PROBLEM 1 Host IP Address 172.30.1.33 Subnet Mask 255.255.255.0 Number of Subnet Bits Number of Subnets Number of Usable Host Bits per Subnet Number of Usable Hosts per Subnet Subnet Address for this IP Address IP Address of First Host on this Subnet IP Address of Last Host on this Subnet Broadcast Address for this Subnet Step 1: Translate Host IP Address and Subnet Mask into binary notation First, list down these values of numbers 128 64 32 16 8 4 2 1 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 you’ve 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 Host IP address 172.30.1.33 and start with the 1 st octet (172) 172 – 128 = 44
Transcript
PROBLEM 1
Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number of Subnet BitsNumber of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 0 1 0 1 1 0 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
6 – 4 = 2
2 – 2 = 0
128 64 32 16 8 4 2 10 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 10 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 10 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
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 10101100 00011110 00000001 00100001Subnet Mask 11111111 11111111 11111111 00000000Subnet Address 10101100 00011110 00000001 00000000 Subnet Counting Host Counting Range range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s in the last 16 bits of the Subnet Mask
1 2 3 4 5 6 7 81 1 1 1 1 1 1 1
Number of 0’s in the last 16 bits of the Subnet Mask
1 2 3 4 5 6 7 80 0 0 0 0 0 0 0
Using the data from the number of 0’s and 1’s in last 16 bits of the Subnet Mask.
2x = 28 = 256 Subnets (Total)
2y – 2 = 28 -2 = 256 – 2 = 254 Hosts (Total)
Determining the Range of Host Addresses available on the Subnet and the Broadcast Address.
172 30 1 255*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Now that we have all the data, fill in the table given.
Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number of Subnet Bits 8Number of Subnets 256Number of Host Bits per Subnet 8Number of Usable Hosts per Subnet 254Subnet Address for this IP Address 172.30.1.0.IP Address of First Host on this Subnet 172.30.1.1IP Address of Last Host on this Subnet 172.30.1.254Broadcast Address for this Subnet 172.30.1.255
PROBLEM 2
Host IP Address 172.30.1.33Subnet Mask 255.255.255.252Number of Subnet BitsNumber of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 0 1 0 1 1 0 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
6 – 4 = 2
2 – 2 = 0
128 64 32 16 8 4 2 10 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 10 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 10 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
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 10101100 00011110 00000001 00100001Subnet Mask 11111111 11111111 11111111 11111100Subnet Address 10101100 00011110 00000001 00100000 Subnet Counting Range Host Counting Range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s and 0’s in the last 16 bits of the Subnet Mask
172 30 1 35*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Now that we have all the data, fill in the table given.
Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number of Subnet Bits 14Number of Subnets 16,384Number of Host Bits per Subnet 2Number of Usable Hosts per Subnet 2Subnet Address for this IP Address 172.30.1.32.IP Address of First Host on this Subnet 172.30.1.33IP Address of Last Host on this Subnet 172.30.1.34Broadcast Address for this Subnet 172.30.1.35
PROBLEM 3
Host IP Address 192.192.10.234Subnet Mask 255.255.255.0Number of Subnet BitsNumber of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 1 0 0 0 0 0 0
So now we know that a decimal number 192 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 (192)
192 – 128 = 64
64 – 64 = 0
128 64 32 16 8 4 2 11 1 0 0 0 0 0 0
Binary notation for the number 192 is 11000000. (128+64=32)
3rd Octet (10)
10 – 8 = 2
2 – 2 = 0
128 64 32 16 8 4 2 10 0 0 0 1 0 1 0
Binary notation for the number 10 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 11 1 1 0 1 0 1 0
Binary notation for the number 234 is 11101010. (128+64+32+8+2=234)
So the IP address of 192.192.10.234 has its binary form equivalent of:
11000000.11000000.00001010.11101010
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 11000000 11000000 00001010 11101010Subnet Mask 11111111 11111111 11111111 00000000Subnet Address 11000000 11000000 00001010 00000000 Subnet Counting Host Counting Range Range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s and 0’s in the last 16 bits of the Subnet Mask
192 192 10 255*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Therefore, the Broadcast Address is 192.192.10.255.
Now that we have all the data, fill in the table given.
Host IP Address 192.192.10.234Subnet Mask 255.255.255.0Number of Subnet Bits 8Number of Subnets 256Number of Host Bits per Subnet 8Number of Usable Hosts per Subnet 254Subnet Address for this IP Address 192.192.10.0.IP Address of First Host on this Subnet 192.192.10.1IP Address of Last Host on this Subnet 192.192.10.254Broadcast Address for this Subnet 192.192.10.255
PROBLEM 4
Host IP Address 172.17.99.71
Subnet Mask 255.255.0.0Number of Subnet BitsNumber of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 0 1 0 1 1 0 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
1 – 1 = 0
128 64 32 16 8 4 2 10 0 0 1 0 0 0 1
Binary notation for the number 17 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 10 1 1 0 0 0 1 1
Binary notation for the number 99 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 10 1 0 0 0 1 1 1
Binary notation for the number 71 is 00100001. (64+4+2+1=71)
So the IP address of 172.30.1.33 has its binary form equivalent of:
10101100.00010001.01100011.01000111
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 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 10 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 10 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 Subnet Mask 255.255.255.252 has its binary form equivalent of:
11111111.11111111.00000000.00000000
Subnet Address
Perform a bit-wise AND (Logical AND Gate) operation on the Host IP address and Subnet Mask (1 AND 1 results in 1; 0 AND anything results in a 0).
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 10101100 00010001 01100011 01000111Subnet Mask 11111111 11111111 00000000 00000000Subnet Address 10101100 00010001 00000000 00000000 Subnet Host Counting Range Counting Range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s and 0’s in the last 16 bits of the Subnet Mask
172 17 0 35*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Therefore, the Broadcast Address is 172.17.255.255.
Now that we have all the data, fill in the table given.
Host IP Address 172.17.99.71Subnet Mask 255.255.0.0Number of Subnet Bits 0Number of Subnets 1Number of Host Bits per Subnet 16Number of Usable Hosts per Subnet 65,534Subnet Address for this IP Address 172.17.0.0.IP Address of First Host on this Subnet 172.17.0.1IP Address of Last Host on this Subnet 172.17.255.254Broadcast Address for this Subnet 172.17.255.255
PROBLEM 5
Host IP Address 192.168.3.219Subnet Mask 255.255.255.0Number of Subnet Bits
Number of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 1 0 0 0 0 0 0
So now we know that a decimal number 192 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
8 – 8 = 0
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
Binary notation for the number 168 is 10101000. (128+32+8=168)
3rd Octet (3)
3 – 2 = 1
1 – 1 = 0
128 64 32 16 8 4 2 10 0 0 0 0 0 1 1
Binary notation for the number 3 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 11 1 0 1 1 0 1 1
Binary notation for the number 219 is 11011011. (128+64+16+8+2+1=219)
So the IP address of 192.168.3.219 has its binary form equivalent of:
11000000.10101000.00000011.11011011
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 11000000 10101000 00000011 11011011Subnet Mask 11111111 11111111 11111111 00000000Subnet Address 11000000 10101000 00000011 00000000 Subnet Counting Host Counting Range Range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s and 0’s in the last 16 bits of the Subnet Mask
192 168 3 255*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Therefore, the Broadcast Address is 192.192.10.255.
Now that we have all the data, fill in the table given.
Host IP Address 192.192.10.234Subnet Mask 255.255.255.0Number of Subnet Bits 8Number of Subnets 256Number of Host Bits per Subnet 8Number of Usable Hosts per Subnet 254Subnet Address for this IP Address 192.168.3.0IP Address of First Host on this Subnet 192.168.3.1IP Address of Last Host on this Subnet 192.168.3.254Broadcast Address for this Subnet 192.168.3.255
PROBLEM 6
Host IP Address 192.168.3.219Subnet Mask 255.255.255.252Number of Subnet Bits
Number of SubnetsNumber of Usable Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this Subnet
Step 1: Translate Host IP Address and Subnet Mask into binary notation
First, list down these values of numbers
128 64 32 16 8 4 2 1
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 you’ve 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 Host 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 11 1 0 0 0 0 0 0
So now we know that a decimal number 192 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
8 – 8 = 0
128 64 32 16 8 4 2 11 0 1 0 1 0 0 0
Binary notation for the number 168 is 10101000. (128+32+8=168)
3rd Octet (3)
3 – 2 = 1
1 – 1 = 0
128 64 32 16 8 4 2 10 0 0 0 0 0 1 1
Binary notation for the number 3 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 11 1 0 1 1 0 1 1
Binary notation for the number 219 is 11011011. (128+64+16+8+2+1=219)
So the IP address of 192.168.3.219 has its binary form equivalent of:
11000000.10101000.00000011.11011011
Subnet Mask
Repeat the same procedure used for the IP address to convert the Subnet 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 11 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 11 1 1 1 1 1 1 1
Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Determine which bits contain network information and host information
Major Divide (M.D) Subnet Divide (S.D)
Host IP Address 11000000 10101000 00000011 11011011Subnet Mask 11111111 11111111 11111111 11111100Subnet Address 11000000 10101000 00000011 11011000 Subnet Counting Range Host Counting Range
Draw the Major Divide (M.D) as a wavy line where the 1’s in the major network mask (The 16 left-most bits) end and the Subnet Divide (S.D) as a straight line where the 1’s in the given Subnet Mask end. From here, we can get the number of subnet bits from the subnet mask by simply counting the number of bits (0 and 1) in the last 16 bits in the Subnet Mask.
Subnet Bits and Hosts Bits per Subnet
Using this formula, 2x where X = the number of 1’s in the last 16 bits of the Subnet Mask, we can get the total number of subnets. For The number of usable Hosts per subnet, use 2 y - 2 where y = the number of 0’s in the last 16 bits of the Subnet Mask.
Number of 1’s and 0’s in the last 16 bits of the Subnet Mask
192 168 3 255*Note that the First, Last and Broadcast Address bits in the fourth octet are used from the Subnet Address’s fourth octet.
For the first host, make all the bits in the host counting range “0” except for the last bit on the right which will be made “1”. This will give you the first host IP Address on the subnet (172.30.1.1).
To find the last host IP Address, make all the bits in the host counting range “1” except for the last bit which will be made “0”. This will give you the last host IP Address on the subnet (172.30.1.254).
Broadcast address can be deduced by making all the bits in the host counting range into “1”. This will give you the result for the Broadcast Address for the subnet (172.30.1.255).
Converting the First Host, Last Host and Broadcast Address to decimal form.
Use the method in converting the Host IP Address and Subnet Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).
Therefore, the Broadcast Address is 192.192.10.219.
Now that we have all the data, fill in the table given.
Host IP Address 192.192.10.234Subnet Mask 255.255.255.252Number of Subnet Bits 14Number of Subnets 16,384Number of Host Bits per Subnet 2Number of Usable Hosts per Subnet 2Subnet Address for this IP Address 192.168.3.216IP Address of First Host on this Subnet 192.168.3.217IP Address of Last Host on this Subnet 192.168.3.218Broadcast Address for this Subnet 192.168.3.219
