001
IP Addressing and Subnetting
Workbook Version 2.0
Instructor’s Edition
11111110
10010101
00011011
11010011
1010100
10001111100
1011100101011100
101100011101001
1011110100011010
00001010010110010
1001010101100111
1111010101000101
1101001101010011
001010010101010
1010101000110010
010101001011000
110101100011010
11010100001011
001010100110
1001010010
10000110
10011000
0101
Class A
Class B Class C Class D Class E
Class A Class B Class C
Class A Class B Class C
1 – 127
128 – 191 192 – 223 224 – 239 240 – 255
IP Address Classes
(Network 127 is reserved for loopback and internal testing)
Leading bit pattern 0
Leading bit pattern 10
Leading bit pattern 110
(Reserved for multicast)
(Reserved for experimental, used for research)
Private Address Space
10.0.0.0 to 10.255.255.255 172.16.0.0 to 172.31.255.255 192.168.0.0 to 192.168.255.255
Default Subnet Masks
255.0.0.0 255.255.0.0 255.255.255.0
Produced by: Robb Jones jonesr@careertech.net and/or Robert.Jones@fcps.org Frederick County Career & Technology Center Cisco Networking Academy
Frederick County Public Schools
Frederick, Maryland, USA
Special Thanks to Melvin Baker and Jim Dorsch
for taking the time to check this workbook for errors,
and to everyone who has sent in suggestions to improve the series.
00000000.00000000.00000000.00000000
Network . Host . Host . Host
10000000.00000000.00000000.00000000
Network . Network . Host . Host
11000000.00000000.00000000.00000000
Network . Network . Network . Host
Workbooks included in the series:
IP Addressing and Subnetting Workbooks ACLs – Access Lists Workbooks
VLSM Variable-Length Subnet Mask Workbooks
Instructors (and anyone else for that matter) please do not post the Instructors version on public websites. When you do this you are giving everyone else worldwide the answers. Yes, students look for answers this way. It also discourages others; myself included, from posting high quality materials.
Inside Cover
Binary To Decimal Conversion
128 64 32 16 8 4 2 1
Answers
Scratch Area
128 126
146
21 119
10010010 146 01110111 119 1 1 1 1 1 1 1 1 255 11000101 197 11110110 246 00010011 19 10000001 129 00110001 49 01111000 120 11110000 240 00111011 59 00000111 7
00011011 27 10101010 170 01101111 111 11111000 248 00100000 32 01010101 85 00111110 62 00000011 3 11101101 237 11000000 192
64 32 146
1
Decimal To Binary Conversion
Use all 8 bits for each problem
128 64 32 16 8 4 2 1 = 255
238 34 123
Scratch Area
11101110
238
-128 34
_________________________________________
110 -32 -64 2 46 -2 -32 0
00100010
_________________________________________
01111011
_________________________________________
00110010
14 -68
_________________________________________ 50
11111111
_________________________________________ 255
-24 -20
11001000
_________________________________________ 200
00001010
_________________________________________ 10
10001010
_________________________________________ 138
00000001
_________________________________________ 1
00001101
_________________________________________ 13
11111010
_________________________________________ 250
01101011
_________________________________________ 107
11100000
_________________________________________ 224
01110010
_________________________________________ 114
11000000
_________________________________________ 192
10101100
_________________________________________ 172
01100100
_________________________________________ 100
01110111
_________________________________________ 119
00111001
_________________________________________ 57
01100010
_________________________________________ 98
10110011
_________________________________________ 179
00000010
_________________________________________ 2 2
Address
10.250.1.1 150.10.15.0 192.14.2.0 148.17.9.1 193.42.1.1 126.8.156.0 220.200.23.1 230.230.45.58 177.100.18.4 119.18.45.0 249.240.80.78 199.155.77.56 117.89.56.45 215.45.45.0 199.200.15.0 95.0.21.90 33.0.0.0 158.98.80.0 219.21.56.0
Address Class Identification Class
__A___ __B___ __C___ __B___ __C___ __A___ __C___ __D___ __B___ __A___ __E___ __C___ __A___ __C___ __C___ __A___ __A___ __B___ __C___
3
Circle the network portion of these addresses:
177.100.18.4 119.18.45.0 209.240.80.78 199.155.77.56 117.89.56.45 215.45.45.0 192.200.15.0 95.0.21.90 33.0.0.0 158.98.80.0 217.21.56.0 10.250.1.1 150.10.15.0 192.14.2.0 148.17.9.1 193.42.1.1 126.8.156.0 220.200.23.1 4
Circle the host portion of these addresses:
10.15.123.50 171.2.199.31 198.125.87.177 223.250.200.222 17.45.222.45 126.201.54.231 191.41.35.112 155.25.169.227 192.15.155.2 123.102.45.254 148.17.9.155 100.25.1.1 195.0.21.98 25.250.135.46 171.102.77.77 55.250.5.5 218.155.230.14 10.250.1.1
Network & Host Identification
Network Addresses
Using the IP address and subnet mask shown write out the network address:
188.10.18.2 255.255.0.0
10.10.48.80 255.255.255.0
192.149.24.191 255.255.255.0
150.203.23.19 255.255.0.0
10.10.10.10 255.0.0.0
186.13.23.110 255.255.255.0
223.69.230.250 255.255.0.0
200.120.135.15 255.255.255.0
27.125.200.151 255.0.0.0
199.20.150.35 255.255.255.0
191.55.165.135 255.255.255.0
28.212.250.254 255.255.0.0
188 . 10 . 0 . 0
_____________________________
10 . 10 . 48 . 0
_____________________________
192 . 149 . 24 . 0
_____________________________
150 . 203 . 0 . 0
_____________________________
10 . 0 . 0 . 0
_____________________________
186 . 13 . 23 . 0
_____________________________
223 . 69 . 0 . 0
_____________________________
200 . 120 . 135 . 0
_____________________________
27 . 0 . 0 . 0
_____________________________
199 . 20 . 150 . 0
_____________________________
191 . 55 . 165 . 0
_____________________________
28 . 212 . 0 . 0
_____________________________
5
Host Addresses
Using the IP address and subnet mask shown write out the host address:
188.10.18.2 255.255.0.0
10.10.48.80 255.255.255.0
222.49.49.11 255.255.255.0
128.23.230.19 255.255.0.0
10.10.10.10 255.0.0.0
200.113.123.11 255.255.255.0
223.169.23.20 255.255.0.0
203.20.35.215 255.255.255.0
117.15.2.51 255.0.0.0
199.120.15.135 255.255.255.0
191.55.165.135 255.255.255.0
48.21.25.54 255.255.0.0
6
0 . 0 . 18 . 2
_____________________________
0 . 0 . 0 . 80
_____________________________
0 . 0 . 0 . 11
_____________________________
0 . 0 . 230 . 19
_____________________________
0 . 10 . 10 . 10
_____________________________
0 . 0 . 0 . 11
_____________________________
0 . 0 . 23 . 20
_____________________________
0 . 0 . 0 . 215
_____________________________
0 . 15 . 2 . 51
_____________________________
0 . 0 . 0 . 135
_____________________________
0 . 0 . 0 . 135
_____________________________
0 . 0 . 25 . 54
_____________________________
Default Subnet Masks
Write the correct default subnet mask for each of the following addresses:
177.100.18.4 119.18.45.0 191.249.234.191 223.23.223.109 10.10.250.1 126.123.23.1 223.69.230.250 192.12.35.105 77.251.200.51 189.210.50.1 88.45.65.35 128.212.250.254 193.100.77.83 125.125.250.1 1.1.10.50 220.90.130.45 134.125.34.9 95.250.91.99
255 . 255 . 0 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 255 . 0 . 0
_____________________________
255 . 255 . 255 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 255 . 255 . 0
_____________________________
255 . 255 . 255 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 255 . 0 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 255 . 0 . 0
_____________________________
255 . 255 . 255 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
255 . 255 . 255 . 0
_____________________________
255 . 255 . 0 . 0
_____________________________
255 . 0 . 0 . 0
_____________________________
7
ANDING With Default subnet masks
Every IP address must be accompanied by a subnet mask. By now you should be able to look at an IP address and tell what class it is. Unfortunately your computer doesn’t think that way. For your computer to determine the network and subnet portion of an IP address it must “AND” the IP address with the subnet mask.
Default Subnet Masks:
Class A Class B Class C
255.0.0.0 255.255.0.0 255.255.255.0
ANDING Equations:
1 AND 1 = 1 1 AND 0 = 0 0 AND 1 = 0 0 AND 0 = 0
Sample:
What you see… IPAddress:
192 . 100 . 10 . 33 What you can figure out in your head…
Address Class: Network Portion: Host Portion:
C
192 . 100 . 10 . 33 192 . 100 . 10 . 33
In order for you computer to get the same information it must AND the IP address with the subnet mask in binary.
Network Host
IPAddress: 11000000.01100100.00001010.00100001 (192.100. 10.33) Default Subnet Mask: 11111111.01111111.11111111.00000000 (255.255.255.0)
AND: 11000000.01100100.00001010.00000000 (192.100. 10 .0) ANDING with the default subnet mask allows your computer to figure out the network
portion of the address.
8
ANDING With Custom subnet masks
When you take a single network such as 192.100.10.0 and divide it into five smaller networks (192.100.10.16, 192.100.10.32, 192.100.10.48, 192.100.10.64, 192.100.10.80) the outside world still sees the network as 192.100.10.0, but the internal computers and routers see five smaller subnetworks. Each independent of the other. This can only be accomplished by using a custom subnet mask. A custom subnet mask borrows bits from the host portion of the address to create a subnetwork address between the network and host portions of an IP address. In this example each range has 14 usable addresses in it. The computer must still AND the IP address against the custom subnet mask to see what the network portion is and which subnetwork it belongs to.
IPAddress: 192 . 100 . 10 . 0 Custom Subnet Mask: 255.255.255.240
AddressRanges:
192.10.10.0 to 192.100.10.15 192.100.10.16 to 192.100.10.31 192.100.10.32 to 192.100.10.47 192.100.10.48 to 192.100.10.63 192.100.10.64 to 192.100.10.79 192.100.10.80 to 192.100.10.95 192.100.10.96 to 192.100.10.111 192.100.10.112 to 192.100.10.127 192.100.10.128 to 192.100.10.143 192.100.10.144 to 192.100.10.159 192.100.10.160 to 192.100.10.175 192.100.10.176 to 192.100.10.191 192.100.10.192 to 192.100.10.207 192.100.10.208 to 192.100.10.223 192.100.10.224 to 192.100.10.239 192.100.10.240 to 192.100.10.255
Network
(Range in the sample below)
Sub Network Host
IPAddress: 11000000.01100100.00001010.00100001(192.100. 10.33) CustomSubnetMask: 11111111.01111111.11111111.11110000(255.255.255.240)
AND: 11000000.01100100.00001010.00100000(192.100. 10 .32)
Four bits borrowed from the host portion of the address for the custom subnet mask.
The ANDING process of the four borrowed bits shows which range of IP addresses this particular address will fall into.
In the next set of problems you will determine the necessary information to determine the correct subnet mask for a variety of IP addresses.
9
How to determine the number of subnets and the number of hosts per subnet
Two formulas can provide this basic information:
Number of subnets = 2 s (Second subnet formula: Number of subnets = 2s – 2) Number of hosts per subnet = 2h – 2
Both formulas calculate the number of hosts or subnets based on the number of binary bits used. For example if you borrow three bits from the host portion of the address use the number of subnets formula to determine the total number of subnets gained by borrowing the threebits. Thiswouldbe23 or2x2x2=8subnets
To determine the number of hosts per subnet you would take the number of binary bits used in the host portion and apply this to the number of hosts per subnet formula If five bits are in the hostportionoftheaddressthiswouldbe25or2x2x2x2x2=32hosts.
When dealing with the number of hosts per subnet you have to subtract two addresses from the range. The first address in every range is the subnet number. The last address in every range is the broadcast address. These two addresses cannot be assigned to any device in the network which is why you have to subtract two addresses to find the number of usable addresses in each range.
For example if two bits are borrowed for the network portion of the address you can easily determine the number of subnets and hosts per subnets using the two formulas.
195. 223 . 50 . 0 0 0 0 0 0 0 0
The number of subnets created by borrowing 2 bits is 22 or 2 x 2 = 4 subnets.
The number of hosts created by leaving 6 bits is 26 – 2 or 2x2x2x 2x2x2=64-2=62 usable hosts per subnet.
What about that second subnet formula: Number of subnets = 2 s – 2
In some instances the first and last subnet range of addresses are reserved. This is similar to the first and last host addresses in each range of addreses.
The first range of addresses is the zero subnet. The subnet number for the zero subnet is also the subnet number for the classful subnet address.
The last range of addresses is the broadcast subnet. The broadcast address for the last subnet in the broadcast subnet is the same as the classful broadcast address.
10
Class C Address unsubnetted:
195. 223 . 50 . 0
195.223.50.0 to 195.223.50.255
Class C Address subnetted (2 bits borrowed): 195. 223 . 50 . 0 0 0 0 0 0
0 0
to 195.223.50.63 to 195.223.50.127 to 195.223.50.191 to 195.223.50.255
(Invalid range) (0) 195.223.50.0 (1) 195.223.50.64
(2) 195.223.50.128 (Invalid range) (3) 195.223.50.192
Notice that the subnet and broadcast addresses match.
The primary reason the the zero and broadcast subnets were not used had to do pirmarily with the broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255 addresses in the classful C address or just the 62 usable addresses in the broadcast range?
The CCNA and CCENT certification exams may have questions which will require you to determine which formula to use, and whehter or not you can use the first and last subnets. Use the chart below to help decide.
When to use which formula to determine the number of subnets
Use the 2s – 2 formula and don’t use the zero and broadcast ranges if…
Usethe2s formulaandusethezeroand broadcast ranges if…
Classful routing is used
Classless routing or VLSM is used
RIP version 1 is used
RIP version 2, EIGRP, or OSPF is used
The no ip subnet zero command is configured on your router
The ip subnet zero command is configured on your router (default setting)
No other clues are given
Bottom line for the CCNA exams; if a question does not give you any clues as to whether or not to allow these two subnets, assume you can use them.
This workbook has you use the number of subnets = 2s formula.
11
Custom Subnet Masks
Problem 1
Number of needed subnets 14
Number of needed usable hosts 14
Network Address 192.10.10.0
Address class ____C______
Default subnet mask ____2__5__5__._2__5__5__._2__5_5___._0______
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___ Total number of subnets _________1_6_________
Total number of host addresses _________1_6_________ Number of usable addresses _________1_4_________ Number of bits borrowed _________4__________
Show your work for Problem 1 in the space below. Number of
2561286432 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64128 256
Number of
192 . 10 . 10 . 0 0 0 0 0 0 0 0
Add the binary value 64 numbers to the left of the line to 32
-2 hosts.
create the custom subnet mask.
14
Subtract 2 for the number of usable hosts.
128 64 32 16 8 4 2 1 – Binaryvalues
128 16 Observe the total number of
+16 240
12
Custom Subnet Masks
Problem 2
Number of needed subnets 1000
Number of needed usable hosts 60
Network Address 165.100.0.0
Address class ____B______
Default subnet mask ____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___ Total number of subnets _______1_,0__2_4________
Total number of host addresses _________6_4_________ Number of usable addresses _________6_2_________ Number of bits borrowed _________1_0_________
Show your work for Problem 2 in the space below.
Number of Hosts –
. 256 128 64 32 16 8 Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
4 2
Number of
128 128 64 +64
32 168
192
64 Observe the total number of -2 hosts.
62 Subtract 2 for the number of usable hosts.
2 1
Add the binary value numbers to the left of the line to create the custom subnet mask.
2 +1 255
4
13
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Problem 3
/26 indicates the total number of bits used for the network and subnetwork portion of the address. All bits remaining belong to the host portion of the address.
Number of Hosts –
4 2
Custom Subnet Masks
Network Address 148.75.0.0 /26 B
Address class __________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 3 in the space below.
. 256 128 64 32 16 8
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
1,024
64
62
10
Add the binary value numbers to the left of the line to create the custom subnet mask.
48
128 128
64 32 16
2 +1 255
+64 192
64 Observe the total number of -2 hosts.
62 Subtract 2 for the number of usable hosts.
1024
-2 Subtract 2 for the total number of
2 1
subnets to get the usable number of 1,022 subnets.
14
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Custom Subnet Masks
Problem 4
Number of needed subnets 6
Number of needed usable hosts 30 Network Address 195.85.8.0
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
C
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
Show your work for Problem 5 in the space below. Number of
2561286432 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64128 256
Number of
195 . 85 . 8 . 0 0 0 0 0 0 0 0
128 64 +32 224
32 8 -2 -2 30 6
15
8
32
30
3
128 64 32 16 8 4 2 1 – Binaryvalues
Custom Subnet Masks
Problem 5
Number of needed subnets 6
Number of needed usable hosts 30
Network Address 210.100.56.0
Address class _______
Number of bits borrowed ___________________
C
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Show your work for Problem 4 in the space below. Number of
8
32
30
3
2561286432 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64128 256
Number of
210 . 100 . 56 . 0 0 0 0 0 0 0 0
128 64 32 16 8 4 2 1 – Binaryvalues
128
64 8 32
+32 -2 -2 224 6 30
16
Custom Subnet Masks
Problem 6
Number of needed subnets 126
Number of needed usable hosts 131,070 Network Address 118.0.0.0
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 6 in the space below.
A
255 . 0 . 0 . 0
Default subnet mask _______________________________
255 . 254. 0 . 0
Custom subnet mask _______________________________
128
131,072
131,070
7
Number of Hosts –
Number of
Subnets – 2 4 8 16 32 64128256 .
. 256128 64 32 16 8
4 2
. 128 64 32 16 8 4
118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32
2 1
128 64 32 168
4 +2 254
17
2 1
128 131,072 -2 -2 126 131,070
16 8 4
.
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
Custom Subnet Masks
Problem 7
Number of needed subnets 2000
Number of needed usable hosts 15
Network Address 178.100.0.0
Address class __________
Number of bits borrowed ___________________
Show your work for Problem 7 in the space below.
B
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
2,048
32
30
11
Number of Hosts –
. 256 128 64 32 16 8 4 2 Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32 168
Number of
24 18 255
2,048 32 -2 -2 +1 2,046 30
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Custom Subnet Masks
Problem 8
Number of needed subnets 3
Number of needed usable hosts 45
Network Address 200.175.14.0
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 8 in the space below.
Number of
Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values 200 . 175 . 14 . 0 0 0 0 0 0 0 0
128 4 64
+64 -2 -2 240 2 62
19
C
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
4
64
62
2
Number of 256 128 64 32 16 8 4 2 – Hosts
Custom Subnet Masks
Problem 9
Number of needed subnets 60
Number of needed usable hosts 1,000 Network Address 128.77.0.0
Address class _______
Number of bits borrowed ___________________
Show your work for Problem 9 in the space below.
B
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 252 . 0
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
64
1,024
1,022
6
Number of Hosts –
. 256 128 64 32 16 8 Subnets – 2 4 8 16 32 64 128 256.
Number of
4 2
1 . 128 64 32 16 8 4
0 . 0 0 0 0 0 0 0 0
1,024 -2 1,022
Binary values – 128 64 32 16 8 4 2 128 . 77 . 0 0 0 0 0 0 0
2 1
128 64 32 168
+4 -2 252 62
64
20
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Custom Subnet Masks
Problem 10
Number of needed usable hosts 60
Network Address 198.100.10.0 C
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 10 in the space below.
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
4
64
62
2
Number of 256 128 64 32 16 8 4 2 – Hosts
Number of
Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values 198 . 100 . 10 . 0 0 0 0 0 0 0 0
128 64 4 +64 -2 -2 192 62 2
21
Custom Subnet Masks
Problem 11
Number of needed subnets 250
Network Address 101.0.0.0 A
Address class _______
Total number of subnets ___________________
65,536
Total number of host addresses ___________________
Number of usable addresses ___________________
255 . 0 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 0 . 0
Custom subnet mask _______________________________
256
65,534
Number of Hosts –
Number of bits borrowed ___________________
Show your work for Problem 11 in the space below.
. 256128 64 32 16 8
4 2
8
Number of
Subnets – 2 4 8 1632 64128256 .
.
. 128 64 32 16 8 4
101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32
16 8 4
2 1
2 1
22
128 64 32 168
24
+1 255
256 65,536 -2 -2 254 65,534
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
Custom Subnet Masks
Problem 12
Number of needed subnets 5
Network Address 218.35.50.0 C
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 12 in the space below.
Number of
Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values 218 . 35 . 50 . 0 0 0 0 0 0 0 0
128
64 64 4
+32 -2 -2 224 62 2
23
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
8
32
30
3
Number of 256 128 64 32 16 8 4 2 – Hosts
Custom Subnet Masks
Problem 13
Number of needed usable hosts 25
Network Address 218.35.50.0 C
Address class _______
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
Show your work for Problem 13 in the space below.
Number of
Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values 218 . 35 . 50 . 0 0 0 0 0 0 0 0
128
64 8 32
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
8
32
30
3
Number of 256 128 64 32 16 8 4 2 – Hosts
24
+32 -2 -2 224 6 30
Custom Subnet Masks
Problem 14
Number of needed subnets 10
Network Address 172.59.0.0 B
Address class _______
Number of bits borrowed ___________________
Show your work for Problem 14 in the space below.
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 240 . 0
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
16
4,096
4,094
4
Number of Hosts –
. 256 128 64 32 16
8
4
2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16
172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32
4,096 -2 4,094
16 +16 -2 240 14
25
8
4
2
1
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Custom Subnet Masks
Problem 15
Number of needed usable hosts 50
Number of Hosts –
. 256 128 64 32 16 8 4 2
Network Address 172.59.0.0 B
Address class _______
Number of bits borrowed ___________________
Show your work for Problem 15 in the space below.
255 . 255 . 0 . 0
Default subnet mask _______________________________
255.255.255. 192
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
1,024
64
62
10
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32 168
4
64 1,024 +1+64-2 -2 255 192 62 1,022
2 128
26
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Custom Subnet Masks
Problem 16
Number of needed usable hosts 29
Network Address 23.0.0.0 A
Address class _______
Number of bits borrowed ___________________
255 . 0 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
Show your work for Problem 16 in the space below.
Number of usable addresses ___________________
524,288
32
30
19
Number of Hosts –
. 256128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 1632 64128256 . Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32
.
. 128 64 32 16 8 4
23 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 +32 224
32 -2 30
524,288 -2 524,286
27
16 8 4
2 1
2 1
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
Subnetting
Problem 1
Number of needed subnets 14
Number of needed usable hosts 14
Network Address 192.10.10.0
Address class ____C______
Default subnet mask ____2__5__5__._2__5__5__._2__5_5___._0______
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___ Total number of subnets _________1_6_________
Total number of host addresses _________1_6_________ Number of usable addresses _________1_4_________ Number of bits borrowed _________4__________
What is the 4th
subnet range? _1__9_2__._1_0__._1_0__._4_8____t_o____1_9__2__._1_0__.1__0__.6__3_______
What is the subnet number
for the 8th subnet? ___1_9_2___. _1_0__.__1_0__._1_1__2___
28
What is the subnet broadcast address for
the13thsubnet? ___1_9_2___._1_0__.__1_0__._2__0_7___
What are the assignable addresses for the 9th
subnet? _1_9__2_._1_0_._1_0__.1_2__9___t_o___1_9_2__.1_0__.1_0__.1__4_2___
Show your work for Problem 1 in the space below. Number of
25612864 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
Number of
192. 10 . 10 . 0 0 0 0 0 0 0 0
128 64 32 16 8 4 2 1 – Binary values
(1) 0
(2) 0
(3) 0
(4) 0
(5) 0 1 0 0 192.10.10.64 (6) 0 1 0 1 192.10.10.80 (7) 0 1 1 0 192.10.10.96 (8) 0 1 1 1 192.10.10.112
0 0 0 192.10.10.0 0 0 1 192.10.10.16 0 1 0 192.10.10.32 0 1 1 192.10.10.48
to 192.10.10.15 to 192.10.10.31 to 192.10.10.47 to 192.10.10.63 to 192.10.10.79 to 192.10.10.95 to 192.10.10.111 to 192.10.10.127 to 192.10.10.143 to 192.10.10.159 to 192.10.10.175 to 192.10.10.191 to 192.10.10.207 to 192.10.10.223 to 192.10.10.239 to 192.10.10.255
(9) 1
(10) 1
(11) 1
(12) 1
(13) 1 1 0 0 192.10.10.192 (14) 1 1 0 1 192.10.10.208 (15) 1 1 1 0 192.10.10.224 (16) 1 1 1 1 192.10.10.240
Custom subnet +16 mask 240
-2 Usable hosts -2 14 14
128 64 32
16 16
0 0 0 192.10.10.128 0 0 1 192.10.10.144 0 1 0 192.10.10.160 0 1 1 192.10.10.176
Usable subnets
The binary value of the last bit borrowed is the range. In this problem the range is 16.
The first address in each subnet range is the subnet number.
The last address in each subnet range is the subnet broadcast address.
29
Subnetting
Problem 2
Number of needed subnets 1000
Number of needed usable hosts 60
Network Address 165.100.0.0
Address class ____B______
Default subnet mask ____2__5__5__._2__5__5__._0___. _0_________
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___ Total number of subnets _______1_,0__2_4________
Total number of host addresses _________6_4_________ Number of usable addresses _________6_2_________ Number of bits borrowed _________1_0_________
What is the 15th
subnet range? _1__6_5__._1_0__0__._3__.1__2_8____t_o____1_6__5__._1_0__0__.3__._1_9__1___
What is the subnet number
forthe6thsubnet? ___1_6_5___._1_0__0__._1__._6__4____
30
What is the subnet broadcast address for
the6thsubnet? ___1_6_5___._1_0__0__._1__._1__2_7___
What are the assignable addresses for the 9th
subnet? _1_6__5_._1_0__0_._2_._1__t_o___1_6__5_._1_0_0__.0__.6__2______
Show your work for Problem 2 in the space below.
31
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
1024 512
65,536
32,768 16,384
8,192
4,096 2048
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
2 1
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
165.100.0.63
The first address in each subnet range is the subnet number.
1 1 . 0 0 165.100.3.0
1 1 . 0 1 165.100.3.64 1 1 . 1 0 165.100.3.128 1 1 . 1 1 165.100.3.192
to 165.100.3.63 to 165.100.3.127 to 165.100.3.191 to 165.100.3.255
The last address in each subnet range is the subnet broadcast address.
64 128 (2) (3) Usable -2 64 (4)
1 0 165.100.0.128 1 1 165.100.0.192
hosts 62 32 (5)
1 . 0 0 165.100.1.0
1 . 0 1 165.100.1.64 1 . 1 0 165.100.1.128 1 . 1 1 165.100.1.192
to 165.100.1.63 to 165.100.1.127 to 165.100.1.191 to 165.100.1.255
168 (6) 128 4 (7)
Custom +64
subnet mask 192 2 (8)
The binary value of the last bit borrowed is the range. In this problem the range is 64.
+1 (9) (10) 255 (11) (12) (13) (14) (15) (16)
1 0 . 0 0 165.100.2.0
1 0 . 0 1 165.100.2.64 1 0 . 1 0 165.100.2.128 1 0 . 1 1 165.100.2.192
to 165.100.2.63 to 165.100.2.127 to 165.100.2.191 to 165.100.2.255
(1023) 1 1 1 1 1 1 (1024) 1 1 1 1 1 1
1 1 . 1 0 165.100.255.128 to 165.100.255.191 1 1 . 1 1 165.100.255.192 to 165.100.255.255
(1)
. 0 165.100.0.0 1 165.100.0.64
to
to 165.100.0.127 to 165.100.0.191 to 165.100.0.255
Down
to
Subnetting
Problem 3
Number of needed subnets 2
Hint: It is possible to borrow one bit to create two subnets.
Network Address 195.223.50.0 C
Address class __________
Number of usable addresses ___________________
Number of bits
What is the 2nd subnet range?
What is the subnet number for the 2nd subnet?
What is the subnet broadcast address for the 1st subnet?
What are the assignable addresses for the 1st subnet?
32
255 . 255 . 255 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
Total number of host addresses ___________________
255 . 255 . 255 . 128
Custom subnet mask _______________________________
borrowed ___________________
195.223.50.128
________________________
195.223.50.127
________________________
2
128
126
1
195.223.50.128 – 195.223.50.255
_______________________________________________
195.223.50.1 – 195.223.50.126
______________________________________
Number of
Show your work for Problem 3 in the space below. Number of
25612864 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values
195. 223 . 50 . 0 0 0 0 0 0 0 0
(1) 0 195.223.50.0 to 195.223.50.127 (2) 1 195.223.50.128 to 195.223.50.255
33
Subnetting
Problem 4
Number of needed subnets 750
34
Network Address 190.35.0.0 B
Address class __________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
What is the 15th
subnet range? _______________________________________________
What is the subnet number for the 13th subnet?
What is the subnet broadcast address for the 10th subnet?
What are the assignable addresses for the 6th subnet?
190.35.3.128 to 190.35.3.191 190.35.3.0
________________________
190.35.2.127
________________________
1,024
64
62
10
190.35.1.65 to 190.35.1.126
______________________________________
Show your work for Problem 4 in the space below.
35
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
190 . 35 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
64 -2 62
128 (14) 11 .
128 64
(1) . (2)
(3)
(4)
0 190.35.0.0 to 1 190.35.0.64 to 1 0 190.35.0.128 to 1 1 190.35.0.192 to 0 0 190.35.1.0 to 0 1 190.35.1.64 to 1 0 190.35.1.128 to 1 1 190.35.1.192 to 0 0 190.35.2.0 to 0 1 190.35.2.64 to 1 0 190.35.2.128 to 1 1 190.35.2.192 to 0 0 190.35.3.0 to 0 1 190.35.3.64 to 1 0 190.35.3.128 to 1 1 190.35.3.192 to
190.35.0.63 190.35.0.127 190.35.0.191 190.35.0.255 190.35.1.63 190.35.1.127 190.35.1.191 190.35.1.255 190.35.2.63 190.35.2.127 190.35.2.191 190.35.2.255 190.35.3.63 190.35.3.127 190.35.3.191 190.35.3.255
(5) 1 . 32 (6) 1 .
168
(7) 1 . (8) 1 . (9) 10 . (10) 1 0 . (11) 1 0 . (12) 1 0 . (13) 11 .
+1 252
24
+64 (15) 11 . 252 (16) 11 .
2 1
Subnetting
Problem 5
Number of needed usable hosts 6
Network Address 126.0.0.0 A
Address class __________
Total number of host addresses ________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
What is the 2nd
subnet range? _______1_2__6__.0__._0__.8_____t_o___1__2__6__.0__._0_._1_5_________
What is the subnet number
for the 5th subnet? _____1_2__6_._0_._0_._3_2_________
36
What is the subnet broadcast address for
the 7th subnet? _____1_2__6_._0_._0_._5_5_________
What are the assignable addresses for the 10th
255 . 0 . 0 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 255 . 248
Custom subnet mask _______________________________
subnet? ______1_2_6__.0__.0__.7_3____t_o___1_2__6_.0__.0__.7__8_____
2,097,152
8
6
21
Show your work for Problem 5 in the space below.
37
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
Number of
Number of – Hosts
. 256 128 64 32 16 8 4 2
Subnets- 248163264128256.
Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32 16 8 4
.
2 1
126. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32 168
128 64 32 16 +8 248
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)
1
0 126.0.0.0 1 126.0.0.8
to 126.0.0.7
to 126.0.0.15 to 126.0.0.23 to 126.0.0.31 to 126.0.0.39 to 126.0.0.47 to 126.0.0.55 to 126.0.0.63 to 126.0.0.71 to 126.0.0.79 to 126.0.0.87 to 126.0.0.95 to 126.0.0.103 to 126.0.0.111 to 126.0.0.119 to 126.0.0.127
+1 255
11 100 100
1 126.0.0.72 0 126.0.0.80 1 126.0.0.88 0 126.0.0.96
8 -26
101 101 110 1 1 0 1 1 1 1 1 1
1 126.0.0.104 0 126.0.0.112
24
11
0 126.0.0.48 1 126.0.0.56 0 126.0.0.64
. 128 64 32 16 8 4 2 1
1 10 10
0 126.0.0.16 1 126.0.0.24 0 126.0.0.32 1 126.0.0.40
1 126.0.0.120
Subnetting
Problem 6
Number of needed subnets 10
Network Address192.70.10.0 C
Address class__________
Total number of subnets___________________
Total number of host addresses___________________
Number of usable addresses___________________
Number of bits
What is the 9th subnet range?
What is the subnet number for the 4th subnet?
What is the subnet broadcast address for the 12th subnet?
What are the assignable addresses for the 10th subnet?
38
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 240
Custom subnet mask_______________________________
borrowed___________________
192.70.10.48
________________________
192.70.10.191
________________________
16
16
14
4
192.70.10.128 to 192.70.10.143
_______________________________________________
192.70.10.145 to 192.70.10.158
______________________________________
Show your work for Problem 6 in the space below. Number of
256 128 64 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
Number of
192 . 70 . 10 . 0 0 0 0 0 0 0 0
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)
128 64 32 16 8 4 2 1 – Binary values
to 192.70.10.15 to 192.70.10.31 to 192.70.10.47 to 192.70.10.63 to 192.70.10.79 to 192.70.10.95 to 192.70.10.111 to 192.70.10.127 to 192.70.10.143 to 192.70.10.159 to 192.70.10.175 to 192.70.10.191 to 192.70.10.0207 to 192.70.10.223 to 192.70.10.239 to 192.70.10.255
0 192.70.10.0 1 192.70.10.16
1 0 192.70.10.32
1 1 192.70.10.48 10 0 192.70.10.64 1 0 1 192.70.10.80 11 0 192.70.10.96 1 1 1 192.70.10.112
1 0 0 0 192.70.10.128 1 0 0 1 192.70.10.144 1 0 1 0 192.70.10.160 1 0 1 1 192.70.10.176 1 1 0 0 192.70.10.192 1 1 0 1 192.70.10.208 1 1 1 0 192.70.10.224 1 1 1 1 192.70.10.240
128 16
+64 -2 240 14
39
Problem 7
What is the subnet number
for the 6th subnet? ________________________
40
What is the subnet broadcast address for
10.1.255.255
Subnetting
Network Address 10.0.0.0 /16 Address class ____A______
Default subnet mask ____2__5__5__._0___. _0__._0_____________ Custom subnet mask ____2__5__5__._2__5__5__._0__.__0_________
Total number of subnets _________2_5__6_______ Total number of host addresses _________6_5__,5__3_6____
Number of usable addresses _________6_5__,5__3_4____ Number of bits borrowed _________8__________
What is the 11th
subnet range? _______________________________________________
10.10.0.0 to 10.10.255.255 10.5.0.0
the 2nd subnet? ________________________
What are the assignable addresses for the 9th
10.8.0.1 to 10.8.255.254
subnet? ______________________________________
Show your work for Problem 7 in the space below.
41
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
65,536 -2 65,534
Number of
Number of Hosts –
. 256 128 64 32
16 8
4 2
Subnets – 2 4 8 1632 64128256 . Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32
.
16 8 4
10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32 168
(5) 1 (6) 1 (7) 1 (8) 1 (9) 10 (10) 10 (11) 10 (12) 10 (13) 11 (14) 11 (15) 11 (16) 1 1
0 0 0 1 1 0 1 1 0 0 0 1
+1 255
24
1 0 1 1 0 0 0 1 1 0 1 1
(1) 0
(2) 1
(3) 1 0
(4) 1 1
10.0.0.0 to 10.1.0.0 to 10.2.0.0 to 10.3.0.0 to 10.4.0.0 to 10.5.0.0 to 10.6.0.0 to 10.7.0.0 to 10.8.0.0 to 10.9.0.0 to 10.10.0.0 to 10.11.0.0 to 10.12.0.0 to 10.13.0.0 to 10.14.0.0 to 10.15.0.0 to
10.0.255.255 10.1.255.255 10.2.255.255 10.3.255.255 10.4.255.255 10.5.255.255 10.6.255.255 10.7.255.255 10.8.255.255 10.9.255.255 10.10.255.255 10.11.255.255 10.12.255.255 10.13.255.255 10.14.255.255 10.15.255.255
2 1
8 4
2 1
. 128 64 32 16
Subnetting
Problem 8
Number of needed subnets 5
Network Address 172.50.0.0 B
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
What is the 4th
subnet range? _1__7_2__._5__0__.9__6__.0_____t_o____1_7__2__.5__0__._1_2__7__.2__5__5___
What is the subnet number
for the 5th subnet? ____1_7__2_._5_0__.1__2_8_._0_______
42
What is the subnet broadcast address for
the 6th subnet? ____1_7__2_._5_0__.1__9_1_._2_5__5____
What are the assignable addresses for the 3rd
255 . 255 . 0 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 224 . 0
Custom subnet mask _______________________________
subnet? _1_7__2_._5_0__.6__4_._1___t_o___1_7_2__.5__0_._9_5__.2__5_4____
8
8,192
8,190
3
Show your work for Problem 8 in the space below.
43
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
8,192 224 -2
8,190
128 64 +32
Number of Hosts –
. 256 128 64 32 16 8 4 2
Number of
Subnets – 2 4 8
16 32 64 128 256.
16 8 4 2 1 . 128 64 32 16 8 4 2 1
Binary values – 128 64 32
172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) (2) (3) (4) (5) (6) (7) (8)
1
0 172.50.0.0 to 1 172.50.32.0 to 0 172.50.64.0 to 1 172.50.96.0 to 0 172.50.128.0 to 1 172.50.160.0 to 0 172.50.192.0 to 1 172.50.224.0 to
172.50.31.255 172.50.63.255 172.50.95.255 172.50.127.255 172.50.159.255 172.50.191.255 172.50.223.255 172.50.255.255
1 1 0 1 0
1 1 1 1
Subnetting
Problem 9
Number of needed usable hosts 28
Network Address 172.50.0.0 B
Address class __________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits
What is the 2nd subnet range?
What is the subnet number for the 10th subnet?
What is the subnet broadcast address for the 4th subnet?
What are the assignable addresses for the 6th subnet?
44
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
borrowed ___________________
172.50.1.32
________________________
172.50.0.127
________________________
2,048
32
30
11
172.50.0.32 to 172.50.0.63
_______________________________________________
172.50.0.161 to 172.50.0.190
______________________________________
Show your work for Problem 9 in the space below.
45
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 64 32 168
128 64 +32 224
(1) . 0 172.50.0.0 to (2) 1 172.50.0.32 to (3) 1 0 172.50.0.64 to (4) 1 1 172.50.0.96 to (5) 1 0 0 172.50.0.128 to (6) 1 0 1 172.50.0.160 to (7) 1 1 0 172.50.0.192 to (8) 1 1 1 172.50.0.224 to (9) 1 . 0 0 0 172.50.1.0 to (10) 1 . 0 0 1 172.50.1.32 to (11) 1 . 0 1 0 172.50.1.64 to (12) 1 0 1 1 172.50.1.96 to (13) 1 . 1 0 0 172.50.1.128 to (14) 1 . 1 0 1 172.50.1.160 to (15) 1 . 1 1 0 172.50.1.192 to (16) 1 . 1 1 1 172.50.1.224 to
172.50.0.31 172.50.0.63 172.50.0.95 172.50.0.127 172.50.0.159 172.50.0.191 172.50.0.223 172.50.0.255 172.50.1.31 172.50.1.63 172.50.1.95 172.50.1.127 172.50.1.159 172.50.1.191 172.50.1.223 172.50.1.255
2 +1 252
4
32 -2 30
2 1
Subnetting
Problem 10
Number of needed subnets 45
Network Address 220.100.100.0 C
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
What is the 5th
subnet range? _2__2_0__._1_0_0__._1_0__0_._1_6____t_o___2_2__0__.1__0_0__.1__0_0__._1_9____
What is the subnet number
for the 4th subnet? _____2_2_0__.1_0__0_.1__0_0_._1_2_____
46
What is the subnet broadcast address for
the 13th subnet? _____2_2_0__.1_0__0_.1__0_0_._5_1_____
What are the assignable addresses for the 12th
255 . 255 . 255 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 255 . 252
Custom subnet mask _______________________________
subnet? __2_2__0_.1_0__0_.1_0__0_._4_5___t_o__2__2_0_._1_0_0_._1_0_0_._4_6__
64
4
2
6
Show your work for Problem 10 in the space below.
47
128 64 32 168
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)
1
0 220.100.100.0 to 220.100.100.3 1 220.100.100.4 to 220.100.100.7
+4 252
11
4 -2
1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1
1 220.100.100.52 to 220.100.100.55 0 220.100.100.56 to 220.100.100.59 1 220.100.100.60 to 220.100.100.63
128643216 8 220 . 100 . 100 . 0 0 0 0 0
4 2 1-Binaryvalues 0 0 0
Number of 256 128 64 32 16 8 4 2 – Hosts
Number of
Subnets – 2 4 8 16 32 64 128 256
1 10 10
0 220.100.100.8 to 220.100.100.11 1 220.100.100.12 to 220.100.100.15 0 220.100.100.16 to 220.100.100.19 1 220.100.100.20 to 220.100.100.23 0 220.100.100.24 to 220.100.100.27
11 1 0 0 1 0 0
1 220.100.100.28 to 220.100.100.31 0 220.100.100.32 to 220.100.100.35 1 220.100.100.36 to 220.100.100.39 0 220.100.100.40 to 220.100.100.43 1 220.100.100.44 to 220.100.100.47 0 220.100.100.48 to 220.100.100.51
Subnetting
Problem 11
Number of needed usable hosts 8,000
48
Network Address 135.70.0.0 B
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
255 . 255 . 0 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 224 . 0
Custom subnet mask _______________________________
What is the 6th
subnet range? _______________________________________________
8
8,192
8,190
3
135.70.160.0 to 135.70.191.255 135.70.192.0
What is the subnet number
for the 7th subnet? ________________________
What is the subnet broadcast address for
135.70.95.255
the 3rd subnet? ________________________
What are the assignable addresses for the 5th
135.70.128.1 to 135.70.159.254
subnet? ______________________________________
Show your work for Problem 11 in the space below.
49
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
135 . 70 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) (2) (3) (4) (5) (6) (7) (8)
1
0 135.70.0.0 to 1 135.70.32.0 to 0 135.70.64.0 to 1 135.70.96.0 to 0 135.70.128.0 to 1 135.70.160.0 to 0 135.70.192.0 to 1 135.70.224.0 to
135.70.31.255 135.70.63.255 135.70.95.255 135.70.127.255 135.70.159.255 135.70.191.255 135.70.223.255 135.70.255.255
1 1 0 1 0
1 1 1 1
8,192 -2 8,190
128 64 +32 224
2 1
Subnetting
Problem 12
Number of needed usable hosts 45
Network Address 198.125.50.0 C
Address class __________
Total number of subnets ___________________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits
What is the 2nd subnet range?
What is the subnet number for the 2nd subnet?
What is the subnet broadcast address for the 4th subnet?
What are the assignable addresses for the 3rd subnet?
50
255 . 255 . 255 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
borrowed ___________________
198.125.50.64
________________________
198.125.50.255
________________________
4
64
62
2
198.125.50.64 to 98.125.50.127
_______________________________________________
198.125.50.129 to 198.125.50.190
______________________________________
Show your work for Problem 12 in the space below. Number of
– Binary values
198.125.50.63 198.125.50.127 198.125.50.191 198.125.50.255
256 128 64 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
Number of
198 . 125 . 50 . 0 0 0 0 0 0 0 0
(1) 0 198.125.50.0 to
(2) 1 198.125.50.64 to
(3) 1 0 198.125.50.128 to (4) 1 1 198.125.50.192 to
128 64 +64 -2 192 62
128 64 32 16 8 4 2 1
51
Problem 13
Subnetting
Network Address 165.200.0.0 /26 B
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
255 . 255 . 0 . 0
Default subnet mask _______________________________
255 . 255 . 255 . 192
Custom subnet mask _______________________________
Total number of subnets ___________________
What is the 10th
subnet range? _______________________________________________
1,024
64
62
10
165.200.2.64 to 165.200.2.127 165.200.2.128
What is the subnet number
for the 11th subnet? ________________________
What is the subnet broadcast address for
the 1023rd subnet? ________________________ What are the assignable
addresses for the 1022nd 165.200.255.65 to 165.200.255.126 subnet? ______________________________________
52
165.200.255.191
Show your work for Problem 13 in the space below.
53
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
165 . 200 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
62
(13)
1 1 . 0 0 1 1 . 0 1 1 1 . 1 0 1 1 . 1 1
(2) 1 64 (3) 1 0
128
32 168
(4) 1 1 (5) 1 0 0 (6) 1 0 1 (7) 1 1 0
24
+1 (8) 1 1 1 252 (9) 1 0 . 0 0
(10) 1 0 . 0 1 64 (11) 1 0 . 1 0
-2 128
(12) 1 0 1 1
+64
252 (14)
(1021) 1 1 1 1 1 1 1 1.0 1 (1022) 1 1 1 1 1 1 1 1 . 1 0 (1023) 11111111.11
165.200.255.64 165.200.155.128 165.200.255.192
to 165.200.255.127 to 165.200.255.191 to 165.200.255.255
(1) . 0
165.200.0.0 165.200.0.64 165.200.0.128 165.200.0.192 165.200.1.0 165.200.1.64 165.200.1.128 165.200.1.192 165.200.2.0 165.200.2.64 165.200.2.128 165.200.2.192 165.200.3.0 165.200.3.64 165.200.3.128 165.200.3.192
to 165.200.0.63 to 165.200.0.127 to 165.200.0.191 to 165.200.0.255 to 165.200.1.63 to 165.200.1.127 to 165.200.1.191 to 165.200.1.255 to 165.200.2.63 to 165.200.2.127 to 165.200.2.191 to 165.200.2.255 to 165.200.3.63 to 165.200.3.127 to 165.200.3.191 to 165.200.3.255
(15) (16)
2 1
Subnetting
Problem 14
Number of needed usable hosts 16
Network Address 200.10.10.0 C
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
What is the 7th 200.10.10.192 to 200.10.10.223 subnet range? _______________________________________________
What is the subnet broadcast address for
What is the subnet number
for the 5th subnet? ________________________
54
What are the assignable addresses for the 6th
200.10.10.161 to 200.10.10.190
255 . 255 . 255 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 255 . 224
Custom subnet mask _______________________________
200.10.10.128 200.10.10.127
the 4th subnet? ________________________
subnet? ______________________________________
8
32
30
3
Show your work for Problem 14 in the space below. Number of
256 128 64 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
Number of
200 . 10 . 10 . 0 0 0 0 0 0 0 0
128 64 32 16 8 4 2 1 – Binary values
128
64 32
+32 -2 224 30
(1) 0 200.10.10.0 to (2) 1 200.10.10.32 to (3) 1 0 200.10.10.64 to (4) 1 1 200.10.10.96 to (5) 10 0 200.10.10.128 to (6) 10 1 200.10.10.160 to (7) 11 0 200.10.10.192 to (8) 11 1 200.10.10.224 to
200.10.10.31 200.10.10.63 200.10.10.95 200.10.10.127 200.10.10.159 200.10.10.191 200.10.10.223 200.10.10.255
55
Problem 15
56
Subnetting
Network Address 93.0.0.0 \19 A
Address class __________
Total number of host addresses ___________________
Number of usable addresses ___________________
Number of bits borrowed ___________________
255 . 0 . 0 . 0
Default subnet mask _______________________________
Total number of subnets ___________________
255 . 255 . 224 . 0
Custom subnet mask _______________________________
What is the 15th
subnet range? _______________________________________________
2,048
8,192
8,190
11
93.1.192.0 to 93.1.223.255 93.1.0.0
What is the subnet number
for the 9th subnet? ________________________
What is the subnet broadcast address for
93.0.223.255
the 7th subnet? ________________________
What are the assignable addresses for the 12th
93.1.96.1 to 93.1.127.254
subnet? ______________________________________
Show your work for Problem 15 in the space below.
57
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
128 64 32
(1) (2) (3)
.
0 93.0.0.0 to 1 93.0.32.0 to 1 0 93.0.64.0 to 1 1 93.0.96.0 to 1 0 0 93.0.128.0 to 1 0 1 93.0.160.0 to 1 1 0 93.0.192.0 to 1 1 1 93.0.224.0 to 0 0 0 93.1.0.0 to 0 0 1 93.1.32.0 to 0 1 0 93.1.64.0 to 0 1 1 93.1.96.0 to 1 0 0 93.1.128.0 to 1 0 1 93.1.160.0 to 1 1 0 93.1.192.0 to
93.0.31.255 93.0.63.255 93.0.95.255 93.0.127.255 93.0.159.255 93.0.191.255 93.0.223.255 93.0.255.255 93.1.31.255 93.1.63.255 93.1.95.255 93.1.127.255 93.1.159.255 93.1.191.255 93.1.223.255
Number of Hosts –
. 256128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 1632 64128256 . Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32
.
16 8 4
93. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
168 (4) 4 (5) 2 (6)
+1 (7) 255 (8)
(9) 1 128 (10) 1 64 (11) 1 +32 (12) 1 224 (13) 1 (14) 1
. . .
8,192 (15) 1 -2
. . .
8,190
2 1
2 1
. 128 64 32 16 8 4
Practical Subnetting 1
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 100% growth in both areas. Circle each subnet on the graphic and answer the questions below.
IP Address 172.16.0.0
F0/0
Marketing 24 Hosts
Router A
S0/0/0
S0/0/1 F0/0
Reasearch 60 Hosts
F0/1
Management 15 Hosts
Router B
Minimum number of subnets needed
Extra subnets required for 100% growth
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses in the largest subnet group
4
Number of addresses needed for 100%growthinthelargestsubnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
60
Address class
B
Custom subnet mask
255.255.224.0
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
58
IP address range for Research
IP address range for Marketing IP address range for Management
IP address range for Router A to Router B serial connection
___1_7_2__.1_6_._0_._0__t_o_1_7_2__.3_1__.2_5__5___ __1__7_2_.1_6__.3_2__.0__t_o__1_7_2_._6_3_._2_5__5__ __1_7__2_.1_6__.6_4__.0__t_o__1_7_2_._9_5_._2_5__5__
__1_7_2__.1_6_._9_6_._0__t_o_1_7_2__.1_2__7_.2_5__5__
_____________________________
_____________________________
_________
4
_+________ 8
_=________ 60
_________
_+________ 120
Show your work for Practical Subnetting 1 in the space below.
59
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
4 x1.04
(1) 0 (2) 1 (3) 1 0 (4) 1 1 (5) 10 0 (6) 10 1 (7) 11 0 (8) 11 1
172.16.0.0 172.16.32.0 172.16.64.0 172.16.96.0 172.16.128.0 172.16.160.0 172.16.192.0 172.16.224.0
to 172.16.31.255 to 172.16.63.255 to 172.16.95.255 to 172.16.127.255 to 172.16.159.255 to 172.16.191.255 to 172.16.223.255 to 172.16.255.255
60 x1.0 60
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
172 . 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
2 1
Practical Subnetting 2
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 135.126.0.0
F0/0
Science Lab 10 Hosts
S0/0/0
S0/0/1
Address class
S0/0/1
F0/0
S0/0/0 F0/1
English Department 15 Hosts
Router A
Router B
Router C
F0/1
Tech Ed Lab 20 Hosts
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 30% growth
255.255.255.224
(Round up to the next whole number)
_________
Total number of subnets needed
=
7
Number of host addresses in the largest subnet group
20
Number of addresses needed for 30% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
_________ _+___6_____
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
60
IP address range for Tech Ed _1_3__5_._1_2_6_._0_.0___to__1_3_5__.1_2__6_.0__.3_1__ IP address range for English _1_3_5_._1_2_6_._0_._3_2__t_o__1_3_5_._1_2_6__.0_._6_3_ IP address range for Science _1_3_5_._1_2_6_._0_._6_4__t_o__1_3_5_._1_2_6_._0_._9_5_
IP address range for Router A
to Router B serial connection 1_3__5_.1_2__6_.0__.9_6__t_o__1_3_5__.1_2__6_.0__.1_2_7_
IP address range for Router A
to Router B serial connection1_3_5_._1_2_6_._0_._1_2_8__t_o__1_3_5_._1_2_6_._0_.1__5_9
+
2
_________
26
B
_____________________________
_____________________________
5
_________
Show your work for Problem 2 in the space below.
61
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(Round up to 2)
5
(1) . (2)
(3) 1
(4) 1
(5) 10
(6) 10
(7) 11
(8) 11
0 135.126.0.0 to 1 135.126.0.32 to 0 135.126.0.64 to 1 135.126.0.96 to 0 135.126.0.128 to 1 135.126.0.160 to 0 135.126.0.192 to 1 135.126.0.224 to 0 135.126.1.0 to 1 135.126.1.32 to 0 135.126.1.64 to 1 135.126.1.96 to 0 135.126.1.128 to 1 135.126.1.160 to 0 135.126.1.192 to 1 135.1261.224 to
135.126.0.31 135.126.0.63 135.126.0.95 135.126.0.127 135.126.0.159 135.126.0.191 135.126.0.223 135.126.0.255 135.126.1.31 135.126.1.63 135.126.1.95 135.126.1.127 135.126.1.159 135.126.1.191 135.126.1.223 135.126.1.255
x.3
1.5
20 x.36
(9) 1 . (10) 1 . (11) 1 . (12) 1 (13) 1 . (14) 1 . (15) 1 . (16) 1 .
00 0 0 0 1 0 1 1 0 1 0 1 1 1 1
2 1
Practical Subnetting 3
Based on the information in the graphic shown, design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the questions below.
F0/0
Router A
F0/1
Marketing 50 Hosts
IP Address 172.16.0.0 S0/0/1
Sales 185 Hosts
Administrative 30 Hosts
F0/0
S0/0/0
Router B
Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 25% growth
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses in the largest subnet group
_____________________________
62
Number of addresses needed for
47
25% growth in the largest subnet (Round up to the next whole number)
_+________ 232
Total number of address needed for the largest subnet
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales
IP address range for Marketing
IP address range for Administrative
IP address range for Router A to Router B serial connection
172.16.0.0 to 172.16.0.255
_________
+1 _________
=5 _________
185
_________
B
255.255.255.0
_____________________________
4
_____________________________
172.16.1.0 to 172.16.1.255
_____________________________
172.16.2.0 to 172.16.2.255
_____________________________
172.16.3.0 to 172.16.3.255
_____________________________
Show your work for Problem 3 in the space below.
63
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
172. 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
56.25 (Round up to 57)
4 (4)
1 1
x..25 (5) 1 (6)
225
(7) (8) (9)1. (10)1.
x.25
(1) . (2) (3)
0 172.16.0.0 to 1 172.16.1.0 to 1 0 172.16.2.0 to 1 1 172.16.3.0 to 0 0 172.16.4.0 to 0 1 172.16.5.0 to 11 0 172.16.6.0 to 11 1 172.16.7.0 to 0 0 0 172.16.8.0 to 0 0 1 172.16.9.0 to (11)101 0 172.16.10.0 to (12)1. 01 1 172.16.11.0 to (13)1. 1 0 0 172.16.12.0 to (14)1. 1 0 1 172.16.13.0 to (15)1. 11 0 172.16.14.0 to (16)1. 11 1 172.16.15.0 to
1172.16.0.255 1172.16.1.255 1172.16.2.255 1172.16.3.255 1172.16.4.255 1172.16.5.255 1172.16.6.255 1172.16.7.255 1172.16.8.255 1172.16.9.255 1172.16.10.255 1172.16.11.255 1172.16.12.255 1172.16.13.255 1172.16.14.255 1172.16.15.255
2
1
Practical Subnetting 4
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 70% growth in all areas. Circle each subnet on the graphic and answer the questions below.
F0/0
S0/0/0
S0/0/1
IP Address 135.126.0.0
Router A
S0/0/1
F0/0
S0/0/0
F0/1 Router C F0/0
Washington D.C. 220 Hosts
Router B
Dallas 150 Hosts
New York 325 Hosts
Minimum number of subnets needed
Extra subnets required for 70% growth
5
Address class
B
Custom subnet mask
255.255.240.0
(Round up to the next whole number)
_________
Total number of subnets needed
=
9
Number of host addresses in the largest subnet group
325
Number of addresses needed for 70% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
_________ _+__2_2__8___
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Washington D. C. _____________________________
64
IP address range for Router A135.126.48.0 to 135.126.63.255 to Router B serial connection _____________________________
IP address range for Router A 135.126.64.0 to 135.126.79.255 to Router C serial connection _____________________________
_____________________________
_____________________________
_________
+
4
_________
553
135.126.0.0 to 135.126.15.255
IP address range for New York _____________________________
135.126.16.0 to 135.126.31.255
135.126.32.0 to 135.126.47.255
IP address range for Dallas _____________________________
Show your work for Problem 4 in the space below.
65
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) . (2)
(3) 1
(4) 1
(5) 1 0
(6) 1 0
(7) 1 1
(8) 1 1
0 135.126.0.0 to 1 135.126.16.0 to 0 135.126.32.0 to 1 135.126.48.0 to 0 135.126.64.0 to 1 135.126.80.0 to 0 135.126.96.0 to 1 135.126.112.0 to 0 135.126.128.0 to 1 135.126.144.0 to 0 135.126.160.0 to 1 135.126.176.0 to 0 135.126.192.0 to 1 135.126.208.0 to 0 135.126.224.0 to 1 135.126.240.0 to
135.126.15.255 135.126.31.255 135.126.47.255 135.126.63.255 135.126.79.255 135.126.95.255 135.126.111.255 135.126.127.255 135.126.143.255 135.126.159.255 135.126.175.255 135.126.191.255 135.126.207.255 135.126.223.255 135.126.239.255 135.126.1255.255
(9) 1 . (10) 1 . (11) 1 . (12) 1 (13) 1 . (14) 1 . (15) 1 . (16) 1 .
0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
2 1
Practical Subnetting 5
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 210.15.10.0
F0/1
Science Room 10 Hosts
F0/0
English classroom 15 Hosts
Tech Ed Lab 18 Hosts
Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 100% growth
(Round up to the next whole number)
Total number of subnets needed
_____________________________
Number of host addresses in the largest subnet group
30
Number of addresses needed for
30
100% growth in the largest subnet (Round up to the next whole number)
_+________
Total number of address needed for the largest subnet
60
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router F0/0 Port _____________________________
IP address range for Router F0/1 Port _____________________________
66
_________
C
255.255.255.192
_____________________________
2
_________
+2 _________
=4 _________
210.15.10.0 to 210.15.10.63
210.15.10.64 to 210.15.10.127
Art Classroom 12 Hosts
Number of
Show your work for Problem 5 in the space below. Number of
256 128 64 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 – Binary values
210. 15 . 10 . 0 0 0 0 0 0 0 0
(1) 0 210.15.10.0 to (2) 1 210.15.10.64 to (3) 1 0 210.15.10.128 to (4) 1 1 210.15.10.192 to
210.15.10.63 210.15.10.127 210.15.10.191 210.15.10.255
67
Practical Subnetting 6
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 20% growth in all areas. Circle each subnet on the graphic and answer the questions below.
S0/0/0
S0/0/1
Science Building 225 Hosts
IP Address 10.0.0.0
S0/0/1
S0/0/1
S0/0/0
Technology Building 320 Hosts
Router A
F0/0
Router B
F0/1
S0/0/0
Art & Drama Router C
Administration 35 Hosts
75 Hosts F0/0
F0/1
Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 20% growth
A
68
(Round up to the next whole number) Total number of subnets needed
+ _________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Technology
IP address range for Science
IP address range for Arts & Drama
IP Address range Administration
IP address range for Router A to Router B serial connection
IP address range for Router A to Router C serial connection
IP address range for Router B to Router C serial connection
_____________________________
255.240.0.0
_____________________________
7
_________
2
9
= _________
10.0.0.0 to 10.15.255.255
_____________________________
10.16.0.0 to 10.31.255.255
_____________________________
10.32.0.0 to 10.47.255.255
_____________________________
10.48.0.0 to 10.63.255.255
_____________________________
10.64.0.0 to 10.79.255.255
_____________________________
10.80.0.0 to 10.95.255.255
_____________________________
10.96.0.0 to 10.111.255.255
_____________________________
Show your work for Problem 6 in the space below.
69
1,048,576 2,097,152
4,194,304
131,072
262,144 524,288
16,384 32,768
8,192 4,096
65,536
512
512 1,024
65,536 32,768
2,048
4,096 8,192
16,384
2,048 1,024
131,072
4,194,304 2,097,152
1,048,576
524,288 262,144
Number of
Number of – Hosts
.2561286432168 4 2
Subnets – 2 4 8 1632 64128256 .
Binaryvalues -128 64 32 16 8 4 2 1 .128 64 32 16 8 4
.
2 1
10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)
1
0 10.0.0.0 1 10.16.0.0
to 10.15.255.255 to 10.32.255.255 to 10.47.255.255 to 10.63.255.255 to 10.79.255.255 to 10.95.255.255 to 10.111.255.255 to 10.127.255.255 to 10.143.255.255 to 10.159.255.255 to 10.175.255.255 to 10.191.255.255 to 10.207.255.255 to 10.223.255.255 to 10.239.255.255 to 10.255.255.255
1 10 10
0 10.32.0.0 1 10.48.0.0 0 10.64.0.0 1 10.80.0.0 0 10.96.0.0
11
1 1 1 0 0 1 0 0
1 10.112.0.0 0 10.128.0.0 1 10.144.0.0 0 10.160.0.0 1 10.176.0.0 0 10.192.0.0
1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1
1 10.208.0.0 0 10.224.0.0 1 10.240.0.0
. 128 64 32 16 8 4 2 1
Practical Subnetting 7
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 125% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 177.135.0.0
S0/0/0 F0/1
Router A
F0/0
S0/0/0
Router B
F0/0
Deployment 63 Hosts
Marketing 75 Hosts
Administration 33 Hosts
Sales 255 Hosts
Research 135 Hosts
Address class
Custom subnet mask
_____________________________
Minimum number of subnets needed
Extra subnets required for 125% growth
(Round up to the next whole number)
_________
Total number of subnets needed
_________
Number of host addresses in the largest subnet group
363
Number of addresses needed for 125% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
_________ _+__4_5__4___
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router A Port F0/0 _____________________________
70
+
5
_________
=
9
817
B
255.255.252.0
_____________________________
4
177.135.0.0 to 177.135.3.255
177.135.4.0 to 177.135.7.255
IP address range for Research _____________________________
177.135.8.0 to 177.135.11.255
IP address range for Router A 177.135.12.0 to 177.135.15.255 to Router B serial connection _____________________________
IP address range for Deployment _____________________________
Show your work for Problem 7 in the space below.
71
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 177.135 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) . (2)
(3) 1
(4) 1
(5) 1 0
(6) 1 0
(7) 1 1
(8) 1 1
0 177.135.0.0 1 177.135.4.0 0 177.135.8.0
to 177.135.3.255 to 177.135.7.255 to 177.135.11.255 to 177.135.15.255 to 177.135.19.255 to 177.135.23.255 to 177.135.27.255 to 177.135.31.255 to 177.135.35.255 to 177.135.39.255 to 177.135.43.255 to 177.135.47.255 to 177.135.51.255 to 177.135.55.255 to 177.135.59.255 to 177.135.63.255
(9) 1 . (10) 1 . (11) 1 . (12) 1 (13) 1 . (14) 1 . (15) 1 . (16) 1 .
0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
1 177.135.20.0 0 177.135.24.0 1 177.135.28.0 0 177.135.32.0 1 177.135.36.0 0 177.135.40.0 1 177.135.44.0 0 177.135.48.0 1 177.135.52.0 0 177.135.56.0 1 177.135.60.0
1 177.135.12.0 0 177.135.16.0
2 1
Practical Subnetting 8
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number subnets, and allow enough extra subnets and hosts for 85% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 192.168.1.0
F0/0
S0/0/0
F0/1
Router A
S0/0/1
F0/0
Router B
New York 8 Hosts
Boston 5 Hosts
Research & Development 8 Hosts
Address class
Custom subnet mask
C
Minimum number of subnets needed
Extra subnets required for 85% growth
(Round up to the next whole number)
3
72
Total number of subnets needed
=
6
Number of host addresses in the largest subnet group
13
Number of addresses needed for 85% growth in the largest subnet
12
(Round up to the next whole number)
Total number of address needed for the largest subnet
_+________ 25
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
_____________________________
_________
+3 _________
_________
_________
IP address range for Router A F0/0 _____________________________
255.255.255.224
_____________________________
192.168.1.0 to 192.168.1.31
192.168.1.32 to 192.168.1.63
IP address range for New York _____________________________
IP address range for Router A
to Router B serial connection _____________________________
192.168.1.64 to 192.168.1.95
Show your work for Problem 8 in the space below. Number of
256 128 64 32 16 8 4 2 – Hosts Subnets – 2 4 8 16 32 64 128 256
Number of
192. 168 . 1 . 0 0 0
0 0 0 0 0
192.168.1.0 192.168.1.32 192.168.1.64 192.168.1.96 192.168.1.128 192.168.1.160 192.168.1.192 192.168.1.224
128 64 32 16 8 4 2 1
– Binary values
to 192.168.1.31 to 192.168.1.63 to 192.168.1.95 to 192.168.1.127 to 192.168.1.159 to 192.168.1.1191 to 192.168.1.223 to 192.168.1.255
(1) (2) (3) (4) (5) (6) (7) (8)
0 1 10 11 1 0 0 1 0 1 1 1 0 1 1 1
73
Practical Subnetting 9
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 148.55.0.0
S0/0/0
S0/0/1
S0/0/1
F0/0
S0/0/0
Router C
S0/0/1
F0/1
Router A
Router B
Dallas 1500 Hosts
F0/0
Router D
S0/0/0
74
_____________________________
Ft. Worth 2300 Hosts
Address class
B
Custom subnet mask
255.255.240.0
Minimum number of subnets needed
Extra subnets required for 15% growth
5
(Round up to the next whole number)
_________
Total number of subnets needed
=
6
Number of host addresses in the largest subnet group
2300
Number of addresses needed for 15% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
_________ _+__3_4__5___
_=________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Ft. Worth IP address range for Dallas
IP address range for Router A to Router B serial connection
IP address range for Router A to Router C serial connection
IP address range for Router C to Router D serial connection
148.55.0.0. to 148.55.15.255
_____________________________
_____________________________
_________
+
1
_________
2645
_____________________________
148.55.16.0. to 148.55.31.255
_____________________________
148.55.32.0. to 148.55.47.255
_____________________________
148.55.48.0. to 148.55.63.255
_____________________________
148.55.64.0. to 148.55.79.255
Show your work for Problem 9 in the space below.
75
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4
148. 55 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) . (2)
(3) 1
(4) 1
(5) 1 0
(6) 1 0
(7) 1 1
(8) 1 1
0 148.55.0.0 to 1 148.55.16.0 to 0 148.55.32.0 to 1 148.55.48.0 to 0 148.55.64.0 to 1 148.55.80.0 to 0 148.55.96.0 to 1 148.55.112.0 to 0 148.55.128.0 to 1 148.55.144.0 to 0 148.55.160.0 to 1 148.55.176.0 to 0 148.55.192.0 to 1 148.55.208.0 to 0 148.55.224.0 to 1 148.55.240.0 to
148.55.15.255 148.55.31.255 148.55.47.255 148.55.63.255 148.55.79.255 148.55.95.255 148.55.111.255 148.55.127.255 148.55.143.255 148.55.159.255 148.55.175.255 148.55.191.255 148.55.207.255 148.55.223.255 148.55.239.255 148.55.255.255
(9) 1 . (10) 1 . (11) 1 . (12) 1 (13) 1 . (14) 1 . (15) 1 . (16) 1 .
0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
2 1
Practical Subnetting 10
Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 110% growth in all areas. Circle each subnet on the graphic and answer the questions below.
Sales
115 Hosts
F0/0
Management 25 Hosts
S0/0/0
Marketing 56 Hosts
F0/0 S0/0/1
Router B
F0/1
Research 35 Hosts
IP Address 172.16.0.0
Router A
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 110% growth
(Round up to the next whole number)
B
76
Total number of subnets needed
=
9
Number of host addresses in the largest subnet group
140
Number of addresses needed for 110% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
_________ _+__1_5_4____
= 294 _________
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales/Managemnt IP address range for Marketing IP address range for Research
IP address range for Router A to Router B serial connection
_____________________________
+
5
_________
_________
255.255.255.240
_____________________________
4
_________
172.16.0.0 to 172.16.15.255
_____________________________
172.16.16.0 to 172.16.31.255
_____________________________
172.16.32.0 to 172.16.47.255
_____________________________
172.16.48.0 to 172.16.63.255
_____________________________
Show your work for Problem 10 in the space below.
77
512 1,024
2,048 4,096
8,192 16,384
32,768 65,536
512
65,536 32,768
16,384
8,192
4,096 2048
1024
Number of Hosts –
. 256 128 64 32 16 8
4 2
Number of
Subnets – 2 4 8 16 32 64 128 256.
Binary values – 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 172.16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(1) . (2)
(3) 1
(4) 1
(5) 1 0
(6) 1 0
(7) 1 1
(8) 1 1
0 172.16.0.0 1 172.16.16.0
to 172.16.15.255 to 172.16.31.255 to 172.16.47.255 to 172.16.63.255 to 172.16.79.255 to 172.16.95.255 to 172.16.111.255 to 172.16.127.255 to 172.16.143.255 to 172.16.159.255 to 172.16.175.255 to 172.16.191.255 to 172.16.207.255 to 172.16.223.255 to 172.16.239.255 to 172.16.255.255
(9) 1 . (10) 1 . (11) 1 . (12) 1 (13) 1 . (14) 1 . (15) 1 . (16) 1 .
0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
1 172.16.112.0 0 172.16.128.0 1 172.16.144.0 0 172.16.160.0 1 172.16.176.0 0 172.16.192.0
0 172.16.32.0 1 172.16.48.0 0 172.16.64.0 1 172.16.80.0 0 172.16.96.0
1 172.16.208.0 0 172.16.224.0 1 172.16.240.0
2 1
IP Address: 0.230.190.192 Subnet Mask: 255.0.0.0
Reference Page Inside Front Cover
IP Address: 192.10.10.1 Subnet Mask: 255.255.255.0
Reference Pages 28-29
IP Address: 245.150.190.10 Subnet Mask: 255.255.255.0
Reference Page Inside Front Cover
IP Address: 135.70.191.255 Subnet Mask: 255.255.254.0
Reference Pages 48-49
IP Address: 127.100.100.10 Subnet Mask: 255.0.0.0
Reference Pages Inside Front Cover
IP Address: 93.0.128.1 Subnet Mask: 255.255.224.0
Reference Pages 56-57
IP Address: 200.10.10.128 Subnet Mask: 255.255.255.224
Reference Pages 54-55
IP Address: 165.100.255.189 Subnet Mask: 255.255.255.192
Reference Pages 30-31
IP Address: 190.35.0.10 Subnet Mask: 255.255.255.192
Reference Pages 34-35
IP Address: 218.35.50.195 Subnet Mask: 255.255.0.0
Reference Page Inside Front Cover
IP Address: 200.10.10.175 /22
Reference Pages 54-55 and/or Inside Front Cover
IP Address: 135.70.255.255 Subnet Mask: 255.255.224.0
Reference Pages 48-49
_T__h_e_n_e_t_w__o_r_k__I_D___c_a_n_n_o_t__b_e__0__. __ ________________________________
_O__K_____________________________ ________________________________
_2__4_5__i_s__r_e_s_e_r_v_e_d___f_o_r___________ _e_x_p_e_r_i_m__e_n_t_a_l_u_s_e_._______________
_T__h_is__i_s__t_h_e__b_r_o_a_d__c_a_s_t__a_d_d_r__e_s_s_ _f_o_r__t_h_i_s__r_a_n_g_e_._________________
_1_2__7__i_s_r__e_s_e_r_v_e_d__f_o_r__l_o_o_p_b_a__c_k__ _t_e_s_t_i_n_g_. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_O__K_____________________________ ________________________________
_T_h_i_s_i_s_t_h_e__s_u_b_n_et__a_d_d_r_e_s_s__fo_r__t_h_e__ _3_r_d__u_s_a_b_le__r_a_n_ge__o_f__2_0_0_._1_0_._1_0_.0____
________________________________ _T_h_is__a_d_d_r_e_s_s_i_s_t_a_k_e_n_f_r_o_m__th_e__f_ir_s_t__
78
Valid and Non-Valid IP Addresses
Using the material in this workbook identify which of the addresses below are correct and usable. If they are not usable addresses explain why.
OK
________________________________
_r_a_ng_e__f_or__t_h_is__s_ub_n_e_t__w_h_ic_h__is__in_v_a_l_id_._ _T_h_i_s__h_a_s__a__c_l_a_s_s__B__s_u_b_n__e_t______
_m_ a_ _s _k_. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _A__c_l_a_s_s__C__a_d__d_r_e_s_s__m__u_s_t__u_s_e__a__
_m_i_n_i_m_ _u_m_ _o_f_ _2_ _4_ _b_i_t_s_. _ _ _ _ _ _ _ _ _ _ _
_T_h_i_s__i_s__a__b_r_o_a_d_c_a__s_t_a__d_d_r_e_s_s__. __ ________________________________
IP Address Breakdown
/24
8+8+8 255.255.255.0 256 Hosts
/25
8+8+8+1 255.255.255.128 128 Hosts
/26
8+8+8+2 255.255.255.192 64 Hosts
/27
8+8+8+3 255.255.255.224 32 Hosts
/28
8+8+8+4 255.255.255.240 16 Hosts
/29
8+8+8+5 255.255.255.248 8 Hosts
/30
8+8+8+6 255.255.255.252 4 Hosts
0-255
0-127
0-63
0-15
0-7
0-3
4-7
8-15
8-11
12-15
16-31
16-23
16-19
20-23
24-31
24-27
28-31
32-47
32-39
32-35
36-39
40-47
40-43
44-47
48-63
48-55
48-51
52-55
56-63
56-59
60-63
64-127
64-79
64-71
64-67
68-71
72-79
72-75
76-79
80-95
80-87
80-83
84-87
88-95
88-91
92-95
96-111
96-103
96-99
100-103
104-111
104-107
108-111
112-127
112-119
112-115
116-119
120-127
120-123
124-127
128-255
128-191
128-143
128-135
128-131
132-135
136-143
136-139
140-143
144-159
144-151
144-147
148-151
152-159
152-155
156-159
160-175
16-167
160-163
164-167
168-175
168-171
172-175
176-191
176-183
176-179
180-183
184-191
184-187
188-191
192-255
192-207
192-199
192-195
196-199
200-207
200-203
204-207
208-223
208-215
208-211
212-215
216-223
216-219
220-223
224-239
224-231
224-227
228-231
232-239
232-235
236-239
240-255
240-247
240-243
244-247
248-255
248-251
252-255
79
Visualizing Subnets Using The Box Method
The box method is the simplest way to visualize the breakdown of subnets and addresses into smaller sizes.
Start with a square. The whole square is a single subnet comprised of 256 addresses.
/24 255.255.255.0 256 Hosts 1 Subnet
Split the box in half and you get two subnets with 128 addresses,
/25 255.255.255.128 128 Hosts 2 Subnets
Divide the box into quarters and you get four subnets with 64 addresses,
/26 255.255.255.192 64 Hosts 4 Subnets
80
Split each individual square and you get eight subnets with 32 addresses,
/27 255.255.255.224 32 Hosts 8 Subnets
Split the boxes in half again and you get sixteen subnets with sixteen addresses,
/28 255.255.255.240 16 Hosts 16 Subnets
The next split gives you thirty two subnets with eight addresses,
/29 255.255.255.248 8 Hosts 32 Subnets
The last split gives sixty four subnets with four addresses each,
/30 255.255.255.252 4 Hosts 64 Subnets
81
Class A Addressing Guide
# of Bits Subnet Total # of Total # of Usable # of CIDR Borrowed Mask Subnets Hosts Hosts
______________________________________________________________________________________________
/8 0 255.0.0.0 1 16,777,216 16,777,214 _____________________________________________________________________________________________
/9 1 255.128.0.0 2 8,388,608 8,388,606 _____________________________________________________________________________________________
/10 2 255.192.0.0 4 4,194,304 4,194,302 __________________________________________________________________________________________________
/11 3 255.224.0.0 8 2,097,152 2,097,150 ______________________________________________________________________________________________
/12 4 255.240.0.0 16 1,048,576 1,048,574 _____________________________________________________________________________________________
/13 5 255.248.0.0 32 524,288 524,286 ________________________________________________________________________________________________
/14 6 255.252.0.0 64 262,144 262,142 ______________________________________________________________________________________________
/15 7 255.254.0.0 128 131,072 131,070 __________________________________________________________________________________________________
/16 8 255.255.0.0 256 65,536 65,534 ___________________________________________________________________________________________________
/17 9 255.255.128.0 512 32,768 32,766 _______________________________________________________________________________________________
/18 10 255.255.192.0 1,024 16,384 16,382 _____________________________________________________________________________________________________
/19 11 255.255.224.0 2,048 8,192 8,190 ______________________________________________________________________________________________
/20 12 255.255.240.0 4,096 4,096 4,094 __________________________________________________________________________________________________
/21 13 255.255.248.0 8,192 2,048 2,046 _________________________________________________________________________________________________
/22 14 255.255.252.0 16,384 1,024 1,022 ________________________________________________________________________________________________
/23 15 255.255.254.0 32,768 512 510 ____________________________________________________________________________________________________
/24 16 255.255.255.0 65,536 256 254 _____________________________________________________________________________________________________
/25 17 255.255.255.128 131,072 128 126 ____________________________________________________________________________________________________
/26 18 255.255.255.192 262,144 64 62 ___________________________________________________________________________________________________
/27 19 255.255.255.224 524,288 32 30 ____________________________________________________________________________________________________
/28 20 255.255.255.240 1,048,576 16 14 ____________________________________________________________________________________________________
/29 21 255.255.255.248 2,097,152 8 6 ________________________________________________________________________________________________
/30 22
# of Bits
CIDR Borrowed
255.255.255.252 4,194,304
Class B Addressing Guide
Subnet Total # of Mask Subnets
4
Total # of Hosts
2
Usable # of Hosts
______________________________________________________________________________________________
/16 0 255.255.0.0 1 65,536 65,534
_____________________________________________________________________________________________
/17 1 255.255.128.0 2 32,768 32,766
_____________________________________________________________________________________________
/18 2 255.255.192.0 4 16,384 16,382
__________________________________________________________________________________________________
/19 3 255.255.224.0 8 8,192 8,190
______________________________________________________________________________________________
/20 4 255.255.240.0 16 4,096 4,094
_____________________________________________________________________________________________
/21 5 255.255.248.0 32 2,048 2,046
________________________________________________________________________________________________
/22 6 255.255.252.0 64 1,024 1,022
______________________________________________________________________________________________
/23 7 255.255.254.0 128 512 510
__________________________________________________________________________________________________
/24 8 255.255.255.0 256 256 254
___________________________________________________________________________________________________
/25 9 255.255.255.128 512 128 126
_______________________________________________________________________________________________
/26 10 255.255.255.192 1,024 64 62
_____________________________________________________________________________________________________
/27 11 255.255.255.224 2,048 32 30
______________________________________________________________________________________________
/28 12 255.255.255.240 4,096 16 14
______________________________________________________________________________________________
/29 13 255.255.255.248 8,192 8 6
________________________________________________________________________________________________
/30 14
# of Bits
CIDR Borrowed
255.255.255.252 16,384
Class C Addressing Guide
Subnet Total # of Mask Subnets
4
Total # of Hosts
2
Usable # of Hosts
______________________________________________________________________________________________
/24 0 255.255.255.0 1 256 254
_____________________________________________________________________________________________
/25 1 255.255.255.128 2 128 126
_____________________________________________________________________________________________
/26 2 255.255.255.192 4 64 62
__________________________________________________________________________________________________
/27 3 255.255.255.224 8 32 30
______________________________________________________________________________________________
/28 4 255.255.255.240 16 16 14
_____________________________________________________________________________________________
/29 5 255.255.255.248 32 8 6
________________________________________________________________________________________________
/30 6 255.255.255.252 64 4 2
82
Inside Cover