ASSIGNMENT SHEET - IP Adress Calculation
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ASSIGNMENT SHEET Task 2 : Challenge PROBLEM 1 Host IP Address 172.30.1.33 Network Mask 255.255.0.0 Network Address Network Broadcast Address Total Number of Host Bits Number of Hosts Network address. Step 1 : Translate Host IP address and network mask into binary notation. 172 30 1 33 IP Address 10101100 00011110 00000001 00100001 Subnet Mask 11111111 11111111 00000000 00000000 255 255 0 0 Remember these value number : 2 0 = 1 2 1 = 2 2 2 = 4 2 3 = 8 2 4 = 16 2 5 = 32 2 6 = 64 2 7 = 128 First, take the IP address 172.30.1.33 and start with the 1 st octect (172) . And here the calculation to find the first octect . 172 – 128 = 44 44 – 32 = 12 12 – 8 = 4 4 – 4 = 0 The number “1” to the values used to subtract the octect and “0” for the ones not used. 128 64 32 16 8 4 2 1 1 0 1 0 1 1 0 0 NOTE : 1 AND 1 results in a 1 ; 0 AND anything else result in a 0.
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Transcript of ASSIGNMENT SHEET - IP Adress Calculation
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ASSIGNMENT SHEET
Task 2 : Challenge
PROBLEM 1
Host IP Address 172.30.1.33
Network Mask 255.255.0.0
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts
Network address.
Step 1 : Translate Host IP address and network mask into binary notation.
172 30 1 33IP Address 10101100 00011110 00000001 00100001
Subnet Mask 11111111 11111111 00000000 00000000
255 255 0 0
Remember these value number :
20 = 1
21 = 2
22 = 423 = 8
24 = 16
25 = 32
26 = 64
27 = 128
First, take the IP address 172.30.1.33 and start with the 1st octect (172) . And here the
calculation to find the first octect .
172 – 128 = 44
44 – 32 = 12
12 – 8 = 4
4 – 4 = 0
The number “1” to the values used to subtract the octect and “0” for the ones not used.
128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 0
NOTE : 1 AND 1 results in a 1 ; 0
AND anything else result in a 0.
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Now , we know that decimal number for first octect of 172 is 10101100 that had been converted
into binary form. To double check the calculation , we take the values that assigned with 1 and add
them all like this : 128+32+8+4 = 172
2nd octect (30)
30 – 16 = 14
14 – 8 = 6
6 – 4 = 2
2 – 2 = 0
128 64 32 16 8 4 2 1
0 0 0 1 1 1 1 0
The Binary notation for the number 30 is 00011110. (16+8+4+2=32)
3rd Octect (1)
1-1 =0
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 1
Binary notation for the number 1 is 00000001.
4th octect (33)
33 – 32 = 1
1 – 1 = 0
128 64 32 16 8 4 2 1
0 0 1 0 0 0 0 1
Binary notation for the number 33 is 00100001. (32+1=33)
The IP address of 172.30.1.33 in binary form is :
10101100.00011110.00000001.00100001
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Network Mask
Repeat the same way that we use for the IP address to convert to the Network Mask into
binary notation.
172 30 1 33IP Address 10101100 00011110 00000001 00100001
Subnet Mask 11111111 11111111 00000000 00000000
255 255 0 0
1st octect (255)
255 – 128 = 127
127 – 64 = 63
63 – 32 = 31
31 – 16 = 15
15 – 8 = 7
7 – 4 = 3
3 – 2 = 1
1 – 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1 = 255)
2nd octect (255)
255 – 128 = 127
127 – 64 = 63
63 – 32 = 31
31 – 16 = 15
15 – 8 = 7
7 – 4 = 3
3 – 2 = 1
1 – 1 = 0
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128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
4th Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
So the Network Mask 255.255.0.0 has its binary form equivalent of:
11111111.11111111.00000000.00000000
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Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask
(1 AND 1 results in 1; 1 AND 0 results in a 0).
IP Address 10101100 00011110 00000001 00100001
Network Mask 11111111 11111111 00000000 00000000
Network Address 10101100 00011110 00000000 00000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
the 1st Octet
128 + 32 + 8 + 4 = 172
2nd Octet
16 + 8 + 4 +2 = 30
3rd Octet
= 0
4th Octet
= 0
Therefore, the Network Address is 172.30.0.0.
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Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network Mask; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 00000000 00000000
Wildcard 00000000 00000000 11111111 11111111
Bit-wise OR operation
Network Address 10101100 00011110 00000000 00000000
Wildcard 00000000 00000000 11111111 11111111
Broadcast Address 10101100 00011110 11111111 11111111
Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 32 + 8 + 4 = 172
2nd Octet
16 + 8 + 4 +2 = 30
3rd Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
4th Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Therefore, the Broadcast Address is 172.30.255.255.
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Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the binary notation), we can
determine the total number of usable hosts for this network. (Use the wildcard in binary
notation)
1 2 3 4 5 6 7 8 9 10 11 12 13
14 15 16
1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1
Host Bits: 16
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
216 - 2 = 65,536 - 2
= 65,534 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
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 = 16
Number of Hosts = 65,534
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PROBLEM 2
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
Network Address
Step 1 : Translate Host IP address and network mask into binary notation.
172 30 1 33IP Address 10101100 00011110 00000001 00100001
Subnet Mask 11111111 11111111 11111111 00000000
255 255 255 0
Remember these value number :
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
Take the IP address 172.30.1.33 and start with the 1st octet (172)
172 - 128 = 44
44 - 32 = 12
12 - 8 = 4
4 - 4 = 0
Assign the number "1" to the values used to subtract the octet and "0" for the ones not used.
128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 0
NOTE : 1 AND 1 results in a 1 ; 0
AND anything else result in a 0.
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So now we know that a decimal number 172 is 10101100 converted to binary form. To
double check, we take the values assigned with 1 and add them together: 128+32+8+4=172
2nd octet (30)
30 - 16 = 14
14 - 8 = 6
6 - 4 = 2
2 - 2 = 0
128 64 32 16 8 4 2 1
0 0 0 1 1 1 1 0
Binary notation for the number 30 is 00011110. (16+8+2=32)
3rd Octet (1)
1 - 1 = 0
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 1
Binary notation for the number 1 is 00000001. (1)
4th octet (33)
33 - 32 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 0 1 0 0 0 0 1
Binary notation for the number 33 is 00100001. (32+1=33)
So the IP address of 172.30.1.33 has its binary form equivalent of:
10101100.00011110.00000001.00100001
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Network Mask
Repeat the same procedure used for the IP address to convert the Network Mask into binary
notation.
1st Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
2nd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
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3rd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111
4th Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
So the Network Mask 255.255.255.0 has its binary form equivalent of:
11111111.11111111.11111111.00000000
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Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask (1
AND 1 results in 1; 1 AND 0 results in a 0).
IP Address 10101100 00011110 00000001 00100001
Subnet Mask 11111111 11111111 11111111 00000000
Network Address 10101100 00011110 00000001 00000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 32 + 8 + 4 = 172
2nd Octet
16 + 8 + 4 +2 = 30
3rd Octet
1
4th Octet
0
Therefore, the Network Address is 172.30.1.0.
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Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network address; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 11111111 00000000
Wildcard 00000000 00000000 11111111 11111111
Bit-wise OR operation
Network Address 10101100 00011110 00000001 00000000
Wildcard 00000000 00000000 00000000 11111111
Broadcast Address 10101100 00011110 00000001 11111111
Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 32 + 8 + 4 = 172
2nd Octet
16 + 8 + 4 +2 = 30
3rd Octet
1
4th Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Therefore, the Broadcast Address is 172.30.1.255.
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Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the binary notation), we can
determine the total number of usable hosts for this network. (Use the Wildcard in binary
notation)
1 2 3 4 5 6 7 8
1 1 1 1 1 1 1 1
Host Bits: 8
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
28 - 2 = 256 - 2
= 254 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
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.
Total Number of Host Bits = 8
Number of Hosts = 254
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PROBLEM 3
Host IP Address 192.168.10.234
Network Mask 255.255.255.0
Network Address
Network Broadcast AddressTotal Number of Host Bits
Number of Hosts
Network address.
Step 1 : Translate Host IP address and network mask into binary notation.
192 168 10 234IP Address 11000000 10101000 00001010 11101010
Subnet Mask 11111111 11111111 11111111 00000000
255 255 255 0
Remember these value number :
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
Take the IP address 172.30.1.33 and start with the 1st octet (172)
192 - 128 = 64
64 - 64 = 0
Assign the number "1" to the values used to subtract the octet and "0" for the ones not used.
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0
So now we know that a decimal number 172 is 11000000 converted to binary form. To
double check, we take the values assigned with 1 and add them together: 128+64=192
NOTE : 1 AND 1 results in a 1 ; 0
AND anything else result in a 0.
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2nd octet (168)
168 - 128 = 40
40 - 32 = 8
8 - 8 = 0
128 64 32 16 8 4 2 1
1 0 1 0 1 0 0 0
Binary notation for the number 30 is 10101000. (16+8+2=32)
3rd Octet (10)
10 - 8 = 2
2 - 2 = 0
128 64 32 16 8 4 2 1
0 0 0 0 1 0 1 0
Binary notation for the number 1 is 00001010. (8 + 2 = 10)
4th octet (234)
234 - 128 = 106
106 - 64 = 42
42 - 32 = 10
10 - 8 = 2
2 - 2 = 0
128 64 32 16 8 4 2 1
1 1 1 0 1 0 1 0
Binary notation for the number 33 is 11101010. (128+64+32+8+2=234)
So the IP address of 172.30.1.33 has its binary form equivalent of:
11000000.10101000.00001010.11101010
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Network Mask
Repeat the same procedure used for the IP address to convert the Network Mask into binary
notation.
1st Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
2nd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
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3rd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111
4th Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
So the Network Mask 255.255.255.0 has its binary form equivalent of:
11111111.11111111.11111111.00000000
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Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask (1
AND 1 results in 1; 1 AND 0 results in a 0).
IP Address 11000000 10101000 00001010 11101010
Subnet Mask 11111111 11111111 11111111 00000000
Network Address 11000000 10101000 00001010 00000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
8 + 2 = 10
4th Octet
0
Therefore, the Network Address is 192.168.10.0.
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Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network Mask; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 11111111 00000000
Wildcard 00000000 00000000 00000000 11111111
Bit-wise OR operation
Network Address 11000000 10101000 00001010 00000000
Wildcard 00000000 00000000 00000000 11111111
Broadcast Address 11000000 10101000 00001010 11111111
Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
8 + 2 = 10
4th Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Therefore, the Broadcast Address is 192.168.10.255.
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Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the binary notation), we can
determine the total number of usable hosts for this network. (Use the Wildcard in binary
notation)
1 2 3 4 5 6 7 8
1 1 1 1 1 1 1 1
Host Bits: 8
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
28 - 2 = 256 - 2
= 254 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
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 = 8
Number of Hosts = 254
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PROBLEM 4
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
Network Address
172 17 99 71IP Address 10101100 00011110 01100011 01000111
Network Mask 11111111 11111111 00000000 00000000
255 255 0 0
Remember these value number :
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
Take the IP address 172.30.1.33 and start with the 1st octet (172)
172 - 128 = 44
44 - 32 = 12
12 - 8 = 4
4 - 4 = 12
Assign the number "1" to the values used to subtract the octet and "0" for the ones not used.
128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 0
NOTE : 1 AND 1 results in a 1 ; 0
AND anything else result in a 0.
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So now we know that a decimal number 172 is 10101100 converted to binary form. To
double check, we take the values assigned with 1 and add them together: 128+32+8+4=172
2nd octet (17)
17 - 16 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 0 0 1 0 0 0 1
Binary notation for the number 30 is 00010001. (16+1=17)
3rd Octet (99)
99 - 64 = 35
35 - 32 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 1 1 0 0 0 1 1
Binary notation for the number 1 is 01100011. (64+32+2+1 = 99)
4th octet (71)
71 - 64 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 1 0 0 0 1 1 1
Binary notation for the number 33 is 01000111. (64+4+2+1=234)
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So the IP address of 172.30.1.33 has its binary form equivalent of:
10101100.00010001.01100011.01000111
Network Mask
Repeat the same procedure used for the IP address to convert the Network Mask into binary
notation.
1st Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
2nd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
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Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
Binary notation for the number 255 is 11111111
4th Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
So the Network Mask 255.255.0.0 has its binary form equivalent of:
11111111.11111111.00000000.00000000
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Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask (1
AND 1 results in 1; 1 AND 0 results in a 0).
IP Address 10101100 00010001 01100011 01000011
Subnet Mask 11111111 11111111 00000000 00000000
Network Address 10101100 00010001 00000000 00000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 32 + 8 +4 = 172
2nd Octet
16 + 1 = 17
3rd Octet
0
4th Octet
0
Therefore, the Network Address is 172.17.0.0.
Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network Mask; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 00000000 00000000
Wildcard 00000000 00000000 11111111 11111111
Bit-wise OR operation
Network Address 10101100 00010001 00000000 00000000
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Wildcard 00000000 00000000 11111111 11111111
Broadcast Address 10101100 00010001 11111111 11111111
Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 32 + 8 +4 = 172
2nd Octet
16 + 1 = 17
3rd Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
4th Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Therefore, the Broadcast Address is 172.17.255.255.
Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the binary notation), we can
determine the total number of usable hosts for this network. (Use the Wildcard in binary
notation)
1 2 3 4 5 6 7 8 9 10 11 12 13
14 15 16
1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1
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Host Bits: 16
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
216 - 2 = 65,536 - 2
= 65,534 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
Host IP Address = 172.17.99.71
Network Mask = 255.255.0.0
Network Address = 192.17.0.0
Network Broadcast Address = 192.168.255.255
Total Number of Host Bits = 16
Number of Hosts 65,534
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PROBLEM 5
Host IP Address 192.168.3.219
Network Mask 255.255.0.0
Network Address
Network Broadcast AddressTotal Number of Host Bits
Number of Hosts
Network Address
192 168 3 219IP Address 11000000 10101000 00000011 11011011
Subnet Mask 11111111 11111111 00000000 00000000
255 255 0 0
Remember these value number :
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25
= 3226 = 64
27 = 128
1
Take the IP address 172.30.1.33 and start with the 1st octet (172)
192 - 128 = 64
64 - 64 = 0
Assign the number "1" to the values used to subtract the octet and "0" for the ones not used.
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0
So now we know that a decimal number 172 is 11000000 converted to binary form. To
double check, we take the values assigned with 1 and add them together: 128+64=192
NOTE : 1 AND 1 results in a 1 ; 0
AND anything else result in a 0.
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2nd octet (168)
168 - 128 = 40
40 - 32 = 8
8 - 8 = 0
128 64 32 16 8 4 2 1
1 0 1 0 1 0 0 0
Binary notation for the number 30 is 10101000. (128+32+8=168)
3rd Octet (3)
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 0 0 0 0 0 1 1
Binary notation for the number 1 is 00000011. (2+1 = 3)
4th octet (219)
219 - 128 = 91
91 - 64 = 27
27 - 16 = 11
11 - 8 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 0 1 1 0 1 1
Binary notation for the number 33 is 11011011. (128+64+16+8+2+1=219)
So the IP address of 172.30.1.33 has its binary form equivalent of:
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11000000.10101000.00000011.11011011
Network Mask
Repeat the same procedure used for the IP address to convert the Network Mask into binary
notation.
1st Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
2nd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
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Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
4th Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 0 is 00000000 (Since there is no number to subtract from 0
so, automatically the answer is 00000000).
So the Network Mask 255.255.0.0 has its binary form equivalent of:
11111111.11111111.00000000.00000000
Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask (1 AND 1 results in 1; 1 AND 0 results in a 0).
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IP Address 11000000 10101000 00000011 11011011
Subnet Mask 11111111 11111111 00000000 00000000
Network Address 11000000 10101000 00000000 00000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
0
4th Octet
0
Therefore, the Network Address is 192.168.0.0.
Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network Mask; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 00000000 00000000
Wildcard 00000000 00000000 11111111 11111111
Bit-wise OR operation
Network Address 11000000 10101000 00000000 00000000
Wildcard 00000000 00000000 11111111 11111111
Broadcast Address 11000000 10101000 11111111 11111111
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Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
4th Octet
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Therefore, the Broadcast Address is 192.168.255.255.
Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the broadcast address the last
octect in binary notation), we can determine the total number of usable hosts for this
network. (Use the Wildcard in binary notation)
1 2 3 4 5 6 7 8 9 10 11 12 13
14 15 16
1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1
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Host Bits: 16
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
216 - 2 = 65,536 - 2
= 65,534 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
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 = 16
Number of Hosts = 65,534
PROBLEM 6
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
Network Address
192 168 3 219IP Address 11000000 10101000 00000011 11011011
Subnet Mask 11111111 11111111 11111111 11100000
255 255 255 224
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Take the IP address 172.30.1.33 and start with the 1st octet (172)
172 – 128 =44
44 - 32 = 12
12 - 8 = 4
4 – 4 = 0
Assign the number "1" to the values used to subtract the octet and "0" for the ones not used.
128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 0
So now we know that a decimal number 172 is 10101100 converted to binary form. Todouble check, we take the values assigned with 1 and add them together: (128+32+8+4 =
172 )
2nd octet (168)
168 - 128 = 40
40 - 32 = 8
8 - 8 = 0
128 64 32 16 8 4 2 1
1 0 1 0 1 0 0 0
Binary notation for the number 30 is 10101000. (128+32+8=168)
3rd Octet (3)
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
0 0 0 0 0 0 1 1
Binary notation for the number 1 is 00000011. (2+1 = 3)
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4th octet (219)
219 - 128 = 91
91 - 64 = 27
27 - 16 = 11
11 - 8 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 0 1 1 0 1 1
Binary notation for the number 33 is 11011011. (128+64+16+8+2+1=219)
So the IP address of 172.30.1.33 has its binary form equivalent of:
11000000.10101000.00000011.11011011
Network Mask
Repeat the same procedure used for the IP address to convert the Network Mask into binary
notation.
1st Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
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Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
2nd Octet (255)
255 - 128 = 127
127 - 64 = 63
63 - 32 = 31
31 - 16 = 15
15 - 8 = 7
7 - 4 = 3
3 - 2 = 1
1 - 1 = 0
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0
Binary notation for the number 255 is 11111111
4th Octet (224)
224 - 128 = 96
96 - 64 = 32
32 - 32 = 0
128 64 32 16 8 4 2 1
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1 1 1 0 0 0 0 0
Binary notation for the number 224 is 11100000
So the Network Mask 255.255.0.0 has its binary form equivalent of:
11111111.11111111.11111111.11100000
Network Address
Perform a bit-wise AND (Logical AND Gate) operation on the IP address and subnet mask (1
AND 1 results in 1; 1 AND 0 results in a 0).
IP Address 11000000 10101000 00000011 11011011
Subnet Mask 11111111 11111111 11111111 11100000
Network Address 11000000 10101000 00000011 11000000
Use the method in converting the IP Address and Network Mask to convert the Network
Address into decimals (Reversed; adding the numbers instead of subtracting).
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 1 1
1 1 0 0 0 0 0 0
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
2 + 1 = 3
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4th Octet
128 + 64 = 192
Therefore, the Network Address is 192.168.3.192.
Broadcast Address
To find the broadcast address, use the network address and the wildcard (inverse of
Network Mask; 1 becomes 0 and 0 becomes 1) and perform a bit-wise OR (Logical OR
Gate) operation (1 OR 1 results in 1; 0 OR 1 results in 1; 0 OR 0 results in a 0).
Network Mask (Original) 11111111 11111111 11111111 11100000
Wildcard 00000000 00000000 00000000 00011111
Bit-wise OR operation
Network Address 11000000 10101000 00000011 11000000
Wildcard 00000000 00000000 00000000 00011111
Broadcast Address 11000000 10101000 00000011 11011111
Use the method in converting the IP Address and Network Mask to convert the Broadcast
Address into decimals (Reversed; adding the numbers instead of subtracting).
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 1 1
1 1 0 1 1 1 1 1
For the 1st Octet
128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
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3rd Octet
2 + 1 = 3
4th Octet
128 + 64 + 16 + 8 + 4 + 2 + 1 = 223
Therefore, the Broadcast Address is 192.168.3.223.
Total Number of Host Bits and Usable Hosts.
By counting the number of host bits (The number of"1's" in the binary notation), we can
determine the total number of usable hosts for this network. (Use the Wildcard in binary
notation)
1 2 3 4 5
1 1 1 1 1
Host Bits: 5
Using the formula 2n - 2 (n = Number of Bits) we can deduce the total number of usable
hosts.
2
5
- 2 = 32 - 2
= 30 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
Host IP Address = 192.168.3.219
Network Mask =255.255.225.224
Network Address = 192.168.3.192
Network Broadcast Address =192.168.3.223
Total Number of Host Bits = 5
Number of Hosts = 30
