answer questions in the documents attached
adedavbc
IPSUBNETTINGWORKBOOK.pdf
11111110
1010100 10001111100
1011100101011100 101100011101001 1011110100011010 00001010010110010 1001010101100111 1111010101000101 1101001101010011 001010010101010 1010101000110010 010101001011000 110101100011010 11010100001011 001010100110 1001010010
10011000 01
01 11010011
10000110
10010101 00011011
001 IP Addressing
and Subnetting
Workbook Version 1.5
Student Name:
IP Address Classes
Class A 1 – 127 (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 00000000.00000000.00000000.00000000
Class B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000.00000000
Class C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000.00000000
Class D 224 – 239 (Reserved for multicast)
Class E 240 – 255 (Reserved for experimental, used for research)
Private Address Space
Class A 10.0.0.0 to 10.255.255.255
Class B 172.16.0.0 to 172.31.255.255
Class C 192.168.0.0 to 192.168.255.255
Default Subnet Masks
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
Network . Host . Host . Host
Network . Network . Host . Host
Network . Network . Network . Host
Inside Cover
Produced by: Robb Jones [email protected]
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.
Workbooks included in the series:
IP Addressing and Subnetting Workbooks ACLs - Access Lists Workbooks
VLSM Variable-Length Subnet Mask IWorkbooks
Binary To Decimal Conversion
128 64 32 16 8 4 2 1 Answers Scratch Area 1 0 0 1 0 0 1 0 146
0 1 1 1 0 1 1 1 119
1 1 1 1 1 1 1 1
1 1 0 0 0 1 0 1
1 1 1 1 0 1 1 0
0 0 0 1 0 0 1 1
1 0 0 0 0 0 0 1
0 0 1 1 0 0 0 1
0 1 1 1 1 0 0 0
1 1 1 1 0 0 0 0
0 0 1 1 1 0 1 1
0 0 0 0 0 1 1 1
00011011
10101010
01101111
11111000
00100000
01010101
00111110
00000011
11101101
11000000
128 16 2
146
64 32 16 4 2 1
119
1
Decimal To Binary Conversion
128 64 32 16 8 4 2 1 = 255 Scratch Area _________________________________________ 238
_________________________________________ 34
_________________________________________ 123
_________________________________________ 50
_________________________________________ 255
_________________________________________ 200
_________________________________________ 10
_________________________________________ 138
_________________________________________ 1
_________________________________________ 13
_________________________________________ 250
_________________________________________ 107
_________________________________________ 224
_________________________________________ 114
_________________________________________ 192
_________________________________________ 172
_________________________________________ 100
_________________________________________ 119
_________________________________________ 57
_________________________________________ 98
_________________________________________ 179
_________________________________________ 2
238 -128 110 -64 46
-32 14 -8 6
-4 2
-2 0
1 1 1 0 1 1 1 0 34 -32
2 -2 0
0 0 1 0 0 0 1 0
Use all 8 bits for each problem
2
Address Class Identification
Address Class
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 _____
A
B
3
Network & Host Identification
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
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
4
5
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
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
0 . 0 . 18 . 2
0 . 0 . 0 . 80
6
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
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 255.0.0.0 Class B 255.255.0.0 Class C 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...
IP Address: 192 . 100 . 10 . 33
What you can figure out in your head...
Address Class: C Network Portion: 192 . 100 . 10 . 33 Host Portion: 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.
ANDING with the default subnet mask allows your computer to figure out the network portion of the address.
1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 0 0 0 0 0 0
(255 . 255 . 255 . 0)
(192 . 100 . 10 . 33)
(192 . 100 . 10 . 0)
Network Host
IP Address: Default Subnet Mask:
AND:
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.
IP Address: 192 . 100 . 10 . 0 Custom Subnet Mask: 255.255.255.240
Address Ranges: 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 (Range in the sample below) 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
In the next set of problems you will determine the necessary information to determine the correct subnet mask for a variety of IP addresses.
1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 0
(255 . 255 . 255 . 240)
(192 . 100 . 10 . 33)
(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.
Sub Network HostNetwork
IP Address: Custom Subnet Mask:
AND:
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 (Second subnet formula: Number of subnets = 2 - 2)
Number of hosts per subnet = 2 - 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 three bits. This would be 2 or 2 x 2 x 2 = 8 subnets
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 host portion of the address this would be 2 or 2 x 2 x 2 x 2 x 2 = 32 hosts.
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.
h
3
ss
5
10
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 The number of subnets created by borrowing 2 bits is 2 or 2 x 2 = 4 subnets.
2
The number of hosts created by leaving 6 bits is 2 - 2 or 2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62 usable hosts per subnet.
6
What about that second subnet formula:
Number of subnets = 2 - 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.
s
11
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195.223.50.0 195.223.50.64 195.223.50.128 195.223.50.192
195.223.50.63 195.223.50.127 195.223.50.191 195.223.50.255
(Invalid range)
(Invalid range)
to to to to
(0) (1) (2) (3)
Class C Address unsubnetted:
195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0
195.223.50.0 to 195.223.50.255
Class C Address subnetted (2 bits borrowed): Notice that the subnet and broadcast addresses match.
Use the 2 - 2 formula and don’t use the zero and broadcast ranges if...
Classful routing is used
RIP version 1 is used
The no ip subnet zero command is configured on your router
Use the 2 formula and use the zero and broadcast ranges if...
Classless routing or VLSM is used
RIP version 2, EIGRP, or OSPF is used
The ip subnet zero command is configured on your router (default setting)
No other clues are given
When to use which formula to determine the number of subnets s s
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.
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 = 2 formula.s
Custom Subnet Masks
Problem 1 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
14 14 192.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of 256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 1 in the space below.
Add the binary value numbers to the left of the line to create the custom subnet mask.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 240
16
16
14
4
Observe the total number of hosts.
Subtract 2 for the number of usable hosts.
16 -2 14
12
128 64 32
+16 240
13
Custom Subnet Masks
Problem 2 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
1000 60 165.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
Show your work for Problem 2 in the space below.
Add the binary value numbers to the left of the line to create the custom subnet mask.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
Observe the total number of hosts.
Subtract 2 for the number of usable hosts.
64 -2 62
128 64 32 16 8 4 2
+1 255
128 64 32 16 8 4 2 1 .....
5 12
Binary values -
Number of Hosts -
128 +64 192
10 2 4
2 0 4 8
4 ,0
96 8,192
16,384
32 ,768
65 ,5
36
5 12
1,0 2 4
2 ,0
4 8
4 ,0
96
8,192
16,384
32 ,768
65 ,5
36
Custom Subnet Masks
Problem 3 Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
148.75.0.0 /26
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 3 in the space below.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
/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.
148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
Add the binary value numbers to the left of the line to create the custom subnet mask.
1024 -2
1,022
Observe the total number of hosts.
Subtract 2 for the number of usable hosts.
64 -2 62
Subtract 2 for the total number of subnets to get the usable number of subnets.
128 64 32 16 8 4 2
+1 255
128 64 32 16 8 4 2 1 .....
5 12
Binary values -
Number of Hosts -
128 +64 192
10 2 4
2 0 4 8
4 ,0
96 8,192
16,384
32 ,768
65 ,5
36
5 12
1,0 2 4
2 ,0
4 8
4 ,0
96
8,192
16,384
32 ,768
65 ,5
36
14
Custom Subnet Masks
Problem 4 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
6 30 210.100.56.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of 256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 4 in the space below.
15
16
Custom Subnet Masks
Problem 5 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
6 30 195.85.8.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of 256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 5 in the space below.
17
Custom Subnet Masks
Problem 6 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
126 131,070 118.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 6 in the space below.
118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
Number of Subnets - 2 4 8 16 32 64 128 256 .
. 256 128 64 32 16 8 4 2
Binary values -128 64 32 16 8 4 2 1 . . . . . 128 64 32 16 8 4 2 1 . . . . . 128 64 32 16 8 4 2 1
Number of Hosts -
5 12
1,0 2 4
2 ,0
4 8
4 ,0
96 8,192
16,384 32
,768 65
,5 36
131,0 72
2 62
,14 4
5 2 4 ,2
88 1,048,5
76 2,097,15
2 4,194,304.
5 12
1,0 2 4
2 ,0
4 8
4 ,0
96
8,192
16,384
32 ,768
65 ,5
36
131,0 72
2 62
,14 4
5 2 4 ,2
88
1,048,5 76
2,097,15 2
4,194,304
18
Custom Subnet Masks
Problem 7 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
2000 15 178.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 7 in the space below.
178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 32 16 8 4 2 1 .....
5 12
Binary values -
Number of Hosts -
10 2 4
2 0 4 8
4 ,0
96 8,192
16,384
32 ,768
65 ,5
36
5 12
1,0 2 4
2 ,0
4 8
4 ,0
96
8,192
16,384
32 ,768
65 ,5
36
19
Custom Subnet Masks
Problem 8 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
3 45 200.175.14.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 8 in the space below.
Custom Subnet Masks
Problem 9 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
60 1,000 128.77.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 9 in the space below.
20
Custom Subnet Masks
Problem 10 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
60 198.100.10.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 10 in the space below.
21
Custom Subnet Masks
Problem 11 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
250 101.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 11 in the space below.
22
Custom Subnet Masks
Problem 12 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
5 218.35.50.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 12 in the space below.
23
Custom Subnet Masks
Problem 13 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
25 218.35.50.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 13 in the space below.
24
Custom Subnet Masks
Problem 14 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10 172.59.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 14 in the space below.
25
Custom Subnet Masks
Problem 15 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
50 172.59.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 15 in the space below.
26
Custom Subnet Masks
Problem 16 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
29 23.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 16 in the space below.
27
_______________________________________________
________________________
________________________
______________________________________
192.10.10.48 to 192.10.10.63
192 . 10 . 10 . 112
192 . 10 . 10 . 207
192.10.10.129 to 192.10.10.142
28
Subnetting
Problem 1 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
14 14 192.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
C
255 . 255 . 255 . 0
255 . 255 . 255 . 240
16
16
14
4
What is the 4th subnet range?
What is the subnet number for the 8th subnet?
What is the subnet broadcast address for
the 13th subnet?
What are the assignable addresses for the 9th
subnet?
Show your work for Problem 1 in the space below.
192.10.10.0 to 192.10.10.15 192.10.10.16 to 192.10.10.31 192.10.10.32 to 192.10.10.47 192.10.10.48 to 192.10.10.63 192.10.10.64 to 192.10.10.79 192.10.10.80 to 192.10.10.95 192.10.10.96 to 192.10.10.111 192.10.10.112 to 192.10.10.127 192.10.10.128 to 192.10.10.143 192.10.10.144 to 192.10.10.159 192.10.10.160 to 192.10.10.175 192.10.10.176 to 192.10.10.191 192.10.10.192 to 192.10.10.207 192.10.10.208 to 192.10.10.223 192.10.10.224 to 192.10.10.239 192.10.10.240 to 192.10.10.255
192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of 256 128 64 32 16 8 4 2 - Hosts
16 -2 14
16 -2 14
128 64 32
+16 240
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.
Custom subnet mask
Usable hostsUsable subnets
29
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
(0) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)
_______________________________________________
________________________
________________________
______________________________________
165.100.3.128 to 165.100.3.191
165 . 100 . 1 . 64
165 . 100 . 1 . 127
165.100.2.1 to 165.100.0.62
30
Subnetting
Problem 2 Number of needed subnets
Number of needed usable hosts Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
1000 60 165.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
What is the 15th subnet range?
What is the subnet number for the 6th subnet?
What is the subnet broadcast address for
the 6th subnet?
What are the assignable addresses for the 9th
subnet?
16 5
. 1
0 0 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 16
5 . 1
0 0 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 16
5 . 1
0 0 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 16
5 . 1
0 0 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 16
5 . 1
0 0 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 16
5 .1
0 0 .0
.0 to
1
6 5
.1 0 0 .0
.6 3
16 5
.1 0 0 .0
.6 4
to
1 6 5
.1 0 0 .0
.1 2 7
16 5
.1 0 0 .0
.1 2
8 to
1
6 5
.1 0 0 .0
.1 91
16 5
.1 0 0 .0
.1 92
to
1 6 5
.1 0 0 .0
.2 5
5 16
5 .1
0 0 .1
.0 to
1
6 5
.1 0 0 .1
.6 3
16 5
.1 0 0 .1
.6 4
to
1 6 5
.1 0 0 .1
.1 2 7
16 5
.1 0 0 .1
.1 2
8 to
1
6 5
.1 0 0 .1
.1 91
16 5
.1 0 0 .1
.1 92
to
1 6 5
.1 0 0 .1
.2 5
5 16
5 .1
0 0 .2
.0 to
1
6 5
.1 0 0 .0
.6 3
16 5
.1 0 0 .2
.6 4
to
1 6 5
.1 0 0 .0
.1 2 7
16 5
.1 0 0 .2
.1 2
8 to
1
6 5
.1 0 0 .0
.1 91
16 5
.1 0 0 .2
.1 92
to
1 6 5
.1 0 0 .0
.2 5
5 16
5 .1
0 0 .3
.0 to
1
6 5
.1 0 0 .3
.6 3
16 5
.1 0 0 .3
.6 4
to
1 6 5
.1 0 0 .3
.1 2 7
16 5
.1 0 0 .3
.1 2
8 to
1
6 5
.1 0 0 .3
.1 91
16 5
.1 0 0 .3
.1 92
to
1 6 5
.1 0 0 .3
.2 5
5
D ow
n to
16 5
.1 0 0 .2
5 5
.1 2
8 t
o
16 5
.1 0 0 .2
5 5
.1 91
16 5
.1 0 0 .2
5 5
.1 92
to
1
6 5
.1 0 0 .2
5 5
.2 5
5
T he
b in
ar y
va lu
e of
t he
la st
b it
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ow ed
is th
e ra
ng e.
In th
is p
ro bl
em th
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ng e
is 6
4.
T he
f irs
t ad
dr es
s in
e ac
h su
bn et
r an
ge is
t he
su bn
et n
um be
r.
T he
la st
a dd
re ss
in e
ac h
su bn
et r
an ge
is t
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bn et
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ad ca
st a
dd re
ss .
31
6 4
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U sa
bl e
ho st
s
12 8
6 4
3 2 16 8 4 2 + 1
2 5
5
12 8
+6 4
19 2
C us
to m
su bn
et m
as k
Show your work for Problem 2 in the space below.
12 8
6 4 3 2 16
8
4
2
1
N um
be r
of S
ub ne
ts -
2
4
8
1
6 3 2
6 4 1
2 8
2 5
6 ..
2 5
6 1
2 8
6 4 3
2 16
8
4 2
12 8
6 4
3 2
1
6
8
4
2
1
.... .
512
B in
ar y
va lu
es -
N um
be r
of H
os ts
-
1024 2048 4,096 8,192
16,38 4
32,76 8
65,5 36
512
1,024
2,048
4,096
8,192
16,38 4
32,76 8
65,5 36
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1
1 1 1 1 0 0 0 0 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
. . . . . . . . . . . . . . .
(0 )
(1 )
(2 )
(3 )
(4 )
(5 )
(6 )
(7 )
(8 )
(9 )
(1 0)
(1 1)
(12 )
(1 3)
(1 4)
(15 )
1
1
1 1
1 1
1
1 1
1 1
1 (1
0 2 2 )
(1 0 2 3 )
_______________________________________________
________________________
________________________
______________________________________
32
Subnetting
Problem 3 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
2 195.223.50.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 3rd 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 3rd
subnet?
Show your work for Problem 3 in the space below.
33
128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of 256 128 64 32 16 8 4 2 - Hosts
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0
_______________________________________________
________________________
________________________
______________________________________
34
Subnetting
Problem 4 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
750 190.35.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
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?
Show your work for Problem 4 in the space below.
35
_______________________________________________
________________________
________________________
______________________________________
36
Subnetting
Problem 5 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
6 126.0.0.0
__________
_______________________________
_______________________________
___________________
________________
___________________
___________________
What is the 2nd subnet range?
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 7th subnet?
What are the assignable addresses for the 10th
subnet?
Show your work for Problem 5 in the space below.
37
_______________________________________________
________________________
________________________
______________________________________
38
Subnetting
Problem 6 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10 192.70.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
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?
Show your work for Problem 6 in the space below.
39
40
Subnetting
Problem 7 Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10.0.0.0 /16
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 11th subnet range?
What is the subnet number for the 6th subnet?
What is the subnet broadcast address for
the 2nd subnet?
What are the assignable addresses for the 9th
subnet?
_______________________________________________
________________________
________________________
______________________________________
Show your work for Problem 7 in the space below.
41
_______________________________________________
________________________
________________________
______________________________________
42
Subnetting
Problem 8 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
5 172.50.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 4th subnet range?
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 6th subnet?
What are the assignable addresses for the 3rd
subnet?
Show your work for Problem 8 in the space below.
43
_______________________________________________
________________________
________________________
______________________________________
44
Subnetting
Problem 9 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
28 172.50.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
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?
Show your work for Problem 9 in the space below.
45
_______________________________________________
________________________
________________________
______________________________________
46
Subnetting
Problem 10 Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
45 220.100.100.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 5th subnet range?
What is the subnet number for the 4th subnet?
What is the subnet broadcast address for
the 13th subnet?
What are the assignable addresses for the 12th
subnet?
Show your work for Problem 10 in the space below.
47
_______________________________________________
________________________
________________________
______________________________________
48
Subnetting
Problem 11 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
8,000 135.70.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 6th subnet range?
What is the subnet number for the 7th subnet?
What is the subnet broadcast address for
the 3rd subnet?
What are the assignable addresses for the 5th
subnet?
Show your work for Problem 11 in the space below.
49
_______________________________________________
________________________
________________________
______________________________________
50
Subnetting
Problem 12 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
45 198.125.50.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
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?
Show your work for Problem 12 in the space below.
51
_______________________________________________
________________________
________________________
______________________________________
52
Subnetting
Problem 13 Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
165.200.0.0 /26
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 10th subnet range?
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
subnet?
Show your work for Problem 13 in the space below.
53
_______________________________________________
________________________
________________________
______________________________________
54
Subnetting
Problem 14 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
16 200.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 7th subnet range?
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 4th subnet?
What are the assignable addresses for the 6th
subnet?
Show your work for Problem 14 in the space below.
55
56
Subnetting
Problem 15 Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
93.0.0.0 \19
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 15th subnet range?
What is the subnet number for the 9th subnet?
What is the subnet broadcast address for
the 7th subnet?
What are the assignable addresses for the 12th
subnet?
_______________________________________________
________________________
________________________
______________________________________
Show your work for Problem 15 in the space below.
57
58
Practical Subnetting 1
F0/1 F0/0
S0/0/0 S0/0/1Router A Router B
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.
Marketing 24 Hosts
Management 15 Hosts
F0/0
Reasearch 60 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 100% growth in the largest subnet
Total number of address needed for the largest subnet
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
IP Address 172.16.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
B 255.255.224.0
4 4
8
60
60
120
+
+
=
=
172.16.0.0 to 172.31.255 172.16.32.0 to 172.63.255 172.16.64.0 to 172.95.255
172.16.96.0 to 172.127.255
59
Show your work for Practical Subnetting 1 in the space below.
17 2
. 1
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 17
2 . 1
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 17
2 . 1
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 17
2 . 1
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 17
2 . 1
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 12
8 6
4 3 2 16
8
4
2
1
N um
be r
of S
ub ne
ts -
2
4
8
1
6 3 2
6 4 1
2 8
2 5
6 ..
2 5
6 1
2 8
6 4 3
2 16
8
4 2
12 8
6 4
3 2
1
6
8
4
2
1
.... .
512
B in
ar y
va lu
es -
N um
be r
of H
os ts
-
1024 2048 4,096 8,192
16,38 4
32,76 8
65,5 36
512
1,024
2,048
4,096
8,192
16,38 4
32,76 8
65,5 36
0 1 0 1 0 1 0 1
1 1 0 0 1 1
1 1 1 1
17 2
.1 6 .0
.0 17
2 .1
6 .3
2 .0
17 2
.1 6 .6
4 .0
17 2
.1 6 .9
6 .0
17 2
.1 6 .1
2 8.
0 17
2 .1
6 .1
6 0 .0
17 2
.1 6 .1
92 .0
17 2
.1 6 .2
2 4 .0
to to to to to to to to
17 2
.1 6 .3
1. 2
5 5
17 2
.1 6 .6
3 .2
5 5
17 2
.1 6 .9
5 .2
5 5
17 2
.1 6 .1
2 7.
2 5
5 17
2 .1
6 .1
5 9.
2 5
5 17
2 .1
6 .1
91 .2
5 5
17 2
.1 6 .2
2 3 .2
5 5
17 2
.1 6 .2
5 5
.2 5
5
(0 )
(1 )
(2 )
(3 )
(4 )
(5 )
(6 )
(7 )
6 0
x1 .0 6 04
x1 .0 4
60
Practical Subnetting 2
F0/0 S0/0/0
S0/0/1Router A
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.
Science Lab 10 Hosts
Tech Ed Lab 20 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 30% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 30% growth in the largest subnet
Total number of address needed for the largest subnet
IP address range for Tech Ed
IP address range for English
IP address range for Science
IP address range for Router A to Router B serial connection
IP address range for Router A to Router B serial connection
IP Address 135.126.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C
English Department 15 Hosts
F0/1
F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
5 2
7
20
6
26
B 255.255.255.224
135.126.0.0 to 135.126.0.31 135.126.0.32 to 135.126.0.63 135.126.0.64 to 135.126.0.95
135.126.0.96 to 135.126.0.127
135.126.0.128 to 135.126.0.159
61
Show your work for Problem 2 in the space below.
13 5
. 12
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 13
5 . 12
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 13
5 . 12
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 13
5 . 12
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 13
5 . 12
6 . 0
0
0
0
0
0
0
0 .
0
0
0
0
0
0
0
0 12
8 6
4 3 2 16
8
4
2
1
N um
be r
of S
ub ne
ts -
2
4
8
1
6 3 2
6 4 1
2 8
2 5
6 ..
2 5
6 1
2 8
6 4 3
2 16
8
4 2
12 8
6 4
3 2
1
6
8
4
2
1
.... .
512
B in
ar y
va lu
es -
N um
be r
of H
os ts
-
1024 2048 4,096 8,192
16,38 4
32,76 8
65,5 36
512
1,024
2,048
4,096
8,192
16,38 4
32,76 8
65,5 36
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
13 5.
12 6.
0. 0
13 5.
12 6.
0. 32
13 5.
12 6.
0. 64
13 5.
12 6.
0. 96
13 5.
12 6.
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8 13
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13 5.
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6. 1.3
2 13
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6. 1.6
4 13
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6. 1.9
6 13
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6. 1.1
28 13
5. 12
6. 1.1
60 13
5. 12
6. 1.1
92 13
5. 12
61 .2
24
to to to to to to to to to to to to to to to to
13 5.
12 6.
0. 31
13 5.
12 6.
0. 63
13 5.
12 6.
0. 95
13 5.
12 6.
0. 12
7 13
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1 13
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13 5.
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5 13
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6. 1.3
1 13
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6. 1.6
3 13
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6. 1.9
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6. 1.1
27 13
5. 12
6. 1.1
59 13
5. 12
6. 1.1
91 13
5. 12
6. 1.2
23 13
5. 12
6. 1.2
55
(0 )
(1 )
(2 )
(3 )
(4 )
(5 )
(6 )
(7 )
(8 )
(9 )
(1 0)
(1 1)
(12 )
(1 3)
(1 4)
(15 )
5 x. 3
1. 5
(R ou
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2 0 x. 3 6
62
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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 25% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 25% growth in the largest subnet
Total number of address needed for the largest subnet
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
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0
Administrative 30 Hosts
Sales 185 Hosts
F0/0
IP Address 172.16.0.0
S0/0/1
Marketing 50 Hosts
F0/1 S0/0/0 Router A
Router B
63
Show your work for Problem 3 in the space below.
64
Practical Subnetting 4
F0/0 S0/0/0 S0/0/1Router A
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.
Dallas 150 Hosts New York
325 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 70% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 70% growth in the largest subnet
Total number of address needed for the largest subnet
IP address range for New York
IP address range for Washington D. C.
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 135.126.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C F0/0 F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Washington D.C. 220 Hosts
65
Show your work for Problem 4 in the space below.
66
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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 100% growth in the largest subnet
Total number of address needed for the largest subnet
IP address range for Router F0/0 Port
IP address range for Router F0/1 Port
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0
Science Room 10 Hosts
Tech Ed Lab 18 Hosts
English classroom 15 Hosts
F0/1
Art Classroom 12 Hosts
IP Address 210.15.10.0
67
Show your work for Problem 5 in the space below.
68
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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 20% growth
Total number of subnets needed
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
IP Address 10.0.0.0
_____________________________
_____________________________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
=
F0/0
S0/0/1 Router A
Administration 35 Hosts
Technology Building 320 HostsF0/0
Router B
S0/0/1
Router C
F0/1
F0/1
S0/0/0
Science Building 225 Hosts
S0/0/0
S0/0/1S0/0/0 Art & Drama
75 Hosts
69
Show your work for Problem 6 in the space below.
70
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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 125% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 125% growth in the largest subnet
Total number of address needed for the largest subnet
IP address range for Router A Port F0/0
IP address range for Research
IP address range for Deployment
IP address range for Router A to Router B serial connection
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Marketing 75 Hosts
IP Address 177.135.0.0
Administration 33 Hosts Sales
255 Hosts
Research 135 Hosts
F0/0 S0/0/0 F0/0
F0/1
S0/0/0 Router A
Router B
Deployment 63 Hosts
71
Show your work for Problem 7 in the space below.
72
Practical Subnetting 8
F0/0 S0/0/0 S0/0/1Router A
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.
New York 8 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 85% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 85% growth in the largest subnet
Total number of address needed for the largest subnet
IP address range for Router A F0/0
IP address range for New York
IP address range for Router A to Router B serial connection
IP Address 192.168.1.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router B
F0/1
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Boston 5 Hosts
Research & Development 8 Hosts
73
Show your work for Problem 8 in the space below.
Practical Subnetting 9
F0/0
S0/0/0 S0/0/1Router A
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.
Dallas 1500 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 15% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 15% growth in the largest subnet
Total number of address needed for the largest subnet
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
IP Address 148.55.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C
F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Router D S0/0/0 S0/0/1
74
Ft. Worth 2300 Hosts
75
Show your work for Problem 9 in the space below.
76
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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 110% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for 110% growth in the largest subnet
Total number of address needed for the largest subnet
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
IP Address 172.16.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0 S0/0/0
S0/0/1Router A F0/0
Router B
F0/1
Sales 115 Hosts
Management 25 Hosts
Research 35 Hosts
Marketing 56 Hosts
77
Show your work for Problem 10 in the space below.
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.
IP Address: 0.230.190.192 ________________________________ Subnet Mask: 255.0.0.0 ________________________________
IP Address: 192.10.10.1 ________________________________ Subnet Mask: 255.255.255.0 ________________________________
IP Address: 245.150.190.10 ________________________________ Subnet Mask: 255.255.255.0 ________________________________
IP Address: 135.70.191.255 ________________________________ Subnet Mask: 255.255.254.0 ________________________________
IP Address: 127.100.100.10 ________________________________ Subnet Mask: 255.0.0.0 ________________________________
IP Address: 93.0.128.1 ________________________________ Subnet Mask: 255.255.224.0 ________________________________
IP Address: 200.10.10.128 ________________________________ Subnet Mask: 255.255.255.224 ________________________________
IP Address: 165.100.255.189 ________________________________ Subnet Mask: 255.255.255.192 ________________________________
IP Address: 190.35.0.10 ________________________________ Subnet Mask: 255.255.255.192 ________________________________
IP Address: 218.35.50.195 ________________________________ Subnet Mask: 255.255.0.0 ________________________________
IP Address: 200.10.10.175 /22 ________________________________ ________________________________
IP Address: 135.70.255.255 ________________________________ Subnet Mask: 255.255.224.0 ________________________________
The network ID cannot be 0.
OK
78
Reference Pages 28-29
Reference Page Inside Front Cover
Reference Pages 48-49
Reference Pages 54-55
Reference Page Inside Front Cover
Reference Page Inside Front Cover
Reference Pages 54-55 and/or Inside Front Cover
Reference Pages 56-57
Reference Pages Inside Front Cover
Reference Pages 34-35
Reference Pages 30-31
Reference Pages 48-49
0-127
128-255
0-3 4-7 8-11
12-15 16-19 20-23 24-27 28-31 32-35 36-39 40-43 44-47 48-51 52-55 56-59 60-63 64-67 68-71 72-75 76-79 80-83 84-87 88-91 92-95 96-99
100-103 104-107 108-111 112-115 116-119 120-123 124-127 128-131 132-135 136-139 140-143 144-147 148-151 152-155 156-159 160-163 164-167 168-171 172-175 176-179 180-183 184-187 188-191 192-195 196-199 200-203 204-207 208-211 212-215 216-219 220-223 224-227 228-231 232-235 236-239 240-243 244-247 248-251 252-255
/30 8+8+8+6
255.255.255.252 4 Hosts
/29 8+8+8+5
255.255.255.248 8 Hosts
/28 8+8+8+4
255.255.255.240 16 Hosts
/27 8+8+8+3
255.255.255.224 32 Hosts
/26 8+8+8+2
255.255.255.192 64 Hosts
/25 8+8+8+1
255.255.255.128 128 Hosts
/24 8+8+8
255.255.255.0 256 Hosts
0-7
8-15
16-23
24-31
32-39
40-47
48-55
56-63
64-71
72-79
80-87
88-95
96-103
104-111
112-119
120-127
128-135
136-143
144-151
152-159
16-167
168-175
176-183
184-191
192-199
200-207
208-215
216-223
224-231
232-239
240-247
248-255
0-15
16-31
32-47
48-63
64-79
80-95
96-111
112-127
128-143
144-159
160-175
176-191
192-207
208-223
224-239
240-255
0-63
64-127
128-191
192-255
0-255
IP Address Breakdown
79
/24 255.255.255.0
256 Hosts 1 Subnet
Start with a square. The whole square is a single subnet comprised of 256 addresses.
Visualizing Subnets Using The Box Method
The box method is the simplest way to visualize the breakdown of subnets and addresses into smaller sizes.
/25 255.255.255.128
128 Hosts 2 Subnets
/26 255.255.255.192
64 Hosts 4 Subnets
80
Split the box in half and you get two subnets with 128 addresses,
Divide the box into quarters and you get four subnets with 64 addresses,
81
/27 255.255.255.224
32 Hosts 8 Subnets
Split each individual square and you get eight subnets with 32 addresses,
/30 255.255.255.252
4 Hosts 64 Subnets
/29 255.255.255.248
8 Hosts 32 Subnets
/28 255.255.255.240
16 Hosts 16 Subnets
Split the boxes in half again and you get sixteen subnets with sixteen addresses,
The next split gives you thirty two subnets with eight addresses,
The last split gives sixty four subnets with four addresses each,
82
# of Bits Borrowed
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Subnet Mask
255.0.0.0 255.128.0.0 255.192.0.0 255.224.0.0 255.240.0.0 255.248.0.0 255.252.0.0 255.254.0.0 255.255.0.0
255.255.128.0 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0 255.255.254.0 255.255.255.0
255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252
Total # of Subnets
1 2 4 8 16 32 64 128 256 512
1,024 2,048 4,096 8,192 16,384 32,768 65,536 131,072 262,144 524,288
1,048,576 2,097,152 4,194,304
Total # of Hosts
16,777,216 8,388,608 4,194,304 2,097,152 1,048,576 524,288 262,144 131,072 65,536 32,768 16,384 8,192 4,096 2,048 1,024 512 256 128 64 32 16 8 4
Usable # of Hosts
16,777,214 8,388,606 4,194,302 2,097,150 1,048,574 524,286 262,142 131,070 65,534 32,766 16,382 8,190 4,094 2,046 1,022 510 254 126 62 30 14 6 2
# of Bits Borrowed
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Subnet Mask
255.255.0.0 255.255.128.0 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0 255.255.254.0 255.255.255.0
255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252
Total # of Subnets
1 2 4 8 16 32 64 128 256 512
1,024 2,048 4,096 8,192 16,384
Total # of Hosts 65,536 32,768 16,384 8,192 4,096 2,048 1,024 512 256 128 64 32 16 8 4
Usable # of Hosts 65,534 32,766 16,382 8,190 4,094 2,046 1,022 510 254 126 62 30 14 6 2
Class C Addressing Guide # of Bits
Borrowed 0 1 2 3 4 5 6
Subnet Mask
255.255.255.0 255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252
Total # of Subnets
1 2 4 8 16 32 64
Total # of Hosts
256 128 64 32 16 8 4
Usable # of Hosts
254 126 62 30 14 6 2
CIDR /8 /9 /10 /11 /12 /13 /14 /15 /16 /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30
CIDR /16 /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30
CIDR /24 /25 /26 /27 /28 /29 /30
Class B Addressing Guide
Class A Addressing Guide
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Inside Cover
