Task 1

PROBLEM 1Host IP Address172.30.1.33

Network Mask255.255.0.0

Network Address

Network Broadcast Address

Total Number of Host Bits

Number of Hosts

Network AddressStep 1: Translate Host IP Address and Network Mask into binary notationFirst, list down these values of numbers1286432168421

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 = 4444 32 = 1212 8 = 44 4 = 0Assign the number 1 to the values used to subtract the octet and 0 for the ones not used.1286432168421

10101100

So now we know that a decimal number 172 is10101100converted 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 = 1414 8 = 6 6 4 = 22 2 = 0 1286432168421

00011110

Binary notation for the number 30 is 00011110. (16+8+2=32)3rd Octet (1)1 1 = 01286432168421

00000001

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

4th octet (33)33 32 = 11 1 = 01286432168421

00100001

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111

Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)2nd Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111

Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)3rd Octet (0)1286432168421

00000000

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)1286432168421

00000000

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 AddressPerform 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 Address10101100000111100000000100100001Subnet Mask11111111111111110000000000000000Network Address10101100000111100000000000000000Use the method in converting the IP Address and Network Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

10101100

00011110

00000000

00000000

For the 1st Octet128 + 32 + 8 + 4 = 1722nd Octet16 + 8 + 4 +2 = 303rd Octet04th Octet0Therefore, the Network Address is 172.30.0.0. Broadcast AddressTo 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)11111111111111110000000000000000Wildcard00000000000000001111111111111111Bit-wise OR operationNetwork Address10101100000111100000000000000000Wildcard00000000000000001111111111111111Broadcast Address10101100000111101111111111111111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

10101100

00011110

11111111

11111111

For the 1st Octet128 + 32 + 8 + 4 = 1722nd Octet16 + 8 + 4 +2 = 303rd Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 2554th Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255Therefore, 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)12345678910111213141516

1111111111111111

Host Bits: 16Using 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 Address172.30.1.33


Network Address172.30.0.0

Network Broadcast Address172.30.255.255.

Total Number of Host Bits16

Number of Hosts65,534



Network Address



Number of Hosts



10101100


2nd octet (30)30 16 = 1414 8 = 6 6 4 = 22 2 = 0 1286432168421

00011110

Binary notation for the number 30 is 00011110. (16+8+2=32)3rd Octet (1)1 1 = 01286432168421

00000001

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

4th octet (33)33 32 = 11 1 = 01286432168421

00100001

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


11111111

Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)3rd Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111

Binary notation for the number 255 is 111111114th Octet (0)1286432168421

00000000


So the Network Mask 255.255.255.0 has its binary form equivalent of:11111111.11111111.11111111.00000000

Network AddressPerform 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 Address10101100000111100000000100100001Subnet Mask11111111111111111111111100000000Network Address10101100000111100000000100000000Use the method in converting the IP Address and Network Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

10101100

00011110

00000001

00000000

For the 1st Octet128 + 32 + 8 + 4 = 1722nd Octet16 + 8 + 4 +2 = 303rd Octet14th Octet0Therefore, the Network Address is 172.30.1.0. Broadcast AddressTo 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)11111111111111111111111100000000Wildcard00000000000000001111111111111111Bit-wise OR operationNetwork Address10101100000111100000000100000000Wildcard00000000000000000000000011111111Broadcast Address10101100000111100000000111111111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

10101100

00011110

00000001

11111111

For the 1st Octet128 + 32 + 8 + 4 = 1722nd Octet16 + 8 + 4 +2 = 303rd Octet14th Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255Therefore, the Broadcast Address is 172.30.1.255.

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)12345678

11111111

Host Bits: 8Using 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 Address172.30.1.33



Network Broadcast Address172.30.1.255.


Number of Hosts254



Network Address



Number of Hosts


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 = 6464 64 = 0Assign the number 1 to the values used to subtract the octet and 0 for the ones not used.1286432168421

11000000

So now we know that a decimal number 172 is11000000converted 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 = 4040 32 = 8 8 8 = 0 1286432168421

10101000

Binary notation for the number 30 is 10101000. (16+8+2=32)3rd Octet (10)10 8 = 22 2 = 01286432168421

00001010

Binary notation for the number 1 is 00001010. (8 + 2 = 10)

4th octet (234)234 128 = 106106 64 = 4242 32 = 1010 8 = 22 2 = 01286432168421

11101010

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


11111111

Binary notation for the number 255 is 11111111. (128+64+32+16+8+4+2+1=255)3rd Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


00000000


So the Network Mask 255.255.255.0 has its binary form equivalent of:11111111.11111111.11111111.00000000

Network AddressPerform 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 Address11000000101010000000101011101010Subnet Mask11111111111111111111111100000000Network Address11000000101010000000101000000000Use the method in converting the IP Address and Network Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

11000000

10101000

00001010

00000000

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet8 + 2 = 104th Octet0Therefore, the Network Address is 192.168.10.0. Broadcast AddressTo 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)11111111111111111111111100000000Wildcard00000000000000000000000011111111Bit-wise OR operationNetwork Address11000000101010000000101000000000Wildcard00000000000000000000000011111111Broadcast Address11000000101010000000101011111111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

11000000

10101000

00001010

11111111

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet8 + 2 = 104th Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255Therefore, the Broadcast Address is 192.168.10.255.


11111111




Network Broadcast Address192.168.10.255


Number of Hosts254



Network Address



Number of Hosts



10101100


2nd octet (17)17 16 = 11 1 = 0 1286432168421

00010001

Binary notation for the number 30 is 00010001. (16+1=17)3rd Octet (99)99 64 = 3535 32 = 33 2 = 11 1 = 01286432168421

01100011

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

4th octet (71)71 64 = 77 4 = 33 2 = 11 1 = 01286432168421

01000111

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


11111111


00000000



00000000



10101100

00010001

00000000

00000000

For the 1st Octet128 + 32 + 8 +4 = 1722nd Octet16 + 1 = 173rd Octet04th Octet0Therefore, the Network Address is 172.17.0.0. Broadcast AddressTo 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)11111111111111110000000000000000Wildcard00000000000000001111111111111111Bit-wise OR operationNetwork Address10101100000100010000000000000000Wildcard00000000000000001111111111111111Broadcast Address10101100000100011111111111111111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

10101100

00010001

11111111

11111111

For the 1st Octet128 + 32 + 8 +4 = 1722nd Octet16 + 1 = 173rd Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 2554th Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255Therefore, the Broadcast Address is 172.17.255.255.


1111111111111111









Network Address



Number of Hosts



11000000


2nd octet (168)168 128 = 4040 32 = 8 8 8 = 0 1286432168421

10101000


00000011

Binary notation for the number 1 is 00000011. (2+1 = 3)4th octet (219)219 128 = 9191 64 = 2727 16 = 1111 8 = 33 2 = 11 1 = 01286432168421

11011011

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


11111111


00000000

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)1286432168421

00000000



11000000

10101000

00000000

00000000

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet04th Octet0Therefore, the Network Address is 192.168.0.0. Broadcast AddressTo 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)11111111111111110000000000000000Wildcard00000000000000001111111111111111Bit-wise OR operationNetwork Address11000000101010000000000000000000Wildcard00000000000000001111111111111111Broadcast Address11000000101010001111111111111111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

11000000

10101000

11111111

11111111

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 2554th Octet128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255Therefore, the Broadcast Address is 192.168.255.255.


1111111111111111









Network Address



Number of Hosts



11000000


2nd octet (168)168 128 = 4040 32 = 8 8 8 = 0 1286432168421

10101000


00000011

Binary notation for the number 1 is 00000011. (2+1 = 3)4th octet (219)219 128 = 9191 64 = 2727 16 = 1111 8 = 33 2 = 11 1 = 01286432168421

11011011

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 MaskRepeat the same procedure used for the IP address to convert the Network Mask into binary notation.1st Octet (255)255 128 = 127127 64 = 6363 32 = 3131 16 = 1515 8 = 77 4 = 33 2 = 11 1 = 01286432168421

11111111


11111111


00000000

Binary notation for the number 255 is 111111114th Octet (224)224 128 = 9696 64 = 3232 32 = 01286432168421

11100000

Binary notation for the number 224 is 11100000So the Network Mask 255.255.0.0 has its binary form equivalent of:11111111.11111111.11111111.11100000 Network AddressPerform 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 Address11000000101010000000001111011011Subnet Mask11111111111111111111111111100000Network Address11000000101010000000001111000000Use the method in converting the IP Address and Network Mask to convert the Network Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

11000000

10101000

00000011

11000000

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet2 + 1 = 34th Octet128 + 64 = 192Therefore, the Network Address is 192.168.3.192. Broadcast AddressTo 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)11111111111111111111111111100000Wildcard00000000000000000000000000011111Bit-wise OR operationNetwork Address11000000101010000000001111000000Wildcard00000000000000000000000000011111Broadcast Address11000000101010000000001111011111Use the method in converting the IP Address and Network Mask to convert the Broadcast Address into decimals (Reversed; adding the numbers instead of subtracting).1286432168421

11000000

10101000

00000011

11011111

For the 1st Octet128 + 64 = 1922nd Octet128 + 32 + 8 = 1683rd Octet2 + 1 = 34th Octet128 + 64 + 16 + 8 + 4 + 2 + 1 = 223Therefore, the Broadcast Address is 192.168.3.223.


11111






Number of Hosts30

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Task 1