PROBLEM 1

Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

172.30.1.33
255.255.0.0

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or
number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
Take the IP address 172.30.1.33 and start with the 1st octet (172)
172 128 = 44
44 32 = 12
12 8 = 4
44=0
Assign the number 1 to the values used to subtract the octet and 0 for the ones
not used.
128
1

64
0

32
1

16
0

8
1

4
1

2
0

1
0

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
64=2
22=0
128
0

64
0

32
0

16
1

8
1

4
1

2
1

1
0

4
0

2
0

1
1

4
0

2
0

1
1

Binary notation for the number 30 is 00011110. (16+8+2=32)

3rd Octet (1)
11=0
128
0

64
0

32
0

16
0

8
0

Binary notation for the number 1 is 00000001. (1)

4th octet (33)

33 32 = 1
11=0
128
0

64
0

32
1

16
0

8
0

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

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
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
74=3
32=1
11=0

4
1

2
1

1
1

128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
0

2
0

1
0

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128
0

64
0

32
0

16
0

8
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
0

64
0

32
0

16
0

8
0

4
0

2
0

1
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).
IP Address

10101100 00011110 00000001 00100001

Subnet 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
1
0
0

64
0
0
0

32
1
0
0

16
0
1
0

8
1
1
0

4
1
1
0

2
0
1
0

1
0
0
0

For 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.

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)
Wildcard

11111111 11111111 00000000 00000000

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
1
0
1
1

64
0
0
1
1

32
1
0
1
1

16
0
1
1
1

8
1
1
1
1

4
1
1
1
1

2
0
1
1
1

1
0
0
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.

Total Number of Host Bits and Usable Hosts.

By counting the number of host bits (The number of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

6
1

7
1

8
1

9
1

10
1

11
1

12
1

13
1

14
1

15
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
Network Mask
Network Address
Network Broadcast Address

172.30.1.33
255.255.0.0
172.30.0.0
172.30.255.255.

16
1

Total Number of Host Bits

Number of Hosts

16
65,534

PROBLEM 2
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

172.30.1.33
255.255.255.0

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or
number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
Take the IP address 172.30.1.33 and start with the 1st octet (172)
172 128 = 44
44 32 = 12
12 8 = 4
44=0
Assign the number 1 to the values used to subtract the octet and 0 for the ones
not used.

128
1

64
0

32
1

16
0

8
1

4
1

2
0

1
0

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
64=2
22=0
128
0

64
0

32
0

16
1

8
1

4
1

2
1

1
0

4
0

2
0

1
1

4
0

2
0

1
1

Binary notation for the number 30 is 00011110. (16+8+2=32)

3rd Octet (1)
11=0
128
0

64
0

32
0

16
0

8
0

Binary notation for the number 1 is 00000001. (1)

4th octet (33)

33 32 = 1
11=0
128
0

64
0

32
1

16
0

8
0

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

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
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
74=3

4
1

2
1

1
1

32=1
11=0
128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
1

2
1

1
1

4
0

2
0

1
0

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (255)
255 128 = 127
127 64 = 63
63 32 = 31
31 16 = 15
15 8 = 7
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

Binary notation for the number 255 is 11111111

4th Octet (0)
128
0

64
0

32
0

16
0

8
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

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
1
0
0
0

64
0
0
0
0

32
1
0
0
0

16
0
1
0
0

8
1
1
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.

Broadcast Address

4
1
1
0
0

2
0
1
0
0

1
0
0
1
0

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)

Wildcard

11111111 11111111 11111111 00000000

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
1
0
0
1

64
0
0
0
1

32
1
0
0
1

16
0
1
0
1

8
1
1
0
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.

4
1
1
0
1

2
0
1
0
1

1
0
0
1
1

Total Number of Host Bits and Usable Hosts.

By counting the number of host bits (The number of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the Wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

6
1

7
1

8
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
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

172.30.1.33
255.255.255.0
172.30.1.0
172.30.1.255.
8
254

PROBLEM 3
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.10.234
255.255.255.0

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or
number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
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
1

64
1

32
0

16
0

8
0

4
0

2
0

1
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

2nd octet (168)

168 128 = 40
40 32 = 8
88=0
128
1

64
0

32
1

16
0

8
1

4
0

2
0

1
0

2
1

1
0

Binary notation for the number 30 is 10101000. (16+8+2=32)

3rd Octet (10)
10 8 = 2
22=0
128
0

64
0

32
0

16
0

8
1

4
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
22=0
128

64

32

16

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

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
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

4
1

2
1

1
1

74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
1

2
1

1
1

4
0

2
0

1
0

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (255)
255 128 = 127
127 64 = 63
63 32 = 31
31 16 = 15
15 8 = 7
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

Binary notation for the number 255 is 11111111

4th Octet (0)
128
0

64
0

32
0

16
0

8
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

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
1
1
0
0

64
1
0
0
0

32
0
1
0
0

16
0
0
0
0

8
0
1
1
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.

4
0
0
0
0

2
0
0
1
0

1
0
0
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)
Wildcard

11111111 11111111 11111111 00000000

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
1
1
0
1

64
1
0
0
1

32
0
1
0
1

16
0
0
0
1

8
0
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.

Total Number of Host Bits and Usable Hosts.

4
0
0
0
1

2
0
0
1
1

1
0
0
0
1

By counting the number of host bits (The number of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the Wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

6
1

7
1

8
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
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.10.234
255.255.255.0
192.168.10.0
192.168.10.255
8
254

PROBLEM 4
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

172.17.99.71
255.255.0.0

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or
number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
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
1

64
0

32
1

16
0

8
1

4
1

2
0

1
0

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
11=0
128
0

64
0

32
0

16
1

8
0

4
0

2
0

1
1

2
1

1
1

Binary notation for the number 30 is 00010001. (16+1=17)

3rd Octet (99)
99 64 = 35
35 32 = 3
32=1
11=0
128
0

64
1

32
1

16
0

8
0

4
0

Binary notation for the number 1 is 01100011. (64+32+2+1 = 99)

4th octet (71)

71 64 = 7
74=3
32=1

11=0
128
0

64
1

32
0

16
0

8
0

4
1

2
1

1
1

Binary notation for the number 33 is 01000111. (64+4+2+1=234)

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
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

4
1

2
1

1
1

15 8 = 7
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
0

2
0

1
0

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128
0

64
0

32
0

16
0

8
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
0

64
0

32
0

16
0

8
0

4
0

2
0

1
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).
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
1
0
0
0
0
0
0
0
st
For the 1 Octet

32
1
0
0
0

16
0
1
0
0

8
1
0
0
0

4
1
0
0
0

2
0
0
0
0

1
0
1
0
0

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)
Wildcard

11111111 11111111 00000000 00000000

00000000 00000000 11111111 11111111

Bit-wise OR operation
Network Address

10101100 00010001 00000000 00000000

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
1
0
1
1

64
0
0
1
1

32
1
0
1
1

16
0
1
1
1

8
1
0
1
1

4
1
0
1
1

2
0
0
1
1

1
0
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 of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the Wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

6
1

7
1

8
1

9
1

10
1

11
1

12
1

13
1

14
1

15
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).

16
1

Now that we have all the data, fill in the table given.
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

172.17.99.71
255.255.0.0
192.17.0.0
192.168.255.255
16
65,534

PROBLEM 5
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.3.219
255.255.0.0

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or
number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
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
1

64
1

32
0

16
0

8
0

4
0

2
0

1
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

2nd octet (168)

168 128 = 40
40 32 = 8
88=0
128
1

64
0

32
1

16
0

8
1

4
0

2
0

1
0

2
1

1
1

Binary notation for the number 30 is 10101000. (128+32+8=168)

3rd Octet (3)
32=1
11=0
128
0

64
0

32
0

16
0

8
0

4
0

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
32=1
11=0
128
1

64
1

32
0

16
1

8
1

4
0

2
1

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
2nd Octet (255)

4
1

2
1

1
1

255 128 = 127

127 64 = 63
63 32 = 31
31 16 = 15
15 8 = 7
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
0

2
0

1
0

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128
0

64
0

32
0

16
0

8
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
0

64
0

32
0

16
0

8
0

4
0

2
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

1
0

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 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
1
1
1
0
0
0
0
0
st
For the 1 Octet

32
0
1
0
0

16
0
0
0
0

8
0
1
0
0

4
0
0
0
0

2
0
0
0
0

1
0
0
0
0

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)
Wildcard

11111111 11111111 00000000 00000000

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

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
1
1
1
1

64
1
0
1
1

32
0
1
1
1

16
0
0
1
1

8
0
1
1
1

4
0
0
1
1

2
0
0
1
1

1
0
0
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 of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the Wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

6
1

7
1

8
1

9
1

10
1

11
1

12
1

13
1

14
1

15
1

Host Bits: 16
Using the formula 2n 2 (n = Number of Bits) we can deduce the total number of
usable hosts.

16
1

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
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.3.219
255.255.0.0
192.168.0.0
192.168.255.255
16
65,534

PROBLEM 6
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.3.219
255.255.255.224

Network Address

Step 1: Translate Host IP Address and Network Mask into binary notation
First, list down these values of numbers
128

64

32

16

In order to remember these values start with the number 1, go from right to left,
and double that number seven times. For example, start with 1 on the right side. For
your next number, double the 1 (1 x 2 = 2). So, 2 is your next number
(remembering to go from right to left). For your third number, double the 2 (2 x 2 =
4); to continue the sequence, double the 4 (4 x 2 = 8). Repeat this process until
youve doubled your original number, seven times. The key to this is that every
single one of the values we put in that row are going to have either number 1 or

number 0 assigned to it. To convert the IP address we will take that string of
numbers and start from left to right this time. Each value in an IP address is an
octet. Subtract each octet with the numbers in the list until there is no remainder
left from the octet.
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
1

64
1

32
0

16
0

8
0

4
0

2
0

1
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

2nd octet (168)

168 128 = 40
40 32 = 8
88=0
128
1

64
0

32
1

16
0

8
1

4
0

2
0

1
0

2
1

1
1

Binary notation for the number 30 is 10101000. (128+32+8=168)

3rd Octet (3)
32=1
11=0
128
0

64
0

32
0

16
0

8
0

4
0

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
32=1
11=0
128
1

64
1

32
0

16
1

8
1

4
0

2
1

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
74=3
32=1
11=0

128
1

64
1

32
1

16
1

8
1

4
1

2
1

1
1

4
1

2
1

1
1

4
0

2
0

1
0

4
0

2
0

1
0

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
74=3
32=1
11=0
128
1

64
1

32
1

16
1

8
1

Binary notation for the number 255 is 11111111.

(128+64+32+16+8+4+2+1=255)
3rd Octet (0)
128
0

64
0

32
0

16
0

8
0

Binary notation for the number 255 is 11111111

4th Octet (224)
224 128 = 96
96 64 = 32
32 32 = 0
128
1

64
1

32
1

16
0

8
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
1
1
0
1

64
1
0
0
1

32
0
1
0
0

16
0
0
0
0

8
0
1
0
0

For the 1st Octet

128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
3rd Octet
2+1=3
4th Octet
128 + 64 = 192
Therefore, the Network Address is 192.168.3.192.

Broadcast Address

4
0
0
0
0

2
0
0
1
0

1
0
0
1
0

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)
Wildcard

11111111 11111111 11111111 11100000

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
1
1
0
1

64
1
0
0
1

32
0
1
0
0

16
0
0
0
1

8
0
1
0
1

For the 1st Octet

128 + 64 = 192
2nd Octet
128 + 32 + 8 = 168
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.

4
0
0
0
1

2
0
0
1
1

1
0
0
1
1

By counting the number of host bits (The number of1s in the binary notation), we
can determine the total number of usable hosts for this network. (Use the Wildcard
in binary notation)
1
1

2
1

3
1

4
1

5
1

Host Bits: 5
Using the formula 2n 2 (n = Number of Bits) we can deduce the total number of
usable hosts.
25 2 = 32 2
= 30 (Maximum number of Hosts).
Now that we have all the data, fill in the table given.
Host IP Address
Network Mask
Network Address
Network Broadcast Address
Total Number of Host Bits
Number of Hosts

192.168.3.219
255.255.225.224
192.168.3.192
192.168.3.223
5
30