CALCULATE IP ADDRESSING

QUESTION

IP Address 172.25.114.250Network Mask 255.255.0.0Network Address 172.25.0.0Network Broadcast Address 172.25.255.255Total Number Of Host Bit 65534Number Of Host 16

ANSWER

172 to Binary Number = 10101100

CalculateAnswerBinary Number

172 2860

86 2430

43 2211

21 2101

10 250

5 221

2 210

1 201

Then , we take the remainder number from down to up .

25 to Binary Number = 00011001. You must put 0 to the front to make it become 8 digit

CalculateAnswerBinary Number

25 2121

12 260

6 230

3 211

1 201

25 2 = 12remainder = 112 2 = 6remainder = 06 2 =3remainder = 03 2 =1remainder = 11 2 =0remainder = 1

Then , we take the remainder number from down to up .

So , the Network Address is 10101100 . 00011001 . 00000000 . 00000000 ( 172 . 25 . 0 . 0 ) .

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 172 . 25 . 255 . 255

Host Bit

Total number of Host Bit .

The total number of Host Bit is 65534 .Formula : 2n - 2216 2 =65536 2 = 65534

Number Of Host

Number of Host is 16 , the solution is like below .

32 16 = 16

Task 2 : Challenge

Problem 1

Host IP Address172.30.1.33

Network Mask255.255.0.0

Network Address172.30.0.0

Network Broadcast Address172.30.255.255

Total Number of Host Bits65534

Number of Hosts16

Answer

172 to Binary Number is = 10101100

CalculateAnswerBinary Number

172 2860

86 2430

43 2211

21 2101

10 250

5 221

2 210

1 201

Then, we take the binary number from down to up .

30 to Binary Number =00011110 You must put 0 to the front to make it become 8 digit

CalculateAnswerBinary Number

30 2150

15 271

7 231

3 211

1 201

30 2 = 15remainder = 015 2 = 7remainder = 17 2 =3remainder = 13 2 =1remainder = 11 2 =0remainder = 1

Then, we take the remainder number from down to up .

So , the Network Address is 10101100 . 00011110 . 00000000 . 00000000 ( 172 . 30 . 0 . 0 )

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 172 . 30 . 255 . 255

Host Bit

Total number of Host Bit .

255=11111111.11111111.00000000.00000000

The total number of Host Bit is 65534 .Formula : 2n - 2216 2 =65536 2 = 65534

Number Of Host

Number of Host is 16 , the solution is like below .

32 16 = 16

Problem 2

Host IP Address172.30.1.33

Network Mask255.255.255.0

Network Address172.30.1.0

Network Broadcast Address172.30.1.255

Total Number of Host Bits254

Number of Hosts24

Answer

172 to Binary Number is = 10101100

CalculateAnswerBinary Number

172 2860

86 2430

43 2211

21 2101

10 250

5 221

2 210

1 201

Then, we take the binary number from down to up .

30 to Binary Number =00011110 You must put 0 to the front to make it become 8 digit

CalculateAnswerBinary Number

30 2150

15 271

7 231

3 211

1 201

30 2 = 15remainder = 015 2 = 7remainder = 17 2 =3remainder = 13 2 =1remainder = 11 2 =0remainder = 1

Then , we take the remainder number from down to up .

CalculateAnswerBinary Number

1 201

Then, You must put 0 to the front to make it become 8 digit So , the Network Address is 10101100 . 00011110 . 00000001 . 00000000 ( 172 . 30 . 1 . 0 )

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 172 . 30 . 1 . 255

Host Bit

Total number of Host Bit .

255=11111111.11111111.11111111.00000000

The total number of Host Bit is 254.Formula : 2n - 228 2 =256 2 = 254

Number Of Host

Number of Host is 24 , the solution is like below .

32 8 = 24

Problem 3

Host IP Address192.168.10.234

Network Mask255.255.255.0

Network Address192.168.10.0

Network Broadcast Address192.168.10.255

Total Number of Host Bits524

Number of Hosts8

Answer

192 to Binary Number is = 11000000

CalculateAnswerBinary Number

192 2960

96 2480

48 2240

24 2120

12 260

6 230

3 211

1 201

Then, we take the binary number from down to up .

168 to Binary Number is = 10001000

CalculateAnswerBinary Number

168 2840

84 2420

42 2210

21 2101

10 250

5 220

2 210

1 201

Then, we take the binary number from down to up .

10 to Binary Number is = 10001000

CalculateAnswerBinary Number

10 250

5 221

2 210

1 201

Then, we take the binary number from down to up . So , the Network Address is 11000000 . 10001000 . 00001010 . 00000000 ( 192 . 168 . 10 . 0 ) Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 192 . 168 . 10 . 255

Host Bit

Total number of Host Bit .

255=11111111.11111111.11111111.00000000

The total number of Host Bit is 254 .Formula : 2n - 228 2 =256 2 = 254

Number Of Host

Number of Host is 24 , the solution is like below .

32 8 = 24

Problem 4

Host IP Address172.17.99.71

Network Mask255.255.0.0

Network Address172.17.0.0

Network Broadcast Address172.17.255.255

Total Number of Host Bits65,534

Number of Hosts16

Answer

172 to Binary Number is = 10101100

CalculateAnswerBinary Number

172 2860

86 2430

43 2211

21 2101

10 250

5 221

2 210

1 201

Then, we take the binary number from down to up .

17 to Binary Number is = 00010001

CalculateAnswerBinary Number

17 281

8 240

4 220

2 210

1 201

Then, we take the binary number from down to up . So , the Network Address is 10101100 . 00010001 . 00000000 . 00000000 ( 172 . 17 . 0 . 0 )

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 172 . 17 . 255 . 255

Host Bit

Total number of Host Bit .

The total number of Host Bit is 510 .Formula : 2n - 2216 2 =65,536 2 = 65,534

Number Of Host

Number of Host is 16 , the solution is like below .

32 16 = 16

Problem 5

Host IP Address192.168.3.219

Network Mask255.255.0.0

Network Address192.168.0.0

Network Broadcast Address192.168.255.255

Total Number of Host Bits524,286

Number of Hosts13

Answer

192 to Binary Number is = 11000000

CalculateAnswerBinary Number

192 2960

96 2480

48 2240

24 2120

12 260

6 230

3 211

1 201

Then, we take the binary number from down to up .

168 to Binary Number is = 10001000

CalculateAnswerBinary Number

168 2840

84 2420

42 2210

21 2101

10 250

5 220

2 210

1 201

Then, we take the binary number from down to up . So , the Network Address is 11000000 . 10001000 . 00000000 . 00000000 ( 172 . 168 . 0 . 0 )

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 192 . 168 . 255 . 255

Host Bit

Total number of Host Bit .

The total number of Host Bit is 65534 .Formula : 2n - 2216 2 =65536 2 = 65534

Number Of Host

Number of Host is 16 , the solution is like below .

32 16 = 16

Problem 6

Host IP Address192.168.3.219

Network Mask255.255.255.224

Network Address192.168.3.192

Network Broadcast Address192.168.3.223

Total Number of Host Bits5

Number of Hosts30

Answer

192 to Binary Number = 11000000

CalculateAnswerBinary Number

192 2960

96 2480

48 2240

24 2120

12 260

6 230

3 211

1 201

Then, we take the binary number from down to up .

168 to Binary Number = 10001000

CalculateAnswerBinary Number

168 2840

84 2420

42 2210

21 2101

10 250

5 220

2 210

1 201

Then, we take the binary number from down to up .

3 to Binary Number =00000011

CalculateAnswerBinary Number

3 211

1 201

Then, we take the binary number from down to up .

219 to Binary Number =11011011CalculateAnswerBinary Number

219 21091

109 2541

54 2270

27 2131

13 261

6 230

3 211

1 201

Then, we take the binary number from down to up .

Network Broadcast Address

Question :

To find the Network Broadcast Address , you need to convert the number of 0 to 1 . The number of Network Broadcast Address is 10101100 . 00011001 . 11111111 , 11111111 . The number of Network Broadcast Address just need to convert the 0 to 1 .

Example :

00000000 . 00000000 = 11111111 . 11111111 . After convert to 1 , you need to convert it to standard number .

Answer :

11111111 to standard number is 255 , the solution is like below .( 1 x 20 ) + ( 1 x 21 ) + ( 1 x 22 ) + ( 1x 23 ) + ( 1 x 24 ) + ( 1 x 25 ) + ( 1 x 26 ) + ( 1 x 27 ) +( 1 x 28 ) =

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

So the standard number of Network Broadcast Address is 192 . 168 . 3 .192

Host Bit

Total number of Host Bit .

The total number of Host Bit is 65534 .Formula : 2n - 225 2 =32 2 = 30

SUBNETTING

Problem 1

IP ADDRESS172.30.1.33

SUBNET MASK255.255.255.0

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS17230133

10101110000111100000000100100001

SUBNET MASK11111111111111111111111100000000

2552552550

Total number of Subnet Mask=11111111.11111111.11111111.00000000= 28 =256Number of Hosts Bits per Subnet=8 bitsNumber of Usable Hosts per Subnet=hosts =28 2 =256-2 =254NETWORK10101110000111100000000100000000

1723010

Network = 10101110.00011110.00000001.00000000 =172.30.1.0BROADCAST =replace the number 0 to number 1 appearing on network addressNETWORK10101110000111100000000100000000

1723010

BROADCAST10101110000111100000000111111111

172301255

First Host = replace the number 0 be the number 1 at 32 bits in the broadcastBROADCAST10101110000111100000000111111111

172301255

FIRST HOST1723011

10101110000111100000000100000001

Last Host = replace the number 0 be the number 1 at 17-31 bits in the first hostFIRST HOST1723011

10101110000111100000000100000001

LAST HOST172301254

10101110000111100000000111111110

RESULTHOST IP ADDRESS172.30.1.33

SUBNET MASK255.255.255.0

NUMBER OF SUBNET BITS8 BITS

NUMBER OF SUBNET28 =256

NUMBER OF HOST BITS PER SUBNET256

NUMBER OF USABLE HOSTS PER SUBNET28 =256-2 =254

SUBNET ADDRESS FOR THIS IP ADDRESS172.30.1.0

IP ADDRESS OF FIRST HOST ON THIS SUBNET172.30.1.1

IP ADDRESS OF LAST HOST ON THIS SUBNET 172.30.1.254

BROADCAST ADRESS FOR THIS SUBNET172.30.1.255

Problem 2IP ADDRESS172.30.1.33

SUBNET MASK255.255.255.252

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS17230133

10101110000111100000000100100001

SUBNET MASK11111111111111111111111111111100

255255255255

Total number of Subnet Mask=11111111.11111111.11111111.11111100= 214 =16384Number of Hosts Bits per Subnet=2 bitsNumber of Usable Hosts per Subnet=hosts =22 2 =4-2 =2NETWORK10101110000111100000000100100000

17230132

Network = 10101110.00011110.00000001.00100000 =172.30.1.32FIRST HOST =replace the number 0 to number 1 appearing on network addressNETWORK10101110000111100000000100100000

17230132

FIRST HOST10101110000111100000000100100001

17230133

LAST HOST = replace the number 0 to 1 at 32 on the host range subnet add except on the last bit remains 0NETWORK10101110000111100000000100000011

17230133

LAST HOST17230134

10101110000111100000000100000001

BROADCAST = replace the number 0 to number 1 appearing on network addressNETWORK1723010

10101110000111100000000100000000

BROADCAST17230135

10101110000111100000000100000011

RESULTHOST IP ADDRESS172.30.1.33

SUBNET MASK255.255.255.252

NUMBER OF SUBNET BITS14 BITS

NUMBER OF SUBNET214 =16384

NUMBER OF HOST BITS PER SUBNET4

NUMBER OF USABLE HOSTS PER SUBNET22 =4-2 =2 host per subnet

SUBNET ADDRESS FOR THIS IP ADDRESS172.30.1.32

IP ADDRESS OF FIRST HOST ON THIS SUBNET172.30.1.33

IP ADDRESS OF LAST HOST ON THIS SUBNET 172.30.1.34

BROADCAST ADRESS FOR THIS SUBNET172.30.1.35

Problem 3IP ADDRESS192.192.10.234

SUBNET MASK255.255.255.0

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS19219210234

11000000110000000000101011101010

SUBNET MASK11111111111111111111111100000000

2552552550

Total number of Subnet Mask=11111111.11111111.11111111.00000000= 28 =256Number of Hosts Bits per Subnet=8 bitsNumber of Usable Hosts per Subnet=hosts =28 2 =256-2 =254NETWORK11000000110000000000101000000000

192192100

Network = 11000000.11000000.00001010.00000000 =192.192.10.234FIRST HOST =replace the number 0 to number 1 appearing on network addressNETWORK11000000110000000000101000000000

192192100

FIRST HOST11000000110000000000101000000001

192192101

LAST HOST = replace the number 0 to 1 at 32 on the host range subnet add except on the last bit remains 0NETWORK11000000110000000000101000000000

192192100

LAST HOST11000000110000000000101011111110

19219210254

BROADCAST = replace the number 0 to number 1 appearing on network addressNETWORK192192100

11000000110000000000101000000000

BROADCAST19219210255

11000000110000000000101011111111

RESULTHOST IP ADDRESS192.192.10.234

SUBNET MASK255.255.255.0

NUMBER OF SUBNET BITS8 BITS

NUMBER OF SUBNET28 =256

NUMBER OF HOST BITS PER SUBNET8

NUMBER OF USABLE HOSTS PER SUBNET28 =256-2 =254 host per subnet

SUBNET ADDRESS FOR THIS IP ADDRESS192.192.10.0

IP ADDRESS OF FIRST HOST ON THIS SUBNET192.192.10.1

IP ADDRESS OF LAST HOST ON THIS SUBNET 192.192.10.254

BROADCAST ADRESS FOR THIS SUBNET192.192.10.255

Problem 4IP ADDRESS172.17.99.71

SUBNET MASK255.255.0.0

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS172179971

10101100000100010110001101000111

SUBNET MASK11111111111111110000000000000000

25525500

Total number of Subnet Mask=11111111.11111111.00000000.00000000= 20 =1Number of Hosts Bits per Subnet=16 bitsNumber of Usable Hosts per Subnet=hosts =216 2 =65536-2 =65534

NETWORK10101100000100010000000000000000

1721700

Network = 10101100.00010001.00000000.00000000 =172.17.0.0FIRST HOST =replace the number 0 to number 1 appearing on network addressNETWORK10101100000100010000000000000000

1721700

FIRST HOST10101100000100010000000000000001

1721701

LAST HOST = replace the number 0 to 1 at 32 on the host range subnet add except on the last bit remains 0NETWORK10101100000100010000000000000000

1721700

LAST HOST10101100000100011111111111111110

172170254

BROADCAST = replace the number 0 to number 1 appearing on network addressNETWORK1721700

10101100000100010000000000000000

BROADCAST17217255255

10101100000100011111111111111111

RESULTHOST IP ADDRESS172.17.99.71

SUBNET MASK255.255.0.0

NUMBER OF SUBNET BITS0 BITS

NUMBER OF SUBNET20 =1

NUMBER OF HOST BITS PER SUBNET16 BITS

NUMBER OF USABLE HOSTS PER SUBNET216 =65536-2 =65534 host per subnet

SUBNET ADDRESS FOR THIS IP ADDRESS172.17.0.0

IP ADDRESS OF FIRST HOST ON THIS SUBNET172.17.0.1

IP ADDRESS OF LAST HOST ON THIS SUBNET 172.17.255.254

BROADCAST ADRESS FOR THIS SUBNET172.17.255.255

Problem 5IP ADDRESS192.168.3.219

SUBNET MASK255.255.255.0

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS1921683219

11000000101010000000001111011011

SUBNET MASK11111111111111111111111100000000

2552552550

Total number of Subnet Mask=11111111.11111111.11111111.00000000= 28 =256Number of Hosts Bits per Subnet=8 bitsNumber of Usable Hosts per Subnet=hosts =28 2 =256-2 =254

NETWORK11000000101010000000001100000000

19216830

Network = 11000000.10101000.00000011.00000000 =192.168.3.0FIRST HOST =replace the number 0 to number 1 appearing on network addressNETWORK11000000101010000000001100000000

19216830

FIRST HOST11000000101010000000001100000001

19216831

LAST HOST = replace the number 0 to 1 at 32 on the host range subnet add except on the last bit remains 0NETWORK11000000101010000000001100000000

19216830

LAST HOST11000000101010000000001111111110

1921683254

BROADCAST = replace the number 0 to number 1 appearing on network addressNETWORK19216830

11000000101010000000001100000000

BROADCAST1921683255

11000000101010000000001111111111

RESULTHOST IP ADDRESS192.168.3.219

SUBNET MASK255.255.255.0

NUMBER OF SUBNET BITS8 BITS

NUMBER OF SUBNET28 =256

NUMBER OF HOST BITS PER SUBNET8 BITS

NUMBER OF USABLE HOSTS PER SUBNET28 =256-2 =254 host per subnet

SUBNET ADDRESS FOR THIS IP ADDRESS192.168.3.0

IP ADDRESS OF FIRST HOST ON THIS SUBNET192.168.3.1

IP ADDRESS OF LAST HOST ON THIS SUBNET 192.168.3.254

BROADCAST ADRESS FOR THIS SUBNET192.17.3.255

Problem 6IP ADDRESS192.168.3.219

SUBNET MASK255.255.255.252

SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534 SUBNETTING Task 1: For a Given IP Address and Subnet Mask, Determine Subnet Information Host IP Address 172.25.114.250 Network Mask 255.255.0.0(/16) Subnet Mask 255.255.255.192(/26) Find: Number of subnet bits 10 Number of subnets 210 = 1024 subnet Number of Host Bits per Subnet 6 Number of usable Hosts per subnet 26= 64 2 = 62 bit per subnet Subnet Address for this IP Address 172.25.114.192 IP Address of first host on this Subnet 172.25.114.193 IP Address of last host on this Subnet 172.25.114.254 Broadcast Address for this Subnet 172.25.114.255 Step 1: Translate host IP Address and subnet mask into binary notation. 172 25 114 250 10101100 00011001 01110010 11111010 IP Address 11111111 11111111 11111111 11000000 Subnet Mask 10101100 00011001 01110010 11000000 Subnet Address 172 25 114 192 Subnet address for this IP address 172.25.114.192Identify the bit ranges for subnet and hosts. IP ADDRESS 172 25 114 250 10101110 00011001 01110010 11111010 SUBNET MASK 11111111 111111111 00000000 00000000 255 255 0 0 SUBNET MASK= 11111111.11111111.00000000.00000000 = Network : 2^8=256 Number of usable host bits per subnet =16 Bits Number os usable host per subnet =host : 2^6=65536-2=65534

IP ADDRESS1921683219

11000000101010000000001111011011

SUBNET MASK11111111111111111111111111111100

255255255252

Total number of Subnet Mask=11111111.11111111.11111111.11111100= 214 =16384Number of Hosts Bits per Subnet=2 bitsNumber of Usable Hosts per Subnet=hosts =22 2 =4-2 =2

NETWORK11000000101010000000001111011000

1921683216

Network = 11000000.10101000.00000011.00000000 =192.168.3.0FIRST HOST =replace the number 0 to number 1 appearing on network addressNETWORK11000000101010000000001111011000

1921683216

FIRST HOST11000000101010000000001111011001

1921683217

LAST HOST = replace the number 0 to 1 at 32 on the host range subnet add except on the last bit remains 0NETWORK11000000101010000000001111011000

1921683216

LAST HOST11000000101010000000001111011010

1921683218

BROADCAST = replace the number 0 to number 1 appearing on network addressNETWORK1921683216

11000000101010000000001111011000

BROADCAST1921683219

11000000101010000000001111011011

RESULTHOST IP ADDRESS192.168.3.219

SUBNET MASK255.255.255.252

NUMBER OF SUBNET BITS14 BITS

NUMBER OF SUBNET214 =16384

NUMBER OF HOST BITS PER SUBNET2 BITS

NUMBER OF USABLE HOSTS PER SUBNET22 =4-2 =2 host per subnet

SUBNET ADDRESS FOR THIS IP ADDRESS192.168.3.216

IP ADDRESS OF FIRST HOST ON THIS SUBNET192.168.3.217

IP ADDRESS OF LAST HOST ON THIS SUBNET 192.168.3.218

BROADCAST ADRESS FOR THIS SUBNET192.168.3.219