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
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