AND 1= 1
1 AND 0 = 0
0 AND 0 = 0
Ip address 11000000 10101000 00000011 11011011
Subnet mask 11111111 11111111 11111111 111 00000
Network address 11000000 10101000 00000011 11011111
192 168 3 223
Network address 11000000 10101000 00000011 11000000
mask 11111111 11111111 11111111 11100000
Broadcast 11000000 10101000 11111111 11111111
192 168 255 255
Total number of network mask is 5
n =5
216 =32
216 - 2 =30
FORMULA :2n - 2
